DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING 1st MID-SEMESTER EXAMINATION [2011-2012, SEMESTER I]
NOTE THE FOLLOWING FOLLOWING 1) Total time for this paper is 120 minutes (2 hours). 2) Total marks is 100 and individual marks are mentioned along side each question. 3) Total number of questions is 4 (with sub-parts) and you are required to answer ALL of them. 4) You are allowed to use the statistical tables and the calculator only. 5) Marks will be there for correct formulation of the problem, rather than only the final answer. Hence step wise marking is also there. 6) Draw diagrams and use the concept of set theory where ever necessary.
Question # 1: [15 + 5 + 5(=25) marks]
(a) Suppose you have a circuit as shown below. The probability of closing of each relay of the circuit is . Assume all the relays of the circuit function independently. Then what is the
probability that current will flow between and .
1
2
A
B
3
4
∩ ∪ ∩ ∩∩ ∪ ∩∩ − ∩ ∩ ∩
Solution # 1 (a) (15 marks)
Let
,
= 1,2, 1,2,3, 3,4 4 denote the event that the relay is closed , and let
current flows between point A and B. Then
= {(
1
2)
(
3
)}. 4 )}.
Hence
= (
1
1
2)
2)
(
+ (
3
3
4 )} 4)
be the event that
This means that only when (i) 1 and 2 or (ii) 3 and 4 or
(iii) 1, 2, 3 and 4 are closed will the current flows.
( ) = {(
{(
1
2)
Page 1 of 100
(
3
4 )}
∩ − ∩− − ∩ ∩ ∩
= (
1
2
+
=
2)
2
+ ( 4
=2
4)
3
2
(
1
2
4)
3
4
(b) An urn has 6 red, 4 white and 5 blue balls. Three balls are drawn successively from the box. Given this information find the probability that the balls are drawn in the order where you have red, followed by white and the last one is blue, (i) with replacement, (ii) without replacement.
Solution # 1 (b) (5marks)
∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩
Let us define the events as, R: drawing the red ball, W: drawing the white ball and B: drawing the blue ball. What we require is ( (i) (
) = ( )× (
| )× ( |
(ii) (
)= ( )× (
| )× ( |
)
)=
6 6+4+5
)=
×
6 6+4+5
4
×
6+4+5
×
4 5+4+5
5 6+4+5
×
=
5 5+3+5
8 225
=
4 91
(c) Find the probability of a 2 turning up at most once in two tosses of a fair die. Remember at most means that it (i.e., number 2) may not come also. You can definitely assume that the rolling of one dice does not affect the outcome of the other dice.
Solution # 1 (c) (5marks)
Let us define
1:
As the event that 2 comes as the face when you toss the first dice.
2:
As the event that 2 comes as the face when you toss the second dice.
∩ ∪ ∩ ∪ ∩ ∪ ∩ ∩ ∪ ∩ ∪ ∩ − ∩∩ ∩ ∩∩ ∩ −− ∩∩ ∩∩ ∩∩ ∩ ∩ ∩ ∩ ∩
Given this we need to find the probability of {(
1
2)
(
1
2)
+ (
2)
1
(
)}. 2 )}.
1
Remember that
Thus: is or and is and {(
1
2)
(
1
2)
(
1
2 )}
= (
2)
1
{(
1
2)
(
1
2 )}
{(
1
2)
(
1
2 )}
{(
+ {(
1
1
2)
2)
Page 2 of 100
(
(
1
1
+ (
1
2 )} 2)
(
1
2 )}
2)
∩ − ∩− − ∩ ∩ ∩
= (
1
2
+
=
2)
2
+ ( 4
=2
4)
3
2
(
1
2
4)
3
4
(b) An urn has 6 red, 4 white and 5 blue balls. Three balls are drawn successively from the box. Given this information find the probability that the balls are drawn in the order where you have red, followed by white and the last one is blue, (i) with replacement, (ii) without replacement.
Solution # 1 (b) (5marks)
∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩
Let us define the events as, R: drawing the red ball, W: drawing the white ball and B: drawing the blue ball. What we require is ( (i) (
) = ( )× (
| )× ( |
(ii) (
)= ( )× (
| )× ( |
)
)=
6 6+4+5
)=
×
6 6+4+5
4
×
6+4+5
×
4 5+4+5
5 6+4+5
×
=
5 5+3+5
8 225
=
4 91
(c) Find the probability of a 2 turning up at most once in two tosses of a fair die. Remember at most means that it (i.e., number 2) may not come also. You can definitely assume that the rolling of one dice does not affect the outcome of the other dice.
Solution # 1 (c) (5marks)
Let us define
1:
As the event that 2 comes as the face when you toss the first dice.
2:
As the event that 2 comes as the face when you toss the second dice.
∩ ∪ ∩ ∪ ∩ ∪ ∩ ∩ ∪ ∩ ∪ ∩ − ∩∩ ∩ ∩∩ ∩ −− ∩∩ ∩∩ ∩∩ ∩ ∩ ∩ ∩ ∩
Given this we need to find the probability of {(
1
2)
(
1
2)
+ (
2)
1
(
)}. 2 )}.
1
Remember that
Thus: is or and is and {(
1
2)
(
1
2)
(
1
2 )}
= (
2)
1
{(
1
2)
(
1
2 )}
{(
1
2)
(
1
2 )}
{(
+ {(
1
1
2)
2)
Page 2 of 100
(
(
1
1
+ (
1
2 )} 2)
(
1
2 )}
2)
∩ ∪ ∩ ∪ ∩ − − − {(
2)
1
(
2)
1
(
2 )}
1
1
=
6
×
5
5
+
6
6
×
1
5
+
6
×
6
5
0
6
0
0+0=
35 36
Question # 2: [15 + 15(=30) marks]
(a) Suppose that the p.d.f (remember what p.d.f means) of a random variable (r.v.), follows:
− − ≤ℎ≤ − ≤ ≤ 1
( )=
36
(9
2)
3
(
Sketch the p.d.f. neatly and determine the values of the following: (i) +1) +1) and (iii)
(
> 2). Here
means probability or
discussed in class.
is as
+3
0
( 1
< 0), (ii)
what we have
≤ − ≤ ≤ ∫− − − − − −
Solution # 2 (a) (15 marks)
To verify the fact that the ( ) is actually a p.d.f let us find (+3) +3) = { ( 3
+3) +3) =
+3 1
3 36
(9
2)
=
1
36
3
9
+3
3
=
3
1
27 27
36
27 3
+3} +3}, i.e., 27
+ 27
3
=1
PDF of the function given as (1/36)*(9-x 2)
0.25
0.2
0.15
) x ( f
0.1
0.05
0 -3
-2
-1
0
1
2
3
X Values
(i)
∫ − − − ≤ ≤ ∫− − − − − − (
(ii) 13 27
< 0) =
( 1
+3 1 (9 0 36
+1) =
2)
+1 1 (9 1 36
=
1
36
− 3
9
2)
+3
=
3 0
=
1
36
9
= 0.48 0.481 1
Page 3 of 100
1
27 27
36
3
3
27 3
+1 1
=
1
36
=
9
9×2 36
1 3
=
1 2
+9
1 3
=
52 108
=
− ≤ − ∫− − − − − −− − −− − −− − ∫ − − − − −−
(iii)
(
(
> 2) = 1
+2 1
=1
3 36
1
=1
=
2)
36
2)
(9
8
18
3
36
1
54
8
36
36
3
3
27
+ 27
+
1
=1 3
81
27
3
3
36
=1
=
8
108
3
9
+2
3
3
1
54
8
36
3
3
=
2
27
+
81
27
3
3
= 0.0741
Note the answer in (iii) could also be obtained by calculating (
+3 1
> 2) = =
+2 36
2)
(9
1
81
27
54
36
3
3
3
=
+
8 3
=
1
3
9
36
8
108
+3
3 +2
=
1
36
27
27 3
18 +
8 3
= 0.074
(b) Suppose that the p.d.f (remember what p.d.f means) of a random variable (r.v.), follows:
( )=
(1
0
−
1
)2
ℎ ≤ 0 <
is as
<1
(i) Find the value of the constant and then sketch the p.d.f. neatly (ii) Find the value of class.
1 2
. Here
means probability or what we have discussed in
≤ ≤ ≤ ≤ ∫ − − − − − − −
Solution # 2 (b) (15 marks)
(i) We know one of the important properties of pdf is ( (0
1) = =
1
0
(1
1 )2
2 × (0
= 1, in case 1
1) = 1, thus 1 2
= , then using simple integration we have
=
1 2
1
Hence the p.d.f. is of the form ( ) = ×
1
1
(1
) = 1. Hence:
)2
= × (1 2
Page 4 of 100
)
1 2
2.4
2.2
2
1.8
1.6 ) x ( f
1.4
1.2
1
0.8
0.6
0.4 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
X Values
(ii)
≤ ∫ 1 2
=
1
1 2
2 0
(1
− − 1 2
)
− √
= 1
1
2
Question # 3: [15 + 15(=30) marks]
(a) Two persons A and B play a gamble with a fair dice. In case numbers 1 or 2 appears, then
gives Rs. 30 to person
, otherwise gets Rs. 12. The two persons play the game (rolling
the dice is considered as a game) 10 times, then find the probability mass function assuming the random variable, , denotes the outcome for person . Find the (i) expected value and (ii) variance of .
Solution # 3 (a) (15 marks)
− − − −
A game is a simple example of ~ ( , ), where
2
= , 6
4
= , 6
is the random
= 10 and
variable which denotes the outcome of the game pertaining to the number being equal to 1 or 2. Hence the p.m.f is given by ( ) =
=
10 10
2 10
4
6
6
.
Now remember that the actual outcome is the outcome from the rolling of the dice, . To this
∑
corresponds different realized values of the net amount gained/lost by , which we denote
using ( )
Utilizing this concept we have: (i) (
)=
=
{ ( )}
10 =0
( ) ( )
Page 5 of 100
∑ − − − − − − − − ∑ − − − − − ∑ − − − 10 =0{12
=
( ) = {12
where as already mentioned
)} ×
30(10
) = { ( )2 } 10 =0{12
=
10 =0{12
30(10
)} ×
10 10
2 10 6
×
4 6
in
= 0, … ,10.
[ { ( )}]2
)}2 ×
30(10
×
×
)} is the amount won/lost by
30(10
each of the outcomes of the dice, starting from
(ii) (
10 10
2 10 6
×
4
10 10
×
2 10 6
×
4 6
2
6
To solve this problem refer to the table given below (refer the last page)
(b) A blood test developed by a pharmaceutical company for detecting a certain disease is 98% effective in detecting the disease, given that the disease is in fact present in the individual being tested. The test yields a false positive result (meaning a person without the disease is in-correctly indicated as having the disease) for any 1% of the disease free persons tested. If an individual is randomly chosen from the population and tested for the disease, and given that 0.1% of the population actually has the disease, then what is the probability that the person tested actually has the disease if the test result is positive (which means that the disease is indicated as being present by the test).
Solution # 3 (b) (15 marks)
Let define the event that the test result is positive and let be the other event which means
that the individual actually has the disease. Then given the data we have: ( | ) = 0.98,
( ) = 0.001 and ( | ( | )=
) = 0.01. We are required to find, ( | ), which is given as:
( | )× ( )
( | )× ( )+
×
=
0.98×0.001 0.98×0.001+0.01×0.999
= 0.089
Question # 4: [15 marks]
Super Computer Company which is a retailer store sells all things related to computer, starting from computer, printer, cartridges, flash drives, CDs, DVDs, etc. The number of computers and printers sold on any given day varies, with the probabilities of the various possible sales outcomes being given below in the following table
Page 6 of 100
Solution # 4 (15 marks) Number of computer sold
0 0 0.03
1
2
3
4
0.03
0.02
0.02
0.01
=0, =0,1,2,3,4
= 0.11
Nu
1 0.02
mbe
0.05
0.06
0.02
0.01
=1, =0,1,2,3,4
= 0.16 r of
2 0.01
0.02
0.10
0.05
0.05
=2, =0,1,2,3,4
prin
= 0.23 3 0.01
ters
0.01
0.05
0.10
0.10
=3, =0,1,2,3,4
= 0.27
sold
4 0.01
0.01
0.01
0.05
0.15
=4, =0,1,2,3,4
= 0.23
1.00
=0,1,2,3,4, =0
0
1.00
= 0.08
=0,1,2,3,4, =1
1
= 0.12
′ ′ ′ ∑∀ ∑∀ ∑∀ ∑∀ −∞ ≤ ≤ ∞ ∑∀ −∞ ≤ ≤ ∞ ∑∀
=0,1,2,3,4, =2
2
= 0.24
=0,1,2,3,4, =3
3
= 0.24
=0,1,2,3,4, =4
Denote the printer with and computer with , then each cell denotes }. Then the
and
togather with
,
{
+ ,
=
=
=
,0
,
+ }=
,
,} =
=
=
,
=
,
=
0,
Page 7 of 100
=
=
and
,
=
. It is
= 1.
and for any fixed let
=
,
=
define the joint distribution of
very easy to verify and also intuitive to see that Now for any fixed , let
,
= 0.32
4
,0
and
0,
=
(ii)
,
. Then (i)
∩ ℎ ℎ ℎ ℎ ℎ ℎ ℎ ℎ ℎℎ Moreover: = , = =
=
=
=
=
, =
( = )
=
,
and
=
=
=
,0
,
=
0,
With
respect
to
(
our
problem
we
)=
,
=
,
,
have:
= 0,1,2,3,4 and
= 0,1,2,3,4, where denotes printer and denotes computer.
Thus we have ,
= (
|
0,
)
,
= (
|
,0
)
With the information given above answer the following (show detailed calculations for each in order to get credit ): (i) What is the probability that more than two computes will be sold on any given day?
=0,1,2,3,4, =3
3
+
=0,1,2,3,4, =4
4
= 0.24 + 0.32 = 0.56
(ii) What is the probability that more than two printers will be sold on any given day? =3, =0,1,2,3,4
3
+
=4, =0,1,2,3,4
4
= 0.27 + 0.23 = 0.50
(iii) What is the probability of selling more than two printers GIVEN that more than two computers are sold?
=3,4
=,3,4
=
(0.10+0.10+0.05+0.15) (0.24+0.32)
∩ ∩ ∩ ∩
=
=3
5 7
=3
+
=3
=4
+
=3
+
=4
=3
+
=4
=4
=
=4
= 0.7143
(iv) What is the probability of selling more than two computers AND more than two printers on a given day?
∩ ∩ ∩ ∩ =3
=3
+
=3
=4
+
=4
=3
+
=4
=4
= 0.10 + 0.10 + 0.05 + 0.15 = 0.40
(v) What is the probability that the company has no sales on a given day? Page 8 of 100
∩ ∩ =0
=0
= 0.03
(vi) Given that the company sells no computers, what is the probability that it sells no printers also on a given day? =0
=0
=
=0
=0
=0
=
0.03 (0.03+0.02+0.01+0.01+0.01 )
Page 9 of 100
=
0.03 0.08
=
3 8
= 0.375
Solution for problem # 3 (a)
− 2
(1
2) 10 9 8 7 6 5 4 3 2 1 0
_
_
0 12 24 36 48 60 72 84 96 108 120
-300 -270 -240 -210 -180 -150 -120 -90 -60 -30 0
_ -300 -258 -216 -174 -132 -90 -48 -6 36 78 120
( ) 0.00 0.00 0.00 0.02 0.06 0.14 0.23 0.26 0.20 0.09 0.02 1
× ( ) -0.005 -0.087 -0.658 -2.829 -7.511 -12.291 -10.925 -1.561 7.023 6.763 2.081 -20
{ ( )}
Using this find
− { ( )} =
20
{ ( )} = 3920 = 4320
−−
( 20)2 = 3920
Page 10 of 100
2
90000 66564 46656 30276 17424 8100 2304 36 1296 6084 14400
× ( ) 1.524 22.545 142.222 492.218 991.459 1106.173 524.408 9.364 252.840 527.529 249.718 4320
{ ( )}2 × ( ) 1.328 19.185 117.104 385.567 713.777 669.166 178.444 50.984 611.809 832.740 339.894 3920
{ ( )}
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING 2nd MID-SEMESTER EXAMINATION [2011-2012, SEMESTER I] NOTE THE FOLLOWING 1) Total time for this paper is 90 minutes (1½ hours). 2) Total marks is 75 and individual marks are mentioned alongside each question. 3) Total number of questions is 3 (with sub-parts) and you are required to answer ALL of them. 4) You are allowed to use the statistical tables and the calculator only. 5) Marks will be there for correct formulation of the problem, rather than only the final answer. Hence step wise marking may be there. 6) Draw diagrams very clearly and legibly , use the concept of set theory where ever necessary, use tables where necessary.
Question # 1: [10 + 15 (=25) marks]
a) Assume you are the shop floor manager of a lathe machine work shop and you know that the working life (in years) of lathe machines follow an exponential distribution with E(X) = 5 years. You also know that any of the lathe machines will survive for at least 2 years. Given this information find out what is the probability that a particular lathe machine will survive for 4 or more years? Draw the pdf and cdf of the exponential distribution very clearly/legibly but separately.
b) Let X be the life in hours of a radio tube which is normally distributed with mean µ = 20 and variance σ2. If a purchaser of such a radio tubes requires that at least 90% of the tubes have life exceeding 150 hours then what is the largest value of σ for which the purchaser is still satisfied?
Solutions # 1 (a): 10 marks
Assume you are the shop floor manager of a lathe machine work shop and you know that the working life (in years) of lathe machines follow an exponential distribution with
( )=5
years. You also know that any of the lathe machines will survive for at least 2 years. Given this information find out what is the probability that a particular lathe machine will survive
Page 11 of 100
for 4 or more years? Draw the pdf and cdf of the exponential distribution very clearly/legibly but separately.
− − ≥ ∞ − − ∞ − − ∫ − − Let
be the random variable which denotes the life of the lathe machines in years, such that
~ ( = 2,
= 5), i.e., ( ) =
(
1
2)
5
5
,
2.
Given this information we are required to find the probability that a particular machine will
survive {
for
> 4} =
4
(
1
or
2)
(
=
5
5 4
more
years
2)
=
5
2 5
which
is
given
by
= 0.67 0.6703 032 2
4
0.16 0.14 0.12 0.1 0.08 f(x)
0.06 0.04 0.02 0 1
3
5
7
9
11 13 13 15 15 17 17 19 1 9 21 21 23 23 25 2 5 27 27 29 29 31 3 1 33 33 35 35 37 3 7 39 39 41 41
Exponential pdf when 0.8
= 2 and
=5
0.7 0.6 0.5 0.4 F(x)
0.3 0.2 0.1 0 1
3
5
7
9
11 13 1 3 15 1 5 17 1 7 19 1 9 21 2 1 23 2 3 25 2 5 27 2 7 29 2 9 31 3 1 33 3 3 35 3 5 37 3 7 39 3 9 41 41
Exponential cdf when
= 2 and
=5
Solutions # 1 (b): 15 marks
Let be the life in hours of a radio tube which is normally distributed with mean µ = 20 and 20 and variance σ2. If a purchaser of such a radio tubes requires that at least 90% of the tubes have
Page 12 of 100
life exceeding 150 hours then what is the largest value of σ for which the purchaser is still satisfied?
As given we consider ~ ( = 20,
−
150 20
= 0.90 from which we have
2)
. Now we have
0.1
= 1.28 =
30
{
> 150} = 0.9 0.90, i.e.,
, which implies
=
30 1.28
−
20
>
= 24
Question # 2: [10 + 15 (=25) marks]
a) The height, X, of boy studying in class II of any school in the city of Kanpur, is normally distributed with mean µ(=125 cms) and variance σ2(=100 cms). We also know the heights (in cms) of 5 such boys who have been selected as the sample are, 120, 100, 110, 140 and 130, then what is the probability that the height of any boy selected at random ((i) from the whole population and (ii) from this t his sample) will be between 120 cms and an d 130 cms.
b) To find whether the lathe machine is working properly you take a sample of 25 finished products and check the dimension of each of the finished product. You find the average dimension of the sample to be 65 mm, while the standard error is 15 mm. Then what would your comment be regarding whether the machine is in order or out of order if you knew that the actual dimension should be 66 mm. Assume α=5%.
Solutions # 2 (a): 10 marks
The height,
, of boy studying in class II of any school in the city of Kanpur, is normally
distributed with mean µ(=125 cms) and variance σ2(=100 cms). We also know the heights (in cms) of 5 such boys who have been selected as the sample are, 120, 100, 110, 140 and 130, then what is the probability that the height of any boy selected at random ((i) from the whole population and (ii) from this t his sample) will be between 120 cms and an d 130 cms.
� ̅ ≤ ≤ −− − ≤ ≤ − − ≤ ≤ ≤ � ≤ − √ − ≤ � ≤ √ − − ≤ � ≤
Consider 5,
5
~ ( = 125,
= 100) and more over
5~
= 125,
2
5
= 20 , where
=
= 120. Using this information we calculate the following:
(i) {120 (ii)
2
130} =
{120
5
120 125
120 125
10
10
130} 130} =
120 125 10 5
1.12} = 0.73 0.7372 72
Page 13 of 100
= { 0.5 120 125
5
10 5
0.5} = 0.38 0.383 3
= { 1.12
5
Solutions # 2 (b): 15 marks
To find whether the lathe machine is working properly you take a sample of 25 finished products and check the dimension of each of the finished product. You find the average dimension of the sample to be 65 mm, while the standard error is 15 mm. Then what would your comment be regarding whether the machine is in order or out of order if you knew that the actual dimension should be 66 mm. Assume α = 5%. 5%.
First let us frame out hypothesis
≠ 0:
= 66 vs 66 vs
0:
= 66
( )
2
2
� − ≥ √ −− � ≤ − √ − � ≥ √ − − ≤ − √ ≥ √ ≤ ≥ Based on the information one would reject 1,
or
= 66 if 66 if |
0:
+
1,
2
|
1,
holds, i.e.,
2
is true.
2
Before we solve the problem we need to find
1,
and the value is 1.711. 1.711. Utilizing the set
2
of values given we have:
65
66
15
25 25
× 1.71 1.711 1 or 65
66 +
15
25 25
× 1.71 1.711 1, i.e., 66
As both of them are false hence we cannot reject
0:
60.867 or 60.867 or 66
71.133
= 66, 66, which means that there is no
significant difference and hence machine is in order.
Question # 3: [10 + 15 (=25) marks]
a) A food inspector examines 10 jars of certain brand of butter and obtained the following percentages of impurities, the th e values of which are, 2.3, 2.3 , 1.9, 2.1, 2.8, 2.3, 3.5, 1.8, 1.4, 1 .4, 2.0 and 2.1. Form a 95% level of confidence for the estimate of the mean of the impurity level,
Page 14 of 100
where you can assume the population distribution of the level of impurity as normal, i.e.,
(
X ~ N µ , σ
2
).
b) Consider we have a biased dice (with six faces, marked 1, 2, 3, 4, 5 and 6), such that P[X=i] ∝ i. First find out the pmf and then draw the pmf as well as the cdf on the same graph very clearly and legibly . After that find the value of E(X) and V(X).
Solutions # 3 (a): 10 marks
A food inspector examines 10 jars of certain brand of butter and obtained the following percentages of impurities, the values of which are, 2.3, 1.9, 2.1, 2.8, 2.3, 3.5, 1.8, 1.4, 2.0 and 2.1. Form a 95% level of confidence for the estimate of the mean of the impurity level, where you can assume the population distribution of the level of impurity as normal, i.e.,
(
)
X ~ N µ , σ 2 .
From the data given one can easily find the following which are:
� − ≤ ≤ � − √ − √ − − √ ≤ ≤ √ 10
= 0.5789, using which our confidence interval is 1,
i.e.,
2.22
+
1,
2
0.5789 10
× 1.833
= (1
̅ = 10,
10
= 2.22,
)
2
2.22 +
0.5789 10
× 1.833 = 0.95
Hence LCL is 1.8445 and UCL is 2.5556
Solutions # 3 (b): 15 marks
Consider we have a biased dice (with six faces, marked 1, 2, 3, 4, 5 and 6), such that P[X=i] ∝ i. First find out the pmf and then draw the pmf as well as the cdf on the same graph very clearly and legibly. After that find the value of E(X) and V(X).
Given the information we have: ( 5 + 6) = 1, thus
=
1
( )=
21
0
, from which we obtain
21
Hence the pmf and cdf are:
= )=
ℎ
= 1,2,3,4,5,6
Page 15 of 100
(1 + 2 + 3 + 4 +
≤ ≤ ⎪ ≤≤ ≤ ⎩ ≥ ∑∑ − − 0 1
21 3
( )=
21 6 21 10 21 15 21
1
<1 <2
2
<3
3
<4
4
<5
5
<6
1
6
( )=
6 =1
( )=
6 =1{
=1×
× ( )= 1×
1
21
1
21
+2×
( )}2 × ( ) =
+4×
2
21
+9×
3
21
2
21
+3×
2)
(
+16×
4
21
3
21
+4×
4 21
+5×
{ ( )}2
+25×
5 21
0.35
+36×
6 21
5 21
−
+6×
6 21
= 4.333
4.3332 = 2.22
0.3 0.25 0.2 0.15 0.1 0.05 0 1
2
3
4
PDF of ( ) =
21
Page 16 of 100
5
6
21/21
15/21
10/21
( ) 6/21 3/21 Left
discontinuous 1/21
0
1
2
3
4
5
6
++++++++++++++++END OF QUESTION PAPER ++++++++++++++++
Page 17 of 100
point
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING FINAL SEMESTER EXAMINATION [2011-2012, SEMESTER I]
NOTE THE FOLLOWING 1) Total time for this paper is 180 minutes (3 hours). 2) Total marks is 100 and individual marks are mentioned alongside each question. 3) Total number of questions is 4 (with sub-parts) and you are required to answer ALL of them. 4) You are allowed to use the statistical tables, formulae sheet and the calculator only. 5) Marks will be there for correct formulation of the problem, rather than only the final answer. Hence step wise marking is also there. 6) Draw diagrams accurately/neatly/legibly, and use the concept of set theory where ever necessary. 7) Remember to write your formulations and do your calculations legibly and clearly.
Question # 1: [15 + 10 (=25) marks]
a) Tarun Goel is a doctor and he measures the weights at birth (in kg.) for 15 babies born in a city hospital. The weights are as given below 2.79
3.01
3.19
3.10
2.25
2.61
3.55
3.82
3.38
2.56
2.16
3.06
3.42
3.51
3.64
Tarun, unfortunately is bad in statistics, hence please help him to find the limits between
which the mean weight at birth for all such babies should lie depending on which he can
prescribe medical care as required for children whose weights are below the lower limit. Consider
= 0.05.
Question # 1: [15 + 10 (=25) marks]
a) Tarun Goel is a doctor and he measures the weights at birth (in kg.) for 15 babies born in a city hospital. The weights are as given below 2.79
3.01
3.19
3.10
2.25
2.61
3.55
3.82
3.38
2.56
2.16
3.06
3.42
3.51
3.64
Tarun, unfortunately is bad in statistics, hence please help him to find the limits between
which the mean weight at birth for all such babies should lie depending on which he can
prescribe medical care as required for children whose weights are below the lower limit. Consider
= 0.05. Page 18 of 100
b) A slip of paper is given to Abhishek Malaviya, who marks it with either a plus (+) or
1
minus – sign and the probability of him writing a plus sign is . Abhishek then passes on 3
the slip to Anveeksha Verma, who may either leave it along or change the sign before passing it on to Kanwardeep Singh. Next, Kanwardeep passes the slip to Monica Agrawal and while doing so Kanwardeep may or may not change the sign. Monica further on passes it on to Raksha Agrawal who further on hands over the slip to Saptarshi Sarkar who again passes it
over to Harshil Shah, who finally gives it to the class representative Deepak Gaur . In handing over the slip to their next partner, Anveeksha, Kanwardeep , Monica, Raksha, Saptarshi and Harshil all of them may or may not change the sign written on the slip, and 2
their respective probabilities of changing the sign is . Deepak sees a plus (+) written on the 3
slip, then find the probability that the sign marked by Abhishek was also plus (+).
Solution # 1 (a): 15 marks Tarun Goel is a doctor and he measures the weights at birth (in kg.) for 15 babies born in a
city hospital. The weights are as given below 2.79
3.01
3.19
3.10
2.25
2.61
3.55
3.82
3.38
2.56
2.16
3.06
3.42
3.51
3.64
Tarun, unfortunately is bad in statistics, hence please help him to find the limits between
which the mean weight at birth for all such babies should lie depending on which he can prescribe medical care as required for children whose weights are below the lower limit. Consider
= 0.05.
� � � −√ − ≤ ≤ � √ − −
In case
is the random variable which denotes the weights of the babies then ~ ( ,
Moreover in case is the sample size then know
~
,
2
but still we are required to find the lower and higher levels of weights given the
From the data
= 15,
15
= 3.07,
= 0.5071,
15
14,0.025
= 2.145
Hence:
C.I. is formulated as
1,
+
2
Page 19 of 100
.
. As per the information we do not
value as 0.01.
Thus
2)
1,
= (1
2
)
� −√ − − � √ − =
=
1,
+
= 3.07
2.145 ×
2
1,
= 3.07 + 2.145 ×
2
√ √
0.5071 15
0.5071 15
= 2.7891 = 3.3509
Solution # 1 (b): 10 marks
A slip of paper is given to Abhishek Malaviya , who marks it with either a plus (+) or minus
1
– sign and the probability of him writing a plus sign is . Abhishek then passes on the slip 3
to Anveeksha Verma, who may either leave it along or change the sign before passing it on to Kanwardeep Singh. Next, Kanwardeep passes the slip to Monica Agrawal and while doing so Kanwardeep may or may not change the sign. Monica further on passes it on to Raksha Agrawal who further on hands over the slip to Saptarshi Sarkar who again passes it over to Harshil Shah, who finally gives it to the class representative Deepak Gaur . In handing over
the slip to their next partner, Anveeksha, Kanwardeep , Monica, Raksha, Saptarshi and Harshil all of them may or may not change the sign written on the slip, and their respective 2
probabilities of changing the sign is . Deepak sees a plus (+) written on the slip, then find 3
the probability that the sign marked by Abhishek was also plus (+).
The slip of paper goes in this way Abhishek Malaviya 1st pass Anveeksha Verma 2nd pass Kanwardeep Singh 3rd pass Monica Agrawal 4th pass Raksha Agrawal 5th pass Saptarshi Sarkar 6th pass Harshil Shah 7th pass Deepak Gaur Page 20 of 100
Define the events as follows
: Abhishek Malaviya wrote a plus (+) sign and ( ) =
1 3
: Abhishek Malaviya wrote a minus – sign and (
: Deepak Gaur sees a plus (+) sign
)=
2 3
Now as Deepak Gaur sees the sign as plus (+) and Abhishek Malaviya also marked a plus (+) sign
Hence we are required to find the following probability { | } =
To find •
No change of sign: The number of ways is 3
•
×
2 0
1 4 3
×
2 2
6 2
×
and the corresponding probability
3
×
2 4 3
1 0 3
×
{
2 6
1 5 3
6 6
6 4
and the corresponding
and the corresponding probability
3
} we note that the following would have happened
×
2 1
6 1
and the corresponding probability
3
Change of sign three times: The number of ways is probability is
•
1 2
Change of sign one time: The number of ways is is
•
3
Change of sign six times: The number of ways is
To find
( )× { | }+
and the corresponding probability is
Change of sign four times: The number of ways is
is
•
6 0
3
probability is •
Change of sign two times: The number of ways is is
( )× { | }
{ | } we note that the following would have happened
1 6
•
1 2 3
×
2 3 3
Change of sign five times: The number of ways is probability is
1 1 3
×
2 5 3
Page 21 of 100
6 3
and the corresponding
6 5
and the corresponding
Utilizing these we have:
{ | }=
1 × 3
6 0
×
1 6 2 0 × + 62 3 3
{ | }=
×
1 × 3 1 4 3
6 0
×
1 6 2 0 1 4 2 2 1 2 2 4 1 0 2 6 × + 62 × × + 64 × × + 66 × × 3 3 3 3 3 3 3 3 2 2 1 2 2 4 1 0 2 6 2 6 1 5 2 1 1 2 2 3 6 6 6 + 4 × × + 6 × × + 1 × × + 3 × × + 65 3 3 3 3 3 3 3 3 3 3
×
1 1 1 1 1 +15×4× 6 +15×16× 6 +64× 6 3 36 3 3 3 1 1 1 1 1 2 1 1 1 +15×4× 6 +15×16× 6 +64× 6 + 6×2× 6 +20×8× 6 +6×32× 6 3 36 3 3 3 3 3 3 3
1+60+240+64
1+60+240+64+2(12+160+192 )
=
365 365+728
=
365
1093
×
1 1 2 5 × 3 3
=
= 0.3339
Question # 2: [15 + 10 (=25) marks]
a) Kaushik Choudhury is a student in the Quantitative Techniques for Decision Making course, which is compulsory for MBA students at IIT Kanpur. He has approached you with
− − ≤≤ ≤
the following problem which he is unable to solve. Help him to draw (accurately, neatly and legibly) both ( ) and ( ) of the function. Remember ( ) is given below.
0 0.2 ( )= 0.7 1
2 0
< 2 <0 <2 2
b) Two experimenters, Navpreet Singh and Nimisha Raveendran, take repeated measurements of the length of a copper wire. On the basis of the data obtained by them, which are given below, test whether Navpreet ′s measurement is more accurate (think what accuracy means here) than Nimisha′s. Consider α = 0.05. Navpreet ′s measurement
Nimisha′s measurement
(in mm)
(in mm)
12.47
12.44
12.06
12.34
11.90
12.13
12.23
12.46
12.77
11.86
12.46
12.39
Page 22 of 100
11.96
12.25
11.98
12.78
12.29
12.22
Solution # 2 (a): 15 marks Kaushik Choudhury is a student in the Quantitative Techniques for Decision Making course,
which is compulsory for MBA students at IIT Kanpur. He has approached you with the following problem which he is unable to solve. Help him to draw (accurately, neatly and
− − ≤≤ ≤
legibly) both ( ) and ( ) of the function. Remember ( ) is given below.
0 0.2 ( )= 0.7 1
2 0
< 2 <0 <2 2
Utilizing the distribution function we have
( )
1.0
0.7
Left point discontinuous
0.2
-2
0
Page 23 of 100
+2
Hence ( ) is
( )
0.5
0.3 0.2
-2
0
+2
Solution # 2 (b): 10 marks
Two experimenters, Navpreet Singh and Nimisha Raveendran, take repeated measurements of the length of a copper wire. On the basis of the data obtained by them, which are given below, test whether Navpreet ′s measurement is more accurate (think what accuracy means here) than Nimisha′s. Consider α = 0.05. Navpreet ′s measurement
Nimisha′s measurement
(in mm)
(in mm)
12.47
12.44
12.06
12.34
11.90
12.13
12.23
12.46
12.77
11.86
12.46
12.39
11.96
12.25
11.98
12.78
12.29
12.22
Given the data let us denote and as the random variables which denote the distribution of
measurement made by Navpreet and Nimisha, such that ~ (
Page 24 of 100
,
2)
and ~ (
,
2)
.
More we also know that: (i)
� � = 8,
=
8
= 12.2675,
� �
= 10,
=
10
= 12.2850,
= 0.178786
= 0.333042 and (ii)
To test the hypothesis or statement that Navpreet ′s measurement is more accurate than Nimisha′s and α = 0.05 our hypothesis is as follows:
− ≤ ≤ ≤ − − − − ≤ − − − 2 0
0:
= 0:
2
2
vs
0:
2 0
0:
2
2
2
The rule is reject
0 if 2
1,
1,1
is true.
Now we have 2 10
0.110917 0.031964
that
2 8
= 0.110917,
0:
= 0.031964,
= 0:
2
7,9,0.05
=
1
3.29
= 0.3040,
2
thus
2
=
0.3040 is FALSE, hence we cannot reject the null hypothesis
= 3.470019
2 0
9,7,1 0.05
1
=
2
, which means that there is significantly no difference in Navpreet ′s
measurements/readings with respect to Nimisha′s measurements/readings. Remember :
9,7,1 0.05
=
1
as
7,9,0.05
, ,1
=
1
,
,
Question # 3: [15 + 10 (=25) marks]
a) Saurabh Awasthi is testing the tensile strength of a particular alloy. The sample average is 13.71, while the standard deviation of the sample is 3.55. What should be the minimum sample size Saurabh should collect such that the confidence interval within which the population mean would lie is 3.14? As Saurabh was finishing his task, Nitin Bharadwaj comes running and says that the standard deviation is not of the sample but of the population. In that case what is the new sample size, considering all other information is correct. How many extra observations did Saurabh already collect or need to collect, based on Nitin′s information? Consider
= 0.05.
Page 25 of 100
b) Mohan Kumar K . is measuring the surface finish of crank shaft his production unit is manufacturing. Depending on the quality of surface finish, the crank shafts can be either good or bad. Mohan checks a sample of size 85 and finds 10 of them to be bad. Help Mohan
to formulate a 95% confidence interval and find for him the lower and upper limits of the confidence interval.
Solution # 3 (a): 15 marks Saurabh Awasthi is testing the tensile strength of a particular alloy. The sample average is
13.71, while the standard deviation of the sample is 3.55. What should be the minimum sample size Saurabh should collect such that the confidence interval within which the population mean would lie is 3.14? As Saurabh was finishing his task, Nitin Bharadwaj comes running and says that the standard deviation is not of the sample but of the population. In that case what is the new sample size, considering all other information is correct. How many extra observations did Saurabh already collect or need to collect, based on Nitin′s information? Consider
= 0.05.
Case I
� − − √ ≤ ≤ � − √ − √ ̅ − √ −√ −√ √ − −√ √ − −√ √ − 1
1 1, 2
1
×
1
1
+
1 1, 2
×
1
, i.e., 2 ×
1
confidence interval. Now from the information we have 2×
1
1,0.025
×
1
1
= 3.14, i.e.,
1 1,0.025 1
=
1 1, 2
3.14
1
×
1
= 13.71,
= 0.44225
3.55×2
Now check the distribution tables and from that we have: 1
1 = 16,
1
1 = 17,
1
1 = 18,
1
= 17,
1
= 18,
1
= 19,
16,0.025
= 2.120, i.e.,
1 1,0.025
=
1
17,0.025
= 2.110, i.e.,
18,0.025
= 2.101, i.e.,
1 1,0.025
=
1
1 1,0.025 1
Page 26 of 100
=
2.120 17
2.110 18
2.101 19
= 0.5142
= 0.4973 = 0.4820
is the length of
1
= 3.55, hence
− − − −
−√ √ −√ √ −√ √ −√ √ � − √ ≤ ≤ � √ √ ̅ √ ≅ 1 = 19,
1
1
19,0.025
1 1,0.025
= 2.093, i.e.,
2.093
=
1
20
=
,
=
,
= .
, i.e.,
, .
, .
=
=
,
=
,
= .
, i.e.,
, .
, .
=
1 = 22,
1
= 20,
1
The value of
= 23,
1 which
22,0.025
= 2.074, i.e.,
1 1,0.025 1
.
= .
.
= .
2.074
=
= 0.4680
23
= 0.4325
satisfies this is between 21 and 22, and we will consider
= 22.
1
Case II
×
2
2
2
2
+
×
2
2
, i.e., 2 ×
2
1
= 13.71,
interval. Now from the information given we have
2
= 1.96, i.e.,
= 3.14,
2
2
= 19.64
is the length of confidence
×
2
= 3.55, hence 2 ×
×
2
20
Hence the extra observations collected is 2
Solution # 3 (b): 10 marks Mohan Kumar K . is measuring the surface finish of crank shaft his production unit is
manufacturing. Depending on the quality of surface finish, the crank shafts can be either good or bad. Mohan checks a sample of size 85 and finds 10 of them to be bad. Help Mohan
to formulate a 95% confidence interval and find for him the lower and upper limits of the confidence interval.
From the information given we have:
̂
using
we
normal
− ≤ −− ≤ − ̂ ̂ − − −
distribution
concept
= 85,
have
=
10 85
= 0.1177,
P
(1
2
̂ − ~
,
(1
)
, such that ( ) =
=
, ( )=
2
Page 27 of 100
=
(1
2
)
=
=
)
(1
75 85
= 0.8823. Now
= (1
), as
2
)
, i.e.,
=
(1
)
Hence the 95% confidence 95% confidence interval is:
̂ − − ≤ ≤ ̂ − − − ≤ ≤ ×
(1
)
+
2
0.1177
×
(1
)
= (1
)
2
1.96 1.96 ×
0.1177×0.8823 0.1177×0.8823
0.1 0.1177 + 1.96 ×
85
0.1177×0.8823 0.1177×0.8823 85
= 0.95 .95
Hence LCL is 0.0492 and 0.0492 and UCL is 0.1826
Question # 4: [15 + 10 (=25) marks]
a) Mayank Singh who is a DJ and the group leader of IIT Kanpur music club is very health conscious about himself and his group members. Remember it is a large music group with many members and all the group members follow a strict diet regime. They are only allowed to take special variety of mixed salad and a specially prepared paneer kofta as advised by their nutritionist Arjun Ravindra Khular . They use at least 800 kgs of this combined special food daily. Yes we do agree the amount is huge but remember the group members like to eat . The nutrient contents and the costs are given below in the chart.
Type of food Mixed salad Paneer kofta
Protein 0.09 0.60
Kg per kg of food Fibre 0.02 0.06
Cost (Rs.) 0.30 0.90
The dietary requirement of this special type of food entails an intake of at least 30% protein and at most 5% fibre. Solve the problem and help Arjun (who has no clue of how to solve an optimization problem) such that the aim to minimize the total cost is met.
b) Sumit Kumar is is a high profile person who has just become the CEO of Ghotala Bank Inc. in India, after completing his MBA from IIT Kanpur with flying colours. He is in the process
Page 28 of 100
of devising a loan policy for his bank and the amount involves a maximum of Rs.120 crores. The following table provides the pertinent data about the available types of loans. Type of Loan Personal Car Home Farm Commercial Note
Interest Rate 0.140 0.130 0.120 0.125 0.100
Bad Debt Ratio 0.10 0.07 0.03 0.05 0.02
•
Bad debts are unrecoverable and produce no interest revenue.
•
Competition with other banks requires that Ghotala Bank Inc. allocates at least 40% of the funds to farm and commercial loans.
•
To assist housing industry in the country home loans must equal at least 50% of the personal, car and home loans. lo ans.
•
The bank also has a stated policy of not allowing the overall ratio of bad debts on all loans to exceed 4%.
Help Sumit to to formulate the optimization problem, given the data.
Solution # 4 (a): 15 marks Mayank Singh who is a DJ and the group leader of IIT Kanpur music club is very health
conscious about himself and his group members. Remember it is a large music group with many members and all the group members follow a strict diet regime. They are only allowed to take special variety of mixed salad and a specially prepared paneer kofta as advised by their nutritionist Arjun Ravindra Khular . They use at least 800 kgs of this combined special food daily. Yes we do agree the amount is huge but remember the group members like to eat . The nutrient contents and the costs are given below in the chart.
Type of food Mixed salad Paneer kofta
Protein 0.09 0.60
Kg per kg of food Fibre 0.02 0.06 Page 29 of 100
Cost (Rs.) 0.30 0.90
The dietary requirement of this special type of food entails an intake of at least 30% protein and at most 5% fibre. Solve the problem and help Arjun (who has no clue of how to solve an optimization problem) such that the aim to minimize the total cost is met.
Let the decision variables be:
1:
2:
Salad Paneer kofta
Hence the optimization problem is as follows
≥ ≥ − ≤ ≤ − ≥ ≥ ≥
min .
s.t.:
(1)
+ .
(2)
+
.
+ .
.
.
(
+
)
(3b)
.
.
+ .
.
.
(
+
(3a)
)
.
,
Page 30 of 100
(4a) (4b)
(5a, 5b)
(Eqn: 5a)
(Eqn: 4)
2
(Eqn: 1) A (Eqn: 3)
B
(Eqn: 2)
0,0
(Eqn: 5b)
1
Figure 1
Find points A and B and find the objective functions at A and B and then find the minimum value •
For A: coordinate is (200,600), for which the objective function is 600
•
For B: coordinate is (470.6,329.4), for which the objective function is 437.64
Hence best possible combination is: (470.6,329.4),
= 437.64
Solution # 4 (b): 10 marks Sumit Kumar is a high profile person who has just become the CEO of Ghotala Bank Inc. in
India, after completing his MBA from IIT Kanpur with flying colours. He is in the process of devising a loan policy for his bank and the amount involves a maximum of Rs.120 crores. The following table provides the pertinent data about the available types of loans.
Page 31 of 100
Type of Loan Personal Car Home Farm Commercial Note
Interest Rate 0.140 0.130 0.120 0.125 0.100
Bad Debt Ratio 0.10 0.07 0.03 0.05 0.02
•
Bad debts are unrecoverable and produce no interest revenue.
•
Competition with other banks requires that Ghotala Bank Inc. allocates at least 40% of the funds to farm and commercial loans. To assist housing industry in the country home loans must equal at least 50% of the
•
personal, car and home loans. The bank also has a stated policy of not allowing the overall ratio of bad debts on all
•
loans to exceed 4%. Help Sumit to formulate the optimization problem, given the data.
Let the decision variables be:
1:
Personal loans
2:
Car loans
3:
Home loans
4:
Farm loans
5:
Commercial loans
Given these we have the optimization problem is such that where we maximize difference between total interest and bad debts, where Total interest is
= 0.14 × 0.9
= 0.126
1
1
+ 0.1209
Bad debts is
+ 0.13 × 0.93 2
2
+ 0.12 × 0.97
+ 0.1164
3
+ 0.11875
3
+ 0.125 × 0.95
4
+ 0.098
Page 32 of 100
5
4
+ 0.1 × 0.98
5
≤ ≥ − − ≤ ≥ − ≤ ≤ − ≤ ≥ = 0.1
1
+ 0.07
+ 0.03
2
3
+ 0.05
4
+ 0.02
5
Hence objective function is total interest – bad debts, hence we have Maximize 0.026
+ 0.0509
1
2
+ 0.0864
3
+ 0.06875
4
+ 0.078
5
s.t.: 1
+
2
4
+
5
+
3
+
4
0.4(
1
+
+
12
5
2
+
3
+
0.6
4
+
5 )
0.6
5
4
(1)
(2a),
i.e., 0.4
+ 0.4
1
0.5(
3
2
1
+ 0.4
+
0.5
1
+ 0.5
0.1
1
+ 0.07
2
+
0.5
2
2
3
0
3)
0
3
+ 0.03
3
+ 0.05
4
+ 0.02
5
0.4(
1
+
2
+
3
+
4
+
(2b)
(3a), i.e., (3b)
5 )
(4a),
i.e.,
0.06
1
1,
2,
+ 0.03
3,
4,
2
0.01
5
0
3
+ 0.01
4
+ 0.02
5
0
++++++++++++++++ END OF QUESTION PAPER ++++++++++++++++
Page 33 of 100
(4b)
(5)
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING 1st QUIZ [2012-2013, SEMESTER I] NOTE THE FOLLOWING 1) Total time for this paper is 60 minutes (1 hour). 2) Total marks is 40 and individual marks are mentioned alongside each question. 3) Total number of questions is 4 (with sub-parts) and you are required to answer ALL of them. 4) Marks will be there for correct formulation of the problem, rather than only the final answer. Hence step wise marking is also there. 6) Draw diagrams and use the concept of set theory where ever necessary.
Question # 1 [10 marks]
Determine the probability of three 6's in five tosses of a fair die.
Solution # 1 [10 marks]
Let the tosses of the die be represented by the five (5) spaces as ____, ____, ____, ____,
____. Then in each of the space we will have the events as 6 or 6 , where 6 means the
′
complement which are 1,2,3,4,5. For example, three (3) 6's and two (2) 6 s can occur as
6,6, 6 , 6 , 6 and corresponding probability is
1 6
1
5
1
5
6
6
6
6
× × × × since the concept of
independence holds true between any two tosses. Now the total number of such placements
′
probability is
5 3
1
1
5
1
5
5!
6
6
6
6
6
3!×2!
× × × × × =
×
1 3 6
×
5 2 6
=
5 3
of these three (3) 6's and the two (2) 6 s can happen in
125
3888
=
5! 3!×2!
ways. Hence the
= 0.03215
Question # 2 [10 marks]
A community has 2% people suffering from tuberculosis. The X-ray test which may be used to detect the disease is also not totally reliable. If a person is affected by tuberculosis then there is 95% chance that the X-ray result will be positive. On the other hand, there may be a false alarm so that there is 1% chance that the X-ray test will show a positive result even when the person is not affected by tuberculosis. Given that a person have a positive X-ray result what is the probability thet he is affected by tuberculosis?
Page 34 of 100
Solution # 2 [10 marks]
Let us define the events as given below: A: The person is affected by tuberculosis B: The X-ray shows a positive result Then AC: The person is not affected by tuberculosis This gives us the values of P(A) = 0.02 and P(AC) = 0.98. Furthermore the event B can be broken down into set which are disjoint in the sense the property of exclusiveness and
∩ ∩ ∩
exhaustiveness will hold, such that ( | ) = 0.95 and ( | ( )= (
Hence
)+ (
) = 0.01.
) = ( | )× ( )+ ( |
)× (
) = 0.95 ×
0.02 + 0.01 × 0.98 = 0.0288
Furthermore the probability that a person has tuberculosis given that he/she has a positive Xray result is given by ( | ) =
(
)
( )
=
( | )× ( ) ( )
=
0.02×0.95 0.0288
= 0.66
Question # 3 [10 marks]
Box I contains 3 red and 2 blue marbles while Box II contains 2 red and 8 blue marbles. A fair coin is tossed. If the coin turns up heads (H) then a marble is chosen from Box I, while if the tail (T) comes then a marble is chosen from Box II. Find the probability that a red marble is chosen.
Solution # 3 [10 marks]
Let R denote that a red marble is chosen , while I and II denote the events that Box I and Box II are chosen respectively. Since red marble can result buy choosing either Box I or Box II, hence we can use the result which is as follows and can be understood from the figure below
Ω Here
A
∩ ∪ ∩ =(
)
(
B
). This can be extended for more than 1 set of B, such that Page 35 of 100
∩ ∩ ∪ ≠ ∩ ∪ ∩ ∪ ∪ ∩ ∪ Ω ⋯ − ∩ −∩∩ ∩ ∩ ∩∩ ∩ ⋯ ⋯ ∩ ∩−− ∩∩−∩∩− − ⋯ ∩− − ∩ ∩ ⋯∩ ∩ ∩∩ ∩ =(
1)
(
2)
(
3)
…. (
), where
=
,
= 1,2, … , and
= 1,2, … , .
= 0 for
Moreover taking probability for A we have ( )= ( .
(
+(
2)
1
+(
2)
+(
)+ (
1
. +( 1)
(
1)
(
1
3)
+
2
1
….
. +(
Hence using this concept we have 2
2+8
=
3)
2
+
)
. +(
3)
+
(
+ (
2)
1
2
)
1
). But note here all the terms after
(
1)
+
) are zeros.
1
3
2
3+2
( )= ( ) ( | )+ ( ) ( | )= ×
1
+ × 2
2 5
Question # 4 [10 marks]
Four letters are written and four matching envelopes are prepared. The letters are placed in the envelopes at random. What is the probability that at least one of the letters is placed into the correct envelope.
Solution # 4 [10 marks]
Suppose A1, A2, A3 and A4 represent the events that the 1st, 2nd , 3rd and the 4th letter are placed in the correct envelopes respectively. Then
∪ ∪ ∪ 1
2
3
4 represents
the event
that at least one of the letters is placed in the right envelope. Furthermore using simple set theoretic notation we can easily visualize the illustration given below such that the following would hold true
Ω
A1
A3
A2
A4
Page 36 of 100
∩∪ ∪− ∪∩ − ∩ − ∩ − ∩ ∩∩ − ∩∩ ∩− ∩ ∩ ∩ ∩ − ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩∩ ∩ ∩ ∩ ∩∩ ∩ ∩ ∩ ∩∩ ∩ ∩ ∩ ∩∩ ∩ ∩ ∩ ∩∩ ∩ ∩ ∩∩ ∩ ∩ ∩ ∩∩ ∩ ∩ ∪ ∪ ∪ − − (
1
(
1
4)
+ (
2
4)
3
4)
(
1
3
= (
1)
+ (
3)
(
2
4)
+ (
2
2)
= (
3)
2
2)
+ (
4)
(
+ ( 4)
3
4)
3
3)
(
4)
(
+ (
1
2
2)
1
1
(
3)
2
3)
1
+ (
1
2
4)
3
Now:
(
1)
•
(
1
2)
= (
1)
× (
2| 1)
= × (
•
(
1
3)
= (
1)
× (
3| 1)
= × (
•
(
1
4)
= (
1)
× (
4| 1)
= × (
•
(
2
3)
= (
2)
× (
3| 2)
= × (
•
(
2
4)
= (
2)
× (
4| 2)
= × (
•
(
3
4)
= (
3)
× (
4| 3)
= × (
•
(
1
2
(
3| 1
(
1
(
4| 1
(
1
(
4| 1
(
2
(
4| 2
(
1
(
3| 1
2)
(
4| 1
2
3)
= × × ×1 =
2
3
4)
= ×4
•
•
•
•
Thus: (
1
= (
3)
2)
2)
3)
3)
2
1
1
1
4
3
2
24
= (
2)
1
1
1
1
1
4
3
2
24
= (
3)
1
1
1
1
1
4
3
2
24
= × × =
4)
3
1
= × × =
4)
3
= (
3)
2
1
1
1
1
4
3
2
24
= × × =
3
4)
2)
1
= × × =
4)
2
= (
= (
1
•
4)
× (
= (
4| 1
=
4
1 4 1 4 1 4
1 4 1 4 1 4
1
1
1
4
3
12
1
1
1
4
3
12
1
1
1
4
3
12
1
1
1
4
3
12
1
1
1
4
3
12
1
1
1
4
3
12
2| 1)
= × =
3| 1)
= × =
4| 1)
= × =
3| 2)
= × =
4| 2)
= × =
4| 3)
= × =
× (
3| 1
2)
= (
1)
× (
2| 1)
×
× (
4| 1
2)
= (
1)
× (
2| 1)
×
× (
4| 1
3)
= (
1)
× (
3| 1)
×
× (
4| 2
3)
= (
2)
× (
3| 2)
×
3)
= (
1
× (
3| 1
1
3)
2
3)
2
= (
1
1
1
1
4
3
2
24
1
1
4
12
× (
×6+
1
24
1)
×4
Page 37 of 100
4| 1
× (
1
24
2
2| 1)
= 0.625
2)
2)
×
×
++++++++++++++++ END OF QUESTION PAPER ++++++++++++++++
Page 38 of 100
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING 1st MID-SEMESTER EXAMINATION [2012-2013, SEMESTER I] NOTE THE FOLLOWING 1) Total time for this paper is 120 minutes (2 hours). 2) Total marks is 100 and individual marks are mentioned alongside each question. 3) Total number of questions is 4 (with sub-parts) and you are required to answer ALL of them. 4) You are allowed to use the statistical tables and the calculator only. 5) Marks will be there for correct formulation of the problem, rather than only the final answer. Hence step wise marking is also there. 6) Draw diagrams and use the concept of set theory where ever necessary.
Question # 1 [10 + 15 (=25 marks)]
(a) The probabilities that Banait Chandan Ambadas will expend a high, medium or low amount of effort in studying for the MBA651 1st mid-semester examination are 50%, 30% and 10%, respectively. Given that Banait expends a high, medium or low amount of effort, the respective conditional probabilities of getting marks of 70 or above in MBA651 1st midsemester examination, are: 90%, 40% and 5% respectively. Find (i) the probability that Banait will get marks of 70 or above in MBA651 course and (ii) that provided he has obtained marks of 70 or above, then what is the probability that Banait spends medium effort for studying for MBA651 1st mid-semester examination.
(b) Consider that there are n persons in the MBA651 class. Then (i) what is the probability that at least two of you will have the same birthday?, (ii) calculate this probability for n = 52 and (iii) how large need n be for this probability to be greater than 0.5? Please do not consider leap year .
Answer # 1 (a) [10 marks]
The probabilities that Banait Chandan Ambadas will expend a high, medium or low amount of effort in studying for the MBA651 1st mid-semester examination are 50%, 30% and 10%, respectively. Given that Banait expends a high, medium or low amount of effort, the respective conditional probabilities of getting marks of 70 or above in MBA651 1st midsemester examination, are: 90%, 40% and 5% respectively. Find (i) the probability that Page 39 of 100
Banait will get marks of 70 or above in MBA651 course and (ii) that provided he has obtained marks of 70 or above, then what is the probability that Banait spends medium effort for studying for MBA651 1st mid-semester examination.
Given the problem let us define the events as follows H: Banait Chandan Ambadas expends high amount of effort M: Banait Chandan Ambadas expends medium amount of effort L: Banait Chandan Ambadas expends low amount of effort A: Banait Chandan Ambadas gets a mark of 70 or above AC: Banait Chandan Ambadas gets a mark of less than 70
( ) = 0.5,
(i) Thus from the data given we know the following:
∩ ∩ ∩ ∩ ∩ ∩
( ) = 0.3,
( ) = 0.1.
We also have ( | ) = 0.90. ( | ) = 0.40 and ( | ) = 0.05 Furthermore:
( )= (
)+ (
)+ (
), as H, M and L are mutually exclusive and
exhaustive events ( )=
(
)
( )
(
× ( )+
)
( )
× ( )+
(
)
( )
× ( )
( ) = ( | )× ( )+ ( | )× ( )+ ( | )× ( ) ( ) = 0.90 × 0.5 + 0.40 × 0.3 + 0.05 × 0.1 = 0.575
(ii)
Given
information
above
we
are
required
∩ ∩∩ ∩ ∩
( | )=
the
(
(
)
( )
Thus ( | ) =
=
)
(
(
)
× ( )
)+ (
= )
)+ (
0.40×0.3
0.90×0.5+0.40×0.3+0.05×0.1
( | )× ( )
to
find
.
( | )× ( )+ ( | )× ( )+ ( | )× ( )
= 0.2087
Answer # 1 (b) [15 marks]
Consider that there are n persons in the MBA651 class. Then (i) what is the probability that at least two of you will have the same birthday?, (ii) calculate this probability for n = 52 and
(iii) how large need n be for this probability to be greater than 0.5? Please do not consider leap year .
Each of you can have your birthday on any one of the 365 days (ignoring leap year). Let be the event that no two persons amongst you have the same birthday. To find this, order the Page 40 of 100
students who are in MBA651 course from 1 to
, such that you may choose a possible
sequence of length of birthdays each chosen as one of the 365 possible dates. There are 365 possibilities for the first element of the sequence, and for each of these choices there are 365
for the second, and so forth. Hence there are 365 number of total outcomes which are possible. But remember that we must find the number of these sequences that have no duplication of birthdays (i.e., no two persons amongst you have the same birthday). For such
a sequence, we can choose any of the 365 days for the first element, then any of the remaining 364 for the second, 363 for the third, and so forth, until we make choices. Thus for the
ℎ
choice, there will be 365
− −
+ 1 possibilities. Hence, the total number of
sequences with no duplications of birthdays (i.e., no two persons amongst you have the same birthday) is {365 × 364 × 363 × … . × (365
+ 1)}.
Assuming that each sequence is equally likely we have where
is the probability that we will have
duplication of their birthdays.
=
365×364×363×….×(365 365
−
+1)
,
number of MBA651 students who have no
(i) Using this, the probability that at least two of the students will have the same birthday is
− − ∗ − − −∗ − −∗
365×364×363×….×(365
given by 1
= 1
(ii) When
= 52, then 1
(iii) Let
=
, then 1
365
= 1
= 1
sheet calculations we find that 1
−
+1)
.
− −
− ≥ ∗
365×364×363×….×(365 52+1) 365 52
365×364×363×….×(365 365
+1)
=1
0.0219 = 0.978
0.5. Using simple Excel
= 0.524305, which means that
= 22.
Question # 2 [15 + 10 (=25 marks)]
(a) Consider three persons Praveen Srinivasan, Tulika Awasthi and Ruchik Sunilbhai Vin are playing a game where a wheel which has numbers marked 1, 2,….., 100 on its circumference is spun. When the wheel stops spinning, the number ′i′, which is closest to a certain pointer (which is fixed and kept next to the wheel) is said to be the outcome for that particular spin. Hence the outcomes are 1, 2,….., 100. Now if the outcome is less than or equal to 40, then Praveen Srinivasan gets Rs. 40 from Tulika Awasthi; if it is more than or equal to 61, then Ruchik Sunilbhai Vin gets Rs. 40 from Tulika Awasthi and if the outcome is anything else then Tulika Awasthi gets Rs. 60 each from Praveen Srinivasan and Ruchik Sunilbhai Vin
Page 41 of 100
respectively. The probability of any particular number being the outcome is uniform. Find the expected value (in Rs.) Tulika Awasthi wins/losses after 3 such games are played, if we consider a game to be the spinning of the wheel only once.
(b) Help Sunny Goyal with the following problem where he has been told to draw F(x), as well as f(x) on the same graph, where:
F ( x) = 0
if x < 0
= 1 − e −α ( x −θ )
if x ≥ 0.
Remember α > 0. Help Sunny calculate f(x) and also obtain P[a ≤ X ≤ b] for any given numbers a and b.
Answer # 2 (a) [15 marks]
Consider three persons Praveen Srinivasan , Tulika Awasthi and Ruchik Sunilbhai Vin are playing a game where a wheel which has numbers marked 1, 2,….., 100 on its circumference
is spun. When the wheel stops spinning, the number , which is closest to a certain pointer (which is fixed and kept next to the wheel) is said to be the outcome for that particular spin. Hence the outcomes are 1,2,…..,100. Now if the outcome is less than or equal to 40, then Praveen Srinivasan gets Rs. 40 from Tulika Awasthi ; if it is more than or equal to 61, then Ruchik Sunilbhai Vin gets Rs. 40 from Tulika Awasthi and if the outcome is anything else
then Tulika Awasthi gets Rs. 60 each from Praveen Srinivasan and Ruchik Sunilbhai Vin respectively. The probability of any particular number being the outcome is uniform. Find the expected value (in Rs.) Tulika Awasthi wins/losses after 3 such games are played, if we consider a game to be the spinning of the wheel only once.
Denote the three persons Praveen Srinivasan , Tulika Awasthi and Ruchik Sunilbhai Vin as PS, TA and RS respectively. Let us consider the problem from TA′s point of view. If the
−
random number denotes the amount of money won/lost by TA, then for = 0,1,2,3, which denotes the number of wins by TA in the three games played, we have Page 42 of 100
= {120
− − ∑ ∑ − − − − − − (3
−
)40}. Furthermore the probability of her losing is
her winning is
=
20
=
80
100
4
= , while the probability of 5
1
= . Thus when 3 games are played it is a simple example of a
100
5
binomial distribution where
= 3,
1
4
= and
= {120
= and
5
5
(3
)40}.
Hence
3 =0
( )= =
4 1 5
120 ×
+ 360 × =
3!×20 53
3 =0{120
()=
3! 3!
3! 3!
×
1 0
×
5
1 3 5
×
4 3 5
(3
)40}
+40×
3!
3
×
2!×1!
3
1 1 5
×
4 2 5
+200×
3!
1!×2!
×
1 2 5
×
4 0
×
5
{ 6 4 + 1 6 + 2 0 + 3} =
24.00
Answer # 2 (b) [10 marks]
Help Sunny Goyal with the following problem where he has been told to draw F(x), as well as f(x) on the same graph, where:
F ( x) = 0
if x < 0
= 1 − e −α ( x −θ )
if x ≥ 0.
Remember α > 0. Help Sunny calculate f(x) and also obtain P[a ≤ X ≤ b] for any given numbers a and b.
−− − − − −− − It is given that: ( ) = 0
(
=1
)
for
for
≥
<0 0
Thus:
( )= =
( )
( )
for
= 0
=
(
)
for
To draw the graph consider for simplicity form 1
and
≥
<0 0
= 0 and
and the graphs are as follows
Page 43 of 100
= 1. Thus the pdf and cdf are of the
0.7 0.6 0.5 0.4 f(x) 0.3
F(x)
0.2 0.1 0 1
3
5
7
9
11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41
≤ ≤ − − −− − To
find
(
)
[
(
)
],
we
≤ ≤ −− − − −−−≤ ≤ [
find
]=1
(
)
, which is also shown in the graph above. We can also find
]=
(
1
)
=
[
Question # 3 [(5 + 5) + 15 (=25 marks)]
(a) Help Tarunava Das solve the following problem, where the probability of closing of the ith relay in the circuits (Fig I and Fig. II), shown below, is given by p1=1/2, p2=1/3, p3=1/4, p4=1/5 and p5=1/6. If all relays function independently , then what is the probability that current flows between A and B for the respective circuits (Fig I and Fig. II).
Page 44 of 100
1
2
3 Fig I
A
B 4
5 1
A
4 Fig II
3 2
B 5
(b) Tanveer Singh Mahajan and his team and Minaz and her team are asked to separately design a new product within a month. From past experience we know that the probability that team Tanveer Singh Mahajan is successful is 2/3, the probability that team Minaz is successful is ½, and the probability that at least one team is successful is 3/4. Assuming that exactly one successful design is produced, what is the probability that it was designed by Minaz and her team?
Answer # 3 (a) [(5 + 5) marks]
Help Tarunava Das solve the following problem, where the probability of closing of the ith relay in the circuits (Fig I and Fig. II), shown below, is given by p1=1/2, p2=1/3, p3=1/4, p4=1/5 and p5=1/6. If all relays function independently , then what is the probability that current flows between A and B for the respective circuits (Fig I and Fig. II).
Page 45 of 100
1
2
3 Fig I
A
B 4
5 1
4
A
Fig II
3
B
2
∩ ∩
5
Consider Figure I
Let
, = 1,2,3,4,5 be the even when the
ℎ
is on and it connects the circuit. Hence current
can flow when we have 1,2,3 are closed, 4,5 are closed and 1,2,3,4,5 are closed, i.e., the events are: 1
3 =
2
1,2,3 are closed
1
2
3
A
B 4
∩ 4
5 =
5
4,5 are closed
1
2
3
A
B 4
∩ ∩ ∩ ∩ 1
2
3
4
5 =
5
1,2,3,4,5 are closed
Page 46 of 100
1
2
3
A
B 4
5
Thus current flows if
∩ ∩∩ ∩ ∪ ∩ ∩ ∩ ∩ − ∩ ∩ − − ℎ {(
3
1
2
4
5)
3)
(
4
5 )}
(
=
1
1
1
1
1
1
1
2
3
4
5
6
2
= × × + ×
=
3)
2
30
720
+
24
1
720
720
=
53
720
(
+
5)
4
1
1
1
1
3
4
5
6
× × × ×
(
1
2
= 0.073611
Consider Figure II
Let
, = 1,2,3,4,5 be the even when the
is on and it connects the circuit. Hence current
can flow when we have the following scenarios
Answer # 3 (b) [15 marks] Tanveer Singh Mahajan and his team and Minaz and her team are asked to separately design
a new product within a month. From past experience we know that the probability that team Tanveer Singh Mahajan is successful is 2/3, the probability that team Minaz is successful is
½, and the probability that at least one team is successful is 3/4. Assuming that exactly one successful design is produced, what is the probability that it was designed by Minaz and her team?
Question # 4 [10 + 15 (=25 marks)]
(a) Sagar Bhardwaj selects a number at random from (1, 2, …. , 100). Given that the number selected by him is divisible by 2, help him to find the probability that it is also divisible by 3 or 5.
(b) Suppose Vinayak Gangwar randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of him getting exactly 2 red cards (i.e., hearts or diamonds)?
Answer # 4 (a)[10 marks]
Page 47 of 100
Sagar Bhardwaj selects a number at random from (1, 2, …. , 100). Given that the number
selected by him is divisible by 2, help him to find the probability that it is also divisible by 3 or 5.
Answer # 4 (b) [15 marks]
Suppose Vinayak Gangwar randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of him getting exactly 2 red cards (i.e., hearts or diamonds)?
This is a hypergeometric experiment in which we know the following: N = 52; since there are 52 cards in a deck k = 26; since there are 26 red cards in a deck. n = 5; since we randomly select 5 cards from the deck. x = 2; since 2 of the cards we select are red.
We plug these values into the hyper-geometric formula as follows which is given as
−− ×
,
= 0,1, … . , . Thus we have
( )=
26 2
× 26 3
52 5
= 0.32513.
++++++++++++++++ END OF QUESTION PAPER ++++++++++++++++
Page 48 of 100
( )=
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING FINAL SEMESTER EXAMINATION [2012-2013, SEMESTER I]
NOTE THE FOLLOWING 1) Total time for this paper (which consists of 3 pages) is 180 minutes (3 hours). 2) Total marks is 100 and individual marks are mentioned alongside each question. 3) Total number of questions is 4 (with sub-parts) and you are required to answer ALL of them. 4) You are allowed to use the statistical tables, formulae sheet and the calculator only. 5) Marks will be there for correct formulation of the problem, rather than only the final answer. Hence step wise marking is also there. 6) Draw diagrams accurately/neatly/legibly, and use the concept of set theory where ever necessary. 7) Remember to write your formulations and do your calculations legibly and clearly.
Question # 1(a): [15 marks]
(a) There are n objects marked 1, 2,…, n and also n places marked 1, 2,…, n. The objects are distributed over these n places, such that one object is allotted to each place. Find the probability that none of the objects occupies the place corresponding to itself, i.e., ith object is not in the ith place. Also find the value when n is very large.
Answer # 1(a): [15 marks]
Let us define Ai the event that the object numbered i occupies the place numbered i, i = 1, 2,…., n. It is quite intuitive and obvious to note that Ai′s, i = 1, 2,…., n are not mutually exclusive events. Now these n objects may distributed over these n places in a total of n! (i.e., n
Pn) ways, which may be assumed to be equally likely. So now we have Ai occurs when the
th
th
i place is occupied by the i object, while the remaining (n-1) places is to be occupied by the
remaining (n-1) objects in any arbitrary order. Hence this total number of ways is (n-1)! (i.e., n-1
Pn-1), hence
(
)=
−
(
1)!
!
=
1
for each i. Again both Ai and A j ( i≠ j) occur when the ith
Page 49 of 100
and the jth places are occupied by the corresponding objects, while the other (n-2) places are
⋂
filled up with the remaining (n-2) objects, so now we have
=
− −
(
2)!
!
1
=
(
,
1)
for each i and j, where i< j.
⋂ ⋂ ⋂ ⋃ ⋃ ⋃ − − − − −⋯ − − − − − ⋃ ⋃ ⋃ − −⋯ − − − ⋃ ⋃ ⋃ − − − ⋯ − − − ⋃ ⋃ ⋃ − ≈ (
In the same logic we have:
1
2
……
2
×
)=
1
!
. Now using the theorem of general
probability we have: (
1
2
)=
……
1
×
1
1
(
1)
+
3
1
×
(
1)(
… . +( 1)(
2)
1)
1
× , which implies we have n number of terms in the above expansion, where !
which has =
(
1
!
1
1
the
required
(
1
2
1)
×
1
!
1
=
)=
……
… . +( 1)(
1
2
which has ( 1)( 2
derive
terms,
which has
(
1)
1)
1
1
1!
2!
+
1
1
number of terms and so on till the last term
… . +( 1)(
3!
Just
as:
note
(
1
that
) = exp( 1)
……
=
number of terms, hence we have:
probability
.
2
1
×
1)
1
× . From which we can easily !
1
2
when
n
…… is
)=1
very
large
1
1
1!
2!
we
+
1
3!
have
0.36788
Note: In case one wants to find the probability that that at least one of the objects occupies
the place corresponding to itself, i.e., ith object is in the ith place then the corresponding
⋃ ⋃ ⋃ − − ≈
probability is given by (
1
2
……
)=1
exp( 1)
0.63212
Question # 1(b): [10 marks]
For the above problem, i.e., # 1 (a), find the probability that there is exactly m number of such matches when we place these n number of objects in these n number of places?
Answer # 1(b): [10 marks]
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First consider the probability that m specified events, marked A1, A2,…, Am (here Ai are defined as we have done for solution # 1(a)) occur while the others do not. Let us furthermore
⋂ ⋂ ⋂ − −∑ − ∑ ∩ ⋯ ⋂ ⋃ −⋃ −−− ∩ −−− −− −−− − −−− −⋯ ∩ − −− −− − −− −⋯ − −− − − − − − −⋯ − −⋯ − − −
define
=
and
=1
given by: (
=
then the probability we are interested to find is
= +1
) = ( ) × ( | ).
Now we know and can also find if required that ( ) = ( | )=1
Now
=
=
= +1
( | )=1
But:
|
|
= +1
=
(
!
, while
+
..
, which would immediately lead us to the fact that:
.
|
1)!
(
Thus:
)!
|
, = +1
= +1
= +1
(
|
, for j = m+1, m+2,…, n
)!
=
(
2)!
(
)!
for j,k = m+1, m+2,…, n and so on.
Hence:
( | )=1
1
(
×
1)!
(
+
)!
×
2
(
2)!
(
)!
Which leads us to: (
)=
1
!
(
)!
1
×(
1)! +
Hence the required probability is the sum of group of m events may be any one of
2
×(
2)!
such probabilities, and since the specified
groups formed.
So we finally have: ×
=
1
!
1
1
!
(
1
1!
)!
+
1
2!
1
×(
+ ( 1)(
1)! +
)
×
2
×(
2)!
1
(
)!
Question # 2(a): [15 marks] Gangaredgy Papaigari & Company manufactures bars of steel and the company claims that
the average breaking strength of its product is not less than 52. Nishu Navneet is the Page 51 of 100
Controller General of quality and to test the breaking strength of this product he samples 15 such bars from the factory of Gangaredgy Papaigari & Company and notes down the breaking strength. Those values are as given as: 51.3 52.1 50.3 50.2 51.9 50.0 52.5 50.7 49.3 49.3 48.3 48.1 48.2 47.8 47.5. Examine if Gangaredgy Papaigari & Company′ s claim is supported by these data.
Answer # 2(a): [15 marks]
As per the information provided in the problem the hypothesis will be framed keeping in mind that we will always try to disapprove the statement made by the person, hence we always try to reject
≥ 0,
which results in the following formulation, i.e.,
≥ 0:
=
0
52. If
that is the case, then we should have the following hypothesis to test, which is: 0:
=
0
vs
52
:
=
sn , is true then we reject n
i.e., if the following, X n < µ 0 − t n −1,α
< 52
0
Now for the sample (here the sample size is 15) given we can easily calculate that
− ∑ −� ∑ − − −− √ − √ 1
=1
= 49.8333 and
= 0.05, i.e., (1
Hence the RHS is
2
=(
1
1)
=1(
1,
×
= 52
=
)2 = 2.7124. For the problem consider
) = 0.95. Hence from the table we have
0
�
1.761 ×
1.6469 15
always less than 51.25117, and as this is true we reject
1,
=
14,0.05
= 1.761.
= 51.25117. Now 49.833 is 0,
i.e., the claim of the
manufacturer Gangaredgy Papaigari & Company is false.
Question # 2(b): [10 marks]
Three plants hereby marked C1, C2 and C3 produce respectively 10, 50 and 40 percent of a company′s output. Although plant C1 is small, Rajat Bhakri who is the general manager and is in charge of all these three plants, firmly believes that this plant produces good quality of products and on an average only 1% of the corresponding products produced are defective. While for plant C2, 97 percent of the goods produced are of good quality, and for plant C3 the percent of bad products produced is 4%. All the products from these three plants are Page 52 of 100
dispatched to the central warehouse for final inspection and subsequent dispatch. Find the probability that if an item or product is picked up at random that it will be defective. Also calculate that if the item is found to be defective, then what is the conditional probability that it was produced in plant C2.
Answer # 2(b): [10 marks]
For this problem we use Bayes theorem. Before that let us define the events as follows:
1=
product is produced in plants # 1
2=
product is produced in plants # 2
3=
product is produced in plants # 3
D = product is defective and it can be produced in one of the three plants Thus with this notational concept we derive the following, given the information from the problem:
∩ ∩ ∩ (
1)
( |
= 0.1, (
1)
2)
= 0.5, (
= 0.01, ( |
2)
= 0.4
= 0.03, ( |
( )= (
Now
3)
1)
+ (
3)
2)
= 0.04
+ (
3)
= 0.1 × 0.01 + 0.5 × 0.03 + 0.4 ×
0.04 = 0.032
Now using Bayes Theorem we have
(
2|
)=
( 2 | )× ( 2 ) ( 1 | )× ( 1 )+ ( 2 | )× ( 2 )+ ( 3 | )× ( 3 )
=
0.5×0.03 0.1×0.01+0.5×0.03+0.4×0.04
= 0.46875
Question # 3(a): [15 marks] x
Consider the logarithmic distribution with the probability mass function (p.m.f), f ( x) = , for x = 1, 2, …….., where 0 < θ < 1. Show that, for this distribution the mean (µ) =
− µ . (1 − θ )
and variance (σ2) = µ
1
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C θ x
C θ
(1 − θ )
∑∞ ∑∞ ⋯ ⋯ ⋯⋯ − − − ∎ ∑∞ − ∑∞ ⋯ −− − − − − − − − ∎
Answer # 3(a): [15 marks]
(i) ( ) =
( )=
=1
= ( +
=1
2
..) =
+
(1 +
+
2
+
..)
Now let us consider +
2
+
..
+
2
+
..
=1+
1
1
=
Hence:
1 (1
) = 1, i.e.,
2)
(ii) ( ) = ( 2)
(
=
=(
1
. Substituting we have ( ) = (
)
1
1
{ ( )}2 . Hence first let us calculate (
2
=1
1
( )=
= ( +2
=1
2
+3
3
2)
)
.
…..) =
(1 + 2 + 3
2
+
..)
Now let us consider 2
2
+4
3
…...
2
+3
3
… ..
=1+2 +3
2
=
Hence:
2)
(
+2
2 (1
=
) = 1+
(1
(1
+
2
+
3
=
)
)2
, such that ( ) =
1
=(
1
2
(1
)
1
=
. Hence )
2
=
1
(1
)2
1
1
Question # 3(b): [10 marks]
Consider the two sided confidence interval for the mean µ when σ is known, which is given
by xn − zα 1
σ
n
≤ µ ≤ xn + zα
2
σ
, where α1+α2=α. If α1=α2=α/2, then we have the usual
n
100(1-α)% confidence interval (C.I.) for µ. In the above formulation when α1≠α2 the interval
σ ( zα + zα ) . Prove that the n
is not symmetric about µ and the length of the interval is L =
1
2
length of the interval L is minimized when α1=α2=α/2.
Answer # 3(b): [10 marks]
Given σ is known we have the confidence interval as given
√ √ 2
=
where α 1 = α 2 = . Hence the actual C.I length, is
1
+
2
2
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̅ − √ ≤ ≤ ̅ − 1
Now to find the minimum length we must have
=
1
√ √ √ 1
1
Thus:
=
1
×1+
+
(remember we could have also differentiated wrt to
2
2
×
= 0, i.e.,
1
2 1
=
2
−
1
(1) Also remember that:
≤ ≤ ∫ − 1
1
2
×
2
=
2 1
( )
1
2
(2)
= 0
Using both the equations (1) and (2) we have
distribution is symmetric. Moreover it is given that
1
, i.e.,
1
=
2
+
2
= , hence
1
1
=
2 as
=
2
=
normal
2
Question # 4(a): [15 marks] Harpreet Singh & Associates produces both interior and exterior paints from two raw
materials M1 and M2. The following table provides the basic data for the problem. Raw Material
Tons of raw material per ton of Exterior Paint Interior Paint 6 4 1 2 4 ton 5
Maximum daily availability (tons) 24 6
M1 M2 Profit per (× ) A marketing survey has revealed that the daily demand for interior paint cannot exceed that
of exterior paint by more than 1 ton, also the maximum daily demand of interior paint is 2 tons. Harpreet Singh & Associates had attended the MBA651 class rarely, hence is unable to solve the problem. Help him to solve this optimization problem using simple method (tableau based solution only). Please note no graphical method solution technique will be considered.
Answer # 4(a): [15 marks]
The starting tableau is as give Step # 1
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Basic
z
x1
x2
s1
s2
s3
s4
Solution
Z
1
-5
-4
0
0
0
0
0
s1
0
6
4
1
0
0
0
24
s2
0
1
2
0
1
0
0
6
s3
0
-1
1
0
0
1
0
1
s4
0
0
1
0
0
0
1
2
Thus the calculations yields
−−
− − − − 2 1
2 5
3 6
3 6
•
(1, 5, 4,0,0,0,0,0)— ( 5) × 0,1, , , 0,0,0,4 = 1,0,
•
(0,6,4,1,0,0,0,24) × = 0,1, , , 0,0,0,4
•
(0,1,2,0,1,0,0,6)— 1 × 0,1, , , 0,0,0,4 = 0,0, ,
•
(0, 1,1,0,0,1,0,1)— ( 1) × 0,1, , , 0,0,0,4 = 0,0, , , 0,1,0,5
•
(0,0,1,0,0,0,1,2)— (0) × 0,1, , , 0,0,0,4 = (0,0,1,0,0,0,1,2)
1 6
−
, , 0,0,0,20
2 1 3 6
2 1
4
1
3 6
3
6
, 1,0,0,2
2 1
5 1
3 6
3 6
2 1 3 6
These calculations would be used for the 2nd tableau which is given below
Step # 2 Basic
z
x1
x2
s1
s2
s3
s4
Solution
Z
1
0
-2/3
5/6
0
0
0
20
x1
0
1
2/3
1/6
0
0
0
4
s2
0
0
4/3
-1/6
1
0
0
2
s3
0
0
5/3
1/6
0
1
0
5
s4
0
0
1
0
0
0
1
2
Thus the calculations yields • • • • •
− −− − − − − − − − − 1,0,
2 5
2
3 6
3
, , 0,0,0,20
2 1
2
3 6
3
0,1, , , 0,0,0,4 4
1
3
6
0,0, ,
× 0,0,1,
× 0,0,1,
4
8 4
5
3 6
3
(0,0,1,0,0,0,1,2)
8 4
1 3
5 1
× 0,0,1,
(1) × 0,0,1,
1 3
3
8 4
2
, , 0,0,
, , 0,0,
3
, 1,0,0,2 × = 0,0,1,
0,0, , , 0,1,0,5
1 3
−
, , 0,0,
1 3
, , 0,0,
8 4
1 3
, ,0,0,
8 4
3 2
3 1
1
1
4
2
3
5
8
4
= 0,1,0, ,
2
2
3 2
= 0,0,0, , 1
= 0,0,0, ,
These would be used for the 3rd tableau which is given below
Page 56 of 100
− − − 4 2
3
3
= 1,0,0, , , 0,0,21
8
3 4
, 0,0,
, 1,0
,0,1,
1 2
5 2
Step # 3 Basic
z
x1
x2
s1
s2
s3
s4
Solution
z
1
0
0
¾
½
0
0
21
x1
0
1
1
¼
-½
0
0
3
x2
0
0
1
-1/8
¾
0
0
3/2
s3
0
0
0
3/8
-5/4
1
0
5/2
s4
0
0
0
1/8
-3/4
0
1
½
Hence the answer is Maximum value is 21,
1
= 3,
3
2
= , 2
1
= 0,
2
= 0,
5
3
= , 2
4
=
1 2
Question # 4(b): [10 marks] Ankita Hemant Sahare Industries, a nationally known manufacturer of clothing, produces
four varieties of ties. One is an expensive, all-silk tie, one is an all-polyester tie, and two are
blends of polyester and cotton. The following table illustrates the cost and availability (per monthly production planning period) of the three materials used in the production process: Material
Cost/meter (XYZ units)
Material
Available/Monthly
(meters)
Silk
21
800
Polyester
6
3,000
Cotton
9
1,600
The firm has some fixed contracts with several major department store chains to supply ties. The contracts require that Ankita Hemant Sahare Industries supply a minimum quality of each tie but allow for a larger demand if Ankita Hemant Sahare Industries chooses to meet that demand, subject to a condition that it should not exceed the maximum requirement. The following table summarizes the contract demand for each of the four styles of ties, the selling price per tie, and the fabric requirements of each variety. Variety
of S.P. per tie
Tie
(XYZ units)
Monthly
Monthly
Material
Type
of
contract
contract
requirement
Material
(Minimum)
(Maximum)
per tie
requirement
All Silk
6.70
6,000
7,000
0.125
100% silk
All polyester
3.55
10,000
14,000
0.080
100% ploy
cotton 4.31
13,000
16,000
0.100
50%
Poly
Page 57 of 100
poly
Blend 1
and
50%
cotton Poly
cotton 4.81
6,000
8,500
0.100
Blend 2
30%
poly
and
70%
cotton Formulate the optimization problem for Ankita Hemant Sahare Industries. You are not required to solve it .
Answer # 4(b): [10 marks]
Let us define the following •
x1= number of all silk ties produced per month
•
x2= number of polyester ties produced per month
•
x3= number of blend 1 poly-cotton ties produced per month
•
x4= number of blend 2 poly-cotton ties produced per month
First the firm must establish the profit per tie
For all silk ties (x1), each requires 0.125 yard of silk, at a cost of 21/yard. Therefore, the
•
cost per tie is 2.62. The selling price per silk tie is 6.70, leaving a net profit of (6.702.62)=4.08 per unit of x1. For all polyester ties (x2), each requires 0.080 yard of polyester, at a cost of 6/yard.
•
Therefore, the cost per tie is 0.48. The selling price per polyester tie is 3.55, leaving a net profit of (3.55-0.48)=3.07 per unit of x2. For all blend 1 poly cotton ties (x3), each requires 0.050 yard of polyester, at a cost of
•
6/yard and 0.050 yard of cotton, at a cost of 9/yard. Therefore, the cost per tie is 0.75. The selling price per polyester tie is 4.31, leaving a net profit of (4.31-0.75)=3.56 per unit of x3. For all blend 2 poly cotton ties (x4), each requires 0.030 yard of polyester, at a cost of
•
6/yard and 0.070 yard of cotton, at a cost of 9/yard. Therefore, the cost per tie is 0.81. The selling price per polyester tie is 4.81, leaving a net profit of (4.81-0.81)=4.00 per unit of x4.
Hence the optimization problem is as follows
≤ : 4.08
. .:
1
0.125
+ 3.07 1
2
+ 3.56
3
+ 4.00
4
800
Page 58 of 100
≤ ≤ ≤≤ ≤≤ ≤≤ ≤≤
0.08
2
+ 0.05
3
0.05
3
+ 0.07
4
6000
+ 0.03
7000
1
2
14000
13000
3
16000
4
3000
1600
10000
6000
4
8500
++++++++++++++++ END OF QUESTION PAPER ++++++++++++++++
Page 59 of 100
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING FIRST EXAMINATION [2013-2014, SEMESTER I]
Question # 1 (a): [5 marks]
Given the data: 23, 45, 65, 78, 76, 23, 9, 12. Find the mean and standard deviation
Answer # 1 (a): [5 marks]
The mean is given by
∑ − ∑ −− ( )=
1
1
=1
= (2 3 + 4 5 + 6 5 + 7 8 + 7 6 + 2 3 + 9 + 2) = 40.125 8
The variance is given by ( )=
(78
1
1
( )}2 = {(23
=1{
40.125)2 + (76
8
−−
40.125)2 + (45
40.125)2 + (23
40.125)2 } = 796.6094
−−
40.125)2 + (9
40.125)2 + (65
−−
40.125)2 +
40.125)2 + (2
Question # 1 (b): [10 marks]
You toss an unbiased coin repeatedly till you get head (H) the first time. Then draw the probability mass function as well as the distribution function (in the same graph) for this problem.
Answer # 1 (b): [10 marks]
As discussed and also derived in class the distribution is called the Geometric Distribution
− ⋯ − ⋯ such that = (1
( )=
~ ( ), where
is the probability of success of getting the head, H, while
) is the corresponding probability of getting the tail, T. Hence , where
= 0,1,
is he randon variable denoting the case where we are
interested to find the number of tails coming before the 1st head. The corresponding ( ) = 1
+1
. Here
=
= 0.5. With
= 0,
,20 (as an example) the pmf and cdf are shown
below
Page 60 of 100
1 0.8 0.6 f(x) 0.4
F(x)
0.2 0 0
5
10
15
20
25
Question # 1 (c): [10 marks]
The probability density function (p.d.f) of he random variable X is given by
( )=
ℎ
√
0 <
0
<4 . Given this find (i) P(X<1/4) and (ii) draw the p.d.f
and c.d.f clearly and legibly on the same graph.
Answer # 1 (c): [10 marks]
∫ √
The pdf is ( ) = have
4 1
0
√
and considering the fundamental properties of distribution function we
= 1, hence
1
= . Now he value of 4
pdf and cdf graphs are as shown below
≤ ∫ √ 1 4
=
1
1 4
1
4 0
1
= . Finally the 4
1 0.9 0.8 0.7 0.6 0.5
f(x)
0.4
F(x)
0.3 0.2 0.1 0 1
3
5
7
9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41
Question # 2 (a): [15 marks]
Page 61 of 100
A and B play a game in which they alternatively toss a pair of dice. The one who is first to get a total of 7 wins the game. Find the probability that (i) the one who tosses first will win the same, (ii) the one who tosses second will win the game. Answer # 2 (a): [15 marks]
(i) The probability of getting a number 7 on a single toss of a pair of dices assumed to be fair) 1
is given by . Remember 7 can be formed by the following combinations which are: (1,6), 6
(2,5), (3,4), (4,3), (,2) and (6,1). If we suppose that A is the first to toss then A will win in any one of the following mutually cases with the following indicated associated probabilities as shown below Case 1: A wins he 1st toss and probability is
1 6 5
5
1
6
6
6
Case 2: A loses on 1st toss, then B loses and then A wins and probability is × ×
Case 3: A loses on 1st toss, then B loses, then again A loses followed by B′s loss and finally 5
5
5
5
1
6
6
6
6
6
A wins and probability is × × × × and so on as it goes till infinity.
1
5
5
1
5
5
5
5
1
6
6
6
6
6
6
6
6
6
Then the probability that A wins is given by + × × + × × × × +
⋯
=
6 11
.
(ii) Now consider B wins (which is to do with the part (ii) of the problem), then the 5
1
5
5
5
6
6
6
6
6
⋯ ∪ ∪ ∪ ≤ 1
5
5
5
5
5
1
6
6
6
6
6
6
6
corresponding probability is: × + × × × + × × × × × +
Question # 2 (b): [15 marks]
Prove for any set of events A1, A2,….., Ak ,
⋯ …+ (
(
1
2
)
…..
=
(
5
11
1)
+ (
2)
+
). Hint: You cannot use Venn diagram or any illustrative/diagrammatic method.
Answer # 2 (b): [15 marks]
Use method of induction and he schematic diagram is shown below First let us illustrate this with the simple Vein diagram for the ease of understanding, but the proof will be given mathematically
Ω
Ai A1
A2 An
A4 An-1
Page 62 of 100
Consider there are only two events, i = 1, 2, namely A1 and A2. In that case we can write A1 ∪ A2 = A1 + ( A2 − A1 ) , where events A1 and ( A2 − A1 ) both belong to A and they are disjoint. Thus we would have: P ( A1 ∪ A2 ) = P ( A1 ) + P ( A2 − A1 ) − P[ A1 ∩ ( A2 − A1 )] .
Hence P ( A1 ∪ A2 ) = P ( A1 ) + P ( A2 − A1 ) ≤ P ( A1 ) + P( A2 ) , since ( A2 − A1 ) ⊂ A2 for the fact that the probability function P is monotone (increasing). Or better still for a clear understanding we can state that 0 ≤ P[ A1 ∩ ( A2 − A1 )] ≤ 1 always holds, hence P ( A1 ∪ A2 ) ≤ P ( A1 ) + P ( A2 ) Now considering the number of events is n, i.e., the events can now be denoted as A1, A2,…., An such that one can write A1 ∪ A2 ∪ ∪ An = ( A1 ∪ A2 ∪ ∪ An−1 ) ∪ An , i.e.,
= ( A1 ∪ A2 ∪ ∪ An−1 ) + { An − ( A1 ∪ A2 ∪ ∪ An−1 )} . Here again we divide
the whole set of events in to separate class of events one being ( A1 ∪ A2 ∪ ∪ An−1 ) and the other being { An − ( A1 ∪ A2 ∪ ∪ An −1 )}. Thus: P( A1 ∪ A2 ∪ ∪ An ) = P ( A1 ∪ A2 ∪ ∪ An −1 ) + P{ An − ( A1 ∪ A2 ∪ ∪ An−1 )}
− P[( A1 ∪ A2 ∪ ∪ An−1 ) ∩ { An − ( A1 ∪ A2 ∪ ∪ An−1 )}] ≤ P( A1 ∪ A2 ∪ ∪ An −1 ) + P{ An − ( A1 ∪ A2 ∪ ∪ An −1 )}
( A1 ∪ A2 ∪ ∪ An−1 ) and { An − ( A1 ∪ A2 ∪ ∪ An−1 )} are disjoint (same logic can be used
here as for i = 1, 2) Hence using similar logic we obtain: P( A1 ∪ A2 ∪ ∪ An ) ≤ P( A1 ∪ A2 ∪ ∪ An−1 ) + P( An )
Similarly we would have Page 63 of 100
P( A1 ∪ A2 ∪ ∪ An ) ≤ P ( A1 ∪ A2 ∪ ∪ An − 2 ) + P ( An −1 ) + P ( An ) P( A1 ∪ A2 ∪ ∪ An ) ≤ P( A1 ∪ A2 ∪ ∪ An−3 ) + P( An −2 ) + P( An−1 ) + P( An ) P( A1 ∪ A2 ∪ ∪ An ) ≤ P ( A1 ∪ A2 ∪ ∪ An −4 ) + P( An−3 ) + P ( An−2 ) + P ( An−1 ) + P( An )
. . n P( A1 ∪ A2 ∪ ∪ An ) ≤ P ( A1 ) + P( A2 ) + + P( An ) , i.e., P Ai ≤ ∑ P( Ai ) i=1 i =1 n
Question # 3: [20 marks]
(i) Find the probability that different given birthdays.
≤ (
365) number of persons selected at random will have
∎
(ii) Determine how many people are required in problem # 3(a) to make the probability of distinct birthdays less than ½.
Answer # 3: [20 marks]
(i) We assume that there are only 365 days in a year and that all birthdays are equally probable (this assumption is not true in reality). Then the first of the people has of course some birthday with probability
365
= 1. If the second person is to have a different birthday
ℎ− ℎ− − ⋯ − − − ⋯ − − − ⋯ − − − − − − −⋯ − − ⋯ − −− − ⋯ − −⋯ − −− −⋯ then it is
364 365
365
. Using this logic the
associated probability is given by
person has a birthday different from the others and the
(365
(
1
+1)
)=
2
365
×
. Thus we have the following
365
365 365
×
365 1 365
×
×
(365
+1)
365
= 1
1
365
×
1
× 1
365
(ii) Denoting the given probability by and taking the natural logarithms we find =
1
1
365
+
1
2
365
+
+
Now we are aware from simple calculus that
1
1
365
(1
we have: = =
1+2+ +(
1)
365
(
1)
730
1 12 +22 + +( 365 2
2
(
1)(2
1)2
1)
12×365 2
Page 64 of 100
)=
2
2
3
3
. Utilizing this
For
≪
If now
− − − − − −− −
365 we neglect the higher terms, hence 1
=
(
1)
730
= (as per the information give) then we get = 2
Solving the quadratic equation in we get, from which we get
2
.
(
1)
730
506 = 0. Thus (
1
=
2
=
0.693.
23)( + 22) = 0
= 23. With this one can conclude that if is larger than 23 then we can
give better than even odds that at least two people will have the same birthday.
++++++++++++++++ END OF QUESTION PAPER ++++++++++++++++
Page 65 of 100
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING FINAL SEMESTER EXAMINATION [2013-2014, SEMESTER I]
Question # 1: [15 + 15 (=30) marks]
(a) As a detective you are searching for clue and it can be presumed that it is equally likely that the crime has been committed by any one of the three persons you are investigating. Let (1
ℎ
−
) denote the probability that the crime has been committed by the
ℎ
person where
= 1,2,3. Then what is the conditional probability that the crime has been committed by the
person, given that investigating, person 1 did not prove fruitful, = 1,2,3?
(b) Suppose
(
is a random variable and its probability mass function is
1
= 2) = and
(
3
1
= 3) = . Given this information first find 6
graph draw both pmf as well as cdf.
Answer # 1: [15 + 15 (=30) marks]
(a) Let
, = 1,2,3, be the event that the
ℎ
(
1
= 1) = , 2
( ) and on the same
person has committed the crime. Moreover let
be the event that after investigating person you pass the verdict that the person is not
guilty. From Bayes’ formula: we obtain
1 1 3 1 1 1 3 +1 3 1 3 1 1 1 3 +1 3
+1
1 3
+1
1 3
1
(b) Given
=
=
(
1
,
1 +2
while
for
(
= 2,3 we
1|
∩
)=
have:
( 1 ) ( )
=
=
∑ ∑ 3 =1
3 =1
( | 1) ( 1 ) ( | ) ( )
( |
)
1
1 +2
1
= 1) = , 2
(
1
= 2) = and 3
(
1
= 3) = the cdf is given below:
Page 66 of 100
6
(
)
=
=
≤ ≤ ≤ ≤ ≤ 0 1
( )=
2 5 6
<1
1
2
2
3
1
. Utilizing this we draw the pmf and cdf as give:
3
1.2 1 0.8 0.6
f(x)
0.4
F(x)
0.2 0 1
2
3
− ℎ
Question # 2: [15 + 15 (=30) marks]
(a) Suppose that
is a continuous random variable whose probability density function is
given by ( )=
(4
2
2)
0 <
<2
0
. Find
( ),
( ) and draw the pdf and cdf on the same
graph.
(b) The time, in hours, it takes to locate and repair an electrical breakdown in a certain
factory is a random variable, say , whose density function is given by
ℎ ( )=
1 0< 0
<1 . If the cost involved for a breakdown of duration is given by
3
,
then what is the expected value for such a breakdown?
Answer # 2: [15 + 15 (=30) marks]
∫ − ∫−∞∞ ∫ −
(a) Form the properties of distribution function we must have: ( ) = 1. From this we obtain
3
= . Now by formulae: (i) 8
Page 67 of 100
( )=
+
2
0
(4
( )
2
=
3
2)
2
8 0
(4
=
2
2 1
3)
= 1, while (ii)
∫−∞∞ − ∫ − − ( )=
+
( )}2 ( )
{
=
2
3
8 0
1)2 (4
(
2
2
3)
=
5
1 0.9 0.8 0.7 0.6 0.5
f(x)
0.4
F(x)
0.3 0.2 0.1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
∫ − ≤ ≤ ∫−∞∞ ∫ − ∫ (b) Letting
=
3
denote the cost, so we first calculate its distribution function (remember it
is the cdf and NOT pdf/pmf) which is as follows: 1 3
=
0
1 3
2 3
for 0
1 3
. Let us now differentiate 1. Thus: ( ) =
( )=
(
3
< )=
<
1 3
=
( ) with respect to to obtain ( ). Thus ( ) =
+
( )
=
1
1
3 0
2 3
=
1
1
3 0
1 3
=
1 4
Question # 3: [20 + 10 + 10(=40) marks]
(a) Civil engineers believe that
, the amount of weight (in units of 1,000 kgs) that a certain
span of a bridge can withstand without structural damage resulting, is normally distributed with mean 400 and standard deviation 40. Suppose that the weight (again, in units of 1,000 kgs) of a car is a random variable with mean 3 and standard deviation .3. How many cars would have to be on the bridge span for the probability of structural damage to exceed 0.1?
(b) The manufacturer of a new fiberglass tyre claims that its average life will be at least 40,000 kms. To verify this claim a sample of 12 tyres is tested, with their lifetimes (in 1,000s
Page 68 of 100
of kms) being as follows: 36.1,40.2,33.8,38.5,42.0,35.8,37.0,41.0,36.8,37.2,33.0 and 36.0. Test the manufacturer’s claim at the 5 percent level of significance.
(c) A sample of 100 transistors is randomly chosen from a large batch and tested to determine if they meet the current standards. If 80 of them meet the standards, then formulate an
approximate 95 percent confidence interval for , the fraction of all the transistors.
⋯ ≥ ℎ ⋯ ∑ ∑ ∑ − − − ∑ − ∑ ⋯− − ≥ ⋯√ −− − ≥ √ − − √ ≤ ≥ Answer # 3: [20 + 10 + 10(=40) marks]
(a) Let
denote the probability of structural damage when there are n cars on the bridge. {
That is
1
+
}, where
+
is the weight of the
follows from the central limit theorem that , hence
=1(
car,
~ (3 , 0.09 ). As
=1
= 1,
, . Now it
is independent of
) is also normal with certain mean and variance.
{
=1(
)} = 3
{
=1(
)} =
400 and
(
)+ (
=1
) = 0.09 + 1600
Utilizing these in the concept of standard normal distribution we obtain: {
1
Hence
+
+
400 3
0.09 +1600
1+
0} =
(3
+
400 )
0.09 +1600
1.28, thus
0 (3
400 )
0.09 +1600
. Thus
= +1.28.
117
(b) To determine whether the foregoing data are consistent with the hypothesis that the mean life is at least 40,000 km, we will test he following hypothesis
� ≥∑ ∑ −� − − √ √ − − − :
4000
1
:
< 4000
1
12 =1{
12 }
=
7.4633. More over we need to use t-distribution and from the table we have
11,0.05
=
From the data we have:
−−
12
=
12
12 =1
vs
= 37.2883, while
2
=
12 1
1.796. Furthermore from the values he t value comes ou to be 3.4448. Since this value of t is less than
the null hypothesis,
11,0.05
=
12(37.2883 40) 7.4633
=
1.796 which gives us the answer that
, is rejected at the 5 percent level of significance
Page 69 of 100
=
− ≤ − − ≤ − ̂ ̂ − − ≤ ≤ ̂
(c) We have the formulae as given:
2
simplified considering the fact that
− − − (1
)
= (1
2
0.8
1.96
0.8×0.2 100
(1
)
= (1
), which ones
2
(1
= yields us
)
+
2
).
Putting
, 0.8 + 1.96
0.8×0.2 100
the
values
gives
= (0.7216,0.878).
++++++++++++++++++++END OF QUESTION PAPER++++++++++++++++++++
Page 70 of 100
us:
MBA651: QUANTITATIVE METHODS FOR DECISION MAKING 1st QUIZ EXAMINATION [2014-2015, SEMESTER I]
Question # 1: [10 + 15 =(25) marks]
(a) A certain brand of cigarette is used by smokers of three states A, B and C. It is estimated that 60% of the sales takes place in state A, 30% in state B and 10% state C. It is also estimated that 70% of the smokers of state A, 5% of the smokers of state B and 50% of the smokers of state C smoke that particular brand. Given a person is smoking that particular brand what are the probabilities that he/she is from states A, B and C respectively?
(a) Let us define the following events which are as stated below
1:
The person belongs to state
2:
The person belongs to state
3:
The person belongs to state
: The person is a smoker of the particular brand of cigarette
Thus:
( |
(
3)
1)
(
= 0.60,
2)
(
= 0.30,
3)
= 0.10,
( |
1)
= 0.70,
( |
2)
= 0.05,
= 0.50,
Hence using Baye’s Theorem we get (
1| ) =
(
=
Similarly:
(
2|
1)
×
( |
1)
+
(
1)
×
( |
1)
(
2)
×
( |
2)
+
0.60 × 0.70
0.60 × 0.70 + 0.30 × 0.05 + 0.10 × 0.50
) = 0.03 and
(
3|
) = 0.10
Question # 1: [10 + 15 =(25) marks]
Page 71 of 100
(
3)
×
= 0.87
( |
3)
(b) A perfect coin and a perfect die are thrown repeatedly in that order. What is the probability of a head appearing before a/an/the (i) six, (ii) even number and (iii) sum is exactly 12.
(b) Part I We shall first obtain the probability of the complementary event, i.e., the probability that no head will appear before the first six. If this complementary event is denoted by may occur in one of the following mutually exclusive forms as stated below
, then
(i) In the first pair of throws tail appears as well as six. (ii) In the first pair of throws tail appears but not six, in the second pair of throws tail appears as well as six. (iii) In the first pair of throws tail appears but not six, similarly in the second pair of throws tail appears but not six, while in the third pair of throws tail appears as well as six. This continues as stated above
ℎ ℎ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ⋯ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ⋯ ⋯ ⋯ − −
Now if
denotes the appearance of tail in the
throw of the coin and
that of six in the
throw of the die, then we may write =(
(
1
)=
1)
+(
1
(
1
1)
2)
2
1
(
+
1
+(
1
2)
2
1
1
2
+
(
3)
3
2
1
2
1
+
.
3
2
3)
+
.
Since all throws are independent hence we have (
)=
(
1)
(
+
(
(
)=
)=
1 2
1
12
×
1
1 6
+ 5
12
1 2
× 1
5 6
× 1
(
1 2
×
1)
1)
1 6
= , hence 7
+
+
(
1 2
×
1
5 6
(
1)
(
)
(
2)
(
5
1
×
( )=
1 2
×
6
1
×
6 7
Page 72 of 100
)
2
2
×
(
2)
(
2)
)
(
3)
(
1 6
+
=
1
12
3)
+
1+
5
12
+
5
12
2
(b) Part II We shall first obtain the probability of the complementary event, i.e., the probability that no
head will appear before the even number. If this complementary event is denoted by then
may occur in one of the following mutually exclusive forms as stated below
,
(i) In the first pair of throws tail appears as well as even number. (ii) In the first pair of throws tail appears but not even number, in the second pair of throws tail appears as well as even number. (iii) In the first pair of throws tail appears but not even number, similarly in the second pair of throws tail appears but not even number, while in the third pair of throws tail appears as well as even number.
This continues as stated above
ℎ ℎ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ⋯ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ∩ ⋯ ⋯ ⋯ − −
Now if
denotes the appearance of tail in the
number in the =(
(
1
)=
throw of the coin and
that of even
throw of the die, then we may write
1)
+(
1
(
1
1)
2)
2
1
(
+
1
+(
1
2)
2
1
1
2
+
(
3)
3
2
1
2
1
+
.
3)
3
2
.
Since all throws are independent hence we have (
)=
(
1)
+
(
(
)=
)=
1 4
1 2
1
×
1 4
1 2
+ 1
1 2
× 1
1 2
×
(
(
1 2
= , hence 3
1)
+
1)
(
1
1
×
2
+
2
(
1)
(
)
(
2)
1
×
( )=
1 2
×
1 2
×
1
)
(
1 2
2 3
(b) Part III Page 73 of 100
×
2
1 2
(
2)
(
2)
)
(
3)
(
×
1 2
+
=
1 4
3)
+
1+
1 4
+
1 4
2
+
We shall have a set of outcomes for exactly 12 as sum the minimum being 6 + 6 and the maximum being when we have 1 + 1+.. 12 times. Using this combination and the fact that we have to formulate the problem as complementary cases we can get the answer.
Question # 2: [10 + 15 =(25) marks]
(a) Show that the distribution function F defined by
F ( x) = 0
if x < 0
= 1 − e −α ( x −θ )
if x ≥ 0.
Here α > 0. With this information sketch the graph of F(x) and f(x) on the same graph , and obtain P[a ≤ X ≤ b] for any given numbers a and b.
(a) We all know the relation between
F ( x) = 0
( ) and
then the diagrams are as follows for
= 1 1.2
if x < 0
−−
if x ≥ 0
(
( )=
if x ≥ 0
In order to facilitate the drawing assume
F ( x) = 0
( )
( ), i.e.,
( ), thus we get
=
( )=0
if x < 0
= 1 − e −α ( x −θ )
− − − )
if x < 0 if x ≥ 0
= 1 (which we assume for simplicity), and
( )=0
if x < 0
( )=
if x ≥ 0
=0
1 0.8 0.6
f(x) F(x)
0.4 0.2 0 0
0.5
1
1.5
2
2.5
3
3.5
− − −− − − ≤ ≤ − − − − −− By formulae we have =
2
3
, if
(
= 2 and
)= ( )
( ) = (1
=3
Page 74 of 100
)
(1
)=
Question # 2: [10 + 15 =(25) marks]
(b) There are four urns containing (i) white,
3 black
and finally (iv)
1 white,
4 white,
1 black,
4 black
(ii)
2 white,
2 black,
(iii)
3
balls respectively. One ball is taken from
each urn at random, then, what is the probability that (i) two balls are white & two balls are black and (ii) one ball is white & three balls are black?
(i) Two balls are white & two balls are black The combination of getting two white and two black is obtained as follows: BBWW, BWBW, BWWB, WWBB, WBWB, WBBW. Now the combination of one is given as
≡ ≡ ≡ ≡ ≡ ≡ 1
1+ 1 1
1+ 1
1
1+ 1
×
× ×
2
2+ 2 2
×
×
2+ 2
2
×
2+ 2
3
3+ 3 3
3+ 3
3
3+ 3
×
× ×
4
4+ 4 4
1
,
1+ 1 1
,
4+ 4
4
1+ 1
1
,
4+ 4
1+ 1
1
Combine them to get the result as: 3
3+ 3
×
1
1+ 1
×
4
+
4+ 4
2
2+ 2
×
1
1+ 1
3
3+ 3
×
×
2
2+ 2
4
4+ 4
×
+
3
3+ 3
1+ 1
×
1
1+ 1
×
×
4
4+ 4
2
2+ 2
2
2+ 2
×
1+ 1
3
2
2+ 2
2
×
2+ 2
3
×
3+ 3
2+ 2
×
1
+
2
×
3+ 3
×
×
2+ 2
3+ 3 3
×
3+ 3
3
×
3+ 3
4
×
2
3
×
4+ 4 4
×
4+ 4
4
×
4+ 4
1
+
4+ 4
×
4
×
3
3+ 3
1+ 1
×
4
4+ 4
×
2
2+ 2
×
+
4
4+ 4
(ii) one ball is white & three balls are black The combination of getting one white and three blacks is obtained as follows: WBBB, BWBB, BBWB, BBBW. Now the combination of one is given as
≡ ≡ 1
1+ 1 1
1+ 1
× ×
2
2+ 2 2
2+ 2
× ×
3
3+ 3 3
3+ 3
× ×
≡ ≡ 4
4+ 4 4
4+ 4
, ,
1
1+ 1 1
1+ 1
Page 75 of 100
× ×
2
2+ 2 2
2+ 2
× ×
3
3+ 3 3
3+ 3
× ×
4
4+ 4 4
4+ 4
Combine them to get the result as: 3
3+ 3
×
4
4+ 4
+
1
1+ 1
×
2
2+ 2
×
3
3+ 3
1
1+ 1
×
4
4+ 4
×
2
2+ 2
+
×
1
1+ 1
3
3+ 3
×
×
2
2+ 2
4
+
4+ 4
×
3
3+ 3
×
1
1+ 1
×
4
4+ 4
++++++++++++++++ END OF QUESTION PAPER ++++++++++++++++
Page 76 of 100
2
2+ 2
×
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING 1st MID-SEMESTER EXAMINATION [2014-2015, SEMESTER I]
Question # 1: [10 + 15 =(25) marks]
(a) Suppose there are three chests and each chest has two drawers. One of the chests has a gold coin in each of its drawers, the next chest has one gold coin in one of its drawer and one silver coin in the other drawer, while the last chest has a silver coin in each of its drawers. One of the chests is drawn at random and then one of its drawers is opened at random and a gold coin is found in that drawer. Then what is the probability that this chest contains a gold coin in its other drawer?
Answer # 1: [10 + 15 =(25) marks]
Let the state of nature/event be denoted as given: 1:
Chest with gold coin in both the drawers
2:
Chest with one gold coin in one drawer and a silver coin in the other drawer
3:
Chest with silver coin in both the drawers
: The coin found in the randomly opened drawer in a randomly drawn chest is gold ( |
Thus: (
3)
1)
= 1,
( |
2)
1
= and 2
( |
3)
1
= . With this information we are required to find 3
formulae:
(
1|
)=
( | 1 )×
( 1 )+
( | 1 )× ( | 2 )×
( 1) ( 2 )+
(
= 0 and also ( |
( | 3 )×
1 ),
( 3
1)
=
(
2)
=
which is given by the
= )
1 3 1 1 1 1 1× + × +0× 3 2 3 3
1×
=
2 3
(b) In a population of ( + 1) individuals, a person, called a progenitor, sends out an email at random to different individuals, each of whom in turn again forwards the email to other individuals and so on. That is at each step, each of the recipients of the email forwards it to of the
other individuals at random. Then what is the probability of the email not being
relayed back to the progenitor even after steps of circulation.
Page 77 of 100
⋯
Answer # 1: [10 + 15 =(25) marks]
In a population of ( + 1) individuals a person, called the progenitor sends out an email at random to different individuals, each of whom in turn again, forwards the email at random to
other individuals and so on. That is at every step, each of the recipients of the email
forwards it to
of the
other individuals at random. We are interested in finding the
probability of the email not relayed back to the progenitor even after steps of circulation. The number of possible recipients for the progenitor is
. The number of possible choices
each one of these recipients has after the first step of circulation is again
, and thus the
number of possible ways this first stage recipients can forward the email equals ×
×
×
. Therefore after the second step of circulation the total number of possible
=
1+
configurations equals
. Now there are
one of whom can forward the email to
×
=
2
many second stage recipients each
possible recipients yielding a possible
third stage recipients after 3 steps of circulation and
1+ + 2
2
many
many total possible
configuration. Proceeding in this manner one can see that after the email has been circulated through
−
ℎ −
1 steps, at the
step of circulation the number of senders equals
1
collectively make
− −− 1
who can
many choices. Thus the total number of possible configurations
after the email has been circulated through steps equals
⋯ − 1+ + 2 + +
1
=
1 1
. Now
the email does not come back does not come back to the progenitor in any of these steps, if
− −
and only if none of, starting from the circulation to the
−
1
recipients after
recipients of the progenitor after the first step of 1 steps of circulation, send to the progenitor, or in
ℎ − − ⋯ − − − − − −− − − − − −− − − − −−− − − − − − −
other words each of these recipients/senders at every step makes a choice of forwarding the email to individuals from a total of
1 instead of the original . Thus the number of
ways the email can get forwarded the second, third, ….., 1
equals
+ 2+
1
=
progenitor remains interest equals (
!(
1)!
1)!
×
)!
!
1 1
1
=
1
1
The number of choices for the
. Thus the number of possible outcomes favorable to the event of 1
×
!(
1
step avoiding the progenitor
1
=
1
1
, yielding the probability of interest as 1
= 1
1
Page 78 of 100
1
=
Question # 2: [10 + 15 =(25) marks
(a) Suppose
(
is a random variable and its probability mass function is
1
= 2) = and 3
1
(
= 3) = . Given this information first find 6
graph draw both pmf as well as the cdf.
Answer # 2: [10 + 15 =(25) marks]
≤ ≤ ≤ ≤ ⎩ ≤
Given
1
(
2
0
1
= 2) = and 3
<1
1
( )=
(
= 1) = ,
2 5 6
1
1
2
2
3
(
(
1
= 1) = , 2
( ) and on the same
1
= 3) = the cdf is given below: 6
. Utilizing this we draw the pmf and cdf as give:
3
1.2 1 0.8 0.6
f(x)
0.4
F(x)
0.2 0 1
2
Answer # 2: [10 + 15 =(25) marks]
(b) A cubic die is thrown number of time, and
3
is the total number of spots shown. Find
≤ ∑ ℎ ∑ (
0.1
),
(
). Also state Chebyshev′s inequality and find an such that
0.1.
−
3.5 >
Answer # 2: [10 + 15 =(25) marks]
Let
=
=1
, where
that the random variables, 1 6
, where
is the number of spots on the
throw. One should remember
, are independent and identically distributed with
= 1,2,3,4,5,6. Hence
( )=
=1
(
Page 79 of 100
1
(
= )=
= ) = (1 + 2 + 3 + 4 + 5 + 6) = 6
− − − − − − − ≥ ≤ − ≥ ≤ − ≤ − ≤ ≤ ≥ 7 2
( )=
,
(
Thus
)=
{|
(
1 6
7
2
7 2
1
2
(
and
)|
(
1
7 2
+ 2
2
35
)=
2
7 2
+ 4
2
7 2
+ 5
2
+ 6
7 2 2
=
35 12
.
from where Chebychev′s inequality is given by
12
1
)}
7 2
+ 3
{|
2 , i.e.,
(
)|
(
2}
1
)
2 2
. This equivalent
form can be derived very simply Now we are given (|
i.e.,
3.5 > 0.1 (
3.5 | > 0.1 )
)
)2
(0.1
0.1
35
=
Comparing equation (1) and (2) we have
12
×
1750
1
0.01
2
=
1750 6
0.1, i.e.,
6
(1) (2)
2920
Question # 3: [10 + 15 =(25) marks
(a) Consider the switching network shown in the figure below. It is equally likely that a switch will or will not work. Find the probability that a closed path will exist between terminals A and B. S1
A
B S3 S2
S4
Answer # 3: [10 + 15 =(25) marks]
ℎ − ℎ
We are given the probabilities of Which means that
(
1
)=
(
1,
2)
2,
3,
(
=
4 as
3)
=
( 1) = (
4)
(
2)
=
(
3)
(
=
4)
1
= . 2
1
= . 2
Now for a closed path to exist between A and B we should have the following: (
)=1
(
)
Now we need to know how many such paths are there such that no connection exits between A and B. They are: •
1 and
1 2
=
1
16
2 are
open, while
3 and
4 are
closed such that the probability is
Page 80 of 100
1 2
1
1
2
2
× × ×
• • •
1,
2 and
3 are
open, while
1,
2 and
4 are
open, while
1,
3 and
4 are
open, while
1,
•
2,
3 and
4 are
1
1
1
1
1
2 1
2 1
2 1
2 1
16 1
2 1
2 1
2 1
2 1
16 1
2
2
2
16
4 is
closed such that the probability is × × × =
3 is
closed such that the probability is × × × =
2 is
closed such that the probability is × × × = 1
1
1
1
2 1
2
2
2
2
16
open such that the probability is × × × =
Hence the total probability that there is no connection between A and B is 1 16
+
1 16
=
5 16
1 16
+
1 16
+
1 16
+
. We simply add are these paths are mutually exclusive
ℎ − (
)=1
5 16
=
11 16
with probability density function ( ) =
(b) The random variable
−− − 1
(
1)!
,
> 0 and
= 1,2, …., is called an Erlang distribution with λ and as the two parameters. Calculate
both ( ) and ( ) and sketch both the pdf and cdf for λ = 2 and
= 3.
Answer # 3: [10 + 15 =(25) marks]
∫∞ − ∫∞ − − ∫∞ −
is the random variable for which the probability density function is given by ( ) =
which is the Erlang distribution. One should remember that ( )= = =
( )
0
0
(
(
∈ℝ +
,
= 1,2, … .. and
−− − 1
(
1)!
> 0.
1)!
1)! 0
Now you can solve this problem using either by parts integration, which is lengthy as we use basic differentiation and integration concepts, else use gamma function formulae.
By parts
− ∫∞ − − − − ∞ ∫∞ − − ( )=
(
=(
1)! 0
1
1)!
0
+
1
0
Page 81 of 100
− − −
=(
1)!
=(
1
2
0
1)!
(
!
=
+1
2
0
1)
2
2
1
×
(
+
0
1
1)!
=(
∞ ∞ − − − − − − ∫∞ − − − − ∞ − − − ∞ − − − − ∞ −⋯ − − ∞ 1
0
1)
2
3
0
0
!
…
+1
0
Using gamma function
Γ ∫∞ − − ∫∞ − − − ∫∞ − − Γ − ∫∞ − − ∫∞ − − − − − − ∫ ∫ −
We know that ( ) = ( )=
=
1)! 0
(
( )=
1
0
( )}2
{
(
1)! 0
=(
×
1)!
Now when
1)!
and hence using this we have
0
( +1)
×
+1
=
(
1)!
×
!
+1
=
( )
+1
=(
1
=
0
{ ( )}2
2
( +1)! +2
2
= 2 and
=
( +1)
2
2
2
=
2
= 3, the density function is of the form
distribution function is ( ) =
0
2
4
2
Let us calculate the distribution ( ) =
0
4
2
2
− ( )=4
2
2
, while its
. Though conceptually to draw the graph very
sketchily we need not derive it but still we want to draw the graph nicely/neatly and legibly. First let us derive the following
∫ −− − − ∫ − − − − − − − − − − − − − − − −⋯ − − − − − − − − − − − − − − − − −⋯ − − − − ( )=
=( =( =(
1
0
(
1)!
1
1)! 0
1
1)!
1
2
1)!
(
1
(
2
2
0
1
….+
1
1
1)(
0
2
(
2)
3
3
1)(
2)
3
1)!
Page 82 of 100
0
3
….
…
(
(
1)!
0
1)!
+0+0+
− − − − − − − − ⋯ − − − −− ⋯ −−− −− ∑− −− ∑ −− =1
….+(
(
1)!
×
1)!
(
=1
1
+(
1)!
1
×
2
2
− − − − −
+(
1)!
×
(
1)(
2)
3
3
+
1)!
1!
+
2 2
2!
+
.+
1
(
1
1)!
1( ) =0 !
= 2 and
( )=1
1
1+
=1
With
×
= 3, we have
2
2 (2 ) =0 !
=1
2
1+2 +
4 2 2!
f(x) F(x)
0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9 2.1 2.3 2.5 2.7 2.9 3.1 3.3 3.5 3.7 3.9 4.1 4.3 4.5 4.7 4.9
Erlang probability and distribution functions for = 2 and
= 3 ale)
++++++++++++++++ END OF QUESTION PAPER ++++++++++++++++
Page 83 of 100
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING 2nd QUIZ EXAMINATION [2014-2015, SEMESTER I]
Question # 1: [15 + 10 =(25) marks]
(a) The data given relates to the per cent shrinkage in nylon fibre as a result of temperature tests at 125°C and 150°C. Test whether the shrinkage at 150°C is greater than that at 125°C and assume shrinkage is normally distributed in each of the two populations pertaining to the two different temperatures. Consider α = 0.10.
125°C 3.48
3.58
3.54
3.61
3.57
3.65
3.49
3.61
3.67
3.62
3.72
3.96
4.01
3.59
3.81
3.45
4.06
3.65
4.08
3.69
3.50
150°C 3.88
�
(a) Consider and are the random variables which denote the shrinkage for 125° and 150° , respectively hence
− − 2 ,150,10
125,12
= 3.58417 and
= 0.045921
�
150,10
= 3.82100. Moreover
2
,125,12
= 0.004954 and
Now our hypothesis is as stated below
− − � −� − − − 0:
0:
=0
vs
0:
>0
=0
vs
0:
>0
= ( ) and
Where as one can easily understand that
=
( ) are the actual or population
means for 125° and 150° respectively. Thus
one
would
reject
+
×
2
150,10
125,12
Moreover under
>
0 we
12+10 2,0.05
must remember
0
,125,12
12
2
+
= 0. Thus:
Page 84 of 100
,150,10
10
is
we
, else accept the hypothesis
have
0.
� −� − 150,10 2
125,12
,125,12
12
2
+
= 3.82100
,150,10
10
=
0.004954 12
3.58417 = 0.2368, +
0.045921 10
−
= 0 (under
0 ),
20,0.05
= 1.725,
= 0.0707. Using these value we have
0.2368 > 0.1220 which is true, hence we reject
0
(b) A manufacturer of bars of steel claims that the average breaking strength of his product is not less than 52. The breaking strength of each bar in a sample of 15 is noted as given as: 51.3 50.2 51.9
50.0
52.5
50.7
49.3
49.3
48.3
48.1 48.2
47.8
52.1
50.3
47.5. Given this
data we are to examine if the manufacturer ′s claim is supported by these data. Assume
= 0.05.
(b) As per the information provided in the problem the hypothesis will be framed keeping in mind that we will always try to disapprove the statement made by the person, hence we always try to reject
≥ ≥ − ∑ −� − − − √ − √ which results in the following formulation, i.e.,
0:
=
0
0,
52. If that is the case, then we should
have the following hypothesis to test, which is: 0:
i.e., if the following, X n
=
0
vs
52
:
=
s < µ 0 − t n −1,α n , is true then we reject n
< 52
0
Now for the sample (here the sample size is 15) given we can easily calculate that 49.8333 and (1
2
=(
1
1)
)2 = 2.7124. For the problem consider
=1(
) = 0.95. Hence from the table we have
1,
reject
×
0,
= 52
1.761 ×
� ∑ −
1.6469 14
1,
=
14,0.05
=
1
=
=1
= 0.05, i.e.,
= 1.761. Hence the RHS is
0
= 51.2249. Now 49.833 is always less 51.2249, hence we
i.e., the claim of the manufacturer is false.
Question # 2: [15 + 10 =(25) marks]
(a) Two experimenters, Shruti Mittal and Parva Goyal, take repeated measurements of the length of a copper wire. On the basis of the data obtained by them, which are given below, test whether Shruti′s measurement is more accurate (think what accuracy means here) than Parva′s. Consider α=0.05. Shruti′s measurement
Parva′s measurement
Page 85 of 100
(in mm)
(in mm)
12.47
12.44
12.06
12.34
11.90
12.13
12.23
12.46
12.77
11.86
12.46
12.39
11.96
12.25
11.98
12.78
12.29
12.22
� �
(b) Given the data let us denote
and
as the random variables which denote the distribution of
measurement made by Shruti and Parva, such that ~ ( More we also know that: (i)
� � =
8
= 12.2675,
= 10,
=
2
,
) and ~ (
= 12.2850,
10
,
2
).
= 0.333042 and (ii)
= 0.178786
= 8,
To test the hypothesis or statement that Shruti′s measurement is more accurate than Parva′s and α = 0.05 our hypothesis is as follows:
− ≤ ≤ ≤ − − − − ≤ − − − 2 0: 0
2
= 0:
2
2 0: 0
vs
0:
2
2
2
The rule is reject
0 if
1,
2
is true.
1,1
Now we have 2 10
= 0.110917,
3.470019 2
2 8
= 0.031964,
9,7,1 0.05
=
1
7,9,0.05
=
1
3.29
2
= 0.3040, thus
0.3040 is FALSE, hence we cannot reject the null hypothesis that
2
=
0.110917
=
0.031964
2 0: 0
= 0:
2
, which means that there is significantly no difference in Shruti′s measurements/readings with
respect to Parva ′s measurements/readings. Remember:
9,7,1 0.05
=
1
7,9,0.05
as
, ,1
=
1
,
,
(b) Ashish Jindal is testing the tensile strength of a particular alloy. The sample average is 13.71, while the standard error is 3.55. What should be the minimum sample size Ashish should collect such that the confidence interval within which the population mean would lie is 3.14? As Ashish was
Page 86 of 100
finishing his task, Ankur Jhavery comes running and says that the value 3.55 if not the standard error but the standard deviation . In that case what is the new sample size, considering all other
information is correct. How many extra observations did Ashish already collect or need to collect, based on Ankur ′s information? Consider α=0.05.
(b) Case I
� − − √ ≤ ≤ � − √ − √ ̅ − √ −√ −√ √ − −√ √ − −√ √ − −√ √ − − −√ √ − −√ √ −√ √ − � − √ ≤ ≤ � √ √ ̅ √ ≅ 1 1, 2
1
1
×
1
1
+
1 1, 2
interval. Now from the information we have 1 1,0.025
3.14, i.e.,
1
=
3.14
3.55×2
1
×
1
1
, i.e., 2 ×
= 13.71,
1 1, 2
1
×
1
is the length of confidence
= 3.55, hence 2 ×
1
1,0.025
×
1 1
=
= 0.44225
Now check the distribution tables and from that we have: 1
1 = 16,
1
= 17,
16,0.025
= 2.120, i.e.,
1 1,0.025 1
1
1 = 17,
1
= 18,
17,0.025
= 2.110, i.e.,
1 1,0.025
1
1 = 18,
1
= 19,
18,0.025
= 2.101, i.e.,
1 1,0.025
1
1 = 19,
1
= 20,
19,0.025
= 2.093, i.e.,
1 1,0.025
1
1
1
1
2.120
=
= =
=
17
2.110 18
2.101 19
2.093 20
=
,
=
,
= .
, i.e.,
, .
, .
=
=
,
=
,
= .
, i.e.,
, .
, .
=
1 = 22,
The value of
1
= 23,
1 which
22,0.025
= 2.074, i.e.,
1 1,0.025 1
=
.
.
2.074 23
= 0.5142
= 0.4973 = 0.4820
= 0.4680 = . = .
= 0.4325
satisfies this is between 21 and 22, and we will consider
1
= 22.
Case II
2
×
2
2
2
+
×
2
from the information given we have i.e.,
2
= 19.64
2
1
, i.e., 2 ×
×
2
= 13.71,
2
is the length of confidence interval. Now
= 3.55, hence 2 ×
×
2
20
Hence the extra observations collected is 2
Page 87 of 100
2
= 3.14,
= 1.96,
2
DEPARTMENT OF INDUSTRIAL & MANAGEMENT ENGINEERING MBA651: QUANTITATIVE METHODS FOR DECISION MAKING FINAL EXAMINATION [2014-2015, SEMESTER I]
NOTE THE FOLLOWING 1) Total time for this paper (which consists of 3 pages) is 180 minutes (3 hours). 2) Total marks is 100 and individual marks are mentioned alongside each question. 3) Total number of questions is 4 (with sub-parts) and you are required to answer ALL of them. 4) You are ONLY allowed to use the calculator and the statistical tables. 5) Marks will be there for correct formulation of the problem, rather than only the final answer. Hence step wise marking is also there. 6) Draw diagrams accurately/neatly/legibly, and use the concept of set theory where ever necessary.
Answer # 1 (a) [10 marks]
(a) Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being equally likely. The first to arrive will wait for 15 minutes and will leave if the other has not yet arrived. What is the probability that they will meet?
Let us use as sample space the unit square, whose elements are the possible pairs of delays for the two of them. Our interpretation of equally likely pairs of delays is to let the probability of a subset of Ω be equal to its area. This probability law satisfies the three probability axioms. The event that Romeo and Juliet will meet is the shaded region as shown in the figure below, and its probability is calculated to be 1
− − 1 2
3
3
1
4
4
2
× ×
3
3
4
4
× × =1
−
9 16
=
7 16
.
Page 88 of 100
¼ ¼
Answer # 1 (b) [15 marks]
Suppose you are on a game show, and you are given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say # 1 [but the door is not opened], and the host, who knows what is behind the doors, opens another door, say # 3, which has a goat. He/She then says to you, "Do you want to pick door # 2?" Is it to your advantage to switch your choice? Justify your answer mathematically.
The decision tree is as follows Total
Stay
Switch
Probability 1 2
Door # 1
1 2
1
Door # 2
3 1
Door # 3
3
1
1
2
6
1
1
2
6
× = × =
Car
Goat
Car
Goat
Goat
Car
Goat
Car
1 3 1 3 1 3
Door # 2
Door # 3
Door # 3 Door # 2
1 3 1 3
×1=
×1 =
1 3 1 3
Tree showing the probability of every possible outcome if the player initially picks Door # 1. Consider the discrete random variables, all taking values in the set of door numbers as {1,2,3}. Page 89 of 100
Let C: Denotes the door behind which you have the car S: Denotes the door selected by the player H: Denotes the door opened by the host Then, if the player initially selects Door # 1, and the host opens Door # 3, then the probability of winning by switching is given by
( = 2|
= 3,
= 1) =
∑ ( =3| =2, =1)× ( =2| =1)
3 =1
( =3| = , =1)× ( = | =1)
=
1 3 1 1× 3
1×
1 1 × 2 3
+
+
1 0× 3
=
2 3
Answer # 2 (a) [10 marks]
(a) Priyanka Chopra is a student in the Quantitative Techniques for Decision Making course, which is compulsory for MBA students at IIT Kanpur. She has approached you with the
− − ≤≤ ≤
following problem which she is unable to solve. Help her to draw (accurately, neatly and legibly) both ( ) and ( ) of the function. Remember ( ) is given below.
0 0.3 ( )= 0.8 1
3 0
< 3 <0 <4 4
Utilizing the distribution function we have
1.0
0.8
( )
Left point discontinuous
0.3
-3
0
Page 90 of 100
+4
Hence ( ) is
( )
0.5
0.3 0.2
-3
0
+4
Answer # 2 (b) [15 marks] A random variable (r.v) X is said to have a Weibull distribution with parameters α and β ( α > 0 and
− −
β > 0) if the pdf of
( ; , ); (ii) ; (iii)
and (iv)
1
( ; , )=
is given by
,
( ; , ) and,
. Draw the graphs of
> 0. Then find (i)
( ; , ) when α = 1.
Can you say something about this distribution which you drew [Hint: This distribution which you just drew has already been studied and discussed in class in details].
It is given that the pdf is ∞
show that
− − ∞
∫ f
X
( x;α , β )dx = ∫
α
β α 0
0
1
( ; , )=
x
α −1
, where
x α exp − dx = 1 . β
α
Put
∞
x = y , β
α −1
which
means
∞
∫ f
X
> 0, hence our first task is to
that
x α α dx = dy . β
∞
Thus
( x;α , β )dx = ∫ e− y dy = (− e− y )0 = 1 , which proves that it is a legitimate pdf.
0
0
(i) Now to find the cdf, we consider: x
∫
F X ( x;α , β ) = f X ( x;α , β )dx = 0
x
α
∫ β
α
0
x
α −1
x α exp − dx β
Page 91 of 100
we
have:
α
x xα −1 Again using the same substitution where = y , which means that α α dx = dy , we have β β F X ( x;α , β ) = (− e
x − y β
α
x α = 1 − exp − β
)
0
(ii) To calculate the mean, i.e.,
= µ , we have
α α ∞ x α x E ( X ) = ∫ xf X ( x;α , β )dx = ∫ x α xα −1 exp − dx = α ∫ xα exp − dx β β 0 β β 0 0
∞
∞
α
α
x xα −1 Substitute = y , which means that α α dx = dy . Thus the expected value is given by β β ∞
1
∫
E ( X ) = β y α e − dy = β Γ1 + y
0
1
, where we use the concept of gamma function which is
α
∞
1 α
Γ( z ) = ∫ e− t t z −1dz . Thus µ = β Γ1 + 0
(iii) To calculate the median, i.e., µ e
∫
= µ , we have
F X (µ e ; α , β ) = f X (µ e ; α , β )dx = 0
x
means that α
α −1 α
β
µ e
α
∫ β
α
x
α −1
0
µ e
dx = dy . Thus we have:
α x α x 1 exp − dx = . Substitute = y , which 2 β β
∫e
− y
dy = (− e
µ e − y β
)
0
0
α
1 2
= , i.e., 1 − e
µ − e β
α
=
1 . Which 2
1
mean µ e = β {loge 2}α . (iv)
To
df X (µ ; α , β ) dx
calculate
=
d α
α
dx β
the
x
α −1
mode,
x α exp− β
i.e.,
= µ ,
we
Thus:
df X (µ ;α , β ) dx
=
α β α
x
α −1
x α α x α d α −1 exp − + α exp − x dx β β β dx d
Page 92 of 100
should
first
find
=−
α β α
x
α −1
x α α α −1 α x α α − 2 exp − α x + α (α − 1) x exp − β β β β 1
(α − 1) β α α −1 µ = , i.e., 1 0 + ( − ) = x α . o α β α α
Equating this to zero yields: − x
α
− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − ≤ − Which mean
=
(
1)
1 2
1
i.e.,
2
i.e.,
i.e., i.e.,
when 1
1
2
2
×
×
2
, provided
( ; , ) 2
=
+
(
1)
2
2
2
+
(
1)
2
2
2
+
(
1)
2
× (2
×(
2)
1) ×
2
1
1
and some similar conditions, thus
+
(
×(
1) × (
(
2
( ; , ) 2
, i.e.,
.
2
=
< 0. First let us find
1)
2)
1
2) . This is always < 0 for the case
1 2
is the mode.
When α = 1, then the pdf is of the form f X ( x;α = 1, β ) =
1 β
x , which is the one β
exp −
x , β
parameter exponential distribution we all know where F X ( x;α = 1, β ) = 1 − exp −
> 0.
If we draw the pdf and cdf of the exponential distribution considering ~ ( = 5,
= 0)
then the graphs are given below.
Page 93 of 100
1 0.9 0.8 0.7 0.6 0.5
f(x)
0.4
F(x)
f X(x),
0.3
FX(x)
0.2 0.1 0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40
Answer # 3 (a): [10 mark]
(a) The Police of Kanpur compiles data on robbery and property crimes and publishes the information for the public to see. A simple random sample of pickpocket offenses yielded the losses, in INR, shown below. Use the data (given below) to find a 95% confidence interval for the mean loss, due to pickpocket offenses. The data is as follows: 447, 207, 627, 430, 883, 313, 844, 253, 397, 214, 217, 768, 1064, 26, 587, 833, 277, 805, 653, 549, 649, 554, 570, 223 and 443.
Let us consider normal distribution of data. From the data we have
= 0.05 and
= 25. We
are required to use the t-distribtion so we note its dof (degree of freedom) as 24 (=25-1). The
̅ √ ̅ ∑
confidence interval for is
±
,
.
2
From the data and the table we have:
∑ −̅ 1
24
25 { =1
25 }
2
25
=
1
25
25 =1
= 513.32,
24,0.025
= 2.064,
24
=
= 262.23, utilizing which we have the lower limit and upper limit
values as 405.07 and 621.57 respectively.
Answer # 3 (b): [15 mark]
(b) A manufacturer of bars of steel claims that the average breaking strength of his product is not less than 52. The breaking strength of each bar in a sample of 15 is noted as given as:
Page 94 of 100
51.3, 52.1, 50.3, 50.2, 51.9, 50.0, 52.5, 50.7, 49.3, 49.3, 48.3, 48.1, 48.2, 47.8 and 47.5. We are to examine if the manufacturer ′s claim is supported by these data. Consider here
= 5%.
As per the information provided in the problem the hypothesis will be framed keeping in mind that we will always try to disapprove the statement made by the person, hence we always try to reject
≥ ≥ − ∑ −� − − − √ − √ which results in the following formulation, i.e.,
0:
=
0,
52. If that is the case, then we should
0
have the following hypothesis to test, which is: 0:
i.e., if the following, X n
=
vs
52
0
:
=
s < µ 0 − t n −1,α n , is true then we reject n
< 52
0
Now for the sample (here the sample size is 15) given we can easily calculate that 49.8333 and (1
2
=
1
(
) = 0.95. Hence from the table we have
1,
reject
×
0,
= 52
1.761 ×
=
)2 = 2.7124. For the problem consider
=1(
1)
� ∑ −
1.6469 14
1,
=
14,0.05
1
=
=1
= 0.05, i.e.,
= 1.761. Hence the RHS is
0
= 51.2249. Now 49.833 is always less 51.2249, hence we
i.e., the claim of the manufacturer is false.
Answers # 4 (a): [10 marks]
(a) A manufacturer produces three types of plastic fixtures. The time required for molding, trimming, and packaging are given below and they are in hours per dozen fixtures. How many fixtures of each type should be produced to obtain maximum profit? Process Molding
Type A 1
Type B
Type C
2
3
Total time available 12000
2 Trimming Packaging Profit
2
2
3
3
1
1
4600
1
1
2400
2
3
2
11
16
15
Page 95 of 100
____
Let number of type A, type B and type C be , and respectively, then the profit function is
≤ ≤ ≥ ≤ ≥ 11 + 16 + 15
s.t:
+2 +
2
+
3 1
+
2
2 3 1 3
, ,
3
12000
2
+ +
4600
1
2400
2
0
Hence adding slacks we have
11 + 16 + 15 + 0
s.t:
+2 +
2
+
3 1
+
2
2 3 1 3
, , ,
3
+
2
+ +
+0
1
+
1
2
+0
3
= 12000
2
= 4600
+
2
1, 2, 3
1
3
= 2400
0
Using this we have the solution as shown: Variables
1
-11
0
2
-16
-15
0
0
( )
3
0
1
1
2
3/2
1
0
0
12000
2
2/3
2/3
1
0
1
0
4600
3
½
1/3
½
0
0
1
2400
Variables
1
2
( )
3
0
-3
8
0
½
1
3/4
½
0
0
6000
2
1/3
0
1/2
-1/3
1
0
600
3
1/3
0
1/4
-1/6
0
1
400
-3
Variables
2
1
0
0
2
96000
3
0
-3/4
13/2
0
0
1
3/8
¾
0
-3/2
5400
0
0
¼
-1/6
1
-1
200
Page 96 of 100
9
( )
99600
1
Variables
0
3/4
-1/2
0
1
0
3
2
1200
( )
3
0
0
6
3
6
100200
0
1
0
1
-3/2
0
5100
0
0
1
-2/3
4
-4
800
1
0
0
0
-3
6
600
Maximum value is 11 × 600 + 16 × 5100 + 15 × 800 = 100200 which is as shown in the last tableau Slacks can be found from the constraints and they are 3
6 0 0 + 2 × 5 1 0 0 + × 8 0 0 + 0 = 12000, i.e., 2
2 3 1 2
2
× 6 0 0 + × 5100 + 800 + 0 = 4600, i.e., 3 1
1
3
2
1
2
= 0 as also shown in the tableau
= 0 as also shown in the tableau
× 6 0 0 + × 5100 + × 8 0 0 + 0 = 2400, i.e.,
3
= 0 as also shown in the tableau
Answers # 4 (b): [15 marks] (b) Greenberg Motors, Inc., manufactures two different electrical motors for sale under contract to Drexel Corp., a well-known producer of small kitchen appliances. Its model GM3A is found in many Drexel food processors, and its model GM3B is used in the assembly of blenders. Three times each year, the procurement officer at Drexel contracts Irwin Greenberg, the founder of Greenberg Motors, to place a monthly order for each of the coming four months. Drexel’s demand for motors varies each month on its own sales forecasts, production capacity, and financial position. Greenberg has just received the January-April order and must begin their own four-month production plan. The demand of motors is shown as Models
January
February
March
April
GM3A+GM3B
8800
8700
8500
8600
One should also remember the following for the company as per policy and they are: (i) The desirability of producing the same number of each motor each month. This simplifies planning and the scheduling of workers and machines, (ii) The necessity to keep down inventory carrying, or holding, cost. This suggests producing in each month only what is needed in that month, (iii) Warehouse limitations that cannot be exceeded without great additional storage costs and (iv) The company’s no-layoff policy, which has been effective in preventing a unionization of the shop. This suggests a minimum production capacity that should be used each month. Moreover production costs are currently $10 per GM3A motor produced and $6 per GM3B unit. A labor agreement going into
Page 97 of 100
effect on March 1 will raise each figure by 10%, however. Furthermore each GM3A motor held in stock costs $0.18/month, and each GM3B has a carrying cost of $0.13/month. Greenberg’s accountants allow monthly ending inventories as an acceptable approximation to the average inventory levels during the month. The carrying cost structure is as follows: Model
January
February
March
April
GM3A
800
700
1000
1100
GM3B
1000
1200
1400
1400
Suppose that Greenberg is starting the new four-month production cycle with a change in design specifications that left no old motors in stock on January 1. Greenberg also want to have on hand an additional 450 GM3As and 300 GM3Bs at the end of April. Storage area of Greenberg Motors can hold a maximum of 3,300 motors of either type (they are similar in size) at any one time. Greenberg has a base employment level of 2.240 labor hrs/month such that capacity is 2,240 hrs/month. In a busy period, though, the company can bring two skilled former employees on board (they are now retired) to increase capacity to 2,560 hrs/month. Each GM3A motor produced requires 1.3 hrs of labor, and each GM3B takes a worker 0.9 hrs to assemble. With this information formulate a LP accordingly, you are NOT required to solve it .
Variables for the problem
XA,i = number of model GM3A motor produced in month i, (i=1, 2, 3, 4 for January to April) XB,i = number of model GM3B motors produced in month i, (i=1, 2, 3, 4 for January to April) IA,i = level of on-hand inventory for GM3A motors at end of month i, (i=1, 2, 3, 4 for January to April) IB,i = level of on-hand inventory for GM3B motors at end of month i, (i=1, 2, 3, 4 for January to April)
Objective function which has production costs
Production costs plus the fact that labor agreement would raise it by 10%, would give us the production cost function as 10XA1+10XA2+11XA3+11XA4+6XB1+6XB2+6.60XB3+6.60XB4
(Eqn 1)
Objective function which has inventory costs
Considering the fact that inventory carrying costs also come we have the inventory carrying cost function as 0.18 I A1+0.18 I A2+0.18 I A3+0.18 I A4+0.13 I B1+0.13 I B2+0.13 I B3+0.13 I B4
Page 98 of 100
(Eqn 2)
Objective function which has both production and inventory costs
Combining (Eqn 1) and (Eqn 2) we have the total cost which is the objective function and it is shown below
10XA1+10XA2+11XA3+11XA4+6XB1+6XB2+6.60XB3+6.60XB4+0.18 I A1+0.18 I A2+0.18 I A3+0.18 I A4+0.13 I B1+0.13 I B2+0.13 I B3+0.13 I B4
(Eqn 3)
Constraints related to inventory quantities
Inventory constraints set the relationship between closing inventory this month, closing inventory last month, this month’s production, and sales this month. Thus: [Inventory at the end of this month]=[Inventory at the end of this month] + [current month’s production]-[sales to Drexel this month]
(Eqn 4)
X A1- I A1=800, which is January demand for GM3A X B1- I B1=1000, which is January demand for GM3B
XA2+IA1-IA2=700, which is February demand for GM3A XB2+IB1-IB2=1200, which is February demand for GM3B XA3+IA2-IA3=1000, which is March demand for GM3A XB3+IB2-IB3=1400, which is March demand for GM3B XA4+IA3-IA4=1100, which is April demand for GM3A XB4+IB3-IB4=1400, which is April demand for GM3B I A4=450, ending inventory for GM3A I B4=300, ending inventory for GM3B
Constraints related to inventory space utilization
[Inventory space utilized by motor type A]+[Inventory space utilized by motor type (Eqn 5)
B]≤[Total space]
IA1+IB1≤3300, total combined inventory space utilization for GM3A and GM3B for January IA2+IB2≤3300, total combined inventory space utilization for GM3A and GM3B for February IA3+IB3≤3300, total combined inventory space utilization for GM3A and GM3B for March IA4+IB4≤3300, total combined inventory space utilization for GM3A and GM3B for April
Constraints related to employment utilization
[Minimum utilization of labor for GM3A and GM3B]≤[Actual utilization of labor for GM3A and GM3B]≤[Maximum utilization of labor for GM3A and GM3B] Page 99 of 100
(Eqn 6)
