2

Q4 (2.3D) The demand for an item over the next our quar ers s , , un s. e price per unit starts at $20 in the first quarter and is ncrease y eac quar er erea er. e supp er can provide no more than 400 units in any quarter. oug we can a e a van age o ower pr ces n early quarters, a storage cost $3.50 is incurred per un quar er. n a on, e max mum num er o units that can be held over from one quarter to the nex canno excee . eve op an an op o p ma schedule for purchasing the item to meet the eman . P K Sahoo, BITS-Hyderabad Campus

1

X i = Production in quarter i Solution: Consider X I i = End inventory for quarter i

Minimize Z= 20X 1 +22X 2 +24X 3 +26X 4 + 3.5(I 1 +I 2 +I 3 ) 1

2

3

4

Demand

300

400

450

250

P K Sahoo, BITS-Hyderabad Campus

2

X i = Production in quarter i Solution: Consider X I i = End inventory for quarter i

Minimize Z= 20X 1 +22X 2 +24X 3 +26X 4 + 3.5(I 1 +I 2 +I 3 ) 1

2

3

4

Demand

300

400

450

250


2

Constraints: X 1 =300+I 1

I 1 +X 2 =400+I 2 I +X =250

= i

, = , , ,

i

, = , ,

I 0 =I 4 =0


3

Sec 2.3.3 Investment Q1. (2.2C) Fox Enterprises is considering six years. The expected (present value) returns and can undertake any of the projects partially or prorate both the return and outlays proportionately. F rm l h r l m LPP n rmin h optimal project mix that maximizes the total r rn I n r h im l fm n P K Sahoo, BITS-Hyderabad Campus

4

Cash outlay ($1000) Project

Year 1

Year 2 Year 3 Year 4 Return ($1000)

1

10.5

14.4

2.2

2.4

32.40

2

8.3

12.6

9.5

3.1

35.80

3

10.2

14.2

5.6

4.2

17.75

.

.

.

.

.

12.3

10.1

8.3

6.3

18.20

.

.

.

.

.

70.0

35.0

20.0

5

Available 60.0 ($1000)


5

Solution: Let xi = Undertaken portion of project i Max z= 32.4x +35.8x +17.75x +14.80x +18.2x +12.35x6 Subject to 10.5x1+8.3x2+10.2x3+7.2x4+12.3x5+9.2x6 ≤ 60 .

1

.

2

.

3

.

4

.

5

.

6

2.2x1+9.5x2+5.6x3+7.5x4+8.3x5+6.9x6 ≤ 35 2.4x1+3.1x2+4.2x3+5x4+6.3x5+5.1x6 ≤ 20 0≤ x j ≤ 1, j=1,2……6 P K Sahoo, BITS-Hyderabad Campus

6

Sec. 2.3.5 Blending and Refining Q4. (2.3 E) A refinery manufactures two grades of et fuel F1 & F2 b blendin four t es of gasoline, A, B, C & D. Fuel F1 uses gasolines A, B C & D in the ratio 1:1:2:4 and fuel F2 uses the ratio 2:2:1:3. The supply limits for A, B, C & D are 1000, 1200, 900 & 1500 bbl/da . The costs er bbl for gasolines A, B, C & D are $120, $90, $100 & $150. Fuels F1 & F2 sell for $200 and $250 er bbl. The minimum demand for F1 & F2 is 200 & 400 bbl/da . Determine the o timal roduction mix for F1 & F2. P K Sahoo, BITS-Hyderabad Campus

7

Solution: Let = o gaso ne n ue X Bi =bbl of gasoline B in fuel i o gaso ne n ue Ci = X Di =bbl of gasoline D in fuel i, i=1,2 ons er Y A =X A1 +X A2 F 1 =X 1 +X B1 +X C1 +X 1 B = B1 + B2 F 2 =X A2 +X B2 +X C2 +X D2 Y C =X C1 +X C2 D = D1 + D2 Ai


8

Max z = 200F1+250F2 –(120YA+90YB+100YC+150YD )

subject to the constraints X A1 =X B1

X A1 =0.5X C1

X A1 =0.25X D1

X A2 =X B2

X A2 =2X C2

X A2 =(2/3)X D2

Y A ≤ 1000, ≥ 200,

Y B ≤ 1200,

Y C ≤ 900,

Y D ≤ 1500

≥ 400


9

Q. Electra produces two types of electric

motors, each on a separate assembly line. The respective daily capacities of the two lines are 150 and 200 motors. Type I motor uses 2 ni f r in l r ni m n n n type II motor uses only 1 unit. The supplier day. The profits per motor of types I and II are an respec ve y. ormu a e e problem as a LPP and find the optimal daily production. P K Sahoo, BITS-Hyderabad Campus

10

motors and x2 type II motors per day. The objective is to find x1 and x2 so as to ax m ze

e pro

u ect to t e constra nts

=

1

2

≤ 150

x1

2 x1 + x2

≤

x1 , x2

0


≥

400 11

above problem. Step 1:Determination of the Feasible solution space The non-negativity restrictions tell that the . Then we replace each inequality constraint by an equality and then graph the resulting line notin that two oints will determine a line uniquely). P K Sahoo, BITS-Hyderabad Campus

12

divides the plane into two half-spaces and a on one a -space e cons ra n w e ≤ and on the other it will be ≥. To determine the “correct” side we choose a reference oint and see on which side it lies. (Normally (0,0) is chosen. If the constraint , , some other point.) Doing like this we would . P K Sahoo, BITS-Hyderabad Campus

13

Step 2:Determination of the optimal solution The determination of the optimal solution re uires the direction in which the ob ective function will increase (decrease) in the case of . find this by assigning two increasing ecreas ng va ues or z an en raw ng e graphs of the objective function for these two values. The optimum solution occurs at a point beyond which any further increase decrease of z will make us leave the feasible space. P K Sahoo, BITS-Hyderabad Campus

14

Graphical solution Maximize z=8 x1+5 x2 Optimum = a (100,200) 0,200) z=1200

(150,100)

z=1800 z=1000 z=400

(150,0)

z=1700


2 x + x x1

≤

400

≤

150

x2 ≤ 200 x1, x2≥ 0 x1 15

.

=

+

Subject to x1 + 3 x2 ≥ 15 2 x1 + 2 x2 ≥ 20 3 x1 + 2 x2 ≥ 24 x1, x2 ≥ 0


16

Graphical Solution

(0,12) z = 26 z = 30 z= 36 z = 39

46

z = 22.5

z = 42

Minimum at (7.5,2.5) P K Sahoo, BITS-Hyderabad Campus

(15,0) 17

Note: Corner point feasible solution: A corner point feasible solution is a solution that lies at the corner point of the feasible solution space.

Q. Maximize z = 2 x1 + x2 Subject to x1 + x2 ≤ 40 4 x1 + x2 ≤ 100 x1, x2 ≥

, P K Sahoo, BITS-Hyderabad Campus

18

,

z maximum at (20,20)

(25,0) P K Sahoo, BITS-Hyderabad Campus

19

Q. Maximize z = 2 Subject to

+ x ≤ 10

x2

2 x1 + 5 x2

≤

≤ 18

x1 + x2 1

x x

60

2

≥

0


20

z is maximum at (13, 5)

Max z = 31

(5,10) (10,8)

(0,10) z = 20 z = 10 z = 4

[1]

(13,5) z = 31 (14.6,0) [4]

[3]


[2] 21

Exceptional Cases Usually a LPP will have a unique optimal so ut on. ut t ere are pro ems w ere t ere may be no solution, may have alternative optimum solutions and “unbounded” solutions. We ra hicall ex lain these cases in the following slides. We note that the the “corners” of the set of all feasible points. P K Sahoo, BITS-Hyderabad Campus

22

n i

r h LPP

Maximize z = 10 x1 + 5 x2

x2

≤

200

2 x1 + x2

≤

400

1 P K Sahoo, BITS-Hyderabad Campus

,

2 23

Graphical solution Maximize z=10 x1+5 x2 2 x + x (100,200) 0,200) z=600

z maximum =2000 at z=1500

z=2000

z=1000 z=400

(150,100)

x1

≤

400

≤

150

x2 ≤ 200 x1, x2≥ 0 x1

(150,0) P K Sahoo, BITS-Hyderabad Campus

24

Thus we see that the objective function z is maximum at the corner 150,200 and also has an alternative optimum solution at , . that z is maximum at each point of the line . has an infinite number of (finite) optimum so ut ons. s appens w en t e o ect ve function is “parallel” to one of the constraint equations. P K Sahoo, BITS-Hyderabad Campus

25

Q. Maximize z = 5 x1 + 7 x2 Subject to 2 x1 -

1

x2 ≤ -1

+ 2 x2 ≤ -1

1,

2

No feasible solution


26

Q. Maximize z = x1 + x2 Subject to z=70

- x1 x2 - 3 x1 + x2 ≤ 30

=

x1, x2 ≥ 0 z=30

unbounded solution


=

27

Section 7.1

Let x = ( x1, x2) and y = ( y1, y2) be two points in t e x1 2 p ane. ny po nt = t 1, t 2 on t e ne segment joining them is of the form t 1 = (1 - λ) x1 + λ y1 2

=

-

x2

y2

for some λ: 0 ≤ λ ≤ 1 λ

= 0 corresponds the ‘left’ endpoint x

λ

= 1 corresponds the ‘right’ endpoint y P K Sahoo, BITS-Hyderabad Campus

28

More generally, let x=( x1, x2, …, xn) and y=( y1, y2, …, yn) be two points in the nim n i n l Th lin m n joining x and y is the set of all points of , 0 ≤ λ ≤ 1 in the sense that the ith coor na e o , name y i s g ven y t i = (1 - λ) xi + λ yi for all i = 1 , 2, …n.


29

Convex sets A subset S in the n-space Vn is called convex if x, y belong to S implies the line se ment oinin them also lies in S.

convex subset

NOT convex


30

In V n, the “half spaces” x=( x1, x2, …, xn) | a1 x1+a2 x2 + …+ an xn ≤ b x= x x

…

a x +a x + …+ a x

“ 1,

2,

b

” …,

n

1 1

2 2

…

n n

. P K Sahoo, BITS-Hyderabad Campus

31

Theorem: The intersection of any number of convex sets is a convex set. Corollary: The set of all feasible solutions o an s a convex su se .

Let S be a convex set. A oint t in S is called an extreme point of S if it is not . P K Sahoo, BITS-Hyderabad Campus

32

In other words, whenever x, y are two points in S, we cannot find a λ: 0 < λ < 1 such that t = (1 - λ) x + λ y Extreme point

Extreme point

Every point on the oun ary s an xtreme point


33

Theorem: Let S be a nonem t closed convex set that is either bounded from . at least one extreme point.


34

2

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