1. Decision variables: There are two products company producing A and B. So Decision Variables are A and B. Maximize Z = 20A+50B Subject to constraints: 1.) A>=80% of A+B A>=0.8A+0.8B 0.2A-0.8B>=0 2.) A<=100 3.) 2A+4B<=240 4.) 8B>=0 Non Negative variables are: A, B>=0 2. Decision variables: There are two Nutrition constituents in the products company producing A and B. so Decision variables are A and B. Maximize Z = 20A+40B Subject to constraints are: Vitamins and proteins in different nutrition Nutrition 1= 36A+6B>=108 or 6A+B>=18 2= 3A+12B<=36 or A+4B>=12 3= 20A+10B<=100 or 2A+B>=10 So LPP formulation is: Maximize Z = 20A+40B Subject to constraints are: 6A+B>=18 A+4B>=12 2A+B>=10 Non Negative variables are: A, B>=0 3. A manufacturer of steel ingots has plants at two cities A and B. The ingots are supplied to four markets C, D, E, and F. the manufacturer wants to know the pattern of shipment from each production plant to each market that would minimize the total transportation cost. You are given the following data on capacity and demand as also the unit transportation costs. Plant Capacities (units each) Origin 1 (plant A) = 3000 Origin 2 (plant B) = 1000 Destination Demands (units each) Destination 1 (market C) = 800 Destination 2 (market D) = 1200 Destination 3 (market E) = 300 Destination 4 (market F) = 1700 Unit Transportation Costs in Rupees: From Origin 1 to Destination 1 =5 From Origin 1 to Destination 2 =8 From Origin 1 to Destination 3 =2 From Origin 1 to Destination 4 =6 From Origin 2 to Destination 1 =7 From Origin 2 to Destination 2 =5 From Origin 2 to Destination 3 = 10 From Origin 2 to Destination 4 =4
Basic Feasible Solution by Vogel’s Approximation Method Plant
Markets C
A
D
5
8
7
Demand
5
3000
3
1 1
1
1000
1
1 -
-
1700
10 1000 1200
800
Col Penalty
6 300
200
Row Penalty
F
2
800 B
Supply E
4
300
1700
2
3
8
2
2 2 2
3 3 -
-
2 2 2
4000
Basic Feasible Solution by Vogel’s Approximation Method Plant
Markets C
A
5
D 8
2
800 B
7
Demand
6
10 1000 1200
F
300
200 5
800
Supply E
3000 1700
4 300
1000 1700
4000
Since the solution is not degenerate: No. of occupied cells = m – n – 1 = 5. We can apply the optimality test. Total transportation cost = 5x800 + 8x200 + 2x300 + 6x1700 + 5x1000 = Rs.21400 Evaluation of unoccupied cells: Cell BC BE BF
Closed Path BC-BD-AD-AC-BC BE-BD-AD-AE-BE BF-BD-AD-AF-BF
Evaluation + 7 – 5 + 8 – 5 = + 10 + 10 – 5 + 8 - 2 = + 11 +4–5+8–6= +1
Opportunity Cost -10 -11 -1
Since the opportunity costs for all unoccupied cells are negative therefore no cost advantage is possible for reallocation and the solution is optimal. The optimum cost is Rs.21400. 4. Solve the following transportation problem for maximization of profit. Warehouse
X Y Z Demand
A 12 8 14 180
Per unit Profit (Rs.) Market B C 18 6 7 10 3 11 320 100
Supply (units) D 25 18 20 400
200 500 300 1000
The given problem is that of maximization and to convert it into a problem of minimization we deduct from the highest cell value i.e. 25 every individual cell value and reconstruct transportation problem as follows and obtain basic feasible solution by VAM:
Warehouse
Market A
B
13
X
Supply (units)
C
7
Row Penalty
D
19
0
15
7
200
7
-
-
-
500
8
8
2
3
300
6
6
3
8
200 17
Y
18 100
11
Z
22
180 180 2
Demand Col Penalty
400 14
20 320 11
6 6 -
5
100 100 1
4 4 4
400 5
1 1 1
1000
2 -
The solution is not degenerate, no. of occupied cells = 6 = m + n – 1. Total profit = 18x200 + 7x100 + 18x400 + 14x180 + 3x20 + 11x100 = Rs.15180. We test this solution for optimality by using MODI test: Warehouse
Market A
B
13
X
17 -10
11
Z Demand vj
15
22
7
14
20 320 7
200
0
500
11
300
15
-4
-4
100
180 180 -4
0 -20
200 18
ui
D
19
-17
Y
C
7
Supply (units)
400 5 6
100 100 -1
400 -4
1000
Since all the ∆ij are not negative, opportunity cost to cell ZD is positive and reallocation through a close path drawn from this cell will offer benefit. Warehouse
Market A
X Y Z Demand
13
B 7
Supply (units) C
19
D 0
200 17 18 + 15 7 100 400 11 22 14 5 180 - 20 100 + 180 320 100 400
200 500 300 1000
Solution is not degenerate and we can go ahead with the optimality test. Warehouse
Market A
X
13
Y
17
B 200 18 -4
11
Demand vj
-14
15
22 320 7
0
500
11
300
9
-4
7 +1
14 -6
200
0
120
180 180 2
ui
D
19
-11
Z
C
7
Supply (units)
380 5
100 100 5
20 400 -4
1000
Cell YC has positive opportunity cost and will be considered for reallocation along a closed path drawn from this cell. Since the solution is not degenerate and we keep continue with the optimality test. Warehouse
Market A
X
B
13
7
19
-15
Y
17
18
11 180 180 2
Demand vj
22
-6
320 7
0
500
11
300
9
-4
7 100
14
200
0
15 120
ui
D
-15
200
-4
Z
C
Supply (units)
-1
100 4
280 5 120 400 -4
1000
All the cells have negative opportunity cost hence no cost advantage is available in reallocation. The solution is optimal. Total optimal profit = 18x200 + 7x120 + 10x100 + 18x280 + 14x180 + 20x120 = 3600 + 840 + 1000 + 5040 + 2520 + 2400 = Rs.15400 5. Maruti Machines Company has plants at Delhi, Kolkata and Mumbai. Its major distribution centres are located at Bangalore and Jaipur. The capacities of the three plants during the next quarter are 1000, 1500 and 1200 machines. The quarterly demand at the two distribution centres is 2300 and 1400 machines. The transportation cost per machine per km is Rs.0.08. The distance in km between the plants and distribution centres is as given below:
Delhi Kolkata Mumbai Bangalore Jaipur
Delhi 3000 2000 3000 500
Kolkata 3000 2500 2000 2500
Mumbai 2000 2500 1000 1500
Bangalore 3000 2000 1000 2500
Jaipur 500 2500 1500 2500 -
Give the minimum transportation cost distribution schedule in case the centre supply from all the sources could pass through any source or distribution, before it is redistributed. For this transshipment problem, buffer stock = total supply = total demand = 3700 units. Adding 3700 units to each supply/demand point, we get the transshipment table as follows and obtain basic feasible solution by using Vogel’s approximation method:
D D
K 240
0
M 160
B 240
J
40 3700 1000 240 0 200 160 200 3700 1100 400 160 200 0 80 120 3700 1200 240 160 80 0 200 3700 40 200 120 200 0 3700
K M B J Demand Column Penalty
3700 40 40 40 40 40 240
3700 160 -
3700 80 80 -
6000 80 80 80 80 40 80
Supply 4700
40
40
40
40
40
40
5200
160
40
40
40
40
40
4900
80
80
40
40
-
-
3700
80
80
200
-
-
-
3700
40
40
40
40
40
-
Row Penalty
5100 40 40 40 40 40 160
Since occupied cells are equal to no. of rows plus no. of columns minus one, solution is not degenerate. We may proceed with the optimality test. Total transportation cost = 3700x0 + 1000x40 + 3700x0 + 1100x160 + 400x200 + 3700x0 + 1200x80 + 3700x0 + 3700x0 = 40000 + 176000 + 80000 + 96000 = Rs.392000 D D K M B J Dem and vj
0 3700 240 -80 160 -240 240 -240 40 -80
K
M
240
B
160 -400
0
240 -240
200 0 80 -320
200
1100 1200
200 -240
5200
160
4900
80
3700
0
3700
-40
0 200
3700
-160
0
400 120
0
120 -440
200
80
ui
1000
-240
3700
-240 160
40
160 -120
3700 200
Supply 4700
J
-160 0 3700
-240
3700
3700
3700
6000
5100
0
-160
-80
0
40
Since all unoccupied cells have opportunity cost either negative or zero the solution is optimal but not unique. The optimal transportation schedule and minimum cost is as under: D K K M Optimal cost =
J B J B
1000 1100 400 1200
1000x40 = 40000 1100x160 = 176000 400x200 = 80000 1200x80 = 96000 Rs.392000
6. A manufacturer of complex electronic equipment has just received a sizeable contract and plans to subcontract part of the job. He has solicited bids for 6 subcontracts from 3 firms. Each job is sufficiently large and any firm can take only one job. The table below shows the bids as well as the cost estimate (in lakhs of rupees) for doing the job internally. Not more than three jobs can be performed internally. Find the optimal assignment that will result in minimum total cost. Firm Jobs 1 2 3 4 5 6 44 67 41 53 48 64 1 46 69 40 45 45 68 2 43 73 37 51 44 62 3 50 65 35 50 46 63 Internal There are 6 subcontracts and three could be undertaken internally, so we can reconstruct the assignment table as under: Firm Jobs 1 2 3 4 5 6 44 67 41 53 48 64 1 46 69 40 45 45 68 2 43 73 37 51 44 62 3 50 65 35 50 46 63 I1 50 65 35 50 46 63 I2 50 65 35 50 46 63 I3 Now it’s a balanced problem. First we do the row treatment: Firm Jobs 1 2 3 4 3 26 0 12 1 6 29 0 5 2 6 33 0 14 3 15 30 0 15 I1 15 30 0 15 I2 15 30 0 15 I3
5 7 5 7 11 11 11
Next, we do the column treatment and draw minimum lines to connect all the zeroes. Firm Jobs 1 2 3 4 5 0 0 0 7 2 1 3 3 0 0 0 2 3 7 0 9 2 3 12 4 0 5 6 I1 12 4 0 5 6 I2 12 4 0 5 6 I3
6 23 28 25 28 28 28
6 0 5 2 5 5 5
Since the minimum lines connecting all the zeroes are less than no. of rows or columns, we improve the assignment table and connect all the zeroes by minimum no. of lines Firm Jobs 1 2 3 4 5 6 0 0 2 7 2 0 1 3 3 2 0 0 5 2 1 5 0 7 0 0 3 10 2 0 3 4 3 I1 10 2 0 3 4 3 I2 10 2 0 3 4 3 I3
Since the minimum lines connecting all the zeroes are less than no. of rows or columns, we improve the assignment table and connect all the zeroes by minimum no. of lines Firm Jobs 1 2 3 4 5 6 0 0 4 7 2 0 1 3 3 4 0 0 5 2 1 5 2 7 0 0 3 8 0 0 1 2 1 I1 8 0 0 1 2 1 I2 8 0 0 1 2 1 I3 Since the minimum lines connecting all the zeroes are less than no. of rows or columns, we improve the assignment table and connect all the zeroes by minimum no. of lines Firm Jobs 1 2 3 4 5 6 0 1 5 7 2 0 1 3 4 5 0 0 5 2 1 6 3 7 0 0 3 7 0 0 0 1 0 I1 7 0 0 0 1 0 I2 7 0 0 0 1 0 I3 Since the minimum lines connecting all the zeroes are equal to no. of rows or columns, we can make the assignment at this stage in the following manner: Firm Jobs 1 2 3 4 5 6 0 1 5 7 2 0 1 3 4 5 0 0 5 2 1 6 3 7 0 3 0 7 0 0 1 0 I1 0 7 0 0 1 0 I2 0 7 0 0 0 1 I3 0 The optimal assignment schedule is: Firm 1 – Subcontract 1 – Cost Rs.44 lakhs Firm 2 – Subcontract 4 – Cost Rs.45 lakhs Firm 3 – Subcontract 5 – Cost Rs.44 lakhs Internally – Subcontract 2 - Cost Rs.65 lakhs Internally – Subcontract 3 – Cost Rs.35 lakhs Internally – Subcontract 6 – Cost Rs.63 lakhs Total optimal cost = Rs.296 lakhs 7. A company has four territories open and four salesmen available for assignment. The territories are not equally rich in their sales potential. It is estimated that a typical salesman operating in each territory would bring in the following annual sales: Territory I II III IV Annual Sales (Rs.) 60000 50000 40000 30000 The four salesmen are also considered to differ in ability; it is estimated that working under the same conditions, their yearly sales would be proportionately as follows: Salesman A B C D Proportion 7 5 5 4 If the criterion is maximum expected total sales, the intuitive answer is to assign the best salesman to the richest territory, the next best salesman to the second richest territory and so on. Verify this answer by the assignment method.
If A is working in territory I his sales will be: 7/21 of sales of territory I i.e. 7/21 x 60000 = Rs.20000, similarly we calculate sales of different salesman working in differtent terrotory and draw an assignment table as follows: (Sales amount in Rs.) Salesman Territory I II III IV A 20000 16667 13333 10000 B 14285 11904 9524 7143 C 14286 11905 9524 7143 D 11429 9524 7619 5714 It is a case of maximization. To convert it from the largest sales figure i.e. Rs.20000. Salesman I A 0 B 5715 C 5714 D 8571
to a case of minimization, deduct every cell value Territory II 3333 8096 8095 10476
Assignment problem is balanced. First do row treatment. Salesman Territory I II A 0 3333 B 0 2381 C 0 2381 D 8571 1905
III 6667 10476 10476 12381
IV 10000 12857 12857 14286
III 6667 4761 4762 3810
IV 10000 7142 7143 5715
Now do the column treatment and draw minimum number of lines to connect all zeroes. Salesman Territory I II III IV A 0 1428 2857 4285 B 0 476 951 1427 C 0 476 952 1428 D 8571 0 0 0 Since minimum no. of lines is less than no. of rows or columns we improve the matrix and draw minimum lines connecting zeroes. Salesman Territory I II III IV A 0 952 2381 3809 B 0 0 475 951 C 0 476 476 952 D 9047 0 0 0 Since minimum no. of lines is less than no. of rows or columns we improve the matrix and draw minimum lines connecting zeroes. Salesman Territory I II III IV A 0 952 2086 3334 B 0 0 0 476 C 0 476 0 477 D 9522 475 0 0
Since minimum no. of lines is equal to no. optimal solution: Salesman I A 00 B 0 C 0 D 9522
of rows or columns we make the assignments for Territory II 952 0 476 475
III 2086 0 0 0
IV 3334 476 477 0
The optimum assignment schedule is: A-I, B-II, C-III and D-IV The optimal sales = 20000.00 + 11904.76 + 9523.80 + 5714.29 = Rs.47142.85 The criterion is maximum expected total sales and the optimum schedule conforms to the intuitive answer that is to assign the best salesman to the richest territory, the next best salesman to the second richest territory and so on. 8. A traveling salesman has planned to visit 5 cities. He would like to start from a particular city, visit each city only once and return to the starting city. The traveling cost in rupees is given in the table below. Find the least cost route. From City To City A B C D E 7 5 3 5 A 7 8 4 3 B 5 8 6 2 C 3 4 6 2 D 5 3 2 2 E This traveling salesman problem can be solved as assignment problem. If the optimal assignment table also satisfies the additional constraints that no city is to be visited twice before completing the tour of all the cities, it is also the optimal solution to the given traveling salesman problem. If it does not, it can be adjusted. The upper bound for solution is from city A to B to C to D to E to A at the cost of 7 + 8 + 6 + 2 + 5 = Rs.28 As going from city A to A, B to B, etc. is not allowed assign a large penalty ∞ to these cells in the above table. The resulting table will have all diagonal elements ∞. Subtract the smallest element of each row from all the elements of the row and, if necessary, the smallest element of each column from all the elements of the column. This yields table no. 1 and no. 2. Also draw minimum no. of lines to connect all the zeroes in Table No. 2: Table No. 1 From City To City A B C D E ∞ 4 2 0 2 A 4 ∞ 5 1 0 B 3 6 ∞ 4 0 C 1 2 4 ∞ 0 D 3 1 0 0 ∞ E Table No. 2 From City To City A B C D E ∞ 3 2 0 2 A 3 ∞ 5 1 0 B 2 5 ∞ 4 0 C 0 1 4 ∞ 0 D 2 0 0 0 ∞ E No. of lines < no. of rows or columns, we improve the matrix: Table No. 3
From City
To City A B C D E ∞ 2 1 0 2 A 3 ∞ 4 1 0 B 4 0 2 4 ∞ C 0 0 3 ∞ 0 D 3 0 0 1 ∞ E No. of lines < no. of rows or columns, we improve the matrix: Table No. 4 From City To City A B C D E ∞ 1 0 0 2 A 2 ∞ 3 1 0 B 1 3 ∞ 4 0 C 0 0 3 ∞ 1 D 3 0 0 2 ∞ E No. of lines < no. of rows or columns, we improve the matrix: Table No. 5 From City To City A B C D E ∞ 1 0 0 3 A 1 ∞ 2 0 0 B 0 2 ∞ 3 0 C 0 0 3 ∞ 2 D 3 0 0 2 ∞ E No. of lines = no. of rows or columns, we can make assignments at this stage. Table No. 6 From City To City A B C D E ∞ 1 0 3 A 0 0 1 ∞ 2 0 B 0 2 ∞ 3 C 0 0 3 ∞ 2 D 0 3 0 2 ∞ E 0 The route for this assignment is: A to C to E to B to D to A and cost is 5 + 2 + 3 + 4 + 3 = Rs.17 This solution fulfills the condition of not visiting twice a city unless all the cities are visited and the solution is less than the upper bound solution hence it is optimal solution. 9. A company has two grades of inspectors, I and II to undertake quality control inspection. At least 1500 pieces must be inspected in an 8-hour day. A Grade I inspector can check 20 pieces in an hour with an accuracy of 96%. A Grade II inspector checks 14 pieces an hour with an accuracy of 92%. Wages of grade I inspector are Rs.5 per hour while those of grade II inspector are Rs.4 per hour. Any error made by an inspector costs Rs.3 to the company. If there are, in all, 10 grade I inspectors and 15 grade II inspectors in the company, find the optimal assignment of inspectors that minimizes the daily inspection cost. Formulate LPP. Solution: Decision Variables: Objective is to minimize the daily inspection cost. No. of inspections done by each Grade I and Grade II inspectors is given hence the number of variables are 2. Let A and B are the number of Grade I and Grade II inspectors respectively who are assigned task of inspection.
Objective Function: Table Showing Inspection Cost Per Inspector Grade I Inspector No. of inspections per hour Daily hours of work Total inspection per day Erroneous inspection per day Wages Cost to the Company for error Daily cost of inspection
Grade II Inspector
20 8 Hours 20x8 = 160 160x4/100 = 6.4 8x5 = Rs.40 6.4x3 = Rs.19.2
14 8 Hours 14x8 = 112 112x8/100 = 8.96 8x4 = Rs.32 8.96x3 = Rs.26.88
40 + 19.2 = Rs.59.2
32 + 26.88 = Rs.58.88
Objective function can be expressed algebraically as follows: Minimize
Z = 59.20A + 58.88B
Constraints: Constraints can be formulated as follows: For no. of inspections constraint: 160A + 112B > 1500 For no. of Grade I inspector constraint: A < 10 For no. of Grade II inspector constraint: B < 15 Hence the LPP is: Minimize Z = 59.20A + 58.88B Subject to: 160A + 112B > 1500 A < 10 B < 15 A, B > 0 10. Decision Variables: The companys objective is to fulfill the requirements of special feed Protein and fiber. So Decision Variables are Symbolized by A and B Objective Function: Minimize Z = 0.3A+0.05B Subject to constraints are: 0.09A+0.02B>=0.3 0.6A+0.06B>=0.9 Non Negative variables are: A, B>=0 Y 2.5 2 1.5 1 0.5 X 0.5
1
1.5
2
2.5
3
3.5
So Minimize Z= 0.3A+0.05B (A, B) (0, 2.5)=0.3(0) + 0.05(2.5) = 0.125 (3.33, 0)= 0.3(3.33) + 0.05(0) = 1 (0.75, 0.4)= 0.3(0.75) + 0.05(0.4) = 0.245 The minimum cost is Rs.0.125*800=Rs. 100 11. The manager of an oil refinery has to decide upon the optimal mix of two possible blending processes of which the inputs and outputs per production run are as follows: Process 1 2
Input Crude A 5 4
Output Crude B 3 5
Gasoline X 5 4
Gasoline Y 8 4
The maximum amounts available of crude A and crude B are 200 units and 150 units respectively. Market requirements show that at least 100 units of gasoline X and 80 units of gasoline Y must be produced. The profits per production run from process 1 and process 2 are Rs.3 and Rs.4 respectively. Formulate the problem as a linear programming problem. Solution: Decision Variables: Profit per production run of Process 1 and Process 2 is given and objective is to maximise the profit of the oil refinery hence there are two decision variables. Let the oil refinery runs A, and B number of runs of Process 1 and Process 2 respectively for attaining the objective. Objective Function: The objective function can be stated algabreically as follows: Maximize Z = 3A + 4B Constraints: For input of crude A availability constraint: 5A + 4B < 200 For input of crude B availability constraint: 3A + 5B < 150 For market requirement constraint of output X: 5A + 4B > 100 For market requirement constraint of output Y: 8A + 4B > 80 Where A, B > 0 Hence the LPP is: Maximize Z = 3A + 4B Subject to: 5A + 4B < 200 3A + 5B < 150 5A + 4B > 100 8A + 4B > 80 A, B > 0
12. Write the dual L.P Problem for the following primal LP Problem: Minimize Z= 4X + Y Subject to: 3X + Y = 2 4X + 3Y>=6 X + 2Y<=3 X, Y>=0
Rewrite the given primal LPP as follows: Minimize Z= 4X + Y Subject to: 3X + Y>= 2 -3X - Y>= -2 4X + 3Y>=6 -X - 2Y>=-3 The dual of the given problem is as follows: Maximize G= 2Y1-2Y2+6Y3-3Y4 Subject to: 3Y1-3Y2+4Y3-Y4<=4 Y1-Y2+3Y3-2Y4<=1 Y1, Y2, Y3, Y4>=0 We can combine third and fourth constraints and the final dual LPP is Maximize G= 2Y+6Y3-3Y4 Subject to: 3Y+4Y3-Y4<=4 Y+3Y3-2Y4<=1 Y with unrestricted sign, Y3, Y4>=0 13. Is 50 20 55 30 35 25 An optimal solution of the transportation problem: A B C A 6 1 9 B 11 5 2 C 10 12 4 Required Units 85 35 50 If not, modify it to obtain the optimal solution.
D 3 8 7 45
Supply 70 55 90 215
Solution:
A
A 6
B
11
C
10
Required Units Vj
85 6
0 5 80
B 1 5 12 35 1
35 1 -7
C 9 2 4 50 -3
-12 50 -3
D 3 8 7 45 3
35 0 10
Supply 70
Ui 0
55
5
90
4
215
Here the solution is not an optimal solution, so we have to modify this solution for getting an optimal solution, No. of occupied cells= no. of Row + no. of Column – 1 6=3+4-1 So the solution is not degenerate, so we can proceed for the optimality test Total transportation cost=(35*1)+(35*3)+(5*11)+(50*2)+(80*10)+(10*7)=1165 A 6
A B C Required Units
11 5 10 + 80 85
B 1 5 + 12
35
C 9 2
50
4
35
50
D 3 + 8
35
7 45
10
Supply 70 55 90 215
Now we have to proceed for the optimality A 6
A B C Required Units
11 10 + 85 85
B 1 5 + 12
30 5
C 9 2 4
35
50
50
D 3 + 8
40
7 45
5
Supply 70 55 90 215
Now the optimal transportation cost is Rs. 1160 (30+120+25+100+850+35) 14. A firm marketing a product has four salesmen S1, S2, S3, and S4. There are three customers C1, C2, and C3. The probability of making a sale to a customer depends upon the salesmancustomer support. The table below represents the probability with which each of the salesman can sell to each of the customers: Customer C1 C2 C3
Salesmen S1 0.7 0.5 0.3
S2 0.4 0.8 0.9
S3 0.5 0.6 0.6
S4 0.8 0.7 0.2
If only one salesman is to be assigned to one customer, what combination of salesmen and customers shall be optimal? Profit obtained by selling one unit to C1 is Rs. 500, to C2 is Rs.450 and to C3 is Rs.540. What is the total expected profit? Solution: Multiply each customers profit value with the probability of each salesman and customer Customer Salesmen S1 S2 S3 S4 C1 0.7*500 0.4*500 0.5*500 0.8*500 C2 0.5*450 0.8*450 0.6*450 0.7*450 C3 0.3*540 0.9*540 0.6*540 0.2*540
We have to assign on salesmen to one customer Since there are four salesman and three customers we have to add one customer (dummy cell) So the matrix (table) will be as follow: Customer Salesmen S1 S2 S3 S4 C1 350 200 250 400 C2 225 360 270 315 C3 162 486 324 108 C4 0 0 0 0 Row Deduction Customer
Salesmen S1
C1 C2 C3 C4
S3
S4
150
S2 00
50
200
0
135
45
90
54
378
216
0
0
0
0
0
Optimal Assignment C1 to S2 =200 C2 to S1 =225 C3 to S4 =108 C4 to S3 =000 Total Expected Profit is Rs. 533 15. Management of a company has decided to have a state-of-the-art fiber-optic network installed to provide high-speed communications between its major centers. The nodes shown in the figure show the geographical layout of the corporation’s major centers. The lines are the potential locations of fiber-optic cables. The number next to each line gives the cost (in lakh rupees) if that particular cable is chosen for installation. Any pair of centers does not need to have a cable directly connecting them in order to take full advantage of the fiber-optic technology for high-speed communications between these centers. All that is necessary is to have a series of cables that connect the centers. The problem is to determine which cables should be installed to minimize the total cost of providing high-speed communications between every pair of centers.
B G
7 2
2 E
4
5
A
5
C 1
4
7
1
3
D
F 4
SOLUTION: 01. If we start from the node A, consider that link which has the least cost. Since A is connected to three different nodes i.e. B, C, and D and the cost of links are 2, 5, and 4 million dollars, we consider connecting A to B. 02. From node B it is possible to connect C and E at respective costs of 2 and 7 million dollars. We select the link BC being less costly. 03. From node C connection is possible to the nodes D, E and F at respective costs of 1, 4 and 3 million dollars. We select the least costly link that is CD. 04. To connect node F, we have links from nodes D, C, and E. But we donot consider link EF for the reason that node E is not yet connected. Out of the remaining two links CF is less costly hence selected. 05. Link FE to connect node E is the preferable choice for the cost advantage. And finally to connect G the obvious choice is link EG. 06. The solution with selected links is graphically shown herebelow. Notice that all the nodes are connected with the cables and the optimal cost is 2 + 2 + 1 + 3 + 1 + 5 = $14 million.
B G 2
2
5 E
A
C 1
D
1
3
F
16. A small furnishing company manufactures tables and chairs. Each chair requires 4 man-hours of labour while each table requires 5 man-hours of labour. If only 80 man-hours are available in a week and the owner of the company would neither hire additional labour nor utilize overtime, then formulate the linear goal programming problem. Both the table and the chair fetch a profit of Rs.100 each. The owner has a target to earn a profit of Rs.2000 per week. Also he would like to supply 10 chairs, if possible per week to a sister concern. Solution: The goals of the company and their assigned priorities are: Priority Goal 1st To avoid hiring extra labour or utilize overtime nd 2 To reach a profit goal of Rs.2000 a week 3rd To supply 10 chairs a week to the sister concern Let X1 and X2 denote the number of chairs and tables to be produced per week. If Ui and Oi represent the amount by which the ith goal is under achieved and over achieved respectively, then the goals (expressed as constraints in order of priority) are: g1: 4X1 + 5X2 + U1 – O1 = 80, g2: 100X1 + 100X2 + U2 – O2 = 2000, g3: X2 + U3 – O3 = 10,
(manhour goal) (profit goal) (chair goal)
To achieve the first goal completely, the overtime O1 must be minimized. Similarly, the second goal will be fulfilled when the underachievement in profit, U2 is minimized and the chair goal will be attained when underachievement U3 is minimized. Therefore, the objective function of the problem may be written as: Minimize Z = {O1, U2, U3} Subject to: 4X1 + 5X2 + U1 – O1 = 80, 100X1 + 100X2 + U2 – O2 = 2000 X2 + U3 – O3 = 10, X1, X2, U1, U2, U3, O1, O2, O3 > 0 17. An airline company decided to expand its operations. The basic issue facing management is whether to purchase more small airplanes to add some new short flights or to start moving into national market by purchasing some large airplanes for new cross-country flights (or both). Many factors will go into management’s final decision, but the most important one is which strategy is likely to be most profitable. How many airplanes of each type should be purchased to maximise the total net annual profit? The following information is available:
Net annual profit per airplane Purchase cost per airplane Maximum purchase quantity
Small Airplane $ 1 million $ 5 million 2
Large Airplane $ 5 million $ 50 million No maximum
Capital Available $ 100 million
18. Answer the following: i. If a primal LP problem has constraints with sign of equality, the dual LP problem will have ____________ variables. Explain the term you write in the blank space. Ans: Unrestricted ii. In a transportation problem, if there are no transport links available between a supply and a demand centre, how are they resolved? Ans: Infinite value iii. How a maximization transport problem is converted into a minimization problem. Ans: Highest value from the matrix- all lowest values iv. What is a degenerate solution in transportation problems? How degeneracy is removed before optimality test is conducted? Ans: No of occupied cells does not eqials to No.of Row+No. of column-1 Degeneracy is removed by putting the Epsilon or Zero value(0) v. At what stage optimal assignments can be made in assignment problems? 19. An auto manufacturer company believes that its most likely customers are high-income women and men. To reach these groups, the company has embarked on an ambitious TV advertising campaign and has decided to purchase 1-minute commercial spots on two types of programmes: comedy shows and cricket games. Each comedy commercial is seen by 7 million high-income women and 2 million high-income men. Each cricket game is seen by 2 million high-income women and 12 million high-income men. A 1-minute comedy ad costs Rs.50000 and a 1-minute cricket ad costs Rs.100000. The company would like the commercials to be seen by at least 28 million high-income women and 24 million highincome men. Use linear programming to determine how the company can meet its advertising requirements at minimum cost. Solve the problem by graphical method. Formulate dual LPP. Solution: Decision Variables: The company has embarked on an ambitious TV advertising campaign and has decided to purchase 1-minute commercial spots on two types of programmes: comedy shows and cricket games so it is symbolized as A and B. Objective Function: Minimize Z= 50000A+100000B Subject to constraints: 7A+2Y>=28 2A+12B>=24 Non Negative Variables: A, B >=0 20. Given below is the unit costs array with supplies and demands. Find the optimal solution. Source I II III Demand Solution:
Sink I 8 9 25
Supply II 10 9 11 32
III 7 4 10 40
IV 6 7 8 23
50 40 30 120
Source I
Sink I 8
II
infinite
III
9
Demand Vj
25
II 10 9
0
25 8
III 7
2 -2
11
4
0 40
10
-2
30 32 10
40 7
IV 6 7 8
23 -4 -1
23 6
Supply
Ui
50
0
40
-3
30
1
120
No. of occupied cell (5) does not equals to 4+3-1. The solution is not degenerate. So Epsilon or Zero had been put at independent cell. Independent cell is that at which closed path had not been made. Transportation cost: 8*25=200 2*10=20 6*23=138 4*40=160 11*30=330 Total transportation cost is Rs. 848 21. Well-done Company has taken the 3rd floor of a multi- storey building for rent with a view to locate one of their zonal offices. There are five main rooms in this floor to be assigned to five managers. Each room has its own advantages and disadvantages. Some have window, some are closer to the washrooms or to the canteen or secretarial pool. The rooms are all of different sizes and shapes. Each of the five managers was asked to rank their room preferences amongst the rooms 301, 302, 303, 304 and 305. Their preferences were recorded in a table as indicated below: Managers M1 302 303 304
M2 302 304 305 301
M3 303 301 304 305 302
M4 302 305 304 303
M5 301 302 304
Solution: Each of the five managers was asked to rank their room preferences amongst the rooms 301, 302, 303, 304 and 305. So the rooms assigned as follows
301 302 303 304 305
M1 M 1 2 3 M
M2 4 1 M 2 3
M3 2 5 1 3 4
M4 M 1 4 3 2
M5 1 2 M 3 M
Row Deduction M1
M2
M3
M4
M5
M
3
1
M
0
0
0
4
1
M
00
1
00
1
M
1
2
301 302 303 304 305
Manager 1 2 3 4 5
0
1
3
M
1
1
0 0
M
Room 302 304 303 305 301
22. The ABC Oil Company has four refineries and the pipelines between these refineries have the following capacities, million of barrels per day:
From Refinery 1 2 3 4
To Refinery 1 --
2 150 --
3 370 190 --
4 240 310 280 --
Determine the maximum possible daily amount of oil which can be piped between refineries 1 to 4, utilizing all possible flow paths. From your results if the flow capacity for the pipeline between refinery 2 and refinery 3 were increased by 150 barrels per day, by how much would the maximum possible flow from refinery 1 to refinery 4 increases? Solve by networking model.
1
370
150
240
2 190+150 3
280
310 4
Since to maximum refinery flow from 1 to 4 is (370+190+310) 870
If we increase 2 to 3 refinary increase by 150 than new path will be 1 to 3 to 2 to 4 (370+190+150+310) 1020 23. A company produces cricket bats. The company has two production lines, the production rate for line 1 is 60 bats per hour and for line 2 it is 75 bats per hour. The company has entered into a contract to supply. The production manager of the company is trying to determine the best daily operation hours for two lines. He sets the priorities to achieve his goals as given below: Goal 1 : Produce and deliver 1250 bats daily Goal 2 : Limit the daily overtime operations of line 2 to 3 hours Goal 3 : Minimize underutilization of the regular daily operation hours of each line Goal 4 : Minimize the daily overtime operation hours of each line as much as possible
24. United Telecom Company is in the process of providing cable service to five newly developed housing areas. The starting point of creating linkages to other five areas will be area A. Cable length required to connect one area with another is given below. Using minimal spanning network find out the minimum cable length required to connect all the areas. A B C D E F A -2 5 7 10 Infinity B 2 -Infinity 4 3 Infinity C 5 Infinity -5 Infinity 10 D 7 4 5 -8 3 E 10 3 Infinity 8 -Infinity F Infinity Infinity 10 3 Infinity --
C
5
10
F E 10
3
5
3 8
B 2 A
4
7 D
The minimum cable length required to connect all the areas: The cable connected A to B to E, B to D, A to C, D to f So the minimum cable length is 17 (2+3+5+4+3) 25. A company has four manufacturing and five warehouses. Each plant manufactures the same product which is sold at different prices in each warehouse area. The cost of manufacturing and cost of raw materials are different in each plant due to various factors. The capacities of the plants are also different. The data are given in the following table. Plant Item 1 2 3 4 Manufacturing cost (Rs. Per unit) 12 10 8 8 Raw material cost (Rs. Per unit) 8 7 7 5 Capacity per unit time 100 200 120 80 The company has five warehouses. The sales prices, transportation costs and demands are given in the following table. Warehouse Transportation cost (Rs.) per unit Selling price per Demand unit Plant 1 Plant 2 Plant 3 Plant 4 A 4 7 4 3 30 80 B 8 9 7 8 32 120 C 2 7 6 10 28 150 D 10 7 5 8 34 70 E 2 5 8 9 30 90 Formulate the problem as a transportation problem to maximize profit. Find the optimum solution. Solution: Profit = Selling price per unit – {Manufacturing cost (Rs. Per unit) + Raw material cost (Rs. Per unit) + Transportation cost (Rs.) per unit}
A B C D E Supply
Plant 1 6 4 6 4 8 100
Plant 2 6 6 4 10 8 200
Plant 3 11 10 7 14 7 120
Cost table (highest- all lowest values) A B C A 8 10 8 0 -4 B
8
C
3
D
0
2 -3
8 4 3
80 E
14
E
14
70 50 -5 -2
10
10 130
7 9 14
-1 -9 10
Plant 4 14 11 5 13 8 80
D 10 4 0 1 14
-8 0 70 -7 -6
Demand 80 120 150 70 90 510
0 0 0 0 0 10
E 6 6 7 6 14
90 -2 -3 -8 -2
Supply 100
Ui 0
200
2
120
-2
80
-8
10
6
Demand 80 120 150 70 90 510 Vj 8 6 8 2 6 Since the solution is Degenerate we have to find the cell from where the loop can not be possible. And consider the value Epsilone or Zero Now the solution is not degenerate so we can proceed for the optimality test. At the cell BA the opportunity cost is positive so we have to relocate that value
A
A 8
B
8
B 10
0
8 +
C
3
D
0
-3
4 3
80 E
14
Demand Vj
80 8
-
0
14 120 6
-4 70 50 -5 -2
C 8
10
D 10
10 - 130 7 -1
4
9
1
14 + 150 8
-9 10
0
14 70 2
-8 0 70 -7 -6
E 6 6 7 6 14 90 6
90 -2 -3 -8 -2
Supply 100
Ui 0
200
2
120
-2
80
-8
10
6
510
If we reallocate Zero than values are same (as it is) so the optimal solution is 10*8 =80 70*8 =560 10*130 =1300 50*4 =200 10*14 =140 90*6 = 540 Total transportation cost is Rs. 2820
26. Mr. Iyer is a salesman and wants to vist six cities, say, 1, 2, 3, 4, 5, and 6, starting with city 1 where he is stationed. The distances between various cities are given in the following table. Solve it as a traveling salesman problem. To City From City 1 2 3 4 5 6 1 -25 18 35 50 39 2 21 -28 16 30 13 3 22 28 -14 16 20 4 35 12 14 -12 12 5 50 30 16 12 -8 6 39 15 20 12 7 -Solution: Upper Bound = 1 to 2 to 3 to 4 to 5 to 6 to 1 25+28+14+12+8+39 =126 Row Deduction: To City From City 1 1 M 2 8 3 12 4 23 5 42 6 32
2 7 M 14 0 22 8
3 0 15 M 2 8 13
4 17 3 0 M 4 5
5 32 17 2 0 M 0
6 21 0 6 0 0 M
Column Deduction To City From City 1 2 3 4 5 6
1 M 0 4 15 34 24
2 7 M 14 0 22 8
3 0 15 M 2 8 13
4 17 3 0 M 4 5
5 32 17 2 0 M 0
6 21 0 6 0 0 M
Distance is 1 to 3 to 4 to 2 to 1 and 5 to 6 to 5 18+21+14+12 8+7 65 The routes we get in the above assignment: 1 to 3 to 4 to 2 to 1 and 5 to 6 to 5 And the cost of these sub-routes 18+21+14+12 8+7 Since the solution for the whole route = 65, it is lower bound Now we break the shorter route into two sub-routes i.e.5 to 6 and 5 to 6 First we evaluate sub-route 5 to 6 by making 6 to 5 unacceptable. To make T to P unacceptable put a large value ∞ of transportation cost in the cell TP in the Table No. 3 and solve it again for assignment: First we do column treatment on Table No. 3 and then draw minimum no. of lines to connect all the zeroes.
To City From City 1 2 3 4 5 6
1 M 0 4 15 34 24
2 7 M 14 0 22 8
3 0 15 M 2 8 13
4 17 3 0 M 4 5
5 32 17 2 0 M M
6 21 0 6 0 0 M
4 17 3 0 M 0 1
5 32 17 2 0 M M
6 25 4 10 4 0 M
4 17 3 0 M 0 0
5 32 17 2 0 M M
6 25 4 10 4 0 M
4 19 5 0 M 2 0
5 32 17 0 0 M M
6 25 4 8 4 0 M
4 19 5 0 M 2 0
5 32 17 0 0 M M
6 25 4 8 4 0 M
Improvement: 1 To City From City 1 2 3 4 5 6
1 M 0 4 15 30 20
2 7 M 14 0 18 4
3 0 15 M 2 4 9
1 M 0 4 15 30 19
2 7 M 14 0 18 3
3 0 15 M 2 4 8
Improvement: 2 To City From City 1 2 3 4 5 6 Improvement: 3 To City From City 1 2 3 4 5 6
1 M 0 2 15 30 17
2 7 M 12 0 18 1
3 0 15 M 2 4 6
From City 1 2 3 4 5 6
1 M 0 2 15 30 17
2 7 M 12 0 18 1
3 0 15 M 2 4 6
To City
The route that we get is the above assignment: 1 to 3 to 5 to 6 to 4 to 2 to 1 And the cost of this route = 18+21+16+12+8+12 = 87
27. A company makes two kinds of leather belts. Belt A is a high quality belt and belt B is of lower quality. The respective profits are Rs.20 and Rs.15 per belt. Each belt of type A requires twice as much time as belt of type B, and if all belts were of type B, the company could make 1000 per day. The supply of leather is sufficient for only 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 per day are available. There are only 700 buckles a day available for belt B. What should be the daily production of each type of belts to maximize profit? Solution: Decision Variables: A company makes two kinds of leather belts. Belt A is a high quality belt and belt B is of lower quality so that are symbolized as A and B. Objective Function: Maximize Z= 20A+15B Subject to Constraint: 2A+B>=1000 A+B<=800 A<=400 B<=700 A, B<=0 28. A manufacturer must produce a product in sufficient quantity to meet contractual sales in the next four months. The production capacity and unit cost of production vary from month to month. The product produced in one month may be held for sales in later months, but at an estimated storage cost of Rs.2.00 per unit per month. No storage cost is incurred for goods sold in the same month in which they are produced. There is no opening inventory and none is desired at the end of the four months. The necessary details are given in the following table. The management wishes to know how much to produce each month in order to minimize total cost. Month Contracted Sales Maximum production Unit cost of production (Rs.) 1 200 400 28 2 300 500 32 3 500 300 30 4 400 500 34 Formulate the problem as a transportation problem by constructing the appropriate cost and requirements table and solve the problem. Solution:
1
1 28
2
M
32
3
M
M
30
4
M
M
M
Demand
200
2 30
200 100
3 32 34
0 400 100
4 34 36 32 34
0 0 200 200
5 0 0 0 0
0 2 -2 300
Supply 400
0
500
2
300
-2
500
0
200 300 500 400 300 1700 28 30 32 34 0 Transportation cost= 5600+6000+3200+13600+3000+6400+6800=44600
1
1 28
2
M
32
3
M
M
30
4
M
M
M
34
Demand
200 28
300 30
500 32
400 34
200
2 30
200 100
3 32 34
4 34
0
0
36
200
5 0
0
32
300
0
0
200
0 400
-2
0
100
300 0
Supply 400
0
500
2
300
-2
500
0
1700
Transportation cost= 5600+6000+3200+6800+9000+13600=44200 29. The personnel manager of ABC Company wants to assign X, Y and Z to regional offices. But the firm has an opening in the Chennai office and would send someone of the three to that branch if it were more economical than a move to Delhi, Mumbai or Kolkata. It will cost Rs.3000, Rs.2100 and Rs.4500 respectively to relocate Mr. X, Mr. Y, and Mr. Z to Chennai. Cost of assigning the three persons to three different offices is given in the following table. What is the optimum assignment of personnel to offices? Persons
Offices Delhi 2400 1500 1500
X Y Z
Mumbai 3300 4800 3000
Kolkata 3600 3900 6900
Solution:
X Y Z Dummy
Delhi 2400 1500 1500 0
Mumbai 3300 4800 3000 0
Kolkata 3600 3900 6900 0
Chennai 3000 2100 4500 0
Delhi 0 0 0 0
Mumbai 900 3300 1500 0
Kolkata 1200 2400 5400 0
Chennai 600 600 3000 0
Row Deduction X Y Z Dummy
There is not having optimal solution so we have to improve it
X Y Z Dummy
Delhi 0 0 0 600
Mumbai 300 2700 900 0
Kolkata 600 1800 4800 0
Chennai 0 0 2400 0
Improvement: 2
X Y Z Dummy
Delhi 0 0 0 900
Mumbai 0 2400 600 0
Kolkata 300 1500 4500 0
Chennai 0 0 2400 300
X to Mumbai = 3300 Y to Chennai = 2100 Z to Delhi = 1500 Total =6900 30. A firm produces Grade-A and Grade-B oil from two raw materials, M1 and M2. The following table provides the basic data of the problem: Tons of Raw Material per Maximum Daily ton of: Availability Grade-A Oil Grade-B Oil (Tons) Raw Material M1 3.6 2.4 14.4 Raw Material M2 0.6 1.2 3.6 Profit per ton (Rs ‘000) 2.5 2.0 A market survey indicated that the daily demand for Grade-A oil be exceed that of Grade-B oil by more than 1 ton. Also the maximum daily demand of Grade-B oil is 2 tons. Formulate this as a linear programming problem Decision Variables: A firm produces Grade-A and Grade-B oil so that are symbolized as A and B. Objective Function: Maximize Z= 2.5A+2.0B Subject to Constraint: 3.6A+2.4B>=14.4 0.6A+1.2B<=3.6 A>=B+1 OR A-B>=1 B<=2 A, B<=0
Ans: 10.5 Maximize Z= 2.5A+2B (1, 0) 2.5(1)+2(0) = 2.5 (3, 1.5) 2.5(3) + 2(1.5) =10.5 (4,0) 2.5(4) + 2(0) = 10 31. Obtain the dual of the LPP given here: Maximize Z = 8X1 + 10X2 + 5X3 Subject to: X1 – X3 < 4 2X1 + 4X2 < 12 X1 + X2 + X3 > 2 3X1 + 2X2 – X3 = 8 X1 , X2 , X3 > 0 Solution: We replace the variable X3 with difference of two non-negative variables, X4 - X5 and rewrite the given primal LPP as follows: Maximise Z = 8X1 + 10X2 - 5X3 Subject to: X1 - X3 < 4 X1 + 4X2 < 12 -X1 - X2 - X3< 2 -3X1 - 2X2 + X3 < -8 3X1 + 2X2 - X3 < 8 X1 X2, X4, X5 > 0
The dual LPP can be written as follows: Minimise G = 4Y1 + 12Y2 + 2Y3 + 8Y4 – 9Y5 Subject to:
Y1 + Y2 - Y3 - 3Y4 + 3Y5 > 8 4Y2 - Y3 - 2Y4 +2Y5 > 10 Y1 - Y3 + Y4 – Y5 > -5 Y1, Y2, Y3, Y4, Y5 > 0
32. A company manufactures two products, radios and transistors, which must be processed through assembly and finishing departments. Assembly has 90 hours available; finishing can handle up to 72 hours of work. Manufacturing one radio requires 6 hours in assembly and 3 hours in finishing. Each transistor requires 3 hours in assembly and 6 hours in finishing. If profit is Rs.120 per radio and Rs.90 per transistor, formulate the linear goal programming problem to realize profit of Rs.2100 This is a single goal (profit of Rs.2100) problem. Let X1 = production of no. of units of radios for achieving the profit objective X2 = production of no. of units of transistors for achieving the profit objective Du = amount by which profit goal is under achieved Do = amount by which profit goal is over achieved The given problem can be expressed as the following goal problem: Minimize Z = Du, (under achievement of the profit target) Subject to: 120X1 + 90X2 + Du - Do = 2100 [Since profit earned + underachievement – overachievement = target profit] 6X1 + 3X2 < 90 [assembly department constraint] 3X1 + 6X2 < 72 [finishing department constraint] X1, X2, Du, Do > 0 33. Solve graphically the following LPP: Maximise Z = 8X + 16Y Subject to: X + Y < 200 X < 125 3X + 6Y < 900 X, Y > 0 Solution: Draw graph by your own Coordinators (0,150) (70,120) (125,70) (125,0)
Z= 8x + 16y 8(0)+16(150) = 2400 8(70)+16(120) = 2480 8(125)+16(70) = 2120 8(125)+16(0) = 1000
34. A manufacturer of ties produces four types of ties. The following table illustrates the cost and availability of the three materials used in the production process: Material
Cost per Meter (Rs.) 1050 300 450
Silk Polyster Cotton
Material availability per month (Meter) 800 3000 1600
The firm has fixed contracts with several major departmental stores to supply ties. The contracts require the manufacturer to supply a minimum quantity of each tie but allow for a larger demand if the manufacturer chooses to meet that demand. The following table summarizes the contract demand for each tie, the selling price per tie and the fabric requirement of each variety. Formulate this as a linear programming problem. Variety of Tie All Silk All Polyster Poly-cotton Blend-1 Poly-cotton Blend-2
S.P. per Tie (Rs.) 335 180
Monthly contract minimum 5500 9500
Material required per tie (Mts) 7000 0.125 14000 0.080
Monthly demand
215
12500
16000
240
5500
8500
Material requirement
100% silk 100% polyster 50% polyster & 0.100 50% cotton 30% polyster & 0.100 70% cotton
Solution: Decision Variables: Company produces four types of products that are silk, Polyster, Poly-cotton Blend-1, Poly-cotton Blend-2 so decision variables are A, B, C, D Objective function: Material * cost Total cost Selling Price - Total Cost A 0.125*1050 131.25 335-131.25 B 0.08*300 24 180-24 C (50% of 0.1*300) + 37.5 215-37.5 (50% of 0.1*450) D (30% of 0.1*300) + 40.5 240-40.5 (70% of 0.1*450) Max Z= 203.75A+156B+177.5C+199.5D Subject to constraints: Monthly contract minimum A>=5500 B>=9500 C>=12500 D>=5500 Monthly demand A<=7000 B<=14000 C<=16000 D<=8500 Material availability Silk: 0.125A<=800 Polyster: 0.08B + 0.05C + 0.03D <=3000 Cotton: 0.05C + 0.07D <=1600 A, B, C, D >=0
Profit 203.75 156 177.5 199.5
35. Given the following traffic flows, in hundreds of cars per hour. Draw the network diagram and find out what is the maximum traffic flow from City 1 to City 7? S. No. 1 2 3 4 5 6 7 8 9 10 11 12
From City To City
Flow
S. No.
From City To City
Flow
1 1 1 2 2 2 3 3 3 4 4 4
4 8 5 0 3 3 0 3 1 3 3 4
13 14 15 16 17 18 19 20 21 22 23 24
5 5 5 5 5 5 6 6 6 7 7 7
1 0 2 0 1 5 1 4 6 2 1 0
2 3 5 1 4 5 1 5 6 2 5 7
1 2 3 4 6 7 3 5 7 4 5 6
1 to 3 to 6 to 5 to 2 to 4 to 7 8+1+4+0+3+4 =20 Maximum Traffic Flow is 20 36. A company produces three products A, B, and C. All these are processed in a central plant. Production of one unit of A needs 2 hours, while production of a unit of B and C requires respectively 3 and 1 hours. The regular plant capacity is 40 hours per week. The marketing department has informed the maximum sales per week of A, B and C is 10, 10 and 12 units respectively. The chief executive of the company has established the following goals according to their importance: a) Avoid any underutilization of production capacity b) Meet the order of ABC stores for 7 units of B and 5 units of C per week. c) Avoid the overtime operation of the plant beyond 10 hours.
d) Achieve the sales goal of A, B and C of 10, 10, and 12 units. e) Minimize the overtime operation as much as possible. Formulate this as a linear goal programming problem. Solution: Let x1, x2, and x3 number of units of products A, B and C are produced for attending different goals. Also let P1, P2, P3, P4 and P5 are the weights attached to the goals where P1> P2 > P3 > P4 > P5. Let dui and doi are underachievement and overachievement of the goals i, where i = 1, 2, 3, 4 and 5. The Company has the following 5 goals: Goal 1: Avoid any underutilization of production capacity Goal 2: Meet the order of ABC stores for 7 units of B and 5 units of C per week Goal 3: Avoid the overtime operation of the plant beyond 10 hours Goal 4: Achieve the sales goal of A, B and C of 10, 10, and 12 units Goal 5: Minimize the overtime operation as much as possible The goal constraints are as follows: 2x1 + 3x2 + x3 + du1 – do1 = 40 [avoid under utilization of capacity] x2 + du2x2 – do2 = 7 and x3 + du2x3 – do2 = 5 [meet the order of ABC stores for 7 units of B and 5 units of C per week] 2x1 + 3x2 + x3 + du3 – do3 = 50 [avoid overtime operation of plant beyond 10 hours] x1 + du4x1 – do4x1 = 10; x2 + du4x2 – do4x2 = 10 and x3 + du4x3 – do4x3 = 12 [Achieve the sales goal of A, B and C of 10, 10, and 12 units] 2x1 + 3x2 + x3 + du5 – do5 = 40 [minimize overtime operation as much as possible] The objective function can be formulated as follows: Minimize Z = P1du1 + P2(du2x2 + du2x3) + P3do3 + P4(do4x1 + do4x2 + do4x3) + P5do5 All the decision variables and deviational variables are non-negative. 37. A firm produces three products A, B, and C. It uses two types of raw materials I and II of which 5000 and 7500 units respectively are available. The raw material requirements per unit of the products are given below: Raw Material Requirement per unit of product A B C I 3 4 5 II 5 3 5 The labour time for each unit of productA is twice as that of product B and three times that of product C. The entire labour force of the firm can produce the equivalent of 3000 units. The minimum demand for the three products is 600, 650 and 500 units respectively. Also, the ratio of the numbers of units produced must be equal to 2:3:4. Assuming the profits per unit of A, B, and C are 50, 50, and 80 rupees respectively, formulate the LPP. Solution: Decision Variables: Objective is to maximize profit. Profit per unit of products A, B, and C is given, hence the decision variables are 3. Let A, B, and C are the number of units of production of product A, B, and C respectively. Objective Function: Objective function can be expressed algebraically as follows: Maximize Z = 50A +540B +80C Constraints: Constraints can be formulated as follows: For raw material P: 3A + 4B + 5C < 5000 For raw material Q: 5A + 3B + 5C < 7500
Product B requires ½ and product C requires 1/3rd the time required for product A. Let t hours is the time required to produce one unit of product A. then t/2, and t/3 are the times in hours to produce B and C and since 1600 units of A will need time 1600t hours, we get the time constraint: tA + t/2B + t/3C < 3000t or A + B/2 + C/3 < 3000 Market demand constraints: A > 600, B > 650 and C >500 Finally, since products A, B, and C are to be produced in the ratio 3: 4: 5. A: B: C : : 2:3:4 Or A/2 = B/3 and B/3 = C/5 Thus there are two additional constraints: 3A – 2B = 0 and4B – 3C = 0 Where A, B, C > 0 Hence the LPP is: Maximize Z = 50A +540B +80C Subject to: 3A + 4B + 5C < 5000 5A + 3B + 5C < 7500 A + B/2 + C/3 < 3000 A > 600, B > 650 and C >500 3A – 2B = 0 and4B – 3C = 0 A, B, C > 0
38. Solve graphically the following LPP. Minimize Z = 3X1 + 5X2 Subject to: - 3X1 + 4X2 < 12 2X1 + 3X2 > 12 2X1 - X2 > - 2 X1,, X2 > 0 39. A company has three plants and four warehouses. The supply and demand in units and the corresponding transportation costs are given. The table has been taken from the solution procedure of the transportation problem: Plants Warehouses Supply I II III IV 5 10 4 5 10 A 10 6 8 7 2 25 B 20 5 4 2 5 7 20 C 5 10 5 Demand 25 10 15 5 55 Answer the following questions, giving brief reasons: a) Is this solution feasible? b) Is this solution degenerate? c) Is this solution optimal? d) Does the problem have more than one optimal solution? If so, show all of them. e) If the cost for the route B-III is reduced from Rs.7 to Rs.6 per unit, what will be the optimal solution? Solution
40. Find the minimal spanning tree of the following network: Arc 1-2 1-3 1-4 2-3 2-5 3-7
Distance 2 9 5 4 8 4
Arc 4-7 3-4 3-5 3-6 5-6 5-8
Distance 8 10 3 7 11 5
Arc 6-7 6-8 6-9 6-10 7-9 8-10 9-10
Distance 8 6 9 4 3 2 3
1 to 2 to 3 to 5, 1 to 4, 9 to 7, 6 to 10, 8 to 10 2+4+3+5+2+4+3 =23 41. A company has just developed a new item for which it proposes to undertake a national TV promotional campaign. It has decided to schedule a series of one-minute commercials during peak audience viewing hours of 1 to 5 p.m. To reach the widest possible audience, the company wants to schedule one commercial on each of the network and to have only one commercial appear during each of the four one-hour time blocks. The exposure ratings for each hour, which represent the number of viewers for Rs.10000 spent, are given below:
Viewing Hours 1-2 p.m. 2-3 p.m. 3-4 p.m. 4-5 p.m.
Network A 27.1 18.9 19.2 11.5
B 18.1 15.5 18.5 21.4
C 11.3 17.1 9.9 16.8
D 9.5 10.6 7.7 12.8
Which network should be established each hour to provide the maximum audience exposure?
42. Two kinds of food for children, F1 and F2 are being considered to be purchased. Food F1 costs Rs. 20 a unit while food F2 is available at Rs.40 per unit. The nutrient contents of these foods are as follows:
Nutrients N1 N2 N3
Nutrient content Food F1 40 3 18
Food F2 20 12 3
The minimum requirements are respectively 200, 36, and 54 units. Formulate this LP. Solve the problem so that costs are minimized satisfying the given constraints. Formulate the dual problem. Mini Z= 20A + 40B Constraints 40A+ 20B >=200 3A +12B >=36 18A + 3B >=54 A, B >=0 Note: sum no. 46, 47, 48, 49, 50 from book N D Vohra
