Chapter 3

CHAPTER 31. Maximise Subject to Z = 8x1 6x2 + 7x3 + 2x4 + 0S1 + 0S 2 + 0S3 MA1 MA2 4x1 3x2 + 6x 3 + x 4 + S 1 + 0S 2 + 0S3 0A1 + 0A2 x1 + 2x2 + 3x3 + x4 + 0S1 + S2 + 0S 3 + 0A1 + 0A2 9x1 5x2 + 7x3 x4 + 0S1 + 0S 2 S3 A1 + 0A2 0x1 + 6x2 + 2x3 + 4x4 + 0S1 + 0S 2 + 0S3 + 0A1 + A2 x1, x2, x3, x4, S1, S 2, S3, A1, A2 2.Basis S1 S2 0 0 x1 3 1 7 0 7

= = = =

40 5 60 47 0

Simplex Tableau 1: Non-optimal Solutionx2 2 4* 14 0 14 S1 1 0 0 36 0 S2 0 1 0 10 0 bi 36 10 Z=0 bi /aij 18 5/2

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 x 2 14 Cj Solution Dj x1 5/2 1/4* 7 0 7/2 x2 0 1 14 5/2 0 S1 1 0 0 31 0 S2 1/2 1/4 0 0 7/2 bi 31 5/2 Z = 35 bi /aij 62/5 10

Simplex Tableau 3: Optimal SolutionBasis S1 0 x1 7 Cj Solution Dj x1 0 1 7 10 0 x2 10 4 14 0 14 S1 1 0 0 6 0 S2 3 1 0 0 7 bi 6 10 Z = 70

3.

For solving the problem, we need to multiply the first constraint by 1 to have a non-negative bi value. With slack variables S1 and S 2, the solution follows.

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22Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 0 0 x1 1 4* 21 0 21 x2 2 3 15 0 15 S1 1 0 0 6 0 S2 0 1 0 12 0 bi 6 12 Z=0 bi /a ij 6 3

Cj Solution Dj

Simplex Tableau 2: Optimal SolutionBasis S1 0 x 1 21 Cj Solution Dj x1 0 1 21 3 0 x2 5/4 3/4 15 0 3/4 S1 1 0 0 3 0 S2 1/4 1/4 0 0 21/4 bi 3 3 Z = 63

4.Basis S1 S2 S3 0 0 0 x1 4 2 8 20 0 20 x2 3 5* 2 30 0 30

Simplex Tableau 1: Non-optimal Solutionx3 1 0 0 5 0 5 S1 1 0 0 0 40 0 S2 0 1 0 0 28 0 S3 0 0 1 0 16 0 bi 40 28 16 Z=0 bi /a ij 40/3 28/5 8

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 S 2 30 S3 0 Cj Solution Dj x1 14/5 2/5 36/5* 20 0 8 x2 0 1 0 30 28/5 0 x3 1 0 0 5 0 5 S1 1 0 0 0 116/5 0 S2 3/5 1/5 2/5 0 0 6 S3 0 0 1 0 24/5 0 bi 116/5 28/5 24/5 Z = 168 bi /a ij 58/7 14 2/3

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23Simplex Tableau 3: Non-optimal SolutionBasis S1 0 x 2 30 x 1 20 Cj Solution Dj x1 0 0 1 20 2/3 0 x2 0 1 0 30 16/3 0 x3 1* 0 0 5 0 5 S1 1 0 0 0 64/3 0 S2 4/9 2/9 1/18 0 0 50/9 S3 7/18 1/18 5/36 0 0 10/9 bi 64/3 16/3 2/3 Z = 520/3 bi /a ij 64/3

Simplex Tableau 4: Non-optimal SolutionBasis x3 5 x 2 30 x 3 20 Cj Solution Dj x1 0 0 1 20 2/3 0 x2 0 1 0 30 16/3 0 x3 1 0 0 5 64/3 0 S1 1 0 0 0 0 5 S2 4/9 2/9 1/18 0 0 10/3 S3 7/18 1/18 5/36* 0 0 5/6 bi 64/3 16/3 2/3 Z = 280 bi /a ij 24/5

Simplex Tableau 5: Optional SolutionBasis x3 5 x 2 30 S3 0 Cj Solution Cj x1 14/5 2/5 36/5 20 0 6 x2 0 1 0 30 28/5 0 x3 1 0 0 5 116/5 0 S1 1 0 0 0 0 5 S2 3/5 1/5 2/5 0 0 3 S3 0 0 1 0 24/5 0 bi 116/5 28/5 24/5 Z = 284

5.

Setting x2 = x3 x4, and multiplying constraint involving negative bi, by 1 the LPP is: Maximise Z = 8x1 4x3 + 4x4 Subject to 4x1 + 5x3 5x 4 20 x1 3x 3 + 3x 4 23 x1, x 3, x 4 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 0 0 x1 4* 1 8 0 8 x3 5 3 4 0 4 x4 5 3 4 0 4 S1 1 0 0 20 0 S2 0 1 0 23 0 bi 20 23 Z=0 b i /a ij 5 23

Cj Solution Dj

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24Simplex Tableau 2: Non-optimal SolutionBasis x1 8 S2 0 Cj Solution Dj x1 1 0 8 5 0 x3 5/4 17/4 4 0 14 x4 5/4 17/4* 4 0 14 S1 1/4 1/4 0 0 2 S2 0 1 0 18 0 bi 5 18 Z = 40 b i /aij 72/17

Simplex Tableau 3: Optimal SolutionBasis x1 x2 Cj Solution Dj 8 4 x1 1 0 8 175/17 0 x3 0 1 4 0 0 x4 0 1 4 72/17 0 S1 3/17 1/17 0 0 20/17 S2 5/17 4/17 0 0 56/17 Z = 1688 17 bi 175/17 72/17

From Table 3, the optimal solution is: x1 = 175/17, x2 = 0, and x3 = 72/17 Accordingly, the solution to the original problem is: x1 = 175/17 and x2 = x3 x4 = 0 72/17 = 72/17 and Z = 8 175/17 4(72/17) = 1688/17 6. From the given information Profit per unit of A = Rs 9.60 (0.5 8 + 0.3 6 + 0.2 4) = Rs 3 Profit per unit of B = Rs 7.80 (0.3 8 + 0.3 6 + 0.4 4) = Rs 2 Now, if x1 and x2 be the output and sales of drugs A and B respectively, the LPP may be stated as follows: Maximise Z = 3x 1 + 2x 2 Subject to 0.5x1 + 0.3x 2 1,600 0.3x1 + 0.3x 2 1,400 0.2x1 + 0.4x 2 1,200 x1, x 2 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 0 0 0 x1 0.5* 0.3 0.2 3 0 3 x2 0.3 0.3 0.4 2 0 2 S1 1 0 0 0 1,600 0 S2 0 1 0 0 1,400 0 S3 0 0 1 0 1,200 0 bi 1,600 1,400 1,200 Z=0 bi /aij 3,200 4,667 6,000

Cj Solution Dj

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25Simplex Tableau 2: Non-optimal SolutionBasis x1 3 S2 0 S3 0 Cj Solution Dj x1 1 0 0 3 3,200 0 x2 0.60 0.12 0.28* 2 0 0.2 S1 2 0.6 0.4 0 0 6 S2 0 1 0 0 440 0 S3 0 0 1 0 560 0 bi 3,200 440 560 Z = 9,600 bi /a ij 5,333 3,667 2,000

Simplex Tableau 3: Optimal SolutionBasis x1 3 S2 0 x2 2 Cj Solution Dj x1 1 0 0 3 2,000 0 x2 0 0 1 2 2,000 0 S1 2.86 0.43 1.43 0 0 5.72 S2 0 1 0 0 0 0 S3 2.14 0.43 3.57 0 0 0.72 bi 2,000 200 2,000 Z = 10,000

7.

From Table 3 it is evident that the optimal product is: drug A, 2,000 units; drug B, 2,000 units for a total profit of Rs 10,000. Let the output of belts type A and type B be x1 and x2 respectively. The LPP is: Maximise Z = 20x1 + 15x2 Total profit Subject to Time availability 2x1 + x 2 1,000 Leather availability x1 + x 2 800 400 x1 Buckle availability x2 700 x1, x 2 0

}

Simplex Tableau 1: Non-optimal SolutionBasis S1 0 S2 0 S3 0 S4 0 Cj Solution Dj x1 2 1 1* 0 20 0 20 x2 1 1 0 1 15 0 15 S1 1 0 0 0 0 1,000 0 S2 0 1 0 0 0 800 0 S3 0 0 1 0 0 400 0 S4 0 0 0 1 0 700 0 bi 1,000 800 400 700 Z=0 bi /a ij 500 800 400

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26Simplex Tableau 2: Non-optimal SolutionBasis S1 0 S2 0 x 1 20 S4 0 Cj Solution Dj x1 0 0 1 0 20 400 0 x2 1* 1 0 1 15 0 15 S1 1 0 0 0 0 200 0 S2 0 1 0 0 0 400 0 S3 2 1 1 0 0 0 20 S4 0 0 0 1 0 700 0 bi 200 400 400 700 Z = 8,000 bi /a ij 200 400 700

Simplex Tableau 3: Non-optimal SolutionBasis x 2 15 S2 0 x 1 20 S4 0 Cj Solution Dj x1 0 0 1 0 20 400 0 x2 1 0 0 0 15 200 0 S1 1 1 0 0 0 0 15 S2 0 1 0 0 0 200 0 S3 2 1* 1 2 0 0 10 S4 0 0 0 1 0 500 0 bi 200 200 400 500 Z = 11,000 bi /a ij 200 400 250

Simplex Tableau 4: Optional SolutionBasis x 2 15 S3 0 x 1 20 S4 0 Cj Solution Dj x1 0 0 1 0 20 200 0 x2 1 0 0 0 15 600 0 S1 1 1 1 1 0 0 5 S2 2 1 1 2 0 0 10 S3 0 1 0 0 0 200 0 S4 0 0 0 1 0 100 0 bi 600 200 200 100 Z = 13,000

8.Basis S1 S2 S3 0 0 0 x1 4 4* 4 3 0 3

Simplex Tableau 1: Non-optimal Solutionx2 3 1 9 2 0 2 S1 1 0 0 0 12 0 S2 0 1 0 0 8 0 S3 0 0 1 0 8 0 bi 12 8 8 b i /a ij 3 2 2

Cj Solution Dj

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27Simplex Tableau 2: Non-optimal SolutionBasis S1 0 x1 3 S3 0 Cj Solution Dj x1 0 1 0 3 2 0 x2 2* 1/4 10 2 0 5/4 S1 1 0 0 0 4 0 S2 1 1/4 1 0 0 3/4 S3 0 0 1 0 0 0 bi 4 2 0 Z=6 b i /a ij 2 8

Simplex Tableau 3: Optional SolutionBasis x2 2 x1 3 S3 0 Cj Solution Dj x1 0 1 0 3 3/2 0 x2 1 0 0 2 2 0 S1 1/2 1/8 5 0 0 5/8 S2 1/2 3/8 6 0 0 1/8 S3 0 0 1 0 20 0 bi 2 3/2 20 0 Z = 8.5

It is evident that the optimal solution contained in Tableau 3 is not degenerate (as none of the basic variables assumes a solution value equal to zero). However, the solution given in Tableau 2 is a degenerate one. The improvement of this solution does not lead to another degenerate solution since the outgoing variable (S1) is not a degenerate variable. The solution is temporarily degenerate, therefore. 9. After introducing necessary variables, the problem is: Maximise Z = 3x1 + 2x2 + 3x3 + 0S1 + 0S2 MA1 Subject to 2x1 + x2 + x 3 + S 1 = 2 3x1 + 4x2 + 2x3 S2 + A1 = 8 x1, x 2, x3, S1, S2, A1 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 0 A1 M x1 2 3 x2 1 4 2 0 2 + 4M x3 1 2 3 0 3 + 2M S1 1 0 0 2 0 S2 0 1 0 0 M A1 0 1 M 8 0 bi 2 8 bi /a ij 2 2

Cj 3 Solution 0 Dj 3 + 3M

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28Simplex Tableau 2: Non-optimal SolutionBasis S1 0 x2 2 Cj Solution Dj x1 5/4 3/4 3 0 3/2 x2 0 1 2 2 0 x3 1/2* 1/2 3 0 2 S1 1 0 0 0 0 S2 1/4 1/4 0 0 1/2 A1 1/4 1/4 M 0 M1/2 bi 0 2 Z=4 bi /aij 0 4

Simplex Tableau 3: Optimal SolutionBasis x3 x2 3 2 x1 5/2 1/2 3 0 7/2 x2 0 1 2 2 0 x3 1 0 3 0 0 S1 2 1 0 0 4 S2 1/2 1/2 0 0 1/2 A1 1/2 1/2 M 0 M1/2 bi 0 2 Z=4

Cj Solution Dj

The solution in Simplex Tableau 3 is optimal. It is unique. The solution is degenerate, however. 10. From the given information, No. of working hours available per machine per month = No. of hours per day No. of days Percentage of effective working. Accordingly, the monthly capacity for the three operations is as follows: X: 3 320 = 960 hours Y: 2 320 = 640 hours Z: 1 320 = 320 hours The LPP with x1, x2, and x3 representing the output of products A, B, and C respectively, may be stated as under: Maximise P = 3x1 + 4x2 + 6x 3 Subject to 4x 1 + x2 + 6x 3 960 5x 1 + 3x2 + x 3 640 x 1 + 2x 2 + 3x 3 320 x1, x 2, x 3 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 0 0 0 x1 4 5 1 3 0 3 x2 1 3 2 4 0 4 x3 6 1 3* 6 0 6 S1 1 0 0 0 960 0 S2 0 1 0 0 640 0 S3 0 0 1 0 320 0 bi 960 640 320 Z=0 b i /a ij 160 640 320/3

Cj Solution Dj

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29Simplex Tableau 2: Non-optimal SolutionBasis S1 0 S2 0 x3 6 Cj Solution Dj x1 2 14/3* 1/3 3 0 1 x2 3 7/3 2/3 4 0 0 x3 0 0 1 6 320/3 0 S1 1 0 0 0 320 0 S2 0 1 0 0 1,600/3 0 S3 2 1/3 1/3 0 0 2 bi 320 1,600/3 320/3 Z = 640 bi /a ij 160 800/7 320

Simplex Tableau 3: Optimal SolutionBasis S1 0 x1 3 x3 6 Cj Solution Dj x1 0 1 0 3 800/7 0 x2 4 1/2 1/2 4 0 1/2 x3 0 0 1 6 480/7 0 S1 1 0 0 0 640/7 0 S2 3/7 3/14 1/14 0 0 3/14 S3 13/7 1/14 5/14 0 0 27/14 bi 640/7 800/7 480/7 Z = 5,280/7

Thus, optimal solution is: product A: 800/7 units, product B: nil, product C = 480/7 units. Total profit = Rs 5,280/7 or Rs 754.29. 11. Let x 1, x2, and x3 represent the daily production of dolls A, B, and C respectively. Using the given information, we may state the LPP as follows: Total Profit Maximise Z = 3x 1 + 5x 2 + 4x 3 Subject to 8 Machine M1 time 2x1 + 3x2 Machine M2 time 2x2 + 5x 3 10 Machine M3 time 3x1 + 2x2 + 4x3 15 x1, x 2, x 3 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 0 0 0 x1 2 0 3 3 0 3 x2 3* 2 2 5 0 5 x3 0 5 4 4 0 4 S1 1 0 0 0 8 0 S2 0 1 0 0 10 0 S3 0 0 1 0 15 0 bi 8 10 15 bi /a ij 8/3 5 15/2

Cj Solution Dj

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30Simplex Tableau 2: Non-optimal SolutionBasis x2 5 S2 0 S3 0 Cj Solution Dj x1 2/3 4/3 5/3 3 0 1/3 x2 1 0 0 5 8/3 0 x3 0 5* 4 4 0 4 S1 1/3 2/3 2/3 0 0 5/3 S2 0 1 0 0 14/3 0 S3 0 0 1 0 29/3 0 bi 8/3 14/3 29/3 bi /a ij 14/15 29/12

Simplex Tableau 3: Non-optimal SolutionBasis x2 5 x3 4 S3 0 Cj Solution Dj x1 2/3 4/15 41/15* 3 0 11/15 x2 1 0 0 5 8/3 0 x3 0 1 0 4 14/5 0 S1 1/3 2/15 2/15 0 0 17/15 S2 0 1/5 4/5 0 0 4/5 S3 0 0 1 0 89/15 0 bi 8/3 14/15 89/15 bi /a ij 4 89/41

Simplex Tableau 4: Optimal SolutionBasis x2 5 x1 0 0 1 3 89/41 0 x2 1 0 0 5 50/41 0 x3 0 1 0 4 62/41 0 S1 15/41 6/41 2/41 0 0 45/41 S2 8/41 5/41 12/41 0 0 24/41 S3 10/41 4/41 15/41 0 0 11/41 bi 50/41 62/41 89/41

x3 4 x1 3 Cj Solution Dj

From Tableau 4, it is evident that optimal daily output of the three type of dolls is: Doll A: 89/41, Doll B: 50/41, Doll C: 62/41 The total profit works out to be Rs 765/41 or Rs 18.66. Also, none of the machines would remain idle. 12. Let x1, x2, and x3 be the output of pistons, rings, and valves respectively. Using the given information, we may state the LPP as follows: Profit Maximise Z = 10x1 + 6x2 + 4x3 Subject to Preparatory work x1 + x2 + x 3 100 Machinng 10x1 + 4x2 + 5x 3 600 Allied 2x1 + 2x2 + 6x 3 300 x1, x 2, x 3 0

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31Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 0 0 0 x1 1 10* 2 10 0 10 x2 1 4 2 6 0 6 x3 1 5 6 4 0 4 S1 1 0 0 0 100 0 S2 0 1 0 0 600 0 S3 0 0 1 0 300 0 bi 100 600 300 bi /a ij 100 60 150

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 x 1 10 S3 0 Cj Solution Dj x1 0 1 0 10 60 0 x2 3/5* 2/5 6/5 6 0 2 x3 1/2 1/2 5 4 0 1 S1 1 0 0 0 40 0 S2 1/10 1/10 1/5 0 0 1 S3 0 0 1 0 180 0 bi 40 60 180 bi /a ij 200/3 150 150

Simplex Tableau 3: Optimal SolutionBasis x2 6 x1 10 S3 0 Cj Solution Dj x1 0 1 0 10 100/3 0 x2 1 0 0 6 200/3 0 x3 5/6 1/6 4 4 0 8/3 S1 5/3 2/3 2 0 0 10/3 S2 1/6 1/6 0 0 0 2/3 S3 0 0 1 0 100 0 bi 200/3 100/3 100

The most profitable mix, therefore, is: Pistons = 100/3, Rings = 200/3 and Valves = 0. The corresponding profit = 10 100/3 + 6 200/3 = Rs 733.33. 13. (a) Let x1, x2, and x3 represent, respectively, the number of units of A, B and C. The linear programming formulation is given here: Maximise Z = 12x1 + 3x2 + x3 Subject to 10x1 + 2x 2 + x3 100 7x1 + 3x2 + 2x 3 77 2x1 + 4x2 + x 3 80 x1, x 2, x 3 0

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32(b)Basis S1 S2 S3 0 0 0 x1 10* 7 2 12 0 12 x2 2 3 4 3 0 3

Simplex Tableau 1: Non-optimal Solutionx3 1 2 1 1 0 1 S1 1 0 0 0 100 0 S2 0 1 0 0 77 0 S3 0 0 1 0 80 0 bi 100 77 80 bi /a ij 10 11 40

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis x1 12 S2 0 S3 0 Cj Solution Dj x1 1 0 0 12 0 0 x2 1/5 8/5* 18/5 3 0 3/5 x3 1/10 13/10 4/5 1 0 1/5 S1 1/10 7/10 1/5 0 0 6/5 S2 0 1 0 0 7 0 S3 0 0 1 0 60 0 bi 10 7 60 bi/aij 50 35/8 50/3

Simplex Tableau 3: Optimal SolutionBasis x 1 12 x2 3 S3 0 Cj Solution Dj x1 1 0 0 12 73/8 0 x2 0 1 0 3 35/8 0 x3 1/16 13/16 17/16 1 0 11/16 S1 3/16 7/16 11/8 0 0 15/16 S2 1/8 5/8 9/4 0 0 3/8 S3 0 0 1 0 177/4 0 bi 73/8 35/8 177/4

The product mix so as to maximise profit is: product A: 73/8 units, product B: 35/8 units and product C: nil. Total profit = Rs 12 73/8 + 3 35/8 = Rs 981/8. (c) From Tableau 3 it is clear that S1 = S2 = 0, while S3 = 177/4. Thus, there is no unused capacity in machine centres X and Y, while in machine centre Z a total of 177/4 hours would be unused. 14. Let the monthly production of the products 5-10-5, 5-5-10, and 20-5-10 be x1, x 2 and x3 kg respectively. The LPP is: Maximise Z = 16x1 + 17x2 + 10x 3 Total profit Subject to 1 x + 1 x + 1 x 100 Material A 20 1 20 2 5 3 1 x + 1 x + 1 x 180 Material B 10 1 20 2 20 3 1 x + 1 x + 1 x 120 Material C 20 1 10 2 10 3 Capacity x1 30 x1, x 2, x 3 0

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33Working notes: Profit per unit is worked out as follows: 5-10-5: 40.50 (0.05 80 + 0.10 20 + 0.05 50 + 0.80 20) = 16 5-5-10: 43 (0.05 80 + 0.05 20 + 0.10 50 + 0.80 20) = 17 20-5-10: 45 (0.20 80 + 0.05 20 + 0.10 50 + 0.65 20) = 10 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 S4 0 0 0 0 x1 1/20 1/10 1/20 1 16 0 16 x2 1/20 1/20 1/10* 0 17 0 17 x3 1/5 1/20 1/10 0 10 0 10 S1 1 0 0 0 0 100 0 S2 0 1 0 0 0 180 0 S3 0 0 1 0 0 120 0 S4 0 0 0 1 0 30 0 bi 100 180 120 30 Z=0 b i/a ij 2000 3600 1200

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 S2 0 x 2 17 S4 0 Cj Solution Dj x1 1/40 3/40 1/2 1* 16 0 15/2 x2 0 0 1 0 17 1,200 0 x3 3/20 0 1 0 10 0 7 S1 1 0 0 0 0 40 0 S2 0 1 0 0 0 120 0 S3 1/2 1/2 10 0 0 0 170 S4 0 0 0 1 0 30 0 bi 40 120 1,200 30 Z = 20,400 bi/a ij 1,600 1,600 2,400 30

Simplex Tableau 3: Optimal SolutionBasis S1 0 S2 0 x 2 17 x 1 16 Cj Solution Dj x1 0 0 0 1 16 30 0 x2 0 0 1 0 17 1185 0 x3 3/20 0 1 0 10 0 7 S1 1 0 0 0 0 157/4 0 S2 0 1 0 0 0 471/4 0 S3 1/2 1/2 10 0 0 0 170 S4 1/40 3/40 1/2 1 0 0 15/2 bi 157/4 471/4 1,185 30 Z = 20,625

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3415. Using the given information about profitability and resources, the LPP may be stated as follows: Maximise Z = 4,000x1 + 2,000x2 + 5,000x3 Revenue Subject to Labour hours 12x1 + 7x 2 + 9x 3 1,260 Wood 22x1 + 18x2 + 16x 3 1,9008 Screws 2x1 + 4x2 + 3x 3 396 x1, x 2, x 3 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 0 0 0 x1 12 22 2 4,000 0 4,000 x2 7 18 4 2,000 0 2,000 x3 9 16 3* 5,000 0 5,000 S1 1 0 0 0 1,260 0 S2 0 1 0 0 19,008 0 S3 0 0 1 0 396 0 bi 1,260 19,008 396 bi /aij 140 1,188 132

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 S2 0 x3 5,000 Cj Solution Dj x1 6* 34/3 2/3 4,000 0 2,000/3 x2 5 10/3 4/3 2,000 0 14,000/3 x3 0 0 1 5,000 132 0 S1 1 0 0 0 72 0 S2 0 1 0 0 16,896 0 S3 3 16/3 1/3 0 0 5,000/3 bi 72 16,896 132 b i /aij 12 1,491 198

Simplex Tableau 3: Optimal SolutionBasis x1 4,000 S2 0 x3 5,000 Cj Solution Dj x1 1 0 0 4,000 12 0 x2 5/6 55/9 17/9 2,000 0 37,000/9 x3 0 0 1 5,000 124 0 S1 1/6 17/9 1/9 0 0 1,000/9 S2 0 1 0 0 16,760 0 S3 1/2 1/3 2/3 0 0 4,000/3 bi 12 16,760 124

(c) From Tableau 3, it is evident that for maximum profit, the company should produce 12 Row boats and 124 Kayaks and no Canoes. The maximum revenue is 4,000 12 + 5,000 124 = 668,000. (d) While labour-hours and screws available are fully used, the wood is not used fully. Its spare capacity is 16,760 board feet. (e) The total wood used to make all of the boats in the optimal solution is 22 12 + 16 124 = 2,248 board feet.

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3516. The information given in the problem is tabulated below: Vehicle Type A B C Tonnage 10 20 18 Average speed (kmph) 35 30 30 Working hours/day 18 18 21 Cost (000 Rs) 80 130 150 Crew 3 6 6 The capacity of a vehicle in tonne-kms per day may be obtained by the product of tonnage, average speed, and working hours per day. This works out to be 10 35 18 = 6,300 for A 20 30 18 = 10,800 for B and 18 30 21 = 11,340 for C. Now x1, x2, and x3 be the number of vehicles purchased of types A, B, and C respectively, the LPP may be expressed as: Capacity Maximise Z = 6,300x1 + 10,800x2 + 11,340x3 Subject to Budget 80x1 + 130x 2 + 150x 3 4,000 x2 + x 3 30 Maintenance x1 + 6x2 + 6x 3 150 Crew 3x1 + x1, x 2, x 3 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 Cj Solution Dj 0 0 0 x1 80 1 3 6,300 0 6,300 x2 130 1 6 10,800 0 10,800 x3 150 1 6* 11,340 0 11,340 S1 1 0 0 0 4,000 0 S2 0 1 0 0 30 0 S3 0 0 1 0 150 0 bi 4,000 30 150 Z=0 bi /a ij 400/15 30 25

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 S2 0 x3 11,340 Cj Solution Dj x1 15 1/2* 1/2 6,300 0 630 x2 20 0 1 10,800 0 540 x3 0 0 1 11,340 25 0 S1 1 0 0 0 250 0 S2 0 1 0 0 5 0 S3 25 1/6 1/6 0 0 189 bi 250 5 25 Z = 283,500 b i /aij 50 10 50

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36Simplex Tableau 3: Optimal SolutionBasis S1 x1 x3 0 6,300 11,340 x1 0 1 0 6,300 10 0 x2 20 0 1 10,800 0 54 x3 0 0 1 11,340 20 0 S1 1 0 0 0 200 0 S2 10 2 1 0 0 1,260 S3 70/3 1/3 1/3 0 0 1,680 bi 200 10 20 Z = 289,800

Cj Solution Dj

From Simplex Tableau 3, it may be observed that the company should buy 10 vehicles of type A and 20 vehicles of type C in order to maximise the capacity. The capacity is 289,800 tonne-km per day. 17.Basis S1 S2 S3 0 0 0 x1 4 3 8* 7 0 7 x2 7 3 3 2 0 2

Simplex Tableau 1: Non-optimal Solutionx3 6 4 4 3 0 3 x4 4 1 2 4 0 4 S1 1 0 0 0 20 0 S2 0 1 0 0 10 0 S3 0 0 1 0 25 0 bi 20 10 25 Z=0 bi /aij 10/3 25/8

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 S2 0 x1 7 Cj Solution Dj x1 0 0 1 7 25/8 0 x2 11/2* 15/8 3/8 2 0 37/8 x3 8 5/2 1/2 3 0 1/2 x4 3 1/4 1/4 4 0 9/4 S1 1 0 0 0 65/2 0 S2 0 1 0 0 5/8 0 S3 1/2 3/8 1/8 0 0 7/8 bi 65/2 5/8 25/8 Z = 175/8 bi /aij 65/11

Simplex Tableau 3: Non-optimal SolutionBasis x2 2 S2 0 x1 7 x1 0 0 1 x2 1 0 0 2 65/11 0 x3 16/11 115/22 23/22 3 0 159/22 x4 6/11 17/22 1/22* 4 0 105/22 S1 2/11 15/44 3/44 0 0 37/44 S2 0 1 0 0 515/44 0 S3 1/11 9/44 7/44 0 0 57/44 bi 65/11 515/44 235/44 Z = 2,165/44 bi /aij 235/2

Cj 7 Solution 235/44 Dj 0

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37Simplex Tableau 4: Optimal SolutionBasis x2 2 S2 0 x4 4 Cj Solution Dj x1 12 17 22 7 0 105 x2 1 0 0 2 70 0 x3 14 23 23 3 0 117 x4 0 0 1 4 235/2 0 S1 1 3/2 3/2 0 0 8 S2 0 1 0 0 205/2 0 S3 2 5/2 7/2 0 0 18 bi 70 205/2 235/2 Z = 610

18. Let the output of desks I, II, III and IV be x1, x2, x3 and x4 respectively. The LPP is: Maximise Z = 9x1 + 20x2 + 15x3 + 40x4 Subject to 4x1 + 9x2 + 7x3 + 10x 4 6,000 x1 + x2 + 3x3 + 40x 4 4,000 x1, x2, x 3, x 4 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 0 0 x1 4 1 9 0 9 x2 9 1 20 0 20 x3 7 3 15 0 15 x4 10 40* 40 0 40 S1 1 0 0 6,000 0 S2 0 1 0 4,000 0 bi 6,000 4,000 Z=0 bi/a ij 600 100

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 x4 0 40 x1 15/4 1/40 9 0 8 x2 35/4* 1/40 20 0 19 x3 25/4 3/40 15 0 12 x4 0 1 40 100 0 S1 1 0 0 5,000 0 S2 1/4 1/40 0 0 1 bi 5,000 100 Z = 4,000 bi /aij 4,000/7 4,000

Cj Solution Dj

Simplex Tableau 3: Optimal SolutionBasis x2 x4 20 40 x1 3/7 1/70 9 0 1/7 x2 1 0 20 4,000/7 0 x3 5/7 2/35 15 0 11/7 x4 0 1 40 600/7 0 S1 4/35 1/350 0 0 76/35 S2 1/35 9/350 bi 4,000/7 600/7

Cj Solution Dj

0 0 Z = 104,000/7 16/35

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3819. Introducing necessary surplus and artificial variables, the problem is: Minimise Z = 6x 1 + 4x2 + 0S1 + 0S2 + MA1 + MA2 Subject to 3x1 + 1/2x2 S 1 + A1 = 12 2x1 + x2 S 2 + A2 = 16 x1, x2, S 1, S 2, A1, A2 0 Simplex Tableau 1: Non-optimal SolutionBasis A1 A2 M M x1 3 2 6 0 6 5M x2 1/2 1 4 0 4 3/2M S1 1 0 0 0 M S2 0 1 0 0 M A1 1 0 M 12 0 A2 0 1 M 16 0 bi 12 16 bi/aij 4 8

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis x1 6 A2 M Cj Solution Dj x1 1 0 6 4 0 x2 1/6 2/3* 4 0 3 2/3M S1 1/3 2/3 0 0 2 2/3M S2 0 1 0 0 M A1 1/3 2/3 M 0 2 + 2/3M A2 0 1 M 8 0 bi 4 8 bi /aij 12

Simplex Tableau 3: Optimal SolutionBasis x1 6 S2 0 Cj Solution Dj x1 1 0 6 8 0 x2 1/2 1 4 0 1 S1 0 1 0 12 0 S2 1/2 3/2 0 0 3 A1 0 1 M 0 M A2 1/2 3/2 M 0 M3 bi 8 12 Z = 48

20. Phase I: Introduce surplus and artificial variables to the given problem, assign unit coefficient to the artificial and zero coefficient to the remaining variables to rewrite the problem as under: Minimise Z = 0x 1 + 0x2 + 0S1 + 0S2 + A1 + A2 Subject to 2x1 + x2 S 1 + A1 = 4 x1 + 7x2 S 2 + A2 = 7 x1, x2, S 1, S 2, A1, A2 0

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39Simplex Tableau 1: Non-optimal SolutionBasis A1 A2 1 1 x1 2 1 0 0 3 x2 1 7* 0 0 8 S1 1 0 0 0 1 S2 0 1 0 0 1 A1 1 0 1 4 0 A2 0 1 1 7 0 bi 4 7 bi /aij 4 1

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis A1 1 x2 0 Cj Solution Dj x1 13/7* 1/7 0 0 13/7 x2 0 1 0 1 0 S1 1 0 0 0 1 S2 1/7 1/7 0 0 1/7 A1 1 0 1 0 0 A2 1/7 1/7 1 3 8/7 bi 3 1 bi /aij 21/13 7

Simplex Tableau 3: Optimal SolutionBasis x1 x2 0 0 x1 1 0 0 21/13 0 x2 0 1 0 10/13 0 S1 7/13 1/13 0 0 0 S2 1/13 14/91 0 0 0 A1 7/13 1/13 1 0 1 A2 1/13 14/91 1 0 1 bi 21/13 10/13

Cj Solution Dj

Phase II: Reconsider Simplex Tableau 3, delete columns headed A1 and A2, and replace the Cj row by the coefficients of the original problem. Apply simplex method. This is shown in Table 4, wherein the solution given is found to be optimal and calls for no revision. Thus, optimal solution is: x1 = 21/13, x2 = 10/13, and Z = 31/13. Simplex Tableau 4: Optimal SolutionBasis x1 x2 1 1 x1 1 0 1 21/13 0 x2 0 1 1 10/13 0 S1 7/13 1/13 0 0 6/13 S2 1/13 14/91 0 0 1/13 bi 21/13 10/13

Cj Solution Dj

21. Phase I: Introduce necessary variables. Assign a coefficient of 0 to each of the decision and surplus variable and 1 to each artificial variable.

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40Minimise Subject to Z = 0x1 + 0x2 + 0x3 + 0S1 + 0S2 + A1 + A2 2x1 + 3x2 + x3 S1 + A1 = 4 3x1 + 2x2 + x3 S2 + A2 = 3 x1, x2, x3, S1, S2, A1, A2 0 Simplex Tableau 1: Non-optimal SolutionBasis A1 A2 1 1 x1 2 3* 0 0 5 x2 3 2 0 0 5 x3 1 1 0 0 2 S1 1 0 0 0 1 S2 0 1 0 0 1 A1 1 0 1 4 0 A2 0 1 1 3 0 bi 4 3 bi /aij 2 1

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis A1 1 x1 0 Cj Solution Dj x1 0 1 0 1 0 x2 5/3* 2/3 0 0 5/3 x3 1/3 1/3 0 0 1/3 S1 1 0 0 0 1 S2 2/3 1/3 0 0 2/3 A1 1 0 1 2 0 A2 2/3 1/3 1 0 1/3 bi 2 1 bi /aij 6/5 3/2

Simplex Tableau 3: Optimal SolutionBasis x2 x1 0 0 x1 0 1 0 1/5 0 x2 1 0 0 6/5 0 x3 1/5 1/5 0 0 0 S1 3/5 2/5 0 0 0 S2 2/5 3/5 0 0 0 A1 3/5 2/5 1 0 1 A2 2/5 3/5 1 0 1 bi 6/5 1/5

Cj Solution Dj

Phase II: Reconsider Simplex Tableau 3. Delete columns headed A1 and A2. Also replace the Cj row by co-efficients of the original problem. Solve by simplex. Simplex Tableau 4: Optimal SolutionBasis x 2 150 x 1 150 Cj Solution Dj x1 0 1 150 1/5 0 x2 1 0 150 6/5 0 x3 1/5 1/5 100 0 40 S1 3/5 2/5 0 0 30 S2 2/5 3/5 0 0 30 bi 6/5 1/5 Z = 210

Optimal solution: x1 = 1/5, x2 = 6/5, Z = 210

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4122. Phase I: Introducing surplus and artificial variables in the given problem, and assigning zero coefficient to each of the decision and surplus variables, and a coefficient of unity to the artificial variables, we get Minimise Z = 0x 1 + 0x2 + 0S1 + 0S2 + A1 + A2 Subject to 20x1 + 30x2 S1 + A1 = 900 40x1 + 30x2 S2 + A2 = 1,200 x1, x2, S 1, S 2, A1, A2 0 Simplex Tableau 1: Non-optimal SolutionBasis A1 A2 Cj Solution Dj 1 1 x1 20 40 0 0 60 x2 30* 30 0 0 60 S1 1 0 0 0 1 S2 0 1 0 0 1 A1 1 0 1 900 0 A2 0 1 1 1,200 0 bi 900 1,200 bi /aij 30 40

Simplex Tableau 2: Non-optimal SolutionBasis x2 0 A2 1 Cj Solution Dj x1 2/3 20* 0 0 20 x2 1 0 0 30 0 S1 1/30 1 0 0 1 S2 0 1 0 0 1 A1 1/30 1 1 0 1 A2 0 1 1 300 0 bi 30 300 bi /aij 45 15

Simplex Tableau 3: Optimal SolutionBasis x2 x1 0 0 x1 0 1 0 15 0 x2 1 0 0 20 0 S1 1/15 1/20 0 0 0 S2 1/30 1/20 0 0 0 A1 1/15 1/20 1 0 1 A2 1/30 1/20 1 0 1 bi 20 15

Cj Solution Dj

Phase II: The Simplex Tableau 3 is reproduced below after replacing the Cj row by the coefficients from the objective function of the original problem and deleting the columns headed by A1 and A2. Then the problem is solved using the simplex method. It may be observed from the table that the solution is an optimal one and no further iterations are called for.

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42Simplex Tableau: Optimal SolutionBasis x2 x1 80 60 x1 0 1 60 15 0 x2 1 0 80 20 0 S1 1/15 1/20 0 0 7/3 S2 1/30 1/20 0 0 1/3 bi 20 15

Cj Solution Dj

23.

(a) Let x1 and x2 be the quantity of Ash Trays and Tea Trays, respectively, produced. The problem is: Profit (in paise) Maximise Z = 20x1 + 30x2 Subject to 10x1 + 20x 2 30,000 Stamping 15x1 + 5x 2 30,000 Forming 10x1 + 8x 2 40,000 Painting x1, x 2 0 (b)Basis S1 S2 S3 0 0 0 x1 10 15 10 20 0 20 x2 20* 5 8 30 0 30

Simplex Tableau 1: Non-optimal SolutionS1 1 0 0 0 30,000 0 S2 0 1 0 0 30,000 0 S3 0 0 1 0 40,000 0 bi 30,000 30,000 40,000 Z=0 bi /a ij 1,500 6,000 5,000

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis x2 30 S2 0 S3 0 Cj Solution Dj x1 1/2 25/2* 6 20 0 5 x2 1 0 0 30 1,500 0 S1 1/20 1/4 2/5 0 0 3/2 S2 0 1 0 0 22,500 0 S3 0 0 1 0 28,000 0 bi 1,500 22,500 28,000 Z = 45,000 bi /a ij 3,000 1,800 14,000/3

Simplex Tableau 3: Optimal SolutionBasis x 2 30 x 1 20 S3 0 Cj Solution Dj x1 0 1 0 20 1,800 0 x2 1 0 0 30 600 0 S1 3/50 1/50 7/25 0 0 7/5 S2 1/25 2/25 12/25 0 0 2/5 S3 0 0 1 0 17,200 0 bi 600 1,800 17,200 Z = 54,000

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43Thus, optimal daily output = Ash Trays: 1,800, Tea Trays: 600. Daily profit = Rs 540 Rs 350 (fixed expenses) = Rs 190. (c) The revised LPP is: Maximise Z = 20x1 + 30x2 Subject to 10x1 + 20x 2 30,000 15x1 + 5x 2 30,000 10x1 + 8x 2 40,000 16x1 + 20x 2 36,000 x1, x 2 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 S4 Cj Solution Dj 0 0 0 0 x1 10 15 10 16 20 0 20 x2 20* 5 8 20 30 0 30 S1 1 0 0 0 0 30,000 0 S2 0 1 0 0 0 30,000 0 S3 0 0 1 0 0 40,000 0 S4 0 0 0 1 0 36,000 0 bi 30,000 30,000 40,000 36,000 Z=0 bi /a ij 1,500 6,000 5,000 1,800

Simplex Tableau 2: Non-optimal SolutionBasis x2 30 S2 0 S3 0 S4 0 Cj Solution Dj x1 1/2 25/2 6 6* 20 0 5 x2 1 0 0 0 30 1,500 0 S1 1/20 1/4 2/5 1 0 0 3/2 S2 0 1 0 0 0 22,500 0 S3 0 0 1 0 0 28,000 0 S4 0 0 0 1 0 6,000 0 bi 1,500 22,500 28,000 6,000 Z = 45,000 bi /aij 3,000 1,800 14,000/3 1,000

Simplex Tableau 3: Optimal SolutionBasis x 2 30 S2 0 S3 0 x 1 20 Cj Solution Dj x1 0 0 0 1 20 1,000 0 x2 1 0 0 0 30 1,000 0 S1 2/15 11/6 3/5 1/6 0 0 2/3 S2 0 1 0 0 0 10,000 0 S3 0 0 1 0 0 22,000 0 S4 1/12 25/12 1 1/6 0 0 5/6 bi 1,000 10,000 22,000 1,000 Z = 50,000

Optimal product mix: Ash Trays = 1,000, Tea Trays = 1,000. Total Profit = Rs = 500 Rs 350 = Rs 150 per day.

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4424. If x1 and x2 be the respective output of products A and B, the LPP is: Maximise Z = 30x1 + 40x2 Subject to 4x1 + 2x 2 100 4x1 + 6x 2 180 x1 + x 2 40 20 x1 x2 10 x1, x 2 0 If we let x2 = 10 + x3, we have the revised problem as: Maximise Z = 30x1 + 40x3 + 400 Subject to 4x1 + 2x3 80; 4x 1+ 6x3 120; x1 + x2 30; x1 20 and x1, x3 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 S4 Cj Solution Dj 0 0 0 0 x1 4 4 1 1 30 0 30 x3 2 6* 1 0 40 0 40 S1 1 0 0 0 0 80 0 S2 0 1 0 0 0 120 0 S3 0 0 1 0 0 30 0 S4 0 0 0 1 0 20 0 bi 80 120 30 20 Z=0 bi /a ij 40 20 20

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 x 3 40 S3 0 S4 0 Cj Solution Dj x1 8/3* 2/3 1/3 1 30 0 10/3 x3 0 1 0 0 40 20 0 S1 1 0 0 0 0 40 0 S2 1/3 1/6 1/6 0 0 0 20/3 S3 0 0 1 0 0 10 0 S4 0 0 0 1 0 20 0 bi 40 20 10 20 Z = 800 bi /aij 15 30 30 20

Simplex Tableau 3: Optimal SolutionBasis x1 30 x3 40 S3 0 S4 0 Cj Solution Dj x1 1 0 0 0 30 15 0 x3 0 1 0 0 40 10 0 S1 3/8 1/4 1/8 3/8 0 0 5/4 S2 1/8 1/4 1/8 1/8 0 0 25/4 S3 0 0 1 0 0 5 0 S4 0 0 0 1 0 5 0 bi 15 10 5 5 Z = 850

The optimal solution is: x1 = 15, x2 = 10 + 10 = 20 and Z = 850 + 400 = 1250.

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4525. Maximise Subject to: Z = 20x1 + 40x 2 Total sales

Raw material 2x1 + 4x 2 100 Sales requirement 8x1 + 24x 2 0 x1, x 2 0 Note: Since the sales volume of product A is required to be at least 60 per cent of the total sales, the constraint may be stated as: 20x 1 0.6 (20x 1 + 40x2), which simplifies to be 8x1 + 24x2 0. Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 0 0 x1 2 8 20 0 20 x2 4 24* 40 0 40 S1 1 0 0 100 0 S2 0 1 0 0 0 bi 100 0 bi /aij 25 0

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 x 2 40 Cj Solution Dj x1 10/3* 1/3 20 0 100/3 x2 0 1 40 0 0 S1 1 0 0 100 0 S2 1/6 1/24 0 0 5/3 bi 100 0 bi /aij 30

Simplex Tableau 3: Optimal SolutionBasis x1 x2 20 40 x1 1 0 20 30 0 x2 0 1 40 10 0 S1 3/10 1/10 0 0 10 S2 1/20 1/40* 0 0 0 bi 30 10 bi /aij 400

Cj Solution Dj

Simplex Tableau 4: Optimal (alternate) SolutionBasis x 1 20 S2 0 Cj Solution Dj x1 1 0 20 50 0 x2 2 40 40 0 0 S1 7/20 4 0 0 7 S2 0 1 0 400 0 bi 50 400

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46The following points may be noted: (i) The solutions given in the first two tables are both degenerate. However, degeneracy here is tempo rary. (ii) In each of tables second and third, only one replacement ratio is considered. The other one involves negative denominator and hence, ignored. (iii) The problem has multiple optimal solutions as shown in tableau 3 and 4. 26. Let x1 and x2 be the output (in tonnes) of the products X and Y respectively. The LPP may be stated as follows: Maximise Z = 80x1 + 120x2 Subject to 20x1 + 50x 2 360 x1 + x 2 9 2 x1 x2 3 As this problem involves lower bounds on the values of x 1 and x2, it can be simplified as follows: Let x1 = 2 + x3 and x2 = 3 + x 4 Substituting these relationships, the given problem may be restated as follows: Maximise Z = 80x3 + 120x4 + 520 Subject to 20x3 + 50x 4 170 x3 + x 4 4 x3, x 4 0 Now, we can solve this problem. The variables S1 and S2 are the slack variables used to convert the inequalities into equations. Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 0 0 x3 20 1 80 0 80 x4 50* 1 120 0 120 S1 1 0 0 170 0 S2 0 1 0 4 0 bi 170 4 bi /a ij 17/5 4

Cj Solution Dj

Z = 0 + 520 = 520

Simplex Tableau 2: Non-optimal SolutionBasis x 4 120 S2 0 Cj Solution Dj x3 2/5 3/5* 80 0 32 x4 1 0 120 17/5 0 S1 1/50 1/50 0 0 12/5 S2 0 1 bi 17/5 3/5 bi /aij 17/2 1

0 3/5 Z = 408 + 520 = 928 0

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47Simplex Tableau 3: Optimal SolutionBasis x 4 120 x3 80 Cj Solution Dj x3 0 1 80 1 0 x4 1 0 120 3 0 S1 1/30 1/30 0 0 4/3 S2 2/3 5/3 0 0 160/3 bi 3 1 Z = 440 + 520 = 960

Thus, optimal solution to the revised problem is: x3 = 1 and x4 = 3. Accordingly, the solution to the original problem may be obtained as follows: Output of X, x1 = 2 + x3 or 2 + 1 = 3 tonnes, Output of Y, x2 = 3 + x4 or 3 + 3 = 6 tonnes, and Total profit = 80 3 + 120 6 = Rs 960. 27. Let the production of I1 and I2 be x 1 and x2 units respectively. The LPP is: Maximise Z = 40x1 + 60x2 Subject to x1 + x 2 40 2x1 + x 2 70 x1 + 3x 2 90 x1, x 2 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 0 0 0 x1 1 2 1 40 0 40 x2 1 1 3* 60 0 60 S1 1 0 0 0 40 0 S2 0 1 0 0 70 0 S3 0 0 1 0 90 0 bi 40 70 90 bi /a ij 40 70 30

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 S2 0 x2 60 Cj Solution Dj x1 2/3* 5/3 1/3 40 0 20 x2 0 0 1 60 30 0 S1 1 0 0 0 10 0 S2 0 1 0 0 40 0 S3 1/3 1/3 1/3 0 0 20 bi 10 40 30 Z = 1,800 bi /a ij 15 24 90

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48Simplex Tableau 3: Optimal SolutionBasis x 1 40 S2 0 x 2 60 Cj Solution Dj x1 1 0 0 40 15 0 x2 0 0 1 60 25 0 S1 3/2 5/2 1/2 0 0 30 S2 0 1 0 0 15 0 S3 1/2 1/2 1/2 0 0 10 bi 15 15 25 Z = 2,100

\ Optimal mix: I1 = 15 and I2 = 25 units. Increase in profit = Rs 2,100 Rs 1,800 = Rs 300. Idle time on machine M2 = 15 hours. 28. Let x1 and x2 be the number of programmes on TV and radio respectively. The problem is: Maximise Z = 5,00,000x1 + 3,00,000x 2 Subject to 50,000x1 + 20,000x 2 2,10,000 x1 3 5 x2 x1, x 2 0 * Let x1 = x1 + 3. The revised problem is: * Maximise Z = 5,00,000 x1 + 3,00,000x2 + 15,00,000 Subject to * 50,000 x 1 + 20,000x 2 60,000 x2 5 * x1 , x 2 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 0 S2 0 Cj Solution Dj* x1

x2 20,000 1 3,00,000 0 3,00,000

S1 1 0 0 60,000 0

S2 0 1 0 5 0

bi 60,000 5 Z = 15,00,000

bi /aij 6/5 -

50,000 0 5,00,000 0 5,00,000

Simplex Tableau 2: Non-optimal SolutionBasis* x1 5,00,000 S2 0 Cj Solution Dj * x1

x2 2/5 1 3,00,000 0 1,00,000

S1 1/50,000 0 0 0 10

S2 0 1 0 5 0

bi 6/5 5 Z = 21,00,000

bi /aij 3 5

1 0 5,00,000 6/5 0

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49Simplex Tableau 4: Optimal SolutionBasis x2 3,00,000 S2 0 Cj Solution Dj* x1

x2 1 0 3,00,000 3 0

S1 1/10,000 1/10,000 0 0 30

S2 0 1 0 2 0

bi 3 2 Z = 24,00,000

5/2 5/2 5,00,000 0 2,50,000

Thus, optimal solution calls for 3 programmes in TV and 3 programmes in Radio. Notice that x1 = x* + 3 1 or 0 + 3 = 3 and x2 = 3. This would imply a total reach of 24,00,000, out of which Type A are 15,90,000 while Type B are 8,10,000. 29. Let x1, x2 and x 3 be the number of advertisements in magazines A, B and C respectively. The problem is: Exposure in 000 Maximise Z = 1,000x1 + 900x2 + 280x3 Subject to Budget 10,000x 1 + 5,000x 2 + 6,000x 3 100,000 2 x1 5 Insertion requirement x2 x3 2 x1, x 2, x 3 0 To simplify the problem, we set x1 = 2 + x4 and x3 + 2 + x5. The revised problem is: Maximise Z = 1,000x4 + 900x2 + 280x5 + 2,560 Subject to 10,000x 4 + 5,000x 2 + 6,000x 5 68,000 5 x2 x4, x 2, x 3 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 0 0 x4 10,000* 0 1,000 0 1,000 x2 5,000 1 900 0 900 x5 6,000 0 280 0 280 S1 1 0 0 68,000 0 S2 0 1 0 5 0 bi 68,000 5 bi /a ij 6.8

Cj Solution Dj

Z = 0 + 2,560 = 2,560

Simplex Tableau 2: Non-optimal SolutionBasis x4 1,000 S2 0 Cj Solution Dj x4 1 0 1,000 6.8 0 x2 1/2 1* 900 0 400 x5 6/10 0 280 0 320 S1 1/10,000 0 0 0 1/10 S2 0 1 bi 6.8 5 b i /aij 13.6 5

0 5 Z = 6,800 + 2,560 = 9,360 0

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50Simplex Tableau 3: Optimal SolutionBasis x4 1,000 x 2 900 Cj Solution Dj x4 1 0 1,000 4.30 0 x2 0 1 900 5 0 x5 6/10 0 280 0 320 S1 1/10,000 0 0 0 1/10 S2 1/2 1 0 0 400 bi 4.30 5 Z = 8,800 + 2,560 = 11,360

Thus, optimal ad-mix is: Magazine A: 2 + 4.30 = 6.30, Magazine B = 5, Magazine C = 2 + 0 = 0. Expected exposure = 11,360 (thousand). Note: A non-integer solution is acceptable in LP. 30.Basis A1 A2 M M x1 20 40* 120 0 120 60M

Simplex Tableau 1: Non-optimal Solutionx2 30 30 160 0 160 6M S1 1 0 0 0 M S2 0 1 0 0 M A1 1 0 M 900 0 A2 0 1 M 1,200 0 bi 900 1,200 bi /aij 45 30

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis A1 M x 1 120 Cj Solution Dj x1 0 1 120 30 0 x2 15* 3/4 160 0 70 15M S1 1 0 0 0 M S2 1/2 1/40 0 0 3 M/2 A1 1 0 M 300 0 A2 1/2 1/40 M 0 3 + M/2 bi 300 30 bi /aij 20 40

Simplex Tableau 3: Optimal SolutionBasis x2 x1 160 120 x1 0 1 120 15 0 x2 1 0 160 20 0 S1 1/15 1/20 0 0 14/3 S2 1/30 1/20 0 0 2/3 A1 1/15 1/20 M 0 M 14/3 A2 1/30 1/20 M 0 M 2/3 bi 20 15 Z = 5,000

Cj Solution Dj

The solution will be unbounded in case the objective function is of maximisation type.

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5131. Let x 1 and x2 respectively be the output of the products A and B. The LPP is: Maximise Z = 10x1 + 12x2 Total Profit Subject to Machine M1 2x1 + 3x 2 1,500 3x1 + 2x 2 1,500 Machine M2 x1 + x 2 1,000 Machine M3 x1, x 2 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 0 0 0 x1 2 3 1 10 0 10 x2 3* 2 1 12 0 12 S1 1 0 0 0 1,500 0 S2 0 1 0 0 1,500 0 S3 0 0 1 0 1,000 0 bi 1,500 1,500 1,000 Z=0 bi /a ij 500 750 1,000

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis x2 12 S2 0 S3 0 Cj Solution Dj x1 2/3 5/3* 1/3 10 0 2 x2 1 0 0 12 500 0 S1 1/3 2/3 1/3 0 0 4 S2 0 1 0 0 500 0 S3 0 0 1 bi 500 500 500 bi /a ij 750 300 1,500

0 500 Z = 6,000 0

Simplex Tableau 3: Optimal SolutionBasis x 2 12 x 1 10 S3 0 Cj Solution Dj x1 0 1 0 10 300 0 x2 1 0 0 12 300 0 S1 3/5 2/5 1/5 0 0 16/5 S2 2/5 3/5 1/5 0 0 6/5 S3 0 0 1 0 400 0 bi 300 300 400 Z = 6,600

Optimal product mix: x1 = 300, x2 = 300. Hours unused on machine M3 = 400. Total Profit = 6,600 + 600 = Rs 7,200. 32. If the output of C1, C2 and C3 be x1, x 2 and x3 respectively, the problem is: Maximise Z = 6x 1 + 3x 2 + 2x 3 Subject to 2x1 + 2x2 + 3x 3 300 2x1 + 2x2 + x 3 120 x1, x 2, x 3 0

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52Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 0 0 x1 2 2* 6 0 6 x2 2 2 3 0 3 x3 3 1 2 0 2 S1 1 0 0 300 0 S2 0 1 0 120 0 bi 300 120 Z=0 b i /aij 150 60

Cj Solution Dj

Simplex Tableau 2: Optimal SolutionBasis S1 0 x1 6 Cj Solution Dj x1 0 1 6 60 0 x2 0 1 3 0 3 x3 2 1/2 2 0 1 S1 1 0 0 180 0 S2 1 1/2 0 0 3 bi 180 60 Z = 360

33.

The optimal solution is to produce only 60 units of C1. The answer would not change by given statement. (a) Since there is no artificial variable in the basis, and all the Cj zj values are 0, the given solution is optimal. The optimal product mix is: x1 = 0, x2 = 8/3 units, and x3 = 56/3 units. (b) The given solution is feasible since it involves no artificial variable in the basis. (c) The problem does not have any alternate optimal solution since none of the non-basic variables, x1, S1, and S2 has Dj = 0. (d) The solution given in the table is not degenerate since none of the basic variables has solution value equal to zero. (e) The values in the given table under column headed x1 are 1/3 and 5/6 corresponding to the variables x2 and x3 respectively. Thus, 1/3 unit of x2 and 5/6 unit of x3 have to be foregone to get one unit of x1. Now, to obtain six units of x1, we have to reduce 6 1/3 = 2 units of x2 and 6 5/6 = 5 units of x3. 34. Let S 1, S2 and A1 be the necessary surplus, slack and artificial variables. Simplex Tableau 1Basis A1 S2 M 0 x1 2 4 x2 5* 1 20 0 20 + 5M S1 1 0 0 0 M S2 0 1 0 28 0 A1 1 0 M 50 0 bi 50 28 b i /aij 10 28

Cj 10 Solution 0 Dj 10 + 2M

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53Simplex Tableau 2Basis x2 20 S2 0 Cj Solution Dj x1 2/5 18/5 10 0 2 x2 1 0 20 10 0 S1 1/5 1/5* 0 0 4 S2 0 1 0 18 0 A1 1/5 1/5 M 0 M 4 bi 10 18 Z = 200 b i /aij 90

Simplex Tableau 3Basis x 2 20 S1 0 Cj Solution Dj x1 4 18 10 0 70 x2 1 0 20 28 0 S1 0 1 0 90 0 S2 1 5 0 0 20 A1 0 1 M 0 M bi 28 90 Z = 560

35.

\ Optimal solution is: x1 = 0, x2 = 28 for Z = 560. With slack, surplus and artificial variables, the problem is: Maximise F = 22x + 30y + 25z + 0S1 + 0S 2 + 0S3 MA1 Subject to 2x + 2y + S 1 = 100 2x + y + z + S 2 = 100 x + 2y + 2z S3 + A1 = 100 x, y, z, S1, S 2, S 3, A1 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 0 S2 0 Ai M Cj Solution Dj x 2 2 1 22 0 22 + M y 2 1 2* 30 0 30 + 2M z 0 1 2 25 0 25 + 2M S1 1 0 0 0 100 0 S2 0 1 0 0 100 0 S3 0 0 1 0 0 M A1 0 0 1 M 100 0 bi 100 100 100 b i /aij 50 100 50

Simplex Tableau 2: Non-optimal SolutionBasis S1 S2 y 0 0 30 x 1 3/2 1/2 22 0 7 y 0 0 1 30 50 0 z 2 0 1 25 0 5 S1 1 0 0 0 0 0 S2 0 1 0 0 50 0 S3 1* 1/2 1/2 0 0 15 A1 1 1/2 1/2 M 0 M 15 bi 0 50 50 bi /aij 0 100 F = 1500

Cj Solution Dj

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54Simplex Tableau 3: Non-optimal SolutionBasis S3 S2 y 0 0 30 x 1 1 1 22 0 8 y 0 0 1 30 50 0 z 2 1* 0 25 0 25 S1 1 1/2 1/2 0 0 15 S2 0 1 0 0 50 0 S3 1 0 0 0 0 0 A1 1 0 0 M 0 M bi 0 50 50 bi /aij 50 F = 1500

Cj Solution Dj

Simplex Tableau 4: Optimal SolutionBasis S3 0 z 25 y 30 Cj Solution Dj x 3 1 1 22 0 33 y 0 0 1 30 50 0 z 0 1 0 25 50 0 S1 0 1/2 1/2 0 0 5/2 S2 2 1 0 0 0 25 S3 1 0 0 0 100 0 A1 1 0 0 M 0 M bi 100 50 50 F = 2,750

Optimal Solution: x = 0, y = 50, z = 50, F= 2,750 36.Basis A1 A2 M M x1 2 4* 40 0 40 + 6M

Simplex Tableau 1: Non-optimal Solutionx2 3 3 35 0 35 + 6M S1 1 0 0 0 M S2 0 1 0 0 M A1 1 0 M 60 0 A2 0 1 M 96 0 bi 60 96 bi /aij 30 24

Cj Solution Dj

Simplex Tableau 2: Non-optimal SolutionBasis A1 M x1 40 Cj Solution Dj x1 0 1 40 24 0 x2 3/2* 3/4 35 0 5+ 3M S1 0 0 0 0 M 10 + M S2 1/2 1/4 0 0 A1 1 0 M 12 0 A2 1/2 1/4 M 0 10 M bi 12 24 bi /aij 8 32

2

2

2

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55Simplex Tableau 3: Non-optimal SolutionBasis x 2 35 x1 0 Cj Solution Dj x1 0 1 40 18 0 x2 1 0 35 8 0 S1 2/3 1/2 0 0 10/3 S2 1/3* 1/2 0 0 25/3 A1 2/3 1/2 M 0 M 10/3 A2 1/3 1/2 M M 25/3 bi 8 18 bi /aij 24 36

Simplex Tableau 4: Non-optimal SolutionBasis S2 0 x 1 40 Cj Solution Dj x1 0 1 40 30 0 x2 3 3/2 35 0 25 S1 2 1/2 0 0 20 S2 1 0 0 24 0 A1 2 1/2 M 0 M 20 A2 1 0 M 0 M bi 24 30 b i /aij 12 60

It may be observed from Simplex Tableau 4 that the solution is not optimal as all Dj values are not less than or equal to zero. However, considering the a ij values of the incoming variable S1, the replacement ratios are both found to be negative. Accordingly, the procedure terminates. This indicates the problem has unbounded solution. 37. Let x1 kg of factor A and x2 kg of factor B are used. The LPP is: Maximise Z = 5x 1 + 6x 2 Subject to x1 + x2 = 5, x1 2, x2 4, and x2 0. Simplex Tableau 1: Non-optimal SolutionBasis A1 M A2 M S2 0 Cj Solution Dj x1 1 1* 0 5 0 5 + 2M x2 1 0 1 6 0 6+M S1 0 1 0 0 0 0 A1 1 0 0 M 5 M A2 0 1 0 M 2 M S2 0 0 1 0 4 0 bi 5 2 4 bi /aij 5 2

Simplex Tableau 2: Non-optimal SolutionBasis A1 M x1 5 S2 0 Cj Solution Dj x1 0 1 0 5 2 0 x2 1* 0 1 6 0 6+M S1 1 1 0 0 0 5+M A1 1 0 0 M 3 0 A2 1 1 0 M 0 5 2M S2 0 0 1 0 4 0 bi 3 2 4 bi /aij 3 4

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56Simplex Tableau 3: Optimal SolutionBasis x2 6 x1 5 S2 0 Cj Solution Dj x1 0 1 0 5 2 0 x2 1 0 0 6 3 0 S1 1 1 1 0 0 1 A1 1 0 1 M 0 M 6 A2 1 1 1 M 0 M + 6 S2 0 0 1 0 1 0 bi 3 2 1 Z = 28

Optimal solution: Factor A = 2 kg, Factor B = 3 kg, Profit = Rs 28. 38. To solve the problem using simplex algorithm, we first introduce the necessary slack, surplus, and artificial variables. The augmented LPP is stated below: Maximise Z = 2x1 + 4x2 + 0S1 + 0S2 MA1 MA2 Subject to = 18 2x1 + x2 + S 1 3x1 + 2x2 S 2 + A1 = 30 A2 = 25 x 1 + 2x2 + x1, x2, S 1, S 2, A1, A2 0 Solution to the problem is contained in tables. Simplex Tableau 1: Non-optimal SolutionBasis S1 0 A1 M A2 M Cj Solution Dj x1 2 3 1 2 0 2 + 4M x2 1 2 2* 4 0 4 + 4M S1 1 0 0 0 18 0 S2 0 1 0 0 0 M A1 0 1 0 M 30 0 A2 0 0 1 M 25 0 bi 18 30 25 bi /a ij 18 15 25/2

Simplex Tableau 2: Non-optimal SolutionBasis S1 0 A1 M x2 4 Cj Solution Dj x1 3/2 2* 1/2 2 0 2M x2 0 0 1 4 25/2 0 S1 1 0 0 0 11/2 0 S2 0 1 0 0 0 M A1 0 1 0 M 5 0 A2 1/2 1 1/2 M 0 2M 2 bi 11/2 5 25/2 bi /aij 11/3 5/2 25

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57Simplex Tableau 3: Optimal SolutionBasis S1 0 x1 2 x2 4 Cj Solution Dj x1 0 1 0 2 5/2 0 x2 0 0 1 4 45/4 0 S1 1 0 0 0 0 0 S2 3/4* 1/2 1/4 0 0 0 A1 3/4 1/2 1/4 M 0 M A2 1/4 1/2 3/4 M 0 M 2 bi 7/4 5/2 45/4 Z = 185

The Simplex Tableau 3 gives optimal solution as x1 = 5/2 and x2 = 45/4, with Z = 185. However, this solution is not unique as a non-basic variable, S2, has D j = 0. An alternate optimal solution is given here. Simplex Tableau 4: Optimal Solution (Alternate)Basis S2 0 x1 2 x2 4 Cj Solution Dj x1 0 1 0 2 11/3 0 x2 0 0 1 4 533/12 0 S1 4/3 2/3 1/3 0 0 0 S2 1 0 0 0 7/3 0 A1 1 0 0 M 0 M A2 1/3 1/3 2/3 M 2 bi 7/3 11/3 533/12 Z = 1085

39.Basis S1 0 S2 0 A1 M A2 M x1 4 6 1 0 x2 2 5 0 1 8 0 8+M

Simplex Tableau 1: Non-optimal SolutionS1 1 0 0 0 0 1,600 0 S2 0 1 0 0 0 3,000 0 S3 0 0 1 0 0 0 M S4 0 0 0 1 0 0 M A1 0 0 1 0 M 300 0 A2 0 0 0 1 M 300 0 bi 1,600 3,000 300 300 bi /aij 400 500 300

Cj 10 Solution 0 Dj 10 + M

Simplex Tableau 2: Non-optimal SolutionBasis S1 S2 x1 A2 0 0 10 M x1 0 0 1 0 10 300 0 x2 2* 5 0 1 8 0 8+M S1 1 0 0 0 0 400 0 S2 0 1 0 0 0 1,200 0 S3 4 6 1 0 0 0 10 S4 0 0 0 1 A1 4 6 1 0 A2 0 0 0 1 M 300 0 bi 400 1,200 300 300 bi /aij 200* 240 300

Cj Solution Dj

0 M 0 0 M M 10

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58Simplex Tableau 3: Non-optimal SolutionBasis x2 S2 x1 A2 8 0 10 M x1 0 0 1 0 10 300 0 x2 1 0 0 0 8 200 0 4 M S1 1/2 5/2 0 1/2 0 0 S2 0 1 0 0 0 200 0 S3 2 4 1 2 0 0 2M 6 S4 0 0 0 1 0 0 M A1 2 4 1 2* M 0 M+6 A2 0 0 0 1 M 100 0 bi 200 200 300 100 bi /aij 50 300 50

Cj Solution Dj

2

Simplex Tableau 4: Non-optimal Solution (Final)Basis x2 S2 x1 A1 8 0 10 M x1 0 0 1 0 10 250 0 x2 1 0 0 01 8 300 0 S1 0 3/2 1/4 1/4 0 0 S2 0 1 0 0 0 0 0 S3 0 0 0 1 0 0 S4 1 2 1/2 1/2 0 0 A1 0 0 0 1 M 50 0 A2 1 2 1/2 1/2 M 0 M 3 bi 300 0 250 50

Cj Solution Dj

- M - 10 4

M M + 3

2

2

In Simplex tableau 4, all D j values are less than, or equal to zero. Hence, the solution is final. However, since an artificial variable is a basic variable, it is not feasible. Thus, the given problem has no feasible solution. 40. With slack variables S1, S2 and S3, the problem may be written as: Maximise Z = 50x1 + 110x2 + 120x3 + 0S1+ 0S2 + 0S3 Subject to 3x1 + 3x2 + 5x 3 + S1 = 100 x1 + 3x2 + 4x 3 + S2 = 80 2x1 + 4x2 + 3x 3 + S3 = 60 x 1, x2, x3, S1, S2, S 3 0 Simplex Tableau 1: Non-optimal SolutionBasis S1 S2 S3 0 0 0 x1 3 1 2 50 0 50 x2 3 3 4 110 0 110 x3 5* 4 3 120 0 120 S1 1 0 0 0 100 0 S2 0 0 1 0 80 0 S3 0 0 1 0 60 0 bi 100 80 60 b i /a ij 20 20 20

Cj Solution Dj

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59Simplex Tableau 2: Non-optimal SolutionBasis x 3 120 S2 0 S3 0 Cj Solution Dj x1 3/5 7/5 1/5 50 0 22 x2 3/5 3/5* 11/5 110 0 38 x3 1 0 0 120 20 0 S1 1/5 4/5 3/5 0 0 24 S2 0 1 0 0 0 0 S3 0 0 1 0 0 0 bi 20 0 0 Z = 2,400 b i /a ij 100/3 0 0

Simplex Tableau 3: Non-optimal SolutionBasis x 3 120 x 2 110 S3 0 Cj Solution Dj x1 2 7/3 16/3* 50 0 200/3 x2 0 1 0 110 0 0 x3 1 0 0 120 20 0 S1 1 4/3 7/3 0 0 80/3 S2 1 5/3 11/3 0 0 190/3 S3 0 0 1 0 0 0 bi 20 0 0 Z = 2,400 b i /a ij 10 0

Simplex Tableau 4: Optimal SolutionBasis x 3 120 x 2 110 x1 50 Cj Solution Dj x1 0 0 1 50 0 0 x2 0 1 0 110 0 0 x3 1 0 0 120 20 0 S1 1/8 5/16 7/16 0 0 5/2 S2 3/8 1/16 11/16 0 0 35/2 S3 3/8 7/16 1/16 0 0 25/2 bi 20 0 0 Z = 2,400

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