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Unformatted text preview: Minimize z = 5 x1 + 7 x2 Subject to the constraints 2 x1 + 3 x2 ≥ 42 3 x1 + 4 x2 ≥ 60 x1 + x2 ≥ 18 x1 , x2 ≥ 0
Ans: x1=12, x2=6 Soln(= Min z) = 102 Introducing the surplus and artificial variables, R1, R2, the LPP is modified as follows: Minimize z = 5 x1 + 7 x2 + MR1 + MR2 + MR3 Subject to the constraints
2 x1 +3 x2 − s1 3 x1 + 4 x2 x1 + x2 + R1 −s2 − s3 +R2 = 42 = 60 + R3 =18 x1 , x2 , s1 , s2 , s3 , R1 , R2 , R3 ≥ 0 Basic z z R1 R2 R3 z x2 R2 R3 x1 x2 s1 5+6M 7+8M M 1 5 7 0 2 3 1 3 4 1 0 1 0 0 1 0 0 s2 M 0 0 1 0 s3 M 0 0 0 1 R1 0 M 1 0 0 R2 0 M 0 1 0 0 0 1 0 R3 Sol 0 120M M 0 0 0 1 42 60 18 0 0 0 1 2M 1−+ 33 7 5M −+ M 33 7 8M M − 33 0 98+8M 0 0 1 14 4 4 0 0 0 2/3 1/3 1/3 1/3 4/3 1/3 0 1 0 0 0 1 1/3 4/3 1/3 Basic z z x2 R2 R3 z x2 s1 R3 x1 1 2M −+ 133 2/3 1/3 1/3
1M + 44 x2 0 1 0 0 0 1 0 0 s1 s2 s3 R1 R2 0 0 1 0
7 5M − 4 4 R3 Sol 7 5M −+ 3 3 M
1/3 4/3 1/3 0 0 1 0 0 1 0 7 8M − M 3 3
0 0 1 1/3 4/3 1/3 0 98+8M 0 0 1 14 4 4 0 0 0 1 0 0 0 7M −+ M  M 44 0 105+3M 0 0 1 15 3 3 3/4 1/4 1/4 1/4 3/4 1/4 0 0 1 0 1 0 1/4 3/4 1/4 Basic z z x2 s1 R3 z x2 s1 x1 1 0 0 0 1 0 0 0 x1 x2 0 1 0 0 0 1 0 0 s1 0 0 1 0 0 0 1 0 s2 s3 M 0 0 1 1 3 1 4 1M + 44
3/4 1/4 1/4 0 0 0 1 7M −+ 44
1/4 3/4 1/4 2 1 1 1 R2 7 5M M 4 − 4 0 1 0 M 0 1 0 1/4 3/4 1/4 R1 R3 0 0 0 1 Sol
105+3M 15 3 3 2M 1M 102 1 1 1 3 1 4 6 0 12 Basic z z x2 s1 x1 z x2 s3 x1 1 0 0 0 1 0 0 0 x1 0 0 0 1 0 0 0 1 x2 0 1 0 0 0 1 0 0 s1 0 0 1 0 1 3 1 4 s2 2 1 1 1 1 2 1 3 s3 1 3 1 4 0 0 1 0 R1 M 0 1 0 1M 3 1 4 R2 R3 Sol 2M 1M 102 1 1 1 1M 2 1 3 3 1 4 6 0 12 2M 102 0 1 0 6 0 12 Basic z z x2 s3 x1 1 0 0 0 x1 0 0 0 1 x2 0 1 0 0 s1 1 3 1 4 s2 1 2 1 3 s3 0 0 1 0 R1 1M 3 1 4 R2 1M 2 1 3 R3 2M 0 1 0 Sol 102 6 0 12 This is the optimal tableau. Optimal solution: x1=12, x2=6 Optimal z = Minimum z = 102 (0, 18) (0, 15) (0, 14) Graphical illustration (12,6) (0, 0) (18, 0) (20, 0) (21, 0) TWO PHASE METHOD The Big M method involves manipulation with small and large numbers and so is not suited to a computer. We now look at the TwoPhase method. As the name suggests, the method consists of two phases: In phase I, we minimise the sum of all the artificial variables subject to the same constraint equations. If the original problem had a feasible solution this new problem will give a solution with all artificial variables zero. Taking this as a starting BFS, we solve the original problem. We illustrate by an example. Consider the LPP: Minimize z = 2 x1 + x2 Subject to the constraints 3 x1 +x2 ≥9 x1 +x2 ≥6 x1 , x2 ≥0 Introducing the surplus and artificial variables, R1, R2, the LPP is same as: Minimize z = 2 x1 + x2
+ R1 =9 Subject to the constraints 3 x1 + x2 − s1 x1 + x2 −s2 + R2 = 6 x1 , x2 , s1 , s2 , R1 , R2 ≥ 0 Phase I: Minimize r = R1 + R2
+ R1 =9 Subject to the constraints 3 x1 + x2 − s1 x1 + x2 − s2 + R2 = 6 x1 , x2 , s1 , s2 , R1 , R2 ≥ 0
We now solve it by Simplex method. Basic r r R1 R2 r x1 R2 r x1 x2 1 0 0 1 0 0 1 0 0 x1 4 0 3 1 0 1 0 0 1 0 x2 2 0 1 1 2/3 1/3 2/3 0 0 1 s1 1 0 1 0 1/3 1/3 1/3 0 1/2 1/2 s2 1 0 0 1 R1 0 1 1 0 R2 0 1 0 1 0 0 1 1 1/2 3/2 Sol. 15 0 9 6 3 3 3 0 3/2 9/2 1 4/3 0 0 1/3 1 1 1/3 1/2 1/2 3/2 1/2 Note that Phase I has ended as min r =0. Phase II Now we solve the original LPP with 3 9 the starting BFS: x1 = , x2 = 2 2 Note that the starting Simplex tableau is same as the last tableau except for the first row which is our zRow. Since R1, R2 have served their purpose (of giving a starting BFS), we suppress their columns. Basic z z x1 x2 1 0 0 x1 0 2 1 0 x2 0 1 0 1 s1 s2 R1 R2 Sol. 15/2 0 3/2 9/2 1/2 1/2 0 0 1/2 1/2 1/2 3/2 This is the optimal tableau. Optimal solution: x1 = 3/2, x2 = 9/2 Optimal value : Min z = 15/2 ...
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This note was uploaded on 02/23/2011 for the course MATH 112 taught by Professor Ritadubey during the Spring '11 term at Amity University.
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