L08_Some Miscellaneous LPPs

L08_Some Miscellaneous LPPs - I n this le ctureweshall look...

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Unformatted text preview: I n this le ctureweshall look at som e m llane LPPs. Each proble will illustrate isce ous m a ce rtain ide which will bee a xplaine whe the d n proble is discusse . m d Problem 6 Problem set 3.4A Page 97 Maximize z = 2 x1 + 4 x2 + 4 x3 − 3 x4 Subject to the constraints x1 + x2 + x3 x1 + 4 x2 x1 , x2 , x3 , x4 ≥ 0 =4 + x4 = 8 Normally, when there are equality constraints, we add artificial variables and use either Big M method or two phase method. But in this problem we already have a 2x2 identity submatrix in the coefficient matrix of the constraint equations. Hence there is no need to add artificial variables. We can start the Simplex algorithm with x3, x4 as basic variables. Basic z z x3 x4 z x3 x2 1 0 0 1 0 0 x1 -1 -2 1 1 2 3/4 1/4 x2 -12 -4 1 4 0 0 1 x3 0 -4 1 0 0 1 0 x4 0 3 0 1 3 -1/4 1/4 Sol -8 0 4 8 16 2 2 This is the Optimal tableau. Optimal Soln: x1=0, x2=2, x3=2, x4=0 Max z = 16 Problem 7 Problem set 3.4A Pages 97-98 Minimize z = 3 x1 + 5 x2 + 3 x3 Subject to the constraints x1 + 4 x2 + x3 2 x1 + x2 x1 , x2 , x3 , x4 ≥ 0 ≥7 + x4 ≥ 10 Introducing surplus variables, the problem is: Minimize z = 3 x1 + 5 x2 + 3 x3 Subject to the constraints x1 + 4 x2 + x3 2 x1 + x2 + x4 − s1 =7 − s2 = 10 x1 , x2 , x3 , x4 , s1 , s2 ≥ 0 Since in this problem as we already have a 2x2 identity submatrix below the variables x3, x4 in the coefficient matrix of the constraint equations, there is no need to add artificial variables. We can start the Simplex algorithm with x3, x4 as basic variables. Basic z z x3 x4 z x2 x4 1 0 0 1 0 0 x1 0 -3 1 2 -7/4 1/4 7/4 x2 7 -5 4 1 0 1 0 x3 0 -3 1 0 -7/4 1/4 -1/4 x4 0 0 0 1 s1 -3 0 -1 0 s2 0 0 0 -1 0 0 -1 Sol. 21 0 7 10 35/4 7/4 33/4 0 -5/4 0 1 -1/4 1/4 This is the Optimal tableau. Optimal Soln: x1=0, x2=7/4, x3=0, x4=33/4 Min z = 35/4 Problem 8 Problem set 3.4A Page 98 Maximize z = x1 + 5 x2 + 3 x3 Subject to the constraints x1 + 2 x2 + x3 = 3 2 x1 − x2 x1 , x2 , x3 ≥ 0 =4 There is a unit vector column (coefficients of x3); so we will not add an artificial variable to the first constraint equation. However, to get a starting BFS (= for getting a 2x2 identity submatrix) we have to add an artificial variable to the second constraint equation. And then we solve it by big M method. Thus we have to Maximize z = x1 + 5 x2 + 3 x3 − MR2 Subject to the constraints x1 + 2 x2 + x3 2 x1 − x2 x1 , x2 , x3 , R2 ≥ 0 =3 + R2 = 4 Basic z z x3 R2 z x3 x1 1 0 0 1 0 0 x1 x2 x3 R2 0 M 0 1 ½ +M -1/2 1/2 Sol -4M 0 3 4 2 1 2 -1-2M -5+M 0 -1 -5 -3 1 2 0 0 1 2 -1 -11/2 5/2 -1/2 1 0 0 1 0 Basic z z x3 x1 z x2 x1 1 0 0 1 0 0 x1 0 0 1 0 0 1 x2 -11/2 5/2 -1/2 0 1 0 x3 0 1 0 11/5 2/5 1/5 R2 ½ +M -1/2 1/2 -3/5 +M -1/5 2/5 Sol 2 1 2 21/5 2/5 11/5 This is the Optimal tableau. Optimal Soln: x1=11/5, x2=2/5, x3=0 Max z = 21/5 Problem 9 Problem set 3.4A Page 98 Use the big M method to show that the following problem has no feasible solution: Maximize z = 2 x1 + 5 x2 Subject to the constraints 3 x1 + 2 x2 ≥ 6 2 x1 + x2 ≤ 2 x1 , x2 ≥ 0 Using surplus, artificial and slack variables, the problem becomes: Maximize z = 2 x1 + 5 x2 − M R1 Subject to the constraints 3 x1 + 2 x2 − s1 + R1 2 x1 + x2 x1 , x2 , s1 , s2 , R1 ≥ 0 We now start the Simplex algorithm. =6 + s2 = 2 Basic z z R1 s2 z R1 x1 1 0 0 1 0 0 x1 x2 s1 M 0 -1 0 M -1 0 R1 0 M 1 0 1 0 s2 0 0 0 1 -3/2 1/2 Sol. - 6M 0 6 2 3 1 -2-3M -5-2M -2 -5 3 2 0 0 1 2 1 -4- M/2 1/2 1/2 0 1+3M/2 2-3M Basic z z R1 x1 z R1 x2 1 0 0 1 0 0 x1 0 0 1 8+M -1 2 x2 -4- M/2 1/2 1/2 0 0 1 s1 M -1 0 M -1 0 R1 s2 Sol. 0 1+3M/2 2-3M 1 0 1 0 -3/2 1/2 -2 1 3 1 2 2 0 5+ 4M 10-2M Thus we have reached the optimal tableau but the artificial variable R1 is present at non-zero level. This means the original LPP has no feasible solution. This can be checked graphically also as the next slide shows. (0,3) (0,2) 3x1 + 2 x2 ≥ 6 (1,0) (2,0) 2 x1 + x2 ≤ 2 One Artificial Variable is enough? Problem 8 Problem Set 3.4B Page 103 Consider the LPP Minimize z = 2 x1 − 4 x2 + 3x3 5 x1 −6 x2 + 2 x3 ≥ 5 −x1 +3 x2 +5 x3 ≥ 8 2 x1 + 5 x2 − 4 x3 ≤ 9 x1 , x2 , x3 ≥ 0 Subject to the constraints Adding surplus and slack variables, the problem becomes: Minimize z = 2 x1 − 4 x2 + 3x3 =5 =8 + s3 =9 Subject to the constraints 5 x1 −6 x2 + 2 x3 − s1 −x1 +3 x2 +5 x3 2 x1 +5 x2 − 4 x3 − s2 x1 , x2 , x3 , s1 , s2 , s3 ≥ 0 Normally we have to add two artificial variables, one to each of the first two constraint equations to get a starting BFS. But by subtracting the first constraint from the second constraint equation, we get the equivalent LPP: Minimize z = 2 x1 − 4 x2 + 3x3 =3 =8 Subject to the constraints −6 x1 + 9 x2 + 3 x3 + s1 − s2 − x1 + 3 x2 + 5 x3 2 x1 + 5 x2 − 4 x3 x1 , x2 , x3 , s1 , s2 , s3 ≥ 0 − s2 + s3 =9 Thus we need to add only one artificial variable to the second constraint equation to get a starting BFS (= to get a 3x3 identity submatrix). Doing this , the LPP is Minimize z = 2 x1 − 4 x2 + 3x3 =3 =8 + s3 =9 Subject to the constraints −6 x1 + 9 x2 + 3 x3 + s1 − s2 − x1 + 3 x2 + 5 x3 2 x1 + 5 x2 − 4 x3 x1 , x2 , x3 , s1 , s2 , s3 ≥ 0 − s2 + R2 We shall solve this by two phase method. Phase I Minimize r = R2 Subject to the constraints −6 x1 + 9 x2 + 3 x3 + s1 − s2 − x1 + 3 x2 + 5 x3 2 x1 + 5 x2 − 4 x3 x1 , x2 , x3 , s1 , s2 , s3 ≥ 0 − s2 + R2 =3 =8 + s3 =9 Basic r r s1 R2 s3 r x3 R2 s3 1 0 0 0 1 0 0 0 x1 -1 0 -6 -1 2 9 -2 9 -6 x2 3 0 9 3 5 -12 3 -12 17 x3 5 0 3 5 -4 0 1 0 0 s1 0 0 1 0 0 -5/3 1/3 -5/3 4/3 s2 -1 0 -1 -1 0 2/3 -1/3 2/3 -4/3 R2 0 -1 0 1 0 0 0 1 0 s3 Sol. 0 0 0 0 1 0 0 0 1 8 0 3 8 9 3 1 3 13 Basic r r x3 R2 s3 r x3 x1 s3 1 0 0 0 1 0 0 0 x1 9 -2 9 -6 0 0 1 0 x2 -12 3 -12 17 0 1/3 -4/3 9 x3 0 1 0 0 0 1 0 0 s1 -5/3 1/3 -5/3 4/3 0 s2 2/3 -1/3 2/3 -4/3 0 R2 0 0 1 0 -1 s3 Sol. 0 0 0 1 0 3 1 3 13 0 5/3 1/3 15 -1/27 -5/27 2/9 0 -5/27 2/27 1/9 0 2/9 -8/9 2/3 1 Phase I has ended. And we now start Phase II with the starting BFS obtained in the optimal tableau of Phase I. Also we do not carry the R2 column as it has served its purpose. Basic z z x3 x1 s3 z x3 x1 x2 1 0 0 0 1 0 0 0 x1 0 -2 0 1 0 0 0 1 0 x2 7/3 4 1/3 -4/3 9 0 0 0 1 x3 s1 s2 R2 s3 Sol. 0 0 0 0 1 17/3 0 5/3 1/3 15 0 -11/27 -13/27 -3 0 0 1 -1/27 -5/27 0 0 0 1 0 0 -5/27 2/27 2/9 -8/9 − − − 231 243 11 243 37 243 − − − 43 243 37 243 14 243 -7/27 16/9 -1/27 10/9 4/27 23/9 1/9 5/3 2/81 -8/81 This is the optimal tableau. And the optimal solution is 23 15 10 x1 = , x2 = , x3 = 9 9 9 16 Optimum z = Minimum z = . 9 It is clear that the same procedure can be applied to more than two ≥ constraints. ...
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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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