Question

) Use the branch-and-bound method to find the optimal solution to the ollowing IP: Minimize 9x1 +13x2 +10x3 +8x4 +8x5 s.t.6x1+3x2+2x3+4x4+7x5240 X131,x221,x322,x421,x5s3 X1, X2, X3, X4, X5 20 integer

I need help on the knapsack lp by using branch and bound

Answer 1

MIN ZE 9x113x2 + 10x3+ 8x4 +8x5 subiect to 6x1 + 3x2 + 2x3 + 4x4 + 7x5 >= 40 x2 >= 1 x3>= 2 x4 1 x5 <- 3 and x1,x2,x3,x4,x50 Solution Solution stpes by BigM method Min Z = 9x1 + 13x2 subject to 10x3 8x4 8x5 6x1 3x2 + 2x3-4x4 7x5 240 2 1 and xx2x3. x420 Solution is and Z,-74 (x1-1, x,-1, x,-2-4-1, xs-3) obtainted by the rounded off solution values. The branch and bound diagram г,-78 ZL = 74 Solution stpes by BiqM method In Sub-problem A, x4must be an integer value, so two new constraints are created, r41 andx4z12 Sub-problem B: Solution is found by adding x4 1 Sub-problem C: Solution is found by addingx422 Solution stpes by BigM method Solution stpes by BigM method Min Z-9x 13 2 10*38*4 8x5 subject to 6x1 3r2 2*3 + 4X4 7x5 240 subject to 6x1 3*2 + 2*3 +4*4 +7t5 240 2 1 2 1 2 1 5 E 3 1 2 345

Solution is Solution is Min Zc - 792-1,x and ZL 248 Min Zg- 2,x4-2,x5-3 and ZL-74 values. obtainted by the rounded off solution obtainted by the rounded off solution x1-1, x2-1, x3" 2, #4-1, x5-3 73 0, x2-1,x3-2, x4-2, xs-3 x1 values. The branch and bound diagram x1 = 1,x2 = 1,Xy = 2, x,--, xs = 3 -78 z, = 74 Solution stpes by BigM method 248 Zg = 79 ZL-73 Solution stpes by BigM method z, = 74 Solution stpes by BigM method In Sub-problem B,x2must be an integer value, so two new constraints are created, x2 1 andx2 2 2 Sub-problem D Solution is found by adding x2 1 Solution stpes by BigM method Min Z = 9x1 + 13x2 + 10x3 + 8x4 + 8xs subject to 6x3x2 + 2*3 +4*4 +7*52 40 Sub-problem E: Solution is found by adding x2 2 2. Solution stpes by BigM method Min Z-9x1 + 13x 10x 8*48xs subject to 6x 3*2 + 2*3 + 4*4 + 7x52 40 ts S 3 S 1 S 1 and x 1, x2, x3, X4, x5 and x,x3,x4x 2 0; Solution is Solution is in and ZL = 84 (x1 = 1, x,-1. x3-3, x4 = 1, x,-3) obtainted by the rounded off solution values This Problem has integer solution, so no further branching is required. and ZL = 78 (x1-0, x2ー2.x,-2.#4-1, x5-3 ) obtainted by the rounded off solution values Zg =-> ZD-84, so no further branching is required

The branch and bound diagram Solution stpes by BigM method 248 Zn = 79 Z,,73 Solution stpes by BigM method = 74 Solution stpes by Biqw me る-84 Solution stpes by BigM method Z, 78 Solution stpes by BigM method In Sub-problem C. xmust be an integer value, so two new constraints are created, x, 0 and x21 Sub-problem F: Solution is found by adding x10. Solution stpes by BigM method Sub-problem G: Solution is found by addingx1 2 1 Solution stpes by BigM method subject to 6*132 + 23 +4*4 75 240 subject to 6x1 + 3x2 + 2x3 4x4 7x5240 *533 Solution is Solution is 558 19 and ZZ-81 (x1-0, x2-1,5-2, x4-3, x5-3) obtainted by the rounded off solution and Z1-74(m-1. x2-1,x,-2.x,-2x,-2)ottainted by the rounded off solution This Problem has integer solution, so no further branching is required. The branch and bound diagram

ZA78 z, 74 Solution stpes by BigM method 248 Z-79 71-73 Solution stpes by BigM method Z74 Solution stpes by BigM method 3 rı = 6',-2, 3-2.4-1,5 3 r1 = 0 x2 = 1 x,-2,,-3,5-3 卲ー81 1-1,2 1,3 2,4 558 2,5-7 1=1 x2 = 1 3 3 4 1 s n 84 7z84 Solution stpes by BigM method 7-78 在-74 Solution stpes by BigM method Solution stpes by BiaM method Solution stpes by BiaM method in Sub-problem G, x x-must be an integer value, so two new constraints are created, x 2 andx 3 Sub-problem H: Solution is found by adding2. Solution stpes by BigM method Min Z = 9x1 + 13x2 10x3 8x4 8Kg subject to Sub-problem Solution is found by adding Solution stpes by BigM method Min Z = 9x1 + 13x2 + 10x3 呂84-8x5 subject to 3. 5s3 *5s2 s23 Solution is Solution is and ZL-82 (х,-I, x2-1.x,-2, x,-2, x5-3 ) obtainted by the rounded off solution and ZL-82 (x1-1. x2-1,#3-2, x4-3-x5-2) obtainted by the rounded off solution | values This Problem has integer solution, so no further branching is required. values Zr- 84> ZF 81, so no further branching is required. The branch and bound diagram Xi-1両-1.x,-2.84-2.x Z,-78

x1 = 1,x2 = 1,#3-2,#4--,#5-3 Z 78 Z, 74 Solution stpes by BigM method 248 Zc 79 Zi 73 Solution stpes by BigM method Z 74 Solution stpes by BigM method 19 ZF-81 Z,-81 Zp 84 Z,-84 Solution stpes by BigM method 558 Z,-78 Solution stpes by BigM method Z,-74 Solution stpes by BigM method Solution stpes by BigM method Z, -82 Z,-82 Solution stpes by BigM method ZH 84 Solution stpes by BigM method The branch and bound algorithm thus terminated and the optimal integer solution is ZF-81 and X1-0, x,-1.x,-2-4-3, x5-3