Question

please help!

Use the Big M method to find the optimal solution to the following LP: max z = x1 + x2 s.t. 2x1 + x2 > 3 3x1 + x2 = 3.5 x1 + x2 = 1 X1, X2 = 0

Answer 1

Manz=2, +22 st M, WO After introducing slack, surplus, artificial variablen's Mem Z=34 + 4 +0.5, +0.52 +0.53-MA, 22, +423 34+4 23.5 4+H2 L1 21+22-S, +A=3 34+H2 +62 = 3.5 24+2 +63=1 subject, ; 4, 4, 5, S2, S3, A;> 0 iteration 11 1 0 -M J 1 ICB XB B 32 A 53 Min Ratio AI XB/a, 3/2=1.5 A, 2 -M3 1 o 0 1 3.5 3 1 0 1 0 S2 3.5 = 1.1667 t=14 1 53 1 1 0 0 M O Z=-3M ZJ-2M-M -M Zj-G/2M-1-M-1 M O 0 Negative minimum Zj-ly is - - 2M-1 and its Column inden is 1, so the entering variable is us. Minimum ratio is 1 and its row inden is, 50: the basis variable is Sz learing The Pivot element is 1.

iteration 2 B 4 1 1 o -M BXB 134 RS, SS3 AI Min Reitio AI -MI 1 -1 - -2 1 S2 9.5 O -210 1 - 3 1 1 0 Z=-Mti - Zg M+lm 0 2M + -M E B |Z- Go M 0 2M+lo all, Zgoly> 0 Hence optimal solution is; 4 =1; w=0 Mom z=1 But the solution is not feasible because the final solution violates the 1st constraint auth73.