the first one is how i took in my class

Post by **answerhappygod** » Wed Dec 15, 2021 10:31 am

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Max Z = 12X1+16X2 St 10X1 +20X2 <= 120 8X1 + 6X2 <= 80 X1, X2 >= 0 SIMPLEX TABLEAU METHOD Max 2 =12X1 +16X2 +0S1 + S2 St 10X1 + 20X2 +51 + OS2= 120 8X1 + 6X2 +0S1 + S2 = 80 Basic variables are 4 (X1, X2, S1 and S2) -as a row Cj = 12 for X1, 16 for X2,0 for S1 and 0 for S2 - as a row Column for solution, i.e bi Column for basic solution, x1, x2, s1, s2 -Cbi Initial table- Cbi C 12 16 0 0 Ratio Basic Variable x1 X2 51 52 (Solution) bi 0 S1 1020 1 0 0 S2 8 80 1 Zj 0 0 0 0 Cj-2 12 16 0 D 2i = Cb1*X1.1 +Cb2"x1,2 = 010) +(098) =0 120 80 0 120/20 = 6 80/8 = 10
Gj Column 1 = 12-0 = 12 Pivot element = 20, then 51 is leaving variable and X2 is entering variable Check for the optimality conditions For maximization problems G-Z) <=0; that it can only have 0 or negative values G-Zi) >=0, that it can only have 0 or positive values For the problem at hand, the optimality condition is not met. Therefore we shall continue to iterate the simplex tableau algorithm 1.. determine the pivot element by finding the pivot row and column, pivot element = 20 First iteration Сьој 12 16 0 Basic Variable x1 X2 이 Ratio $1 S2 bi (Solution) 4 0 1/200 -2/51 4/50 4/50 6 32 0 16 X2 1/2 L 0 S2 Z; 8 16 C-2) 4 10 Zx1 = (16*1/2) +(0*4) = 8 Zo = (16*1) +(0*0) = 16 Zs: = (16*1/20) +(0-1) 4/5
Z2 = (16*0) +(0*1) = 0 New value = Old value corresponding pivot column value). corresponding pivot row value perot element X1.S2 = 8- (8) . (10) = 4 20 X2,928- (8) - (20 20 = 0 S1, S2 = 0 - (8) -(1) 20 =-2/5 SZ, S2 = 1 (8) -(0) = 1 20 62,52 = 80 - (8) . (120) = 32 Since the optimality condition is not met, we shall iterate again 24 iteration Cbij 12 16 0 0 Basic Variable X1 X2 si S2 bi (Solution) Ratio 6/1/2 =12 32/4 =8 = 16 X2 0 S2 Z] C-2 1/2 1 4 10 8 16 4 10 1/2010 -2/51 4/5 0 -4/50 6 32 0 이 Cbici 12 16 0 Basic Variable X1 X2 51 S2 Ratio bi (Solution)
16 X2 0 1 1/10 1/8 2 12 X1 10 -1/10 1/48 zj 12 16 2/5 1 128 C:21 12/511 S2 will be leaving the tableau while X1 will be entering the basic solution 0 (1/2) (0) = 1 x1.x2 = 12-(0/2) = 6) = 0 X2, X2 = 1 - - (43)*(?). (C) (1) S1,x2 = 1/20 - - 1/10 X2.52 = 0 - -1/8 b2, X2-6- -(4/2) = 82) - (() = 2 II
Question 2 Min Z=2X1 - 3X2 +6X3 St 3X1 - 2X2 +2X3 <=7 -2X1 - 4X2 <= 12 4X1 + 3X2 + 8X3 <= 10 X1, X2,X3 >=0