Consider the following: Maximize Z = 21X1 + 9X2 + 4X3 subject to: 2X1 + X2 + X3
ID: 3282362 • Letter: C
Question
Consider the following:
Maximize Z = 21X1 + 9X2 + 4X3
subject to:
2X1 + X2 + X3 <= 31 (resource constraint 1)
3X1 + 2X2 + X3 <= 60 (resource constraint 2)
X1 + 2X2 + X3 >= 50 (requirement constraint)
X1 >= 0
X2 >= 0
The simplex method yields the following final set of equations
Z + (1/2)X3 + (2/3)X4 + X6 = 291
X1 + (1/3)X3 + (2/3)X4 + (1/3)X6 = 4
X2 + (1/3)X3 - (1/3)X4 - (2/3)X6 = 23
X5 - (2/3)X3 - (4/3)X4 + (1/3)X6 = 2
Where X4 is the slack variable for resource constraint 1, X5 is the slack variable for resourse constraint 2, and X6 is the slack variable for the requirement constraint.
a) What is the optimal solution, the maximum profit, the marginal values of resources 1 & 2, and marginal cost of the requirements?
b) How much can the coefficient of X2 in the objective function vary without affaecting the optimal solution?
c) By how much would the profit be increased if 5 more units of resourse 1 where available? What would be the new solution?
Explanation / Answer
(a)
Maximize Z = 21X1 + 9X2 + 4X3
2X1 + X2 + X3 <= 31
3X1 + 2X2 + X3 <= 60
X1 + 2X2 + X3 >= 50
X1 >= 0
Tableau #1
x1 x2 x3 s1 s2 s3 s4 s5 z
2 1 1 1 0 0 0 0 0 31
3 2 1 0 1 0 0 0 0 60
1 2 1 0 0 -1 0 0 0 50
1 0 0 0 0 0 -1 0 0 0
0 1 0 0 0 0 0 -1 0 0
-21 -9 -4 0 0 0 0 0 1 0
Tableau #2
x1 x2 x3 s1 s2 s3 s4 s5 z
2 0 1 1 0 0 0 1 0 31
3 0 1 0 1 0 0 2 0 60
1 0 1 0 0 -1 0 2 0 50
1 0 0 0 0 0 -1 0 0 0
0 1 0 0 0 0 0 -1 0 0
-21 0 -4 0 0 0 0 -9 1 0
Tableau #3
x1 x2 x3 s1 s2 s3 s4 s5 z
3/2 0 1/2 1 0 1/2 0 0 0 6
2 0 0 0 1 1 0 0 0 10
1/2 0 1/2 0 0 -1/2 0 1 0 25
1 0 0 0 0 0 -1 0 0 0
1/2 1 1/2 0 0 -1/2 0 0 0 25
-33/2 0 1/2 0 0 -9/2 0 0 1 225
Tableau #4
x1 x2 x3 s1 s2 s3 s4 s5 z
3/2 0 1/2 1 0 1/2 0 0 0 6
2 0 0 0 1 1 0 0 0 10
1/2 0 1/2 0 0 -1/2 0 1 0 25
-1 0 0 0 0 0 1 0 0 0
1/2 1 1/2 0 0 -1/2 0 0 0 25
-33/2 0 1/2 0 0 -9/2 0 0 1 225
Tableau #5
x1 x2 x3 s1 s2 s3 s4 s5 z
1 0 1/3 2/3 0 1/3 0 0 0 4
0 0 -2/3 -4/3 1 1/3 0 0 0 2
0 0 1/3 -1/3 0 -2/3 0 1 0 23
0 0 1/3 2/3 0 1/3 1 0 0 4
0 1 1/3 -1/3 0 -2/3 0 0 0 23
0 0 6 11 0 1 0 0 1 291
Optimal Solution:(maximum profit) z = 291;
x1 = 4, x2 = 23, x3 = 0
marginal value of resource 1 =11 ( value of s1 column in bottom row)
marginal value of resource 2 =0 ( value of s2 column in bottom row)
marginal cost of requirement =1 (s3)
(b) Let 9+ be the profit margin on x1. The reduced cost remains constant for =0 so according to that the coefficient hase to be unchanged ,so that the optimal basis remains unchange
(c)
Maximize Z = 21X1 + 9X2 + 4X3
2X1 + X2 + X3 <= 36
3X1 + 2X2 + X3 <= 60
X1 + 2X2 + X3 >= 50
X1 >= 0
Tableau #1
x1 x2 x3 s1 s2 s3 s4 s5 z
2 1 1 1 0 0 0 0 0 36
3 2 1 0 1 0 0 0 0 60
1 2 1 0 0 -1 0 0 0 50
1 0 0 0 0 0 -1 0 0 0
0 1 0 0 0 0 0 -1 0 0
-21 -9 -4 0 0 0 0 0 1 0
Tableau #2
x1 x2 x3 s1 s2 s3 s4 s5 z
2 0 1 1 0 0 0 1 0 36
3 0 1 0 1 0 0 2 0 60
1 0 1 0 0 -1 0 2 0 50
1 0 0 0 0 0 -1 0 0 0
0 1 0 0 0 0 0 -1 0 0
-21 0 -4 0 0 0 0 -9 1 0
Tableau #3
x1 x2 x3 s1 s2 s3 s4 s5 z
1.5 0 0.5 1 0 0.5 0 0 0 11
2 0 0 0 1 1 0 0 0 10
0.5 0 0.5 0 0 -0.5 0 1 0 25
1 0 0 0 0 0 -1 0 0 0
0.5 1 0.5 0 0 -0.5 0 0 0 25
-16.5 0 0.5 0 0 -4.5 0 0 1 225
Tableau #4
x1 x2 x3 s1 s2 s3 s4 s5 z
1.5 0 0.5 1 0 0.5 0 0 0 11
2 0 0 0 1 1 0 0 0 10
0.5 0 0.5 0 0 -0.5 0 1 0 25
-1 0 0 0 0 0 1 0 0 0
0.5 1 0.5 0 0 -0.5 0 0 0 25
-16.5 0 0.5 0 0 -4.5 0 0 1 225
Tableau #5
x1 x2 x3 s1 s2 s3 s4 s5 z
0 0 0.5 1 -0.75 -0.25 0 0 0 3.5
1 0 0 0 0.5 0.5 0 0 0 5
0 0 0.5 0 -0.25 -0.75 0 1 0 22.5
0 0 0 0 0.5 0.5 1 0 0 5
0 1 0.5 0 -0.25 -0.75 0 0 0 22.5
0 0 0.5 0 8.25 3.75 0 0 1 307.5
Optimal Solution: z = 307.5; x1 = 5, x2 = 22.5, x3 = 0