1. Given the following original problem: max Z = 6x1 + 4x2 + 5x3 s.t. x1 + 2x2 +
ID: 1231733 • Letter: 1
Question
1. Given the following original problem:
max Z = 6x1 + 4x2 + 5x3
s.t. x1 + 2x2 + x3 <= 430
3x1 + 2x3 <= 520
x1 + 4x2 <= 600
x1, x2, x3 >= 0
Answer the following questions given only the following information from SOME ITERATION of the simplex tableau, with s4, s5, and s6 representing the slack variables for constraints 1, 2, and 3, respectively.
(a) Fill in the values in this tableau (which may or may not be optimal), given only the information below.
Basic Z x1 x2 x3 s4 s5 s6 RHS
Z
x3 1.5 -0.25 -0.75
x1 -1 0.5 0.5
x2 0.25 -0.125 0.125
(b) Find and show the optimal solution (Z, x1, x2, x3)?
Explanation / Answer
I changed Z to P, x1=x, x2=y, x3=z to be less confusing
your table should be:
0 0 0 1 1.5 -0.25 -0.75
0 1 0 0 -1 0.5 0.5
0 0 1 0 0.25 -0.125 0.125
your optimal solution should be Z=P=1640, x=x1=0, y=x2=85 and z=x3=260
your first iteration should be
x y z s1 s2 s3 s4 s5 s6 p
1 2 1 1 0 0 0 0 0 0 430
3 0 2 0 1 0 0 0 0 0 520
1 4 0 0 0 1 0 0 0 0 600
-1 0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-6 -4 -5 0 0 0 0 0 0 1 0
second iteration should be
x y z s1 s2 s3 s4 s5 s6 p
1 2 1 1 0 0 0 0 0 0 430
3 0 2 0 1 0 0 0 0 0 520
1 4 0 0 0 1 0 0 0 0 600
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 -1 0 0 0 0 0 1 0 0
-6 -4 -5 0 0 0 0 0 0 1 0
your third iternation should be
x y z s1 s2 s3 s4 s5 s6 p
1 2 1 1 0 0 0 0 0 0 430
3 0 2 0 1 0 0 0 0 0 520
1 4 0 0 0 1 0 0 0 0 600
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-6 -4 -5 0 0 0 0 0 0 1 0
you keep going and the last interation should be
x y z s1 s2 s3 s4 s5 s6 p
1.5 0 1 0 0.5 0 0 0 0 0 260
2 0 0 -2 1 1 0 0 0 0 260
-0.25 1 0 0.5 -0.25 0 0 0 0 0 85
-1 0 0 0 0 0 1 0 0 0 0
-0.25 0 0 0.5 -0.25 0 0 1 0 0 85
1.5 0 0 0 0.5 0 0 0 1 0 260
0.5 0 0 2 1.5 0 0 0 0 1 1640
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