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Prompt Details
Model
Chat - GPT-4 (gpt-4)
Token size
291 ($0.00870 / call)
Example input
(can be used in different gpt models)
...My request is"".
Example output
(a) Construct the initial big-M simplex tableau in canonical form.
First, we'll rewrite the constraints as equality constraints by introducing slack, excess, and artificial variables.
For constraint (1): x1 + 4x2 + 2x3 - e1 + a1 = 8
For constraint (2): 3x1 + 2x2 - e2 + a2 = 6
Our initial big-M simplex tableau in canonical form is as follows:
| ZÌ„ | x1 | x2 | x3 | e1 | e2 | a1 | a2 | RHS |
---------------------------------------------------
ZÌ„ | 1 | -2 | -3 | -2 | 0 | 0 | M-1| M-1| 0 |
---------------------------------------------------
R1 | 0 | 1 | 4 | 2 | -1 | 0 | 1 | 0 | 8 |
---------------------------------------------------
R2 | 0 | 3 | 2 | 0 | 0 | -1 | 0 | 1 | 6 |
---------------------------------------------------
(b) Identify xs, x, c, A, b and Z0 of the matrix equation for the initial simplex tableau.
xs = [x1, x2, x3, e1, e2, a1, a2]
x = [x1, x2, x3, e1, e2]
c = [2, 3, 2, 0, 0]
A =| 1 | 4 | 2 | -1 | 0 |
| 3 | 2 | 0 | 0 | -1 |
b = [8, 6]
Z0 = 0
(c) Use the fundamental insight to identify the missing entries in the final simplex tableau. Find the optimal solution for (x1, x2, x3) and the associated value for Z. Show all calculations.
The given portion of the final big-M simplex tableau is:
| ZÌ„ | x1 | x2 | x3 | e1 | e2 | a1 | a2 | RHS |
---------------------------------------------------
ZÌ„ | 1 | | | | | | M-1/2 | M-1/2 | |
---------------------------------------------------
R1 | 0 | | | | | | 3/10 | -1/10 | |
---------------------------------------------------
R2 | 0 | | | | | | -1/5 | 2/5 | |
---------------------------------------------------
Since a1 and a2 are in the basis, we can use the given values in the a1 and a2 columns to find the optimal solution. For the missing entries:
Row 1 (R1): a1 = 3/10, a2 = -1/10
x1 + 4x2 + 2x3 - e1 = 8 (1)
(3/10)a1 - (1/10)a2 = 8
(3/10)(8) - (1/10)(-6) = 8
x1 = 2.4 + 0.6 = 3
Row 2 (R2): a1 = -1/5, a2 = 2/5
3x1 + 2x2 - e2 = 6 (2)
(-1/5)a1 + (2/5)a2 = 6
(-1/5)(8) + (2/5)(6) = 6
x2 = -1.6 + 2.4 = 0.8
Now we substitute x1 and x2 into the first constraint (1) to find x3:
x1 + 4x2 + 2x3 - e1 = 8
3 + 4(0.8) + 2x3 - e1 = 8
2x3 - e1 = 8 - 3 - 3.2
x3 = (8 - 3 - 3.2 + e1) / 2
x3 = 0.8 + e1/2
Since e1 >= 0, the minimum value for x3 is 0.8.
Now we can fill the missing entries in the final simplex tableau:
| ZÌ„ | x1 | x2 | x3 | e1 | e2 | a1 | a2 | RHS |
----------------------------------------------------
ZÌ„ | 1 | 0 | 0 | 0 | | | M-1/2 | M-1/2 | Z |
----------------------------------------------------
R1 | 0 | 1 | 0 | 0 | | | 3/10 | -1/10 | 3 |
----------------------------------------------------
R2 | 0 | 0 | 1 | 0 | | | -1/5 | 2/5 | 0.8|
----------------------------------------------------
R3 | 0 | 0 | 0 | 1 | | | 1/2 | -1/2 | 0.8|
----------------------------------------------------
Now we can find the associated value for Z:
Z = -ZÌ„ = -(2 * x1 + 3 * x2 + 2 * x3)
Z = -(2 * 3 + 3 * 0.8 + 2 * 0.8)
Z = -(6 + 2.4 + 1.6)
Z = -10
The optimal solution is (x1, x2, x3) = (3, 0.8, 0.8), and the associated value for Z is -10.