Reference no: EM13707120
Question 1:
For the following tableau:
a. Label the rows with the variable names in the current solution.
b. State the current solutions variable name and value and objective function value
c. Perform one simplex method iteration showin p all values for ratios and row o erations for row 0 and pivot row.
|
Basic
|
Z
1
|
x1
0
|
x2
-2/3
|
x3
5/6
|
x4
|
x5
0
|
x6
0
|
r.h.s.
20 tig.
4
|
|
|
0
|
1
|
213
|
1/6
|
0
|
0
|
0
|
2
|
|
|
0
|
0
|
1 1/3
|
- 1/6
|
1
|
0
|
0
|
2
|
|
|
0
|
0
|
1 2/3
|
1/6
|
0
|
1
|
0
|
2
|
|
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
|
d. What variables are in the new solution?
e. Is the solution optimal and why?
Question 2:
TOYCO assembles three types of toys-trains, trucks, and cars-using three operations. The daily limits on the available times for the three operations are 430,460, and 420 minutes, respec¬tively, and the revenues per unit of toy train. tructand car are $3, 52, and $5.respectively.The as¬sembly times per train at the three operations are 1, 3, and 1 minutes, respectively. The corresponding times per train and per car are (2,0,4) and (1,2,0) minutes (a zero time indicates that the operation is not used).
Letting x1, x2, and x3 represent the daily number of units assembled of trains, trucks, and cars, respectively, the associated LP model is given as:
Maximize z = 3x1 + 2x2 + 5x3
subject to
x1 + 2x2 + x3 ≤ 430 (Operation 1)
3x1 + 2x3 ≤ 460 (Operation 2)
x1 + 4x2 ≤ 420 (Operation 3)
x1, x2, x3 ≥ 0
Using x4. x5, and x6 as the slack variables for the constraints of operations 1,2, and 3, respectively, the optimum tableau is
|
Basic
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
Solution
|
|
z
|
4
|
0
|
0
|
1
|
2
|
0
|
1350
|
|
x2
|
|
1
|
0
|
1/2
|
-1/4
|
0
|
100
|
|
-1/4
|
|
X3
|
3/2
|
0
|
1
|
0
|
1/2
|
0
|
230
|
|
X6
|
2
|
0
|
0
|
-2
|
1
|
1
|
20
|
a. What is the production plan? Number of trains =
Number of trucks =
Number of cars =
b. What are the shadow prices for one more minute of time at each operation?
c. If one more unit of time in operation 2 were available, what would be the new value of the objective function?
d. what is the basis matrix of the current solution?
e. Verify the current solution by calculating x = B-1b and labeling the x variables.
F. FOR EXTRA CREDIT calculate the new solution (x's and z with 461 minutes available at operation 2).
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