Question: Problem 1: Set up the initial simplex tableau and work through one iteration (in TABULAR form). Show the values of x 1 , x 2
Problem 1: Set up the initial simplex tableau and work through one iteration (in TABULAR form). Show the values of x1, x2 and z after one iteration.
Maximize Z = 3x1 + 2 x2,
Subject to x1 4
x1 + 3x2 15
2x1 + x2 10
and x1 0, x2 0.
Problem 2: The following tableau represents a specific simplex iteration. All variables are nonnegative. The tableau is not optimal for either a maximization or a minimization problem. Thus, when a nonbasic variable enters the solution, it can either increase or decrease z or leave it unchanged, depending on the parameters of the nonbasic variable.
| Basic | X1 | X2 | X3 | X4 | X5 | X6 | X7 | X8 | RHS |
| Z | 0 | -5 | 0 | 4 | -1 | -10 | 0 | 0 | 620 |
| X8 | 0 | 3 | 0 | -2 | -3 | 1 | 5 | 1 | 12 |
| X3 | 0 | 1 | 1 | 3 | 1 | 0 | 3 | 0 | 6 |
| X1 | 1 | -1 | 0 | 0 | 6 | -4 | 0 | 0 | 0 |
- categorize the variables as basic and nonbasic and provide the current values of all the variables.
- Suppose that the problem is of the maximization type: identify the nonbasic variables that have the potential to improve the value of z. if such a variable enters the basic solution, determine the associated leaving variable., if any, and the associated change in z. Do not use the row operations.
- Which nonbasic variable(s) will not cause a change in the value of z when selected to enter the solution? Why?
Problem 3: Consider the graphical representation of the following LP problem in which X1 and X2 are the number of desks and file cabinets produced per week, respectively:
Max z = 3X1 + 2X2
S.t. 2X1 + X2 < 100 (cutting operation constraint)
X1 + X2 < 80 (welding operation constraint)
X1 < 35 (woodwork constraint)
X1, X2 > 0
a) Using the graphical representation, Find the optimal solution, and answer the following questions:
b) Let c2 be the contribution to profit by each file cabinet. Currently c2 =2. For what range of c2 values does the current solution remains optimal?
c) Let b3 be the number of available woodwork hours. Currently b3 = 35. For what range of b3 values does the current solution remains optimal?
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