Exam 17: the Simplex Solution Method

You are offered the chance to obtain more space. The offer is for 15 units and the total price is 1500. What should you do?

The simplex method is a general mathematical solution technique for solving nonlinear programming problems.

Given the following linear programming problem: maximize 4+3 subject to 4+3\leq23 5-\leq5 ,\geq0 What are the basic variables in the initial tableau?

The basic feasible solution in the initial simplex tableau is the origin where all decision variables equal:

The linear programming problem whose output follows determines how many red nail polishes, blue nail polishes, green nail polishes, and pink nail polishes a beauty salon should stock. The objective function measures profit; it is assumed that every piece stocked will be sold. Constraint 1 measures display space in units, constraint 2 measures time to set up the display in minutes. Constraints 3 and 4 are marketing restrictions. MAX    100x1 + 120x2 + 150x3 + 125x4 Subject to    1. x1 + 2x2 + 2x3 + 2x4 ? 108       2. 3x1 + 5x2 + x4 ? 120       3. x1 + x3 ? 25       4. x2 + x3 + x4 > 50       x1, x2, x3, x4 ? 0 Optimal Solution: Objective Function Value = 7475.000 Variable Value Reduced Costs X1 8 0 X2 0 5 X3 17 0 X4 33 0 Constraint Slack/Surplus Dual Prices 1 0 75 2 63 0 3 0 25 4 0 -25 Objective Coefficient Ranges Variable Lower Limit Current Value Upper Limit X1 87.5 100 none X2 none 120 125 X3 125 150 162 X4 120 125 150 Right Hand Side Ranges Constraint Lower Limit Current Value Upper Limit 1 100 108 123.75 2 57 120 none 3 8 25 58 4 41.5 50 54 -How much space will be left unused?

The simplex method can be used to solve quadratic programming problems.

Multiple optimal solutions cannot be determined from the simplex method.

At the initial basic feasible solution at the origin, only ________ variables have a value greater than zero.

The last step in solving a linear programming model manually with the simplex method is to convert the model into standard form.

Given the following linear programming problem: maximize 4+3 subject to 4+3\leq23 5-\leq5 ,\geq0 What is the optimal value of this objective function?

The simplex method ________ guarantee integer solutions.

Whereas the maximization primal model has ≤ constraints, the ________ dual model has ≥ constraints.

The quantity values on the right-hand side of the primal inequality constraints are the ________ coefficients in the dual.

To determine the sensitivity range for the coefficient of a variable in the objective function, calculations are performed such that all values in the cj - zj row are ________.

A basic feasible solution satisfies the model constraints and has the same number of variables with non-negative values as there are constraints.

Given the following linear programming problem: maximize 4+3 subject to 4+3\leq23 5-\leq5 ,\geq0 What are the Cj values for the basic variables?

The basic feasible solution in the initial simplex tableau is the origin where all decision variables equal zero.

A basic feasible solution satisfies the model constraints and has the same number of variables with negative values as there are constraints.

The dual form of a linear program is used to determine how much one should pay for additional resources.

The linear programming problem whose output follows determines how many red nail polishes, blue nail polishes, green nail polishes, and pink nail polishes a beauty salon should stock. The objective function measures profit; it is assumed that every piece stocked will be sold. Constraint 1 measures display space in units, constraint 2 measures time to set up the display in minutes. Constraints 3 and 4 are marketing restrictions. MAX    100x1 + 120x2 + 150x3 + 125x4 Subject to    1. x1 + 2x2 + 2x3 + 2x4 ? 108       2. 3x1 + 5x2 + x4 ? 120       3. x1 + x3 ? 25       4. x2 + x3 + x4 > 50       x1, x2, x3, x4 ? 0 Optimal Solution: Objective Function Value = 7475.000 Variable Value Reduced Costs X1 8 0 X2 0 5 X3 17 0 X4 33 0 Constraint Slack/Surplus Dual Prices 1 0 75 2 63 0 3 0 25 4 0 -25 Objective Coefficient Ranges Variable Lower Limit Current Value Upper Limit X1 87.5 100 none X2 none 120 125 X3 125 150 162 X4 120 125 150 Right Hand Side Ranges Constraint Lower Limit Current Value Upper Limit 1 100 108 123.75 2 57 120 none 3 8 25 58 4 41.5 50 54 -By how much will the second marketing restriction be exceeded?