Exam 18: Simplex-Based Sensitivity Analysis and Duality

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For this optimal simplex tableau, the right-hand sides for the two original \ge constraints were 300 and 250. The problem was a minimization. Basis -100 -110 -95 0 0 -95 0 -1.5 1 1 -1.5 75 -100 1 2 0 -1 1 50 -100 -57.5 -95 5 42.5 -12125 - 0 -52.5 0 -5 -42.5 a.What would the new solution be if the right-hand side value in the first constraint had been 325? b.What would the new solution be if the right-hand side value for the second constraint had been 220?

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a.x3 = 50, x1 = 75, Z = 12250
b.x3 = 30, x1 = 80, Z = 10850

The ranges for which the right-hand side values are valid are the same as the ranges over which the dual prices are valid.

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The range of optimality for a basic variable defines the objective function coefficient values for which the variable will remain part of the current optimal basic feasible solution.

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Given the following linear programming problem Max 10x1 + 12x2 s.t. 1x1 + 2x2 \ge 40 5x1 + 8x2 \le 160 1x1 + 1x2 \le 40 x1, x2 \ge 0 the final tableau is Basis 10 12 0 0 0 12 0 1 -2.5 -.5 0 20 10 1 0 4 1 0 0 0 0 0 -1.5 -5 1 20 10 12 10 4 0 240 - 0 0 -10 -4 0 a.Find the range of optimality for c1 and c2. b.Find the range of feasibility for b1, b2, and b3. c.Find the dual prices.

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The range of feasibility indicates right-hand side values for which

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The linear programming problem Max 6x1 + 2x2 + 3x3 + 4x4 s.t. x1 + x2 + x3 + x4 \le 100 4x1 + x2 + x3 + x4 \le 160 3x1 + x2 + 2x3 + 3x4 \le 240 x1, x2, x 3, x4 \ge 0 has the final tableau Basis 6 2 3 4 0 0 0 2 0 1 1/2 0 3/2 0 -1/2 30 6 1 0 0 0 -1/3 1/3 0 20 4 0 0 1/2 1 -1/6 -1/3 1/2 50 6 2 3 4 2.33 .67 1 380 - 0 0 0 0 -2.33 -.67 -1 Fill in the table below to show what you would have found if you had used The Management Scientist to solve this problem. LINEAR PROGRAMMING PROBLEM MAX 6X1+2X2+3X3+4X4 S.T. 1) 1X1 + 1X2 + 1X3 + 1X4 < 100 2) 4X1 + 1X2 + 1X3 + 1X4 < 160 3) 3X1 + 1X2 + 2X3 + 3X4 < 240 OPTIMAL SOLUTION Objective Function Value == Variable Value Reduced Cost X1 - - X2 - - X3 - - X4 - - Constraint Slack/Surplus Dual Price 1 - - 2 - - 3 - -  OBJECTIVE COEFFICIENT RANGES \text { OBJECTIVE COEFFICIENT RANGES } Variable Lower Limit Current Value Upper Limit 1 - - - 2 - - - 3 - - - 4 - - -  RIGHT HAND SIDE RANGES \text { RIGHT HAND SIDE RANGES } Constraint Lower Limit Current Value Upper Limit 1 - - - 2 - - - 3 - - -

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Write the dual of the following problem Min Z = 2x1 -3x2 + 5x3 s.t. -3x1 + 2x2 + 5x3 \ge 7 2x1 -x3 \ge 5 4x 2 + 3x3 \ge 8.

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The range of optimality is calculated by considering changes in the cj - zj value of the variable in question.

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Explain the simplex tableau location of the dual constraint for each type of constraint.

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If the dual price for b1 is 2.7, the range of feasibility is 20 \le b1 \le 50, and the original value of b1 was 30, which of the following is true?

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If the simplex tableau is from a maximization converted from a minimization, the signs and directions of the inequalities that give the objective function ranges will need to be adjusted to apply to the original coefficients.

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When sensitivity calculations yield several potential upper bounds and several lower bounds, how is the range determined?

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A linear programming problem with the objective function 3x1 + 8x2 has the optimal solution x1 = 5, x2 = 6. If c2 decreases by 2 and the range of optimality shows 5 \le c2 \le 12, the value of Z

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For the following linear programming problem Max Z -2x1 + x2 - x3 s.t. 2x1 + x2 \le 7 1x1 + x2 + x3 \ge 4 the final tableau is Basis -2 1 -1 0 0 - 0 1 0 -1 1 1 -1 3 1 2 1 0 1 0 0 7 2 1 0 1 0 0 7 - -4 0 1 -1 0 - a.Find the range of optimality for c1, c2 , c3.c4, c5 , and c6. b.Find the range of feasibility for b1, and b2.

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The dual variable represents

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Dual prices and ranges for objective function coefficients and right-hand side values are found by considering

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The entries in the associated slack column of the final tableau indicate the changes in the values of the current basic variables corresponding to a one-unit increase in the right-hand side.

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The primal problem is Min 2x1 + 5x2 + 4x3 s.t. x1 + 3x2 + 3x3 \ge 30 3x1 + 7x2 + 5x3 \ge 70 x1, x2, x3 \ge 0 The final tableau for its dual problem is Basis 30 70 0 0 0 70 0 1 3/4 0 -1/4 1/2 0 0 0 -3/2 1 -1/2 0 30 1 0 -5/4 0 3/4 1/2 30 70 15 0 5 50 - 0 0 -15 0 -5 Give the complete solution to the primal problem.

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For an objective function coefficient change outside the range of optimality, explain how to calculate the new optimal solution. Must you return to the (revised) initial tableau?

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The improvement in the value of the optimal solution per-unit increase in a constraint's right-hand side is

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