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Question
In solving a minimization problem, artificial variables are assigned a ________ in the objective function to eliminate them from the final solution.Answer
This answer is hidden. It contains 9 characters.
Related questions
Q:
At the initial basic feasible solution at the origin, only slack variables have a value greater than:A) 0B) 1C) -1D) 1 or -1
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The basic feasible solution in the initial simplex tableau is the origin where all decision variables equal:A) oB) 1C) -1D) 1 or -1
Q:
A(n) ________ problem can be identified in the simplex procedure when it is not possible to select a pivot row.
Q:
A(n) ________ maximization linear programming problem has an artificial variable in the final simplex tableau where all cj - zj values are less than or equal to zero.
Q:
________ variables are added to constraints and represent unused resources.
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Final tableaus cannot be used to conduct sensitivity analysis.
Q:
At the initial basic feasible solution at the origin, only slack variables have a value greater than 1.
Q:
The basic feasible solution in the initial simplex tableau is the origin where all decision variables equal zero.
Q:
Row operations are used to solve simultaneous equations where equations are multiplied by constants and added or subtracted from each other.
Q:
The last step in solving a linear programming model manually with the simplex method is to convert the model into standard form.
Q:
The ________ form of a linear program is used to determine how much one should pay for additional resources.
A) standard
B) primal
C) feasible
D) dual
E) simplex
Q:
In the simplex procedure, if cj - zj = 0 for a non-basic variable, this indicates that:
A) the solution is infeasible.
B) the solution is unbounded.
C) there are multiple optimal solutions.
D) the formulation is incorrect.
Q:
The method of Lagrange multipliers can only be used in solving nonlinear programming problems with one equality constraint.
Q:
Lagrange multiplier is another name for shadow price.
Q:
In the method of Lagrange multipliers, the model constraints are multiplied by Lagrange multipliers and subtracted from the objective function.
Q:
The slope of a curve at its highest point equals 0.
Q:
When the branch and bound approach is applied to an integer programming problem, it is used in conjunction with the normal non-integer solution approach.
Q:
The branch and bound method is a solution approach that partitions the feasible solution space into smaller subsets of solutions.
Q:
In an assignment problem:
A) one agent can do parts of several tasks.
B) one task can be done by only one agent.
C) each agent is assigned to its own best task.
D) all of the above
Q:
A transportation table contains an optimal solution when all the empty cell evaluations are:
A) positive or 0.
B) negative or 0.
C) positive.
D) negative.
Q:
An assignment problem is a special form of transportation problem where all supply and demand values equal:A) -1 or 1.B) -1.C) 0D) 1
Q:
Use the following table and the information to answer the questions below.
In the following transportation matrix, all cells except 2C have been evaluated and show no improvement. Evaluate that cell. What is the reduced cost if we decide to allocate to cell 2C?
A) -2
B) -1
C) 0
D) +1
E) +2
Q:
The transportation method is a linear programming technique where linearity can be found in which of the following ways?
A) The cost of goods shipped from any source to any destination is a linear function of quantity shipped.
B) The total cost associated with a given plan is a linear function of shipping costs.
C) Cell evaluations assume linear movements through the matrix.
D) The cost of goods shipped from any source to any destination is a linear function of the cost per unit.
Q:
If we are solving a 0-1 integer programming problem, the constraint x1 + x2 = 1 is a mutually exclusive constraint.
Q:
The divisibility assumption is violated by integer programming.
Q:
In a 0-1 integer model, the solution values of the decision variables are 0 or 1.
Q:
In a total integer model, all decision variables have integer solution values.
Q:
The landlord ran the model in Excel and received the sensitivity report for the constraints as contained in the table. Which of the following statements is correct?FinalShadowConstraintAllowableAllowableCellNameValuePriceR.H. SideIncreaseDecrease$B$12Total Time Weeks4.0013.6441.13.1E-16$E$10Total Cost3000.000.0130005503.1E-13$I$4wood floors Included?1.0067.2711.27E-164.6E-01$I$5kitchen tile Included?1.0025.4517.63E-172.8E-01$I$6back door Included?0.000.0011.00E+301.0E+00$I$7garage door opener Included?1.0011.8215.09E-161.0E+00A) If there are 2 more weeks available for work, it would be possible to increase rent by $27.28 per weekB) Given the scenario, it is evident that a constraint is missing.C) If the landlord installed two garage doors, rent could be increased by $23.64.D) Given the scenario, it is evident that the objective function is misstated.
Q:
The landlord uses the following scheme for decision variables:x1 = contractor works on wood floorsx2 = landlord works on wood floorsx3 = contractor works on kitchen tilex4 = contractor works on kitchen tilex5 = contractor works on back doorx6 = contractor works on back doorx7 = contractor works on garage doorx8 = contractor works on garage doorWhich of these constraints would not be appropriate for this scenario?A) 2700x1 + 400x2 + 2500x3 + 1000x4 + 600x5 + 250x6 + 350x7 + 400x8≤ 3000B) 1x1 + 2x2 + 1.5x3 + 3x4 + 5x5 + 1x6 + 0.25x7 + 0.5x8≤ 4C) x3 + x4 = 1D) x1, x2, x3, x4, x5, x6 , x7, x8≥0 and integer
Q:
The landlord uses the following scheme for decision variables:x1 = contractor works on wood floorsx2 = landlord works on wood floorsx3 = contractor works on kitchen tilex4 = contractor works on kitchen tilex5 = contractor works on back doorx6 = contractor works on back doorx7 = contractor works on garage doorx8 = contractor works on garage doorWhat should the objective function be?A) Min Z = 2700x1 + 400x2 + 2500x3 + 1000x4 + 600x5 + 250x6 + 350x7 + 400x8B) Min Z = 2700x1 - 400x2 + 2500x3 - 1000x4 + 600x5 - 250x6 + 350x7 - 400x8C) Max Z = 100x1 + 100x2 + 80x3 + 80x4 + 15x5 + 15x6 + 20x7 + 20x8D) Max Z = 100x1 - 100x2 + 80x3 - 80x4 + 15x5 - 15x6 + 20x7 - 20x8