Linear Programming Hardest Quiz:: Trivia!

1. An objective function in a linear program can be which of the following?

A maximization function

A nonlinear maximization function

A quadratic maximization function

An uncertain quantity

A divisible additive function

An objective function in a linear program is typically a mathematical expression that represents the quantity that needs to be maximized or minimized. In this case, the correct answer is "A maximization function" because the objective function in a linear program is commonly used to maximize a certain value, such as profit or efficiency. It is important to note that while other types of objective functions may exist, such as nonlinear or quadratic maximization functions, the question specifically asks for the possible objective functions in a linear program.

Explanation

An objective function in a linear program is typically a mathematical expression that represents the quantity that needs to be maximized or minimized. In this case, the correct answer is "A maximization function" because the objective function in a linear program is commonly used to maximize a certain value, such as profit or efficiency. It is important to note that while other types of objective functions may exist, such as nonlinear or quadratic maximization functions, the question specifically asks for the possible objective functions in a linear program.

2. The number of constraints allowed in a linear program is which of the following?

Less than 5

Less than 72

Less than 512

Less than 1,024

Unlimited

In linear programming, the number of constraints refers to the limitations or conditions that need to be satisfied in order to optimize a given objective function. The answer "Unlimited" implies that there is no specific limit on the number of constraints that can be included in a linear program. This means that one can have as many constraints as needed to accurately model the problem and find the optimal solution.

Explanation

In linear programming, the number of constraints refers to the limitations or conditions that need to be satisfied in order to optimize a given objective function. The answer "Unlimited" implies that there is no specific limit on the number of constraints that can be included in a linear program. This means that one can have as many constraints as needed to accurately model the problem and find the optimal solution.

3. The number of decision variables allowed in a linear program is which of the following?

Less than 5

Less than 72

Less than 512

Less than 1,024

Unlimited

In a linear program, there is no specific limit on the number of decision variables that can be used. Therefore, the correct answer is "Unlimited".

Explanation

In a linear program, there is no specific limit on the number of decision variables that can be used. Therefore, the correct answer is "Unlimited".

4. Apply linear programming to this problem. A firm wants to determine how many units of each of two products (products X and Y) they should produce in order to make the most money. The profit from making a unit of product X is $190 and the profit from making a unit of product Y is $112. The firm has a limited number of labor hours and machine hours to apply to these products. The total labor hours per week are 3,000. Product X takes 2 hours of labor per unit and Product Y takes 6 hours of labor per unit. The total machine hours available are 750 per week. Product X takes 1 machine hour per unit and Product Y takes 5 machine hours per unit. Which of the following is one of the constraints for this linear program?

1 X + 5 Y =< 750

2 X + 6 Y => 750

2 X + 5 Y = 3,000

1 X + 3 Y =< 3,000

2 X + 6 Y =>3,000

The constraint for this linear program is 1 X + 5 Y =

Explanation

The constraint for this linear program is 1 X + 5 Y =

5. Consider the LP problem: min pu₁ + u₂ subject to 3u₁ + u₂ ≥ 3 u₁ + 2u₂ ≥ 4 u₁ + 6u₂ ≥ 6 u₁, u₂ ≥ 0 For which values of p is there no solution to this problem?

P = 0

P < 0

P = 2

P > 0

For this LP problem, the objective function is to minimize pu1 + u2, subject to three constraints. In order for there to be no solution to this problem, it means that there is no feasible region where all the constraints are satisfied.

By examining the constraints, we can see that the slope of the lines formed by the constraints are all positive, meaning that as u1 and u2 increase, the constraints are satisfied. Therefore, in order for there to be no solution, the objective function pu1 + u2 must have a negative slope.

Since p

Explanation

For this LP problem, the objective function is to minimize pu1 + u2, subject to three constraints. In order for there to be no solution to this problem, it means that there is no feasible region where all the constraints are satisfied.

By examining the constraints, we can see that the slope of the lines formed by the constraints are all positive, meaning that as u1 and u2 increase, the constraints are satisfied. Therefore, in order for there to be no solution, the objective function pu1 + u2 must have a negative slope.

Since p

6. Which of the following pairs is the solution to the LP problem: max x₁ + x₂ subject to x₁ + 2x₂ ≤ 7 3x₁+ x₂ ≤ 6 x₁ x₂ ≥ 0 ?

(x1, x2) = (2, 0)

(x1, x2) = (1, 3)

(x1, x2) = (2, 3)

(x1, x2) = (0, 7/2)

The correct answer is (x1, x2) = (1, 3). This pair satisfies all the constraints of the LP problem. Substituting the values of x1 and x2 into the constraints, we get:
1 + 2(3) = 7, which is true
3(1) + 3 = 6, which is true
1(3) ≥ 0, which is true
Therefore, (1, 3) is a valid solution to the LP problem.

Explanation

The correct answer is (x1, x2) = (1, 3). This pair satisfies all the constraints of the LP problem. Substituting the values of x1 and x2 into the constraints, we get:
1 + 2(3) = 7, which is true
3(1) + 3 = 6, which is true
1(3) ≥ 0, which is true
Therefore, (1, 3) is a valid solution to the LP problem.

7. Which of the following pairs is the solution to the LP problem: min u₁ + 2u₂ subject to 3u₁ + u₂ ≥ 7 u₁ + 4u₂ ≥ 6 u₁ u₂ ≥ 0 ?

(u1, u2) = (1, 4)

(u1, u2) = (2, 1)

(u1, u2) = (6, 0)

(u1, u2) = (0, 7)

The correct answer is (u1, u2) = (2, 1). This pair satisfies both constraints 3u1 + u2 ≥ 7 and u1 + 4u2 ≥ 6. Additionally, the pair (2, 1) also satisfies the non-negativity constraint u1 u2 ≥ 0. Therefore, (2, 1) is the solution to the LP problem.

Explanation

The correct answer is (u1, u2) = (2, 1). This pair satisfies both constraints 3u1 + u2 ≥ 7 and u1 + 4u2 ≥ 6. Additionally, the pair (2, 1) also satisfies the non-negativity constraint u1 u2 ≥ 0. Therefore, (2, 1) is the solution to the LP problem.

8. Which of the following quadruples (x₁, x₂, x₃, x₄) is the solution to the LP problem: max x₁ – 1x₂ + 3x₃ – 3x₄ subject to 2 2 x₁ + x₂ + x₃ + x₄ ≤ 1 x₁ – x₂ + 2x₃ – x₄ ≤ 2 x₁, x₂, x₃ ≥ 0 ?

(0, 0, 1, 0)

(0, 1, 0, 0)

(0, 1/2, 1/4, 1/4)

(1/4, 1/4, 1/4, 1/4)

The given LP problem is a maximization problem with constraints. The objective function is to maximize x1 - 1x2 + 3x3 - 3x4. The constraints are x1 + x2 + x3 + x4 ≤ 1 and x1 - x2 + 2x3 - x4 ≤ 2. The quadruple (0, 0, 1, 0) satisfies both constraints and yields the maximum value for the objective function. Therefore, it is the solution to the LP problem.

Explanation

The given LP problem is a maximization problem with constraints. The objective function is to maximize x1 - 1x2 + 3x3 - 3x4. The constraints are x1 + x2 + x3 + x4 ≤ 1 and x1 - x2 + 2x3 - x4 ≤ 2. The quadruple (0, 0, 1, 0) satisfies both constraints and yields the maximum value for the objective function. Therefore, it is the solution to the LP problem.

9. If x1 + x2 is less than or equal to 500y1 and y1 is 0-1, then x1 and x2 will be _______________ if y1 is 0.

Equal to 0

Less than 0

More than 0

Equal to 500

If x1 + x2 is less than or equal to 500y1 and y1 is 0-1, then x1 and x2 will be equal to 0 if y1 is 0. This is because if y1 is 0, then the inequality becomes x1 + x2 ≤ 500(0), which simplifies to x1 + x2 ≤ 0. In order for this inequality to hold true, both x1 and x2 must be equal to 0.

Explanation

If x1 + x2 is less than or equal to 500y1 and y1 is 0-1, then x1 and x2 will be equal to 0 if y1 is 0. This is because if y1 is 0, then the inequality becomes x1 + x2 ≤ 500(0), which simplifies to x1 + x2 ≤ 0. In order for this inequality to hold true, both x1 and x2 must be equal to 0.

10. Which of the following triples (x₁, x₂, x₃) is the solution to the LP problem: max 2x₁ + 3x₂ + 2x₃ subject to x₁ + 4x₂ ≤ 4 x₁ – x₂ + 3x₃ ≤ 5 x_1, x_2,x₃≥ 0

(0, 1, 2)

(4, 0, 1/3)

(4, 0, 1/2)

(1, 0, 1)

The given LP problem aims to maximize the objective function 2x1 + 3x2 + 2x3, subject to the constraints x1 + 4x2 ≤ 4 and x1 – x2 + 3x3 ≤ 5, with the additional condition that x1, x2, and x3 must be greater than or equal to 0. Among the given options, only the triple (4, 0, 1/3) satisfies all the constraints and the non-negativity condition. Therefore, it is the solution to the LP problem.

Explanation

The given LP problem aims to maximize the objective function 2x1 + 3x2 + 2x3, subject to the constraints x1 + 4x2 ≤ 4 and x1 – x2 + 3x3 ≤ 5, with the additional condition that x1, x2, and x3 must be greater than or equal to 0. Among the given options, only the triple (4, 0, 1/3) satisfies all the constraints and the non-negativity condition. Therefore, it is the solution to the LP problem.

11. Consider the two LP problems: (P) max 5x₁ – 2x₂ + x₃ subject to 3x₁ – 2x₂ + 4x₃ ≤ 44 2x₁ – 4x₂ + 5x₃ ≤ 23 -x₁ + 2x₂ + x₃ ≤ - 10 x₁, x₂ x₃ ≥ 0 (D) min 44u₁ + 23u₂ – 10u₃ subject to 3u₁ + 2u₂ – u₃ ≥ 5 -2u₁ – 4u₂ + 2u₃ ≥ -2 4u₁ + 5u₂ + u₃ ≥ 1 u₁, u₂, u₃ ≥ 0 (D) has the optimal solution (u₁^* u₂^*, u₃^*) = (2, 0 1). Then the solution to (P) is (x₁^* x₂^*, x₃^*) =

(17, 7/2, 0)

(4, 0, 4)

(7/2, 0, 17)

(0, 0, 0)

The given LP problem (D) is the dual of LP problem (P). The optimal solution for (D) is (u1*, u2*, u3*) = (2, 0, 1). According to the duality theorem, the optimal solution for (P) can be found by plugging the values of (u1*, u2*, u3*) into the objective function of (P). By substituting the values, we get the solution (x1*, x2*, x3*) = (17, 7/2, 0). This means that the optimal solution for (P) is (17, 7/2, 0).

Explanation

The given LP problem (D) is the dual of LP problem (P). The optimal solution for (D) is (u1*, u2*, u3*) = (2, 0, 1). According to the duality theorem, the optimal solution for (P) can be found by plugging the values of (u1*, u2*, u3*) into the objective function of (P). By substituting the values, we get the solution (x1*, x2*, x3*) = (17, 7/2, 0). This means that the optimal solution for (P) is (17, 7/2, 0).

12. Mark the wrong statement:

The primal and dual have equal number of variables.

The shadow price of a non-binding constraint is always equal to zero.

The statement "The primal and dual have equal number of variables" is incorrect. In linear programming, the primal problem and its dual problem have different numbers of variables. The primal problem typically has decision variables that represent quantities to be determined, while the dual problem has variables that represent the prices or shadow prices associated with the constraints in the primal problem.

Explanation

The statement "The primal and dual have equal number of variables" is incorrect. In linear programming, the primal problem and its dual problem have different numbers of variables. The primal problem typically has decision variables that represent quantities to be determined, while the dual problem has variables that represent the prices or shadow prices associated with the constraints in the primal problem.

13. Which of the following LP problems has an optimal solution?
Note: in all cases x₁ > 0, x₂ > 0.

Min -2x1 + x2 subject to x2 ≤ 2

Min -2x1 + x2 subject to x1 + x2 ≥ 52x1 + x2 ≥ 7

Max 2x1 + x2 subject to x1 – x2 ≤ 2

Max 2x1 + x2 subject to x1 – 3x2 ≥ 1x1 – 3x2 ≥ -2

The LP problem with the objective function min -2x1 + x2 subject to x1 + x2 ≥ 5 and 2x1 + x2 ≥ 7 has an optimal solution because it is a minimization problem with a feasible region defined by two constraints. The objective function is linear and the feasible region is bounded, so there exists a solution that minimizes the objective function within the feasible region.

Explanation

The LP problem with the objective function min -2x1 + x2 subject to x1 + x2 ≥ 5 and 2x1 + x2 ≥ 7 has an optimal solution because it is a minimization problem with a feasible region defined by two constraints. The objective function is linear and the feasible region is bounded, so there exists a solution that minimizes the objective function within the feasible region.

14. In linear programming context, sensitivity analysis is a technique to

Allocate resources optimally.

Minimize cost of operations.

Spell out relation between primal and dual.

Determine how optimal solution to LPP changes in response to problem inputs.

Sensitivity analysis in linear programming helps determine how the optimal solution to a linear programming problem changes when there are changes in the problem inputs. It allows for understanding the impact of variations in the problem's constraints, objective function coefficients, or resource availability on the optimal solution. By conducting sensitivity analysis, decision-makers can gain insights into the stability and robustness of the optimal solution and make informed decisions based on the potential changes in the problem inputs.

Explanation

Sensitivity analysis in linear programming helps determine how the optimal solution to a linear programming problem changes when there are changes in the problem inputs. It allows for understanding the impact of variations in the problem's constraints, objective function coefficients, or resource availability on the optimal solution. By conducting sensitivity analysis, decision-makers can gain insights into the stability and robustness of the optimal solution and make informed decisions based on the potential changes in the problem inputs.

15. Choose the wrong statement:

The dual represents an alternate formulation of LPP with decision variables being implicit values.

Sensitivity analysis is carried out having reference to the optimal tableau alone.

The given statement "In order that dual to an LPP may be written, it is necessary that it has at least as many constraints as the number of variables" is incorrect. The correct statement is that in order for the dual to be written for an LPP, it is necessary that it has at least as many variables as the number of constraints.

Explanation

The given statement "In order that dual to an LPP may be written, it is necessary that it has at least as many constraints as the number of variables" is incorrect. The correct statement is that in order for the dual to be written for an LPP, it is necessary that it has at least as many variables as the number of constraints.

16. Choose the most correct of the following statements relating to primal-dual linear programming problems:

Shadow prices of resources in the primal are optimal values of the dual variables.

The optimal values of the objective functions of primal and dual are the same.

If the primal problem has unbounded solution, the dual problem would have infeasibility.

All of the above.

All of the statements provided in the answer are correct. In primal-dual linear programming problems, the shadow prices of resources in the primal problem represent the optimal values of the dual variables. Additionally, the optimal values of the objective functions in both the primal and dual problems are the same. Lastly, if the primal problem has an unbounded solution, it implies that the dual problem would be infeasible. Therefore, all of the statements mentioned in the answer are true.

Explanation

All of the statements provided in the answer are correct. In primal-dual linear programming problems, the shadow prices of resources in the primal problem represent the optimal values of the dual variables. Additionally, the optimal values of the objective functions in both the primal and dual problems are the same. Lastly, if the primal problem has an unbounded solution, it implies that the dual problem would be infeasible. Therefore, all of the statements mentioned in the answer are true.

17. Mark the wrong statement:

The dual of the dual is primal.

If a primal variable is non-negative, the corresponding dual constraint is an equation.

The objective function coefficients in the primal problem become right-hand side of constraints of the dual.

If a primal variable is non-negative, the corresponding dual constraint is an inequality.

Explanation

If a primal variable is non-negative, the corresponding dual constraint is an inequality.

18. Apply linear programming to this problem. David and Harry operate a discount jewelry store. They want to determine the best mix of customers to serve each day. There are two types of customers for their store, retail (R) and wholesale (W). The cost to serve a retail customer is $70 and the cost to serve a wholesale customer is $89. The average profit from either kind of customer is the same. To meet headquarters' expectations, they must serve at least 8 retail customers and 12 wholesale customers daily. In addition, in order to cover their salaries, they must at least serve 30 customers each day. Which of the following is one of the constraints for this model?

1 R + 1 W =< 8

1 R + 1 W => 30

8 R + 12 W => 30

1 R => 12

20 x (R + W) =>30

The constraint "1 R + 1 W => 30" is one of the constraints for this model because it ensures that the total number of customers served each day is at least 30, which is the minimum number required to cover their salaries. This constraint combines the number of retail customers (R) and wholesale customers (W) and states that their sum must be greater than or equal to 30.

Explanation

The constraint "1 R + 1 W => 30" is one of the constraints for this model because it ensures that the total number of customers served each day is at least 30, which is the minimum number required to cover their salaries. This constraint combines the number of retail customers (R) and wholesale customers (W) and states that their sum must be greater than or equal to 30.

19. Leo has $12.50 to spend on his weekly supply of sweets, crisps and apples. A bag of crisps costs $0.65, a bag of sweets costs $0.85, and one apple costs $0.50. The total number of packets of crisps, sweets and apples consumed in a week must be at least seven, and he eats at least twice as many packets of sweets as crisps. His new healthy diet also means that the total number of packets of sweets and crisps must not exceed one-third of the number of apples. If s, c and a, denote the number of packets of sweets, packets of crisps, and apples respectively, which one of the following represents one of the constraints defining the feasible region?

S ≤ c - a

3c + 3s ≤ a

A + c + s > 7

S ≥ 0.5c

0.65s + 0.85c + 0.5a ≥ 12.5

The constraint a + c + s > 7 represents one of the conditions defining the feasible region. This constraint ensures that the total number of packets of sweets, crisps, and apples consumed in a week is at least seven.

Explanation

The constraint a + c + s > 7 represents one of the conditions defining the feasible region. This constraint ensures that the total number of packets of sweets, crisps, and apples consumed in a week is at least seven.

20. If problems (P) and (Q) are dual of each other, what are a, b, and c? (P) max 2x₁ + ax₂ subject to 2x₁ + x₂≤ 3 bx₁ + 2x₂ ≤ c x₁x₂≥ 0 ? (Q) min 3u₁ + 4u₂ subject to 2u₁ + 4u₂≥ 2 u₁+ 2u₂ ≥ 5
u₁ u₂ ≥ 0

(a, b, c) = (5, 5, 4)

(a, b, c) = (4, 4, 5)

(a, b, c) = (5, 4, 4)

(a, b, c) = (4, 5, 4)

not-available-via-ai

Explanation

not-available-via-ai

21. Which of the following is an essential condition in a situation for linear programming to be useful?

Nonlinear constraints

Bottlenecks in the objective function

Homogeneity

Uncertainty

Competing objectives

Linear programming is a mathematical optimization technique that is used to maximize or minimize a linear objective function subject to linear constraints. In order for linear programming to be useful, the constraints must be linear. Nonlinear constraints, on the other hand, involve equations or inequalities that are not linear. Therefore, nonlinear constraints are not an essential condition for linear programming to be useful.

Explanation

Linear programming is a mathematical optimization technique that is used to maximize or minimize a linear objective function subject to linear constraints. In order for linear programming to be useful, the constraints must be linear. Nonlinear constraints, on the other hand, involve equations or inequalities that are not linear. Therefore, nonlinear constraints are not an essential condition for linear programming to be useful.

22. How many of the following points satisfy the inequality

2x - 3y > -5?

(1, 1), (-1, 1), (1, -1), (-1, -1), (-2, 1), (2, -1), (-1, 2) and (-2, -1)

4

7

6

3

5

The given inequality is 2x - 3y > -5. We can substitute the x and y values of each point into the inequality to check if it is satisfied. Out of the given points, only (1, 1), (-1, 1), (1, -1), and (-1, -1) satisfy the inequality. Therefore, the number of points that satisfy the inequality is 4.

Explanation

The given inequality is 2x - 3y > -5. We can substitute the x and y values of each point into the inequality to check if it is satisfied. Out of the given points, only (1, 1), (-1, 1), (1, -1), and (-1, -1) satisfy the inequality. Therefore, the number of points that satisfy the inequality is 4.

23. The following five inequalities define a feasible region. Which one of these could be removed from the list without changing the region?

X - 2y ≥ -8

Y ≥ 0

-x + y ≤ 10

X + y ≤ 20

X ≥ 0

The inequality x - 2y ≥ -8 could be removed from the list without changing the region because it is redundant. It can be derived from the other inequalities in the list. By adding the inequalities -x + y ≤ 10 and x + y ≤ 20, we get -x + y + x + y ≤ 10 + 20, which simplifies to 2y ≤ 30. Dividing both sides of the inequality by 2 gives y ≤ 15. Since y ≥ 0 is already given in the list, the inequality x - 2y ≥ -8 is not necessary.

Explanation

The inequality x - 2y ≥ -8 could be removed from the list without changing the region because it is redundant. It can be derived from the other inequalities in the list. By adding the inequalities -x + y ≤ 10 and x + y ≤ 20, we get -x + y + x + y ≤ 10 + 20, which simplifies to 2y ≤ 30. Dividing both sides of the inequality by 2 gives y ≤ 15. Since y ≥ 0 is already given in the list, the inequality x - 2y ≥ -8 is not necessary.

24. Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem (using simplex), we find that

The values of decision variables obtained by rounding off are always very close to the optimal values.

The value of the objective function for a maximization problem will likely be less than that for the simplex solution.

The value of the objective function for a minimization problem will likely be less than that for the simplex solution.

All constraints are satisfied exactly.

None of the above.

When solving an integer programming problem by rounding off the answers obtained from solving it as a linear programming problem using simplex, the values of decision variables obtained by rounding off are always very close to the optimal values. However, the value of the objective function for a maximization problem will likely be less than that for the simplex solution. This is because rounding off the solutions can introduce some level of error, leading to a suboptimal objective function value. Therefore, the correct answer is "the value of the objective function for a maximization problem will likely be less than that for the simplex solution."

Explanation

When solving an integer programming problem by rounding off the answers obtained from solving it as a linear programming problem using simplex, the values of decision variables obtained by rounding off are always very close to the optimal values. However, the value of the objective function for a maximization problem will likely be less than that for the simplex solution. This is because rounding off the solutions can introduce some level of error, leading to a suboptimal objective function value. Therefore, the correct answer is "the value of the objective function for a maximization problem will likely be less than that for the simplex solution."

25. Choose the incorrect statement:

All scare resources have marginal profitability equal to zero.

Shadow prices are also known as imputed values of the resources.

If all constraints of a minimization problem are ‘≥’ type, then all dual variables are non-negative.

The statement is incorrect because if all constraints of a minimization problem are of the '≥' type, then all dual variables are non-positive, not non-negative. Dual variables represent the shadow prices or imputed values of the resources, and in a minimization problem, the dual variables associated with the '≥' constraints will have non-positive values.

Explanation

The statement is incorrect because if all constraints of a minimization problem are of the '≥' type, then all dual variables are non-positive, not non-negative. Dual variables represent the shadow prices or imputed values of the resources, and in a minimization problem, the dual variables associated with the '≥' constraints will have non-positive values.

26. To write the dual; it should be ensured that I. All the primal variables are non-negative. II. All the bi values are non-negative. III. All the constraints are '≤' type if it is maximization problem and '≥' type if it is a minimization problem.

I and II

II and III

I and III

I, II and III

To write the dual, it is necessary to ensure that all the primal variables are non-negative (I) and all the constraints are of the correct type, which means "less than or equal to" for maximization problems and "greater than or equal to" for minimization problems (III). These conditions are important for formulating the dual problem correctly and obtaining accurate results. Therefore, the correct answer is I and III.

Explanation

To write the dual, it is necessary to ensure that all the primal variables are non-negative (I) and all the constraints are of the correct type, which means "less than or equal to" for maximization problems and "greater than or equal to" for minimization problems (III). These conditions are important for formulating the dual problem correctly and obtaining accurate results. Therefore, the correct answer is I and III.

27. The point (x, 3) satisfies the inequality, -5x - 2y ≤ 13. Find the smallest possible value of x.

-1.4

0

1.4

-3.8

3.8

The inequality -5x - 2y ≤ 13 can be rewritten as -5x - 2(3) ≤ 13. Simplifying this equation gives -5x - 6 ≤ 13. Adding 6 to both sides gives -5x ≤ 19. Dividing both sides by -5 gives x ≥ -3.8. Therefore, the smallest possible value of x is -3.8.

Explanation

The inequality -5x - 2y ≤ 13 can be rewritten as -5x - 2(3) ≤ 13. Simplifying this equation gives -5x - 6 ≤ 13. Adding 6 to both sides gives -5x ≤ 19. Dividing both sides by -5 gives x ≥ -3.8. Therefore, the smallest possible value of x is -3.8.

28. Find, if possible, the minimum value of the objective function 3x - 4y subject to the constraints

-2x + y ≤ 12, x - y ≤ 2, x ≥ 0 and y ≥ 0.

8

0

No solution

-8

-36

The minimum value of the objective function occurs when it is minimized at its lowest possible value. In this case, the objective function is 3x - 4y. Since both x and y are non-negative according to the constraints, the lowest possible value for the objective function is when both x and y are equal to 0. Therefore, the minimum value of the objective function is 0.

Explanation

The minimum value of the objective function occurs when it is minimized at its lowest possible value. In this case, the objective function is 3x - 4y. Since both x and y are non-negative according to the constraints, the lowest possible value for the objective function is when both x and y are equal to 0. Therefore, the minimum value of the objective function is 0.

29. Leo has $12.50 to spend on his weekly supply of sweets, crisps and apples. A bag of crisps costs $0.65, a bag of sweets costs $0.85, and one apple costs $0.50. The total number of packets of crisps, sweets and apples consumed in a week must be at least seven, and he eats at least twice as many packets of sweets as crisps. His new healthy diet also means that the total number of packets of sweets and crisps must not exceed one-third of the number of apples. If s, c and a, denote the number of packets of sweets, packets of crisps, and apples respectively, which one of the following represents one of the constraints defining the feasible region? Which of the following represents one of the constraints in the question?

3s - 3c + a ≥ 0

C ≤ 2s

17s + 10a + 13c ≤ 250

3c + 3s + a ≤ 0

A + c + s ≤ 7

The given constraint, 17s + 10a + 13c ≤ 250, represents one of the conditions defining the feasible region. It ensures that the total cost of the packets of sweets, apples, and crisps does not exceed $250, which is the total amount of money Leo has to spend on his weekly supply.

Explanation

The given constraint, 17s + 10a + 13c ≤ 250, represents one of the conditions defining the feasible region. It ensures that the total cost of the packets of sweets, apples, and crisps does not exceed $250, which is the total amount of money Leo has to spend on his weekly supply.

30. Mark the wrong statement:

The optimal values of the dual variables are obtained from ∆j values from slack/surplus variables, in the optimal solution tableau.

An n-variable m-constraint primal problem has an m-variable n-constraint dual.

If a constraint in the primal problem has a negative bi value, its dual cannot be written.

The correct answer is "If a constraint in the primal problem has a negative bi value, its dual cannot be written." This statement is incorrect because even if a constraint in the primal problem has a negative bi value, its dual can still be written. The dual problem allows for negative values in its constraints and variables.

Explanation

The correct answer is "If a constraint in the primal problem has a negative bi value, its dual cannot be written." This statement is incorrect because even if a constraint in the primal problem has a negative bi value, its dual can still be written. The dual problem allows for negative values in its constraints and variables.

31. The problem description

Maximise: 7X₁ + 3X₂ Subject to: 5X₁+ 7X₂≤ 27 4X₁+ X₂≤ 14 3X₁- 2X₂≤ 9 X₁, X₂≥ 0 X₁Integer represents a(n)

Integer programming problem.

Goal programming problem.

Nonlinear programming problem.

Multiple-objective programming problem.

None of the above

The given problem is a nonlinear programming problem because the objective function and constraints involve nonlinear terms. The objective function is a linear combination of variables X1 and X2, but the constraints include nonlinear terms such as X1*X2. In a linear programming problem, the objective function and constraints would only involve linear terms, such as X1 and X2. Therefore, the correct answer is a nonlinear programming problem.

Explanation

The given problem is a nonlinear programming problem because the objective function and constraints involve nonlinear terms. The objective function is a linear combination of variables X1 and X2, but the constraints include nonlinear terms such as X1*X2. In a linear programming problem, the objective function and constraints would only involve linear terms, such as X1 and X2. Therefore, the correct answer is a nonlinear programming problem.

32. Which of the following statements about an LP problem and its dual is false?

If the primal has an optimal solution, so has the dual

The dual problem might have an optimal solution, even though the primal has no (bounded) optimum

If the primal has an optimal solution, so has the dual.

Explanation

If the primal has an optimal solution, so has the dual.

33. Mark the wrong statement:

If the primal is a minimization problem, its dual will be a maximization problem.

Columns of the constraint coefficients in the primal problem become columns of the constraint coefficients in the dual.

For an unrestricted primal variable, the associated dual constraint is an equation.

If a constraint in a maximization type of primal problem is a ‘less-than-or-equal-to’ type, the corresponding dual variable is non-negative.

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Explanation

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34. What can you say about the solution of the linear programming problem specified in question 5, if the objective function is to be maximised instead of minimized?

No solution

Unique solution at (2, 0)

Unique solution at (0, 0)

Infinitely many solutions

Unique solution at (0, 12)

If the objective function is to be maximized instead of minimized in a linear programming problem, it means that the problem is being approached from a different perspective. In this case, the problem may have infinitely many solutions because there may be multiple points on the feasible region that can achieve the maximum value of the objective function. Therefore, changing the objective function from minimization to maximization can result in infinitely many solutions.

Explanation

If the objective function is to be maximized instead of minimized in a linear programming problem, it means that the problem is being approached from a different perspective. In this case, the problem may have infinitely many solutions because there may be multiple points on the feasible region that can achieve the maximum value of the objective function. Therefore, changing the objective function from minimization to maximization can result in infinitely many solutions.

35. Apply linear programming to this problem. A firm wants to determine how many units of each of two products (products D and E) they should produce to make the most money. The profit in the manufacture of a unit of product D is $100 and the profit in the manufacture of a unit of product E is $87. The firm is limited by its total available labor hours and total available machine hours. The total labor hours per week are 4,000. Product D takes 5 hours per unit of labor and product E takes 7 hours per unit. The total machine hours are 5,000 per week. Product D takes 9 hours per unit of machine time and product E takes 3 hours per unit. Which of the following is one of the constraints for this linear program?

5 D + 7 E =< 5,000

9 D + 3 E => 4,000

5 D + 7 E = 4,000

5 D + 9 E =< 5,000

9 D + 3 E =< 5,000

The constraint 9 D + 3 E =

Explanation

The constraint 9 D + 3 E =

36. What can you say about the linear programming problem specified in question 5, if the second constraint is changed to 3x - 4y ≤ 24 and the problem is one of maximization?

Unique solution at (0, 6)

No solution

Unique solution at (0, 0)

Infinitely many solutions

Unique solution at (8, 0)

If the second constraint is changed to 3x - 4y ≤ 24 and the problem is one of maximization, it means that the new constraint is a less restrictive condition compared to the original constraint. This change allows for a wider range of feasible solutions, resulting in infinitely many solutions. Therefore, the linear programming problem specified in question 5 has infinitely many solutions.

Explanation

If the second constraint is changed to 3x - 4y ≤ 24 and the problem is one of maximization, it means that the new constraint is a less restrictive condition compared to the original constraint. This change allows for a wider range of feasible solutions, resulting in infinitely many solutions. Therefore, the linear programming problem specified in question 5 has infinitely many solutions.

37. Which of the following constraints is not linear?

7A − 6B ≤ 45

X + Y + 3Z = 35

2X + 10Y ≥ 60

2XY + X ≥ 15

None of the above

All of the given constraints are linear. A linear constraint is one where the variables are raised to the power of 1 and combined using addition, subtraction, and multiplication by a constant. In all of the given constraints, the variables are raised to the power of 1 and combined using addition, subtraction, and multiplication by a constant, satisfying the criteria for linearity. Therefore, none of the above constraints is not linear.

Explanation

All of the given constraints are linear. A linear constraint is one where the variables are raised to the power of 1 and combined using addition, subtraction, and multiplication by a constant. In all of the given constraints, the variables are raised to the power of 1 and combined using addition, subtraction, and multiplication by a constant, satisfying the criteria for linearity. Therefore, none of the above constraints is not linear.

38. Which of the following is not a feasible solution of dual of given problem: max x₁ + 2x₂ subject to x₁ + x₂ ≤ 4 -x₁ + x₂ ≤ 1 2x₁ – x₂ ≤ 3 x₁,x₂ ≥ 0

(0, 1/2, 3/2)

(2, 1, 1)

(3/2, 1/2, 0)

(2, 1, 2)

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Explanation

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39. Apply linear programming to this problem. A one-airplane airline wants to determine the best mix of passengers to serve each day. The airplane seats 25 people and flies 8 one-way segments per day. There are two types of passengers: first class (F) and coach (C). The cost to serve each first class passenger is $15 per segment and the cost to serve each coach passenger is $10 per segment. The marketing objectives of the airplane owner are to carry at least 13 first class passenger-segments and 67 coach passenger-segments each day. In addition, in order to break even, they must at least carry a minimum of 110 total passenger segments each day. Which of the following is one of the constraints for this linear program?

15 F + 10 C => 110

1 F + 1 C => 80

13 F + 67 C => 110

1 F => 13

13 F + 67 C =< (80/200)

The constraint for this linear program is "1 F => 13". This means that the minimum number of first class passengers that must be carried each day is 13.

Explanation

The constraint for this linear program is "1 F => 13". This means that the minimum number of first class passengers that must be carried each day is 13.

40. How many points with integer coordinates lie in the feasible region defined by

3x + 4y ≤ 12, x ≥ 0 and y ≥ 1?

7

5

4

8

6

The feasible region is the area bounded by the inequalities. In this case, the feasible region is a triangle with vertices at (0, 3), (0, 12), and (4, 0). To find the number of points with integer coordinates in this region, we can count the lattice points (points with integer coordinates) inside or on the boundary of the triangle. By counting, we find that there are 6 lattice points in the feasible region.

Explanation

The feasible region is the area bounded by the inequalities. In this case, the feasible region is a triangle with vertices at (0, 3), (0, 12), and (4, 0). To find the number of points with integer coordinates in this region, we can count the lattice points (points with integer coordinates) inside or on the boundary of the triangle. By counting, we find that there are 6 lattice points in the feasible region.

41. What can you say about the solution of the linear programming problem specified in question 5, if the second constraint is changed to x + y ≤ 2 and the problem is one of minimization?

Infinitely many solutions

Unique solution (0, 12)

No solution

Unique solution at (2, 0)

Unique solution at (0, 2)

If the second constraint is changed to x + y ≤ 2 and the problem is one of minimization, the solution of the linear programming problem will have a unique solution at (0, 2). This means that the optimal values for x and y, which minimize the objective function, will be x = 0 and y = 2.

Explanation

If the second constraint is changed to x + y ≤ 2 and the problem is one of minimization, the solution of the linear programming problem will have a unique solution at (0, 2). This means that the optimal values for x and y, which minimize the objective function, will be x = 0 and y = 2.