EXERCISE 1

We consider the following LP model:

Subject to:

The associated dual model (D) is:

Subject to:

Subject to:

Subject to:

Subject to:

EXERCISE 2

Using the Excel program, the following PL problem has been solved:

The sensitivity report that Solver provides is:

Variable cells
			Final	reduced	Aim	Permissible	Permissible
	Cell	Name	worth	Coste	coefficient	increase	Reduce
	$C$10	X1 X1	100	0	10	10	1E+30
	$C$11	X2 X1	0	-20	20	20	1E+30
	$C$12	X3 X1	100	0	40	1E+30	20
Restrictions
			Final	reduced	Aim	Permissible	Permissible
	Cell	Name	worth	Coste	coefficient	increase	Reduce
	$H$5	<= FORMULA	400	0	500	1E+30	100
	$H$6	<= FORMULA	300	30	300	100	50
	$H$7	>= FORMULA	200	-20	200	20	50

Using the information that appears in the table provided by Solver, answer the following questions:

Question 1: The optimal value of the objective function of your dual model is:

1. -800
2. -400
3. 5000
4. We do not have enough data to determine the optimal value of the objective function of the dual problem.

Question 2: Defining ℎ1, ℎ2 y 𝑒3 and respectively as the slack (and exceedance) variables of constraints 1, 2 and 3. What are the basic variables in the optimal solution?

1. and .
2. and .
3. and .
4. None of the above.

Question 3: If the independent term of the third constraint decreases the objective function by 10 units:

1. Increases by 200 units.
2. Decreases by 200 units.
3. Decreases by 50 units.
4. It is not modified, since it is a secondary variable and therefore its value is zero.

Question 4: If the independent term of the second constraint increases the objective function by 10 units:

1. Increases by 200 units.
2. Decreases by 300 units.
3. Increases by 300 units.
4. It is not modified, since its associated variable is a secondary variable and therefore its value is zero.

Question 5: If the independent term of the first restriction increases the objective function by 5 units:

1. Increases by 2000 units.
2. Increases by 50 units.
3. It is not modified, since its associated variable is a basic variable and therefore the value of  is zero.
4. It is not modified, since its associated variable  is a secondary variable and therefore its value is zero.
   

EXERCISE 3

The NADALSA company prepares, among other products, Christmas batches. This year it has decided to manufacture four types of batches, which are made up of the following products:

Products	Lot 1	Lot 2	Lot 3	Lot 4	Availability
Digging	0	1	2	2	300 bottles
Muscatel	1	1	1	1	120 bottles
Nougats	1	2	3	3	500 nougat bars
Wafers	1	1	1	1	300 boxes of wafers
Iberian sausages	0	0	0	1	15 sausages
Benefit	8	15	19	28

It is assumed that all production is sold. The PL modelwhich allowsTo maximize profit, society is as follows:

  [Benefit]

Subject to:

  [Digging]

  [Muscatel]

  [Nougats]

 [Waffles]

 [Iberian sausages]

Where  is the number of lots of the type  produced. Below is the optimal table obtained:

			8	15	19	28	0	0	0	0	0
	0	60	-2	-1	0	0	1	-2	0	0	0
	19	105	1	1	1	0	0	1	0	0	-1
	0	140	-2	-1	0	0	0	-3	1	0	0
	0	180	0	0	0	0	0	-1	0	1	0
	28	fifteen	0	0	0	1	0	0	0	0	1
Z=2415			11	4	0	0	0	19	0	0	9

Where   is the constraint clearance = 1, … 5.

Observation: Although it is an Integer Programming (PLEP), we will assume that the variables are continuous in order to apply the Simplex algorithm.

Question 1: This problem has 4 decision variables and 5 slack variables. What variables will its dual have in its extended form if we want to solve it using the big M method?

1. 4 decision variables, 5 excess and 5 artificial variables.
2. 5 decision variables, 5 excess and 5 artificial variables.
3. 5 decision variables, 4 excess and 4 artificial variables.
4. 5 decision variables and 4 slack variables.

Question 2: We want to increase profits by 20% this Christmas. How much do we have to increase the units of Iberian sausages available to reach this goal?

1. We cannot know without solving the dual to calculate the shadow price.
2. 86.25
3. 483
4. 53.66

Question 3: Returning to the initial approach. Due to a problem with a supplier, the company has 10 fewer bottles of muscat. What effect will it have on the objective function?

1. This Christmas profits will be reduced by almost 8%.
2. It will not affect the objective function in any way since it is a secondary variable.
3. The objective function will be reduced by 19 monetary units.
4. The objective function will be reduced by 40 monetary units.

Question 4: Performing a post-optimal analysis of this Linear Program we have obtained the following Simplex table:

			8	15	19	28	0	0	0	0	0
	0	-140	-2	-1	0	0	1	-2	0	0	0
	19	105	1	1	1	0	0	1	0	0	-1
	0	-160	-2	-1	0	0	0	-3	1	0	0
	0	380	0	0	0	0	0	-1	0	1	0
	28	15	0	0	0	1	0	0	0	0	1
			11	4	0	0	0	19	0	0	9

We can deduce that:

1. It meets the feasibility criterion, but not the optimality criterion.
2. It meets the optimality criterion, but not the feasibility criterion.
3. The solution is optimal and feasible in this iteration.
4. It has multiple solutions.

Question 5: Applies an iteration (pivot) of the Simplex-Dual algorithm, indicating in detail the operations performed. ¿What is the incoming variable?

1. Enters .
2. Enters .
3. Enters .
4. None of the above.

Question 6:What is the outgoing variable?

1. Comes out .
2. Comes out .
3. Comes out .
4. None of the above.

Question 7:What is the value of the objective function in the next iteration?

1. 1775
2. 2415
3. 2775
4. 3865

Question 8:What is the value of the optimal solution?

1. It does not have an optimal solution since it is not feasible.
2. and the objective function is .
3. and the objective function is .
4. and the objective function is 3865.

Question 9:Returning to the post-optimal analysis performed in question 4, The base matrix  of the system that allows the simplex table to be translated into the formula , with 𝑏 the vector of independent terms, the vector of basic variables and  the vector of secondary variables. The matrix is:

Question 10: What are the values ​​of the independent terms b linked to resource availability that we used when performing the post-optimal analysis above (Question 4)? It is recommended to use the base matrix B to arrive at the result.