Chapter-3 (1)

Exercise

9

Solving Linear Programming

incorrect statement.

______1. An iterative process is a procedure, which is repeated over and over following a random pattern.

______ 2. An inequality relationship may be converted into an equality by the addition of a slack variable.

______ 3. An infeasible solution is characterized as one where at least one constraint is violated.

______ 4. The Cj column in a maximization simplex table indicates the profit per unit for only those variables in the product mix.

______ 5. The significance of the quantity column in the simplex table is that it always shows the right hand side values of the constraints.

______ 6. Artificial, slack, and surplus variables are added to inequality constraints in order to convert them to equalities. Artificial variables are also added to constraints which were equalities to begin with.

incorrect statement.

______ 7. There are always more variables in a simplex problem than there are constraints.

______ 8. If an artificial variable appears in the product mix of a table, which is optimal, it is likely that somebody has defined two or more

incompatible constraints.

______ 9. Shadow prices are identified with variables, which are included in the product mix.

______ 10. For any basic variable in the product mix of a simplex table, the value of Cj – Zj in the column headed by the basic variables will always be negative in a maximization problem and positive in a minimization problem.

incorrect statement. (Answer is in red letter)

______1. An iterative process is a procedure, which is repeated over and over following a random pattern. True

______ 2. An inequality relationship may be converted into an equality by the addition of a slack variable. True

______ 3. An infeasible solution is characterized as one where at least one constraint is violated. True

______ 4. The Cj column in a maximization simplex table indicates the profit per unit for only those variables in the product mix. True

______ 5. The significance of the quantity column in the simplex table is that it always shows the right hand side values of the constraints. True

______ 6. Artificial, slack, and surplus variables are added to inequality constraints in order to convert them to equalities. Artificial variables are also added to constraints which were equalities to begin with. True

incorrect statement.

______ 7. There are always more variables in a simplex problem than there are constraints. False

______ 8. If an artificial variable appears in the product mix of a table, which is optimal, it is likely that somebody has defined two or more

incompatible constraints. False

______ 9. Shadow prices are identified with variables, which are included in the product mix. True

______ 10. For any basic variable in the product mix of a simplex table, the value of Cj – Zj in the column headed by the basic variables will always be negative in a maximization problem and positive in a minimization problem. True

______ 1. Which of the following is not true for the simplex procedures? a. Each iteration results a new solution, which is as good as or better than

the previous solution.

b. The procedure indicates when the optimum solution has been reached. c. The procedure ensures that the optimum solution is reached in the

minimum

number of iteration.

d. All of the above choices are correct.

______ 2. If for a given solution, a slack variable is equal to zero then, a. the solution is not feasible

b. the solution is optimal

c. the entire amount of resource has been used d. all of the above choices are correct

______ 3. A variable, which is not included in the product mix column for a given solution is:

a. always equal to zero b. never equal to zero c. called a basic variable

d. none of the above choices are correct

______ 4. With simplex, the initial solution contains only: a. slack variable in the product mix

b. artificial variables in the product mix

c. slack and artificial variables in the product mix d. real variables in the product mix

______ 5. The procedure for solving a minimization problem with simplex is exactly the same as solving a maximization problem but with one exception, which is:

a. the identification of the replaced row b. the identification of the optimum column c. the computation of the Cj – Zj values

just two resources (2 slack variables), then the combination that will yield the optimal profit is: a. produce all 100 products if both resources are used to full capacity

b. produce, at most 2 products c. produce only one product

d. none of the above choices are correct

______ 7. In converting > constraint to equality one will a. add an artificial variable

b. add a slack variable

c. subtract a surplus variable and add an artificial variable d. subtract a surplus variable

______ 8. In solving maximization problem using simplex method, the optimum column is obtained by getting the column with

a. the largest positive value in the Cj – Zj row b. the largest negative value in the Cj – Zj row c. zero value in the Cj – Zj row

______ 9. Which of the following in a simplex table indicates that an optimal solution for maximization problem has been formed?

a. All the Cj – Zj values are negative or zero. b. All the Cj – Zj values are positive or zero.

c. There are no more slack variables in the product mix. d. All the Cj – Zj values are zeros.

______ 10. Which of the following must equal to zero? a. Basic variables

b. Solution mix variables c. Non-basic variables

__b____ 1. Which of the following is not true for the simplex procedures? a. Each iteration results a new solution, which is as good as or better than

the previous solution.

b. The procedure indicates when the optimum solution has been reached.

c. The procedure ensures that the optimum solution is reached in the minimum number of iteration.

d. All of the above choices are correct.

__c____ 2. If for a given solution, a slack variable is equal to zero then, a. the solution is not feasible

b. the solution is optimal

c. the entire amount of resource has been used d. all of the above choices are correct

__d____ 3. A variable, which is not included in the product mix column for a given solution is:

a. always equal to zero b. never equal to zero c. called a basic variable

d. none of the above choices are correct

__c____ 4. With simplex, the initial solution contains only: a. slack variable in the product mix

b. artificial variables in the product mix

c. slack and artificial variables in the product mix d. real variables in the product mix

__d____ 5. The procedure for solving a minimization problem with simplex is exactly the same as solving a maximization problem but with one exception, which is:

a. the identification of the replaced row b. the identification of the optimum column c. the computation of the Cj – Zj values

__d___ 6. In solving a problem where 100 different products (100 real variables) are produced using just two resources (2 slack variables), then the combination that will yield the optimal profit is:

a. produce all 100 products if both resources are used to full capacity b. produce, at most 2 products

c. produce only one product

d. none of the above choices are correct

___C___ 7. In converting > constraint to equality one will a. add an artificial variable

b. add a slack variable

c. subtract a surplus variable and add an artificial variable d. subtract a surplus variable

__b____ 8. In solving maximization problem using simplex method, the optimum column is obtained by getting the column with

a. the largest positive value in the Cj – Zj row b. the largest negative value in the Cj – Zj row c. zero value in the Cj – Zj row

___a___ 9. Which of the following in a simplex table indicates that an optimal solution for maximization problem has been formed?

a. All the Cj – Zj values are negative or zero. b. All the Cj – Zj values are positive or zero.

c. There are no more slack variables in the product mix. d. All the Cj – Zj values are zeros.

___d___ 10. Which of the following must equal to zero? a. Basic variables

b. Solution mix variables c. Non-basic variables

C. Add the appropriate slack, surplus, and artificial variables.

Use the value of 10 for the objective function coefficient of

artificial variables. Do not solve.

1. Maximize: Z = 4x₁ + 5x₂ subject to: 3x₁+ 6x₂ ≤ 36 4x₁ + 2x₂ ≤ 36 x₁ + x2 ≤ 8 x1,x₂ ≥ 0 2. Maximize: Z = 6x + 3y subject to: 2x+ 4y ≤ 30 4x + 2y ≤ 30 2x + 3y ≥ 18 x₁,x₂ ≥ 0

C. Adding the appropriate slack, surplus, and artificial

variables.

C. Adding the appropriate slack, surplus, and artificial variables.

C. Add the appropriate slack, surplus, and artificial variables.

Use the value of 10 for the objective function coefficient of

artificial variables. Do not solve.

3. Minimize: C = 8x + 4y

subject to: 10x+ 25y ≥ 100 y ≤ 4 x,y ≥ 0 4. Minimize: C = 4x + 5y subject to: x+ 6y = 12 x ≤ 4 y≤ 8 y ≥ 5 x,Y ≥ 0

C. Adding the appropriate slack, surplus, and artificial variables.

3. Minimize: C = 8x + 4y

subject to: 10x+ 25y ≥ 100 y ≤ 4 x,y ≥ 0 C = 8x + 4y + 0S₁ + 0S₂ + 10 A₁ 10x+ 25y + 1S₁ + 0S₂ + 1A₁ = 100 0x + y - 0S₁ + 1S₂ + 0A₁ = 4

C. Adding the appropriate slack, surplus, and artificial variables.

C. Add the appropriate slack, surplus, and artificial variables.

Use the value of 10 for the objective function coefficient of

artificial variables. Do not solve.

5. Minimize: C = 0.05x + 0.06y subject to: 5x+ 25y ≥ 50

25x + 10y ≥ 100 10x + 10y ≥ 60 36x + 20y ≥ 180 x,y ≥ 0

6. Maximize: Z = 4A + 5B subject to: 3A+ 4B ≤ 150 A+2B ≤ 60 A ≥ 25 A,B ≥ 0

C. Adding the appropriate slack, surplus, and artificial variables.

C = 0.05x + 0.06y + 0S₁ +0S₂ + 0S₃ + 0S₄ +10A₁ + 10A₂ + 10A₃ +10A₄ 5x+ 25y - 1S₁ +0S₂ + 0S₃ + 0S₄ +1A₁ + 0A₂ + 0A₃ +0A₄ = 50

25x + 10y + 0S₁ -1S₂ + 0S₃ + 0S₄ +0A₁ + 1A₂ + 0A₃ +0A₄ = 100 10x + 10y + 0S₁ +0S₂ - 1S₃ + 0S₄ +0A₁ + 0A₂ + 1A₃ +0A₄ = 60

C. Adding the appropriate slack, surplus, and artificial variables.

6. Maximize: Z = 4A + 5B subject to: 3A+ 4B ≤ 150 A+2B ≤ 60 A ≥ 25 A,B ≥ 0 Z = 4A + 5B +

0S

+0S

+ 0S

1S

+0S

+ 0S

+ 0A =

0S

+ 1S

+ 0S

+ 0A =

0S

+0S

- 1S

+ 1A =

C. Add the appropriate slack, surplus, and artificial variables.

Use the value of 10 for the objective function coefficient of

artificial variables. Do not solve.

7. Minimize: C = 2x₁ + 5x₂ subject to: 3x₁+ 2x₂ = 30 x₁ ≤ 5 x₂ ≥ 10 x₁,x₂ ≥ 0 8. Minimize: C = 2x₁ + 5x₂ + 9x₃ subject to: 2x₁+ 2x₂ + x₃ = 20 2x₁+ 4x₂ + 5x₃ ≤ 20 x₁ + 2x₂ ≥ 6 x₂ ≥ 10 x₁,x_2, x₃ ≥ 0

C. Adding the appropriate slack, surplus, and artificial variables.

+ 0A

+ 0A

C. Adding the appropriate slack, surplus, and artificial variables.

C = 2x₁ + 5x₂ + 9x₃ + 0S₂ + 0S₃ + 0S₄ + 10A₁ + 10A₃ + 10A₄ 2x₁+ 2x₂ + x₃ + 0S₂ + 0S₃ + 0S₄ + 1A₁ + 0A₃ + 0A₄ = 20

2x₁+ 4x₂ + 5x₃ + 1S₂ + 0S₃ + 0S₄ + 0A₁ + 0A₃ + 0A₄ = 20 x₁ + 2x₂ +0x₃ + 0S₂ - 1S₃ + 0S₄ + 0A₁ + 1A₃ + 0A₄ = 6

C. Add the appropriate slack, surplus, and artificial variables.

Use the value of 10 for the objective function coefficient of

artificial variables. Do not solve.

9. Maximize: Z = 8x₁ + 6x₂ + 2x₃ subject to: 3x₁+ 2x₂ + 6 ≤ 24 4x₁+ 3x₂ + 6x₃ = 48 x₁,x_2, x₃ ≥ 0 10. Minimize: C = 4x + 5y+ 3z subject to: 4x+ 2y + 3z= 100 x≥ 20 y ≤ 6 z ≥ 8 x,y_, z ≥ 0

C. Adding the appropriate slack, surplus, and artificial variables.

C. Adding the appropriate slack, surplus, and artificial variables.

C = 4x + 5y + 3z + 0S₂ + 0S₃ + 0S₄ + 10A₁ +10A₂ +10A₄ 4x + 2y + 3z + 0S₂ + 0S₃ + 0S₄ + 1A₁ + 0A₂ + 0A₄ = 100 1x + 0y + 0z- 1S₂ + 0S₃ + 0S₄ + 0A₁ +1A₂ +0A₄ = 20 0x + 1y + 0z+ 0S₂ + 1S₃ + 0S₄ + 0A₁ +0A₂ +0A₄ = 6 0x + 0y + 1z+ 0S₂ + 0S₃ - 1S₄ + 0A₁ +0A₂ +1A₄ = 8 10. Minimize: C = 4x + 5y+ 3z subject to: 4x+ 2y + 3z= 100 x≥ 20 y ≤ 6 z ≥ 8 x,y_, z ≥ 0

C. Add the appropriate slack, surplus, and artificial variables.

Use the value of 10 for the objective function coefficient of

artificial variables. Do not solve.

11. Minimize: C = 6x₁ + 16x₂ subject to: x₁+ x₂ = 400 x₁= 150 x₂ ≥ 200 x₁,x_2, ≥ 0 12. Minimize: C = 1000x + 1500y subject to: 20x+ 20y ≥ 160 30x+ 60y ≥ 300 x ≥ 2 x,y_, ≥ 0

C. Adding the appropriate slack, surplus, and artificial variables.

C = 6x₁ + 16x₂ + 0S₃ + 10A₁ + 10A₂ + 10A₃ 1x₁+ 1x₂ + 0S₃ + 1A₁ + 0A₂ + 0A₃ = 400 1x₁+ 0x₂ + 0S₃ + 0A₁ + 1A₂ + 0A₃ = 150 0x₁+ 1x₂ - 1S₃ + 0A₁ + 0A₂ + 1A₃ = 200 11. Minimize: C = 6x₁ + 16x₂ subject to: x₁+ x₂ = 400 x₁= 150 x₂ ≥ 200 x₁,x_2, ≥ 0

C. Adding the appropriate slack, surplus, and artificial variables.

C = 1000x + 1500y+ 0S₁ + 0S₂ + 0S₃ + 10A₁ +10A₂ +10A₃ 20x + 20y - 1S₁ + 0S₂ + 0S₃ + 1A₁ +0A₂ +0A₃ = 160

30x + 60y + 0S₁ - 1S₂ + 0S₃ + 0A₁ +1A₂ +0A₃ = 300

1x + 0y + 0S₁ + 0S₂ - 1S₃ + 0A₁ +0A₂ +1A₃ = 2 12. Minimize: C = 1000x + 1500y

subject to: 20x+ 20y ≥ 160 30x+ 60y ≥ 300 x ≥ 2 x,y_, ≥ 0

C. Add the appropriate slack, surplus, and artificial variables.

Use the value of 10 for the objective function coefficient of

artificial variables. Do not solve.

13. Minimize: C = 10x₁ + 20x₂ subject to: x₁+ x₂ ≥ 10 3x₁ + x₂ ≤ 12 x₁,x_2, ≥ 0 14. Minimize: C = 120x₁ + 100x₂ subject to: 2x₁+ x₂ ≤ 18 5x₁+ 4x₂ ≥ 60 x₁,x_2, ≥ 0

C. Adding the appropriate slack, surplus, and artificial variables.

C. Adding the appropriate slack, surplus, and artificial variables.

C. Add the appropriate slack, surplus, and artificial variables.

Use the value of 10 for the objective function coefficient of

artificial variables. Do not solve.

15. Minimize: Z = 2x₁ + 1x₂ subject to: 2x₁+ 3x₂ ≥ 12 3x₁+ 4x₂ ≤ 21 6x₁+ 5x₂ ≤ 30 x₁,x_2, ≥ 0

C. Adding the appropriate slack, surplus, and artificial variables.

D. Solve the following problems using simplex method:

1. Ignacio Furniture Co. makes two types of playhouses: a standard model and a deluxe model. The playhouse are sold to the independent dealers at a profit of P200/standard and P300/ deluxe. A standard requires 30 man-hours for assembly, and 20 man-hours for painting and finishing and 10 man-hours for inspecting. A deluxe requires 75 hours for assembly, 25 man-hours for painting and finishing, and five (5) man-man-hours for inspecting. A production run generally has 15000 man-hours available for assembly, 6500 man-hours available for painting and finishing and 2500 man-hours for inspecting. Determine the maximum profit and optimal values of the decisions and slack variables.

Z = 200x₁ + 300x₂ + 0S₁ + 0S₂ + 0S₃ 30x₁+ 75x₂ + 1S₁ + 0S₂ + 0S₃ = 15,000 20x₁+ 25x₂ + 0S₁ + 1S₂ + 0S₃ = 6,500 10x₁+ 5x₂ + 0S₁ + 0S₂ + 1S₃ = 2,500 1 Maximize: Z = 200x₁ + 300x₂ subject to: Assembly 30x₁+ 75x₂ ≤ 15,000 Painting/finishing 20x₁+ 25x₂ ≤ 6,500 inspecting 10x₁+ 5x₂ ≤ 2,500 x₁,x_2, ≥ 0

D. Solve the following problems using simplex method:

Let x₁ = standard model x₂ = deluxe model

Initial Tableau

200 300 0 0 0

Cont. Soln. Qty. X₁ X₂ S₁ S₂ S₃ ETR

0 S₁ 15,000 30 75 1 0 0

0 S2 6,500 20 25 0 1 0

0 S3 2,500 10 5 0 0 0

Zj 0 0 0 0 0 0

Initial Tableau

200 300 0 0 0

Cont. Soln. Qty. X₁ X₂ S₁ S₂ S₃ ETR

0 S₁ 15,000 30 75 1 0 0 200

0 S2 6,500 20 25 0 1 0 260

0 S3 2,500 10 5 0 0 0 500

Zj 0 0 0 0 0 0

Second Tableau

200 300 0 0 0

Cont. Soln. Qty. X₁ X₂ S₁ S₂ S₃ ETR

300 X₁ 200 2/5 1 1/75 0 0

0 S2 1,500 10 0 -1/3 1 0

0 S3 1,500 8 0 -1/15 0 0

Zj 60,000 120 300 4 0 0

Second Tableau

200 300 0 0 0

Cont. Soln. Qty. X₁ X₂ S₁ S₂ S₃ ETR

300 X₁ 200 2/5 1 1/75 0 0 500

0 S2 1,500 10 0 -1/3 1 0 150

0 S3 1,500 8 0 -1/15 0 0 187.5

Zj 60,000 120 300 4 0 0

Third (Final) Tableau

200 300 0 0 0

Cont. Soln. Qty. X₁ X₂ S₁ S₂ S₃ ETR

300 X₁ 140 0 1 11/75 -1/25 0

200 X₁ 150 1 0 -1/30 1/10 0

0 S3 300 0 0 13/5 -1/80 0

Zj 72,000 200 300 112/3 8 0 Cj - Zj -72,000 0 0 -112/3 -8 0

The solution is optimal: X

= 140 Max Z = 200x

+ 300x

X

= 150 = 72,000

D. Solve the following problems using simplex method:

2. To make one unit of product x requires three (3) minutes in Dept. I and one (1) minute in Dept. II. One unit of product y requires four (4) minutes in Dept.I and two (2) minutes in Dept.II. Profit contribution is P5/unit of x and P8/unit of y. It is required that at least 25 units of x be made to maximize profit if Dept.I and II have 150 and 160 minutes available respectively. What is the maximum profit?

Z = 5x + 8y + 0S₁ + 0S₂ + 0S₃ + 10A₃ 3X+ 4Y + 1S₁ + 0S₂ + 0S₃ +0 A₃ = 150 1X+ 2Y + 0S₁ + 1S₂ + 0S₃ + 0A₃ = 160 1X+ 0Y + 0S₁ + 0S₂ - 1S₃ + 1A₃ = 25 2 Maximize: Z = 5x + 8y subject to: time in dept. 1 3x+ 4y ≤ 150 time in dept .2 1x+ 2y ≤ 160 capacity x ≥ 25 x,y ≥ 0

Initial Tableau

5 8 0 0 0 10

Cont. Soln. Qty. X Y S₁ S₂ S₃ A₃ ETR

0 S₁ 150 3 4 1 0 0 0

0 S₂ 160 1 2 0 1 0 0

10 A₃ 25 1 0 0 0 -1 1

Zj 250 10 0 0 0 -10 10

Initial Tableau

5 8 0 0 0 10

Cont. Soln. Qty. X Y S₁ S₂ S₃ A₃ ETR

0 S₁ 150 3 4 1 0 0 0 50

0 S₂ 160 1 2 0 1 0 0 160

10 A₃ 25 1 0 0 0 -1 1 25

Zj 250 10 0 0 0 -10 10

Second (Final) Tableau

5 8 0 0 0 10

Cont. Soln. Qty. X Y S₁ S₂ S₃ A₃ ETR

0 S₁ 75 0 4 1 0 -3 -3

0 S₂ 135 0 2 0 1 1 -1

5 X 25 1 0 0 0 -1 1

Zj 125 5 0 0 0 -5 5

Cj - Zj -125 0 8 0 0 5 5

The solution is optimal: X = 25 Max Z = 5X + 8Y

Y= 0 = 125

D. Solve the following problems using simplex method:

3. A manufacturer of commercial chemical has an order for a certain mixture containing of three ingredients x1, x2, x3, which costs P8, P7, and P4 per kilo, respectively. The following are specifications: a. It cannot contain more than 35 kilos of x.

b. It must contain at least 15 kilos of x2.

c. It cannot contain more than 40 kilos of X3. d. The weight of the mixture must be 120 kilos.

Find a mixture of the three ingredients, which satisfies his customer’s requirements and still yields the minimum total cost of raw materials.

Z = 8x₁ + 7x₂ + 4x₃ + 0S₁ + 0S₂ + 0S₃ + 10A₂ + 10A₄ 1x₁+ 1x₂ + 1x₃ + 1S₁ + 0S₂ + 0S₃ + 0A₂ +0A₄ = 35 0x₁+ 1x₂ + 0x₃ + 0S₁ - 1S₂ + 0S₃ +1A₂ +0A₄= 15 0x₁+ 0x₂ + 1x₃ + 0S₁ + 0S₂ + 1S₃ +0A₂ + 0A₄ = 40 1x₁+ 1x₂ + 1x₃ + 0S₁ + 0S₂ + 0S₃ +0A₂ + 1A₄ = 120 3. Minimize: Z = 8x₁ + 7x₂ + 4x₃ subject to: content of X x₁+ x₂ + x₃ ≤ 35 content of X2 x₂ ≥ 15 content of X3 x₃ ≤ 40 weight x₁+ x₂ + x₃ = 120 x₁,x_2, x₃≥ 0

Initial Tableau

8 7 4 0 0 0 10 10

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 A2 A4 ETR}

0 S₁ 35 1 1 1 1 0 0 0 0 10 A₂ 15 0 1 0 0 -1 0 1 0 0 S₃ 40 0 0 1 0 0 1 0 0 10 A₄ 120 1 1 1 0 0 0 0 1 Zj 1350 10 20 10 0 -10 0 10 10 Cj - Zj -1350 -2 -13 -6 0 10 0 0 0

Initial Tableau

8 7 4 0 0 0 10 10

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 A2 A4 ETR}

0 S₁ 35 1 1 1 1 0 0 0 0 35 10 A₂ 15 0 1 0 0 -1 0 1 0 15 0 S₃ 40 0 0 1 0 0 1 0 0 0 10 A₄ 120 1 1 1 0 0 0 0 1 120 Zj 1350 10 20 10 0 -10 0 10 10 Cj - Zj -1350 -2 -13 -6 0 10 0 0 0

Second Tableau

8 7 4 0 0 0 10 10

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 A2 A4 ETR}

0 S₁ 20 1 0 1 1 2 0 -1 0 7 X₂ 15 0 1 0 0 -1 0 1 0 0 S₃ 40 0 0 1 0 0 1 0 0 10 A₄ 105 1 0 1 0 1 0 -1 1 Zj 1155 10 7 10 0 3 0 -3 10 Cj - Zj -1155 -2 0 -6 0 -3 0 7 0

Second Tableau

8 7 4 0 0 0 10 10

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 A2 A4 ETR}

0 S₁ 20 1 0 1 1 2 0 -1 0 20 7 X₂ 15 0 1 0 0 -1 0 1 0 0 0 S₃ 40 0 0 1 0 0 1 0 0 40 10 A₄ 105 1 0 1 0 1 0 -1 1 105 Zj 1155 10 7 10 0 3 0 -3 10 Cj - Zj -1155 -2 0 -6 0 -3 0 7 0

Third (Final) Tableau

8 7 4 0 0 0 10 10

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 A2 A4 ETR}

4 X₃ 20 1 0 1 1 2 0 -1 0 7 X₂ 15 0 1 0 0 -1 0 1 0 0 S₃ 20 1 0 0 -1 -2 1 1 0 10 A₄ 85 0 0 0 -1 -1 0 0 1 Zj 1035 4 7 4 -6 -9 0 3 10 Cj - Zj -1035 4 0 0 6 9 0 7 0

The solution is optimal: X₃ = 20 X₁ = 0 Min Z = 8X₁ + 7X₂ + 4X₃ + 10A₄ X₂= 15 A₄ = 85 = 1035

D. Solve the following problems using simplex method:

4. The Jay Gee Bull Ranch has plans to sell cattle to a large meat packaging plant. The manager of the ranch believes that each cattle should receive a minimum of 240 oz. of nutritional ingredients A and a minimum of 150 oz. of nutritional ingredients B each week. There are two grains available which contain both types of nutritional ingredients. One bag of grain 1 contains 12 oz of ingredients A and 25 oz of ingredients B and costs P140/bag; one bag of grain 2 contains 20 oz of ingredients A and five (5) oz. of ingredients B and costs P120/bag. Determine the optimum mixture of grains and the associated minimum cost.

Z = 140x + 120y + 0S₁ + 0S₂ + 1000A₁ + 1000A₂ 12x + 20y -1S₁ + 0S₂ + 1A₁ + 0A₂ = 35 25x + 5y + 0S₁ - 1S₂ + 0A₁ + 1A₂ = 35 4. Minimize: Z = 140x + 120y subject to: Ingredients A 12X + 20Y ≥ 240 Ingredients B 25X + 5Y ≥ 150 x,y ≥ 0

D. Solve the following problems using simplex method:

Let X = grain 1 Y = grain 2

Initial Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR

1000 A₁ 240 12 20 -1 0 1 0

1000 A₂ 150 25 5 0 -1 0 1

Zj 390000 37000 25000 -1000 -1000 1000 1000 Cj - Zj -390000 -36860 -24880 1000 1000 0 0

Initial Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR

1000 A₁ 240 12 20 -1 0 1 0 20

1000 A₂ 150 25 5 0 -1 0 1 6

Zj 390000 37000 25000 -1000 -1000 1000 1000 Cj - Zj -390000 -36860 -24880 1000 1000 0 0

Second Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR 1000 A₁ 168 0 88/5 -1 12/25 1 -12/25

140 X 6 1 1/5 0 -1/25 0 1/25

Zj 168840 140 18300 -1000 4794.4 1000 -4794.4 Cj - Zj -168840 0 -18180 1000 -4794.4 0 5794.4

Second Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR 1000 A₁ 168 0 88/5 -1 12/25 1 -12/25 840/88

140 X 6 1 1/5 0 -1/25 0 1/25 30

Zj 168840 140 18300 -1000 4794.4 1000 -4794.4 Cj - Zj -168840 0 -18180 1000 -4794.4 0 5794.4

Third Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR 120 Y 105/11 0 1 -5/88 3/110 5/88 -3/110

140 X 45/11 1 0 1/88 -1/22 -1/88 1/22 Zj 1145.45 140 120 5.23 -3.09 5.23 -4794.4 Cj - Zj -1145.55 0 0 -5.23 3.09 994.8 6.36

Third Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR 120 Y 105/11 0 1 -5/88 3/110 5/88 -3/110 -168

140 X 45/11 1 0 1/88 -1/22 -1/88 1/22 360 Zj 1145.45 140 120 5.23 -3.09 5.23 -4794.4

Fourth Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR

120 Y 30 5 1 0 -22/110 0 22/110

0 S₂ 360 88 0 1 -4 -1 4

Zj 3,600 600 120 0 -24 0 24

Fourth Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR

120 Y 30 5 1 0 -22/110 0 22/110 0

0 S₂ 360 88 0 1 -4 -1 4 4.09

Zj 3,600 600 120 0 -24 0 24

Fifth (Final) Tableau

140 120 0 0 1000 1000

Cont. Soln. Qty. X Y S₁ S₂ A₁ A₂ ETR 120 Y 105/11 0 1 -5/88 3/110 -5/88 -3/110

140 X 45/11 1 0 1/88 -4/88 1/88 4/88 Zj 1718.18 140 120 -5.23 -3.09 5.23 3.09 Cj - Zj -1718.18 0 0 5.23 3.09 994.77 _996.91

The solution is optimal: Y = 105/11 Min Z = 140X + 120Y X= 45/11 = 1718.18

D. Solve the following problems using simplex method:

5. A manufacturer makes two products; picnic tables and benches, which must be processed through two machine centers. MC1 has up to 60 hours available. MC2 can handle up 48 hours of work. Each picnic table requires four (4) hours in MC1 and two (2) hours in MC2. Each bench takes two (2) hours in MC1 and four(4) hours in MC2. If profit is P80/picnic table and P60/bench, determine the best possible contribution of picnic tables and benches to produce and sell in order to realize the maximum profit.

Z = 80x + 60y + 0S₁ + 0S₂ 4X+ 2Y + 1S₁ + 0S₂ = 60 2X+ 4Y + 0S₁ + 1S₂ = 48 5. Maximize: Z = 80x + 60y subject to: MC 1 4x+ 2y ≤ 60 MC 2 2x+ 4y ≤ 48

D. Solve the following problems using simplex method:

Let X = picnic table Y = bench

Initial Tableau

80 60 0 0

Cont. Soln. Qty. X Y S₁ S₂ ETR

0 S₁ 60 4 2 1 0

0 S₂ 48 2 4 0 1

Zj 0 0 0 0 0

Initial Tableau

80 60 0 0

Cont. Soln. Qty. X Y S₁ S₂ ETR

0 S₁ 60 4 2 1 0 15

0 S₂ 48 2 4 0 1 24

Zj 0 0 0 0 0

Second Tableau

80 60 0 0

Cont. Soln. Qty. X Y S₁ S₂ ETR

80 X 15 1 1/2 1/4 0

0 S₂ 18 0 3 -1/2 1

Zj 1200 80 40 20 0

Second Tableau

80 60 0 0

Cont. Soln. Qty. X Y S₁ S₂ ETR

80 X 15 1 1/2 1/4 0 7.5

0 S₂ 18 0 3 -1/2 1 6

Zj 1200 80 40 20 0

Third Tableau

80 60 0 0

Cont. Soln. Qty. X Y S₁ S₂ ETR

80 X 4 1 0 1/3 -1/6

60 Y 8 0 1 -1/6 1/3

Zj 800 80 60 16.67 6.67 Cj - Zj -800 0 0 -16.67 -6.67

The solution is optimal: X = 4 Max Z = 80X + 60Y y = 8 = 800

D. Solve the following problems using simplex method:

6. Mr. Winston Pe can produce in his shop any mix of three products x,y, and z. The prices and variable cost per unit are:

Product Price/unit Variable cost/unit X P20 P11

Y 12 8 Z 8 6

Winston processes three product in each of the three departments I, II, III. The time requirements per department are as follows:

80 hours per day is available at each of the department. Develop the set of constraints and the objective function to maximize profit. How much is the maximum profit?

Z = 9x + 4y + 2z + 0S₁ + 0S₂ + 0S₃ 2X+ 4Y + 3z + 1S₁ + 0S₂ + 0S₃ = 80 X+ 6Y + 5z + 0S₁ + 1S₂ + 0S₃ = 80 7X+ 2Y + 9z + 0S₁ + 0S₂ + 1S₃ = 80 6. Maximize: Z = 9x + 4y + 2z subject to: Dept. 1 2x+ 4y + 3y ≤ 80 Dept. 2 x+ 6y + 5y ≤ 80 Dept. 3 7x + 2y + 9z ≤ 80

Initial Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 80 2 4 3 1 0 0

0 S₂ 80 1 6 5 0 1 0

0 S₂ 80 7 2 9 0 0 1

Zj 0 0 0 0 0 0 0

Initial Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 80 2 4 3 1 0 0 40

0 S₂ 80 1 6 5 0 1 0 80

0 S₂ 80 7 2 9 0 0 1 11.43

Zj 0 0 0 0 0 0 0

Second Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 400/7 0 24/7 3/7 1 0 -2/7 0 S₂ 480/7 0 40/7 26/7 0 1 -1/7

9 X 80/7 1 2/7 9/7 0 0 1/7

Zj 102.86 9 2.57 11.57 0 0 1.29 Cj - Zj -102.86 0 1.43 -9.57 0 0 -1.29

Second Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 400/7 0 24/7 3/7 1 0 -2/7 133.3 0 S₂ 480/7 0 40/7 26/7 0 1 -1/7 18.46

9 X 80/7 1 2/7 9/7 0 0 1/7 8.89

Zj 102.86 9 2.57 11.57 0 0 1.29 Cj - Zj -102.86 0 1.43 -9.57 0 0 -1.29

Third Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 160/3 -1/3 10/3 0 1 0 -1/3 0 S₂ 4940/63 -26/9 308/63 _{0 0 1} -35/63

2 Z 80/9 7/9 2/9 1 0 0 1/9

Zj 17.78 1.56 0.44 2 0 0 0.22 Cj - Zj -17,78 7.44 3.56 0 0 0 -0.22

Third Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 160/3 -1/3 10/3 0 1 0 -1/3 -160 0 S₂ 4940/63 -26/9 308/63 _{0 0 1} -35/63 _-27.25

2 Z 80/9 7/9 2/9 1 0 0 1/9 11.43 Zj 17.78 1.56 0.44 2 0 0 0.22

Fourth Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 1200/21 0 72/21 3/7 1 0 -6/21

0 S₂ 40 0 360/63 _{26/7 0 1} -9/63

9 X 80/7 1 2/7 9/7 0 0 1/7

Zj 102.86 9 2.57 11.57 0 0 1.29 Cj - Zj -102.86 0 1.43 -9.57 0 0 -1.29

Fourth Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 1200/21 0 72/21 3/7 1 0 -6/21 21.05

0 S₂ 40 0 360/63 _{26/7 0 1} -9/63 ₇

9 X 80/7 1 2/7 9/7 0 0 1/7 31.5

Zj 102.86 9 2.57 11.57 0 0 1.29 Cj - Zj -102.86 0 1.43 -9.57 0 0 -1.29

Fifth (Final) Tableau

9 4 2 0 0 0

Cont .

Soln. Qty. X Y Z S₁ S₂ S₃ ETR

0 S₁ 232/7 0 0 -9/5 1 -3/5 -13/35

4 Y 7 0 1 _{13/20 0 7/40} -1/40

9 X 66/7 1 0 11/7 0 -1/20 19/140

Zj 112.85 9 4 16.74 0 0.25 1.12 Cj - Zj -112.86 0 0 -14,74 0 -0.25 -1.12 The solution is optimal: Y= 7 Max Z = 9X + 4Y + 2Z = 112.86 X = 66/7

D. Solve the following problems using simplex method:

7. The poultry farmer must supplement the vitamins in the feeds he buys. He is considering two supplements, each of which contains the feed required but in different amounts. He must meet or exceed the minimum vitamin requirements.

The vitamin content per gram of the supplements is given in the following table:

Vitamin Supplement 1 supplement 2 1 2 1

2 2 9 3 2 3

Supplement 1 costs P5 per gram and supplement 2 costs P4 per gram. The feed must contain at least 12 units of vitamin 1, 36 units of vitamin 2, and 24 units of vitamin 3. Determine the combination that has the minimum cost.

Z = 5x + 4y + 0S₁ + 0S₂ + 0S₃ +10A₁ + 10A₂ + 10A₃ 2X + 1Y - 1S₁ + 0S₂ + 0S₃ +1A₁ + 0A₂ + 0A₃ = 12 2X + 9Y + 0S₁ - 1S₂ + 0S₃ +0A₁ + 1A₂ + 0A₃ = 36 2X + 3Y + 0S₁ + 0S₂ - 1S₃ +0A₁ + 0A₂ + 1A₃ = 24 7. Minimize: Z = 5x + 4y subject to: vita 1 2x+ 1y ≥ 12 vita 2 2x+ 9y ≥ 36 vita 3 2x + 3y ≥ 24

D. Solve the following problems using simplex method:

Let X = supplement 1 Y = supplement 2

Initial Tableau

5 4 0 0 0 10 10 10

Con t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

10 A₁ 12 2 1 -1 0 0 1 0 0 12 10 A₂ 36 2 9 0 -1 0 0 1 0 4 10 A₃ 24 2 3 0 0 -1 0 0 1 8 Zj 720 60 130 -10 -10 -10 10 10 10 Cj - Zj -0 -55 -126 10 10 10 0 0 0

Second Tableau 5 4 0 0 0 10 10 10 Cont. Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR 10 A₁ 8 16/9 _{0 -1 1/9 0 1 -1/9 0} 4 Y 4 2/9 1 0 -1/9 0 0 1/9 0 10 A₃ 12 4/3 0 0 1/3 -1 0 -1/3 1 Zj 216 32 4 -10 4 -10 10 -4 10 Cj - Zj -216 -27 0 10 -4 10 0 14 0

Second Tableau 5 4 0 0 0 10 10 10 Co nt. Sol n. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR 10 A₁ 8 16/9 _{0 -1 1/9 0 1 -1/9 0 4.5} 4 Y 4 2/9 1 0 -1/9 0 0 1/9 0 18 10 A₃ 12 4/3 0 0 1/3 -1 0 -1/3 1 9 Zj 216 32 4 -10 4 -10 10 -4 10 Cj - Zj -216 -27 0 10 -4 10 0 14 0

Third Tableau

5 4 0 0 0 10 10 10

Con t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

5 X 9/2 1 _{0 -9/16} 1/16 ₀ 9/16 _{-1/16 0} 4 Y 3 0 1 1/8 -1/8 0 -1/8 1/8 0 10 A₃ 6 0 0 3/4 1/4 -1 -3/4 -1/4 ₁ Zj 94.5 5 4 0.188 2.31 -10 -5.19 -2.31 10 Cj - Zj -94.5 0 0 -0.188 _{-2.31 10 15.19 12.31 0}

Third Tableau

5 4 0 0 0 10 10 10

Con t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

5 X 9/2 1 _{0 -9/16} 1/16 ₀ 9/16 _{-1/16 0 72} 4 Y 3 0 1 1/8 -1/8 0 -1/8 1/8 0 -24 10 A₃ 6 0 0 3/4 1/4 -1 -3/4 -1/4 _{1 24} Zj 94.5 5 4 0.188 2.31 -10 -5.19 -2.31 10 Cj - Zj -94.5 0 0 -0.188 _{-2.31 10 15.19 12.31 0}

Fourth Final Tableau 5 4 0 0 0 10 10 10 Cont. Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR 5 X 3 1 _{0 -3/4} 0 _1/4 3/4 _{0 -1/4} 4 Y 6 0 1 1/2 0 -1/2 -1/2 0 1/2 0 S₂ 24 0 0 3 1 -4 -3 -1 ₄ Zj 39 5 4 -1.75 0 -0.75 1.75 0 0.75 Cj - Zj -39 0 0 1.75 _{0 0.75 8.25 0 9.25}

The solution is optimal: Y= 6 Min Z = 5X + 4Y = 39 X = 3

D. Solve the following problems using simplex method:

8. The Ajax Manufacturing Company makes three products, x₁, x₂, and x₃. The profit per unit for each is as follows: x1, P2; x2, P4; and x3, P3. The three products pass through three manufacturing centers as part of the manufacturing process. Product x₁ requires 3 hours in center 1, two (2) hours in center 2, and one (1) hour in center 3; and product x₂ requires 4 hours in center 1, 1 hour in center 2, and 3 hours in center 3; and product x₃ requires 2 hours in each of the 3 centers. Each center has time available as follows; center 1, 60 hours; center 2, 40 hours; and center 3, 80 hours. Determine the optimum product mix for next week production schedule.

Z = 2x₁ + 4x₂ + 3x₃ + 0S₁ + 0S₂ + 0S₃ 3x₁+ 4x₂ + 2x₃ + 1S₁ + 0S₂ + 0S₃ = 60 2x₁+ 1x₂ + 2x₃ + 0S₁ + 1S₂ + 0S₃ = 40 1x₁+ 3x₂ + 2x₃ + 0S₁ + 0S₂ + 1S₃ = 80 8. Maximize: Z = 2x₁ + 4x₂ + 3x₃ subject to: Center 1 3x₁+ 4x₂ + 2x₃ ≤ 60 Center 2 2x₁+ 1x₂ + 2x₃ ≤ 40 Center 3 1x₁+ 3x₂ + 2x₃ ≤ 80 x₁,x_2, x₃≥ 0

Initial Tableau

2 4 3 0 0 0

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 ETR}

0 S₁ 60 3 4 2 1 0 0

0 S₂ 40 2 1 2 0 1 0

0 S₃ 80 1 3 2 0 0 1

Zj 0 0 0 0 0 0 0

Initial Tableau

2 4 3 0 0 0

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 ETR}

0 S₁ 60 3 4 2 1 0 0 15

0 S₂ 40 2 1 2 0 1 0 40

0 S₃ 80 1 3 2 0 0 1 26.67

Zj 0 0 0 0 0 0 0

Second Tableau

2 4 3 0 0 0

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 ETR} 4 X₂ 15 3/4 1 2/4 1/4 0 0

0 S₂ 25 1/4 0 2/4 -1/4 1 0 0 S₃ 35 -5/4 0 2/4 -3/4 0 1

Zj 60 3 4 2 1 0 0

Second Tableau

2 4 3 0 0 0

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 ETR} 4 X₂ 15 3/4 1 2/4 1/4 0 0 30 0 S₂ 25 1/4 0 2/4 -1/4 1 0 50 0 S₃ 35 -5/4 0 2/4 -3/4 0 1 70

Zj 60 3 4 2 1 0 0

Third (Final) Tableau

2 4 3 0 0 0

Cont. _{Soln. Qty. X1 X2 X3 S1 S2 S3 ETR} 3 X₃ 30 3/2 2 1 1/2 0 0

0 S₂ 10 -1/2 -1 0 -1/2 1 0

0 S₃ 20 -2 -1 0 -1 0 1

Zj 90 4.5 6 3 1.5 0 0 Cj - Zj -90 -2.5 -2 0 -1.5 0 0

The solution is optimal: X₃= 30 Max Z = 2X₁ + 4X₂ + 3X₃ = 90 S₂ = 10

D. Solve the following problems using simplex method:

9. The dean of the university, College of B.A, plans the course offerings for the second semester. Student demands make it necessary to offer at least 30 undergraduate and 20 graduate courses in the term. Faculty contracts also dictate that at least 60 courses be offered in total. Each undergraduate course taught costs the college an average of P2,500,000 in faculty wages, and each graduate course costs P3,000,000. How many undergraduate and graduate courses should be taught in the second semester so that total faculty salaries are kept to a minimum?

Z = 2.5Mx + 3My + 0S₁ + 0S₂ + 0S₃ +10MA₁ + 10MA₂ + 10MA₃ 1X + 0Y - 1S₁ + 0S₂ + 0S₃ +1A₁ + 0A₂ + 0A₃ = 30 0X + 1Y + 0S₁ - 1S₂ + 0S₃ +0A₁ + 1A₂ + 0A₃ = 20 1X + 1Y + 0S₁ + 0S₂ - 1S₃ +0A₁ + 0A₂ + 1A₃ = 60 9. Minimize: Z = 2.5Mx + 3My subject to: Course offerings x ≥ 30 y ≥ 20 x + y ≥ 60

D. Solve the following problems using simplex method:

Let X = undergraduate Y = graduate

Initial Tableau

2.5M 3M 0 0 0 10M 10M 10M Con

t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

10M _{A1 30 1 0 -1 0 0 1 0 0} 10M _{A2 20 0 1 0 -1 0 0 1 0} 10M _{A3 60 1 1 0 0 -1 0 0 1} Zj 1100M 20M 20M -10M -10M -10M 10M 10M 10M Cj - Zj -1100M -17.5M _{-17M 10M 10M 10M 0 0 0}

Initial Tableau

2.5M 3M 0 0 0 10M 10M 10M Con

t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

10M _{A1 30 1 0 -1 0 0 1 0 0 30} 10M _{A2 20 0 1 0 -1 0 0 1 0 0} 10M _{A3 60 1 1 0 0 -1 0 0 1 60} Zj 1100M 20M 20M -10M -10M -10M 10M 10M 10M Cj - Zj -1100M -17.5M _{-17M 10M 10M 10M 0 0 0}

Second Tableau

2.5M 3M 0 0 0 10M 10M 10M Con

t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

2.5M _{X 30 1 0 -1 0 0 1 0 0} 10M _{A2 20 0 1 0 -1 0 0 1 0} 10M _{A3 30 0 1 1 0 -1 -1 0 1} Zj 575M 2.5M 20M 7.5M -10M -10M -7.5M _{10M 10M} Cj - Zj -575M 0 _{-17M -7.5M 10M 10M 2.5M 0 0}

Second Tableau

2.5M 3M 0 0 0 10M 10M 10M Con

t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

2.5M _{X 30 1 0 -1 0 0 1 0 0 0} 10M _{A2 20 0 1 0 -1 0 0 1 0 20} 10M _{A3 30 0 1 1 0 -1 -1 0 1 30} Zj 575M 2.5M 20M 7.5M -10M -10M -7.5M _{10M 10M} Cj - Zj -575M 0 _{-17M -7.5M 10M 10M 2.5M 0 0}

Third Tableau

2.5M 3M 0 0 0 10M 10M 10M Con

t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

2.5M _{X 30 1 0 -1 0 0 1 0 0} 3M _{Y 20 0 1 0 -1 0 0 1 0} 10M _{A3 10 0 0 1 1 -1 -1 -1 1} Zj 235M 2.5M 3M 7.5M 7M -10M -7.5M _{-7M 10M} Cj - Zj -235M 0 _{0M -7.5M -7M 10M 17.5M 17M 0}

Third Tableau

2.5M 3M 0 0 0 10M 10M 10M Con

t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

2.5M _{X 30 1 0 -1 0 0 1 0 0 -2.5} 3M _{Y 20 0 1 0 -1 0 0 1 0 0} 10M _{A3 10 0 0 1 1 -1 -1 -1 1 10} Zj 235M 2.5M 3M 7.5M 7M -10M -7.5M _{-7M 10M} Cj - Zj -235M 0 _{0M -7.5M -7M 10M 17.5M 17M 0}

Fourth Tableau

2.5M 3M 0 0 0 10M 10M 10M Con

t.

Soln. Qty. X Y S₁ S₂ S₃ A₁ A₂ A₃ ETR

2.5M _{X 40 1 0 0 1 -1 0 -1 1} 3M _{Y 20 0 1 0 -1 0 0 1 0} 0 _{S1 10 0 0 1 1 -1 -1 -1 1} Zj 160M 2.5M 3M 0 -0.5M -2.5M 0 _{0.5M 2.5M} Cj - Zj -160M 0 _{0 0 0.5M 2.5M 10M 9.5M 7.5M}

The solution is optimal: X= 40 Min Z = 2.5MX + 3MY = 160M Y = 20 Where: M= million pesos