Maximization simplex method

Solving LPP by Simplex Method

Example 1: Clear Shampoo Geena is the Production Manager of Clear Shampoo

and she wants to determine the production mix that will result in maximum profit. She is able to determine the data necessary for her to make a decision. It will take 5 minutes to mix 1 case of Men’s shampoo and 10 minutes to mix 1 case of Women’s shampoo and she has a total of 100 mins available for the mixing process per day. It will take 7 minutes to bottle 1 case of Men’s shampoo and 7 minutes to bottle 1 case of Women’s shampoo and she has a total of 84 mins available for the bottling process per day. It will take 9 minutes to pack 1 case of Men’s shampoo and 5 minutes to pack 1 case of Women’s shampoo and she has a total of 90 mins available for the packing process per day. She will earn 6 dollars for every case of Men’s shampoo produced and 8 dollars for every case of Women’s shampoo produced.

How many cases of Men’s shampoo and Women’s shampoo should be produced per day to maximize profits? How much is the total profit?

Table:

Process/Product

Mins per case Sign Mins per day

Men Women

Mixing 5 10 100

Bottling 7 7 84

Packing 9 5 90

Profit per case 6 8 = Max

Solution:STEP 1: Develop the initial tableau (Table 1).

1. Set up the variables.

x = number of cases of Men’s shampoo to be produced per dayy = number of cases of Women’s shampoo to be produced per day= slack 1 or unused minutes of constraint A (mixing)= slack 2 or unused minutes of constraint B (bottling)= slack 3 or unused minutes of constraint C (packing)

2. Set up the objective function where 0 is the assigned profit per unit of a slack variable:

Maximize P = 6x + 8y + + +

3. Convert the constraints into equalities.

Constraint A (mixing)5x + 10y 1005x + 10y + + + = 100

Constraint B (bottling)7x + 7y 847x+ 7y + + += 84

Constraint C (packing)9x + 5y 909x + 5y + + += 90

4. Compute the values:

= sumproduct (basic column, variable column) 0 = 0(5) +0(7) + 0(9)0 = 0(10) + 0(7) + 0(5)0 = 0(1) + 0(0) + 0(0)0 = 0(0) + 0(1) + 0(0)0 = 0(0) + 0(0) + 0(1)0 = 0(100) + 0(84) + 0(90) = 0 dollar of total profit for this solution

5. Compute the (- ) values:

- = row 6 = 6 – 0 8 = 8 – 0 0 = 0 – 0 0 = 0 – 0 0 = 0 – 0

6. Determine the maximum positive (- ) value:

8 = +maximum (6, 8, 0, 0, 0) = pivot column is y

Table 1: Initial Tableau

Basic

6 8 0 0 0 QuantitySolution x y

0 5 10 1 0 0 100

0 7 7 0 1 0 84

0 9 5 0 0 1 90

Gross

0 0 0 0 0 0

NET - 6 8 0 0 0 TotalProfitMax +? Yes/No No Yes No No No

STEP 2: Determine the pivot row

1. Compute the quantity ratio ():= Q/pivot column10 = 100/1012 = 84/718 = 90/5

2. Determine the minimum positive quantity ratio

10 = +minimum (10, 12, 18) = pivot row is

3. Determine the pivot number: 10 = intersection of pivot column and pivot row

Table 1.1: First Pivot Row

SolutionMix

Pivot col y

Quantity

Min +?Yes/No

10 100 10 Yes

7 84 12 No

5 90 8 No

Pivot row

STEP 3: Develop the second tableau (Table 2)

1. Replace the pivot row with the pivot column in solution column:

Exit = Enter= 8 y

2. Replace the row of the initial tableau with the y row in the second tableau: y row = row of the initial tableau/pivot number 0.5 = 5/10 0 = 0/10

1 = 10/10 0 = 0/100.1 = 1/10 10 = 100/10

3. Replace the row of the initial tableau with new values in the second tableau: New row = old row – (number in old row and pivot column)(y row)

3.5 = 7 – 7(0.5) 0 = 7 – 7(1.0)-0.7 = 0 – 7(0.1) 1 = 1 – 7(0) 0 = 0 – 7(0) 14 = 84 – 7(10)

4. Replace the row of the initial tableau with new values in the second tableau: New row = old row – (number in old row and pivot column)(y row)

6.5 = 9 – 5(0.5) 0 = 5 – 5(1.0)-0.5 = 0 – 5(0.1) 0 = 0 – 5(0) 1 = 1 – 5(0) 40 = 90 – 5(10)


= sumproduct (basic column, variable column) 4 = 8(0.5) +0(3.5) + 0(6.5) 8 = 8(1) + 0(0) + 0(0)0.8 = 8(0.1) + 0(-0.7) + 0(-0.5) 0 = 8(0) + 0(1) + 0(0) 0 = 8(0) + 0(0) + 0(1)80 = 8(10) + 0(14) + 0(40) = 80 dollar of total profit for this solution


- = row 2 = 6 – 4 0 = 8 – 8 -0.8 = 0 – 0.8 0 = 0 – 0 0 = 0 – 0


2 = +maximum (2, 0, -0.8, 0, 0) = pivot column is x

Table 2: Second Tableau:

Basic

6 8 0 0 0 QuantitySolution x y

8 y 0.5 1 0.1 0 0 10

0 3.5 0 -0.7 1 0 14

0 6.5 0 -0.5 0 1 40

Gross 4 8 0.8 0 0 80

NET - 2 0 -0.8 0 0 TotalProfitMax +? Yes/No Yes No No No No

STEP 4: Determine the pivot row

1. Compute the quantity ratio ():= Q/pivot column20 = 10/0.5 4 = 14/3.56.15385 = 40/6.5

2. Determine the minimum positive quantity ratio

4 = +minimum (20, 4, 6.15385) = pivot row is

3. Determine the pivot number: 3.5 = intersection of pivot column and pivot row

Table 2.1: Second Pivot Row

SolutionMix

Pivot col x

Quantity

Min +?Yes/No

y 0.5 10 20 No

3.5 14 4 Yes

6.5 40 6.15385 No

STEP 5: Develop the third tableau (Table 4).

1. Replace the pivot row with the pivot column in solution column:

Exit = Enter= 6x

2. Replace the row of the second tableau with the x row in the third tableau: x row = row of the second tableau/pivot number 1 = 3.5/3.5 0.28571 = 1/3.5

0 = 0/3.5 0 = 0/3.5-0.2 = -0.7/3.5 4 = 14/3.5

3. Replace the y row of the second tableau with new values in the third tableau: New y row = old y row – (number in old y row and pivot column)(x row)

0 = 0.5 – 0.5(1) 1 = 1 – 0.5(0) 0.2 = 0.1 – 0.5(-0.2)

-0.14286 = 0 – 0.5(0.28571) 0 = 0 – 0.5(0) 8 = 10 – 0.5(4)

4. Replace the row of the second tableau with new values in the third tableau: New row = old row – (number in old row and pivot column)(x row)

0 = 6.5 – 6.5(1) 0 = 0 – 6.5(0) 0.8 = -0.5 – 6.5(-0.2)

-1.85714 = 0 – 6.5(0.28571) 1 = 1 – 6.5(0) 14 = 40 – 6.5(4)


= sumproduct (basic column, variable column) 6 = 8(0) + 6(1) + 0(0) 8 = 8(1) + 6(0) + 0(0) 0.4 = 8(0.2) + 6(-0.2) + 0(0.8) 0.57143 = 8(-0.14286) + 6(0.28571) + 0(-1.85714) 0 = 8(0) + 6(0) + 0(1) 88 = 8(8) + 6(4) + 0(14) = 88 dollar of total profit for this solution


- = row 0 = 6 – 6 0 = 8 – 8 -0.4 = 0 – 0.4 -0.57143 = 0 – 0.57143 0 = 0 – 0


None = +maximum (0, 0, -0.4, -0.57143, 0) = final tableau is reached

8. Determine the final solution: y = 8 (8 cases of Women’s shampoo should be produced per day)x = 4 (4 cases of Men’s shampoo should be produced per day)= 14 (14 minutes of packing capacity are unused per day) = 88 (88 dollars is the total profit per day)

Table 4: Third Tableau

Thus, Geena should produce 4 cases of Men’s shampoo and 8 cases of Women’s shampoo per day at a total profit of 88 dollars, with 14 minutes of unused packing capacity.

Basic

6 8 0 0 0 Quantity

Solution x y

8 y 0 1 0.2 -0.14286

0 8

6 x 1 0 -0.2 0.28571 0 4

0 0 0 0.8 -1.85714

1 14

Gross

6 8 0.4 0.57143 0 88

NET - 0 0 -0.4 -0.57143

0 TotalProfit

Max +? Yes/No None

Example 2: NescafeHarold is the Plant Supervisor of Nescafe and he want

to determine the production mix that will result to maximum profit. He is able to determine the data necessary for him to make a decision. It will take 9 minutes to roast 1 case of Classic Coffee and 7 minutes to roast 1 case of Decaf Coffee and he has a total of 126 minutes available for the roasting process per day. It will take 7 minutes to grind 1 case of Classic Coffee and 8 minutes to grind 1 case of Decaf Coffee and he has a total of 112 minutes available for the grinding process per day. It will take 6 minutes to pack 1 case of Classic Coffee and 11 minutes to pack 1 case of Decaf Coffee and he has a total of 132 minutes available for the packing process per day. He will earn 6 dollars for every case of Classic Coffee produced and 4 dollars for every case of Decaf Coffee produced.

How many cases of Classic Coffee and Decaf Coffee should be produced per day to maximize profits? How much is the total profit?

Table: Nescafe

Process/Product

Mins per case Sign Mins per dayClassic Decaf

Roasting 9 7 126

Grinding 7 8 112

Packing 6 11 132


Example 3: Century Tuna Iola is the Operations Director of Century Tuna and she wants to determine the production mix that will result to maximum profit. She is able to determine the data necessary for her to make a decision. It will take 5 minutes to cook 1 case of Regular Tuna and 12 minutes to cook 1 case of Spicy Tuna and she has a total of 120 minutes available for the cooking process per day. It will take 10 minutes to can 1 case of Regular Tuna and 5 minutes to can 1 case of Spicy Tuna and she has a total of 100 minutes available for the canning process per day. It will take 7 minutes to label 1 case of Regular Tuna and 6 minutes to label 1 case of Spicy Tuna and she has a total of 84 minutes available for the labeling process per day. She will earn 15 dollars for every case of Regular Tuna produced and 10 dollars for every case of Spicy Tuna produced.

How many cases of Regular Tuna and Spicy Tuna should be produced per day to maximize profits? How much is the total profit?

Table: Century Tuna

Process/Product

Mins per case Sign Mins per dayRegular Spicy

Cooking 5 12 120

Canning 10 5 100

Labeling 7 6 84


References:1) Quantitative Techniques for Business by Priscilla S. Altares, et.,al.2) Quantitative Techniques in Management

by Jonathan B. Cabero, et., al.3) Quantitative Decision Models

by Edwin J. Loma4) Quantitative Business Method

by David Anderson

Maximization simplex method

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