Minimization model by simplex method

Solving Minimization Model by Simplex Method

Example 1: KraftJacob is the Purchasing Manager of Kraft

Foods and he wants to determine the supply mix that will result on minimum cost. He is able to determine the data necessary for him to make a decision. A gallon of Alaska milk can produce 5 cases of cheese, 7 cases of butter, and 9 cases of cream. A gallon of Nestle milk can produce 11 cases of cheese, 8 cases of butter, and 4 cases of cream. He must produce at least 110 cases of cheese, 112 cases of butter, and 72 cases of cream per day. Alaska milk costs 50 dollars per gallon while Nestle costs 55 dollars per gallon.

How many gallons of Alaska milk and Nestle milk should be purchase per day to minimize costs? How much is the total cost?

Table: Kraft

Product/Supplier

Cases per Gal Sign Cases/Day

Alaska Nestle

Cheese 5 11 110

Butter 7 8 112

Cream 9 4 72

Cost/gal 50 55 = Min

Minimize E = 50x + 55y

Subject to:5x + 11y 110 7x + 8y 112 9x + 4y 72 x, y 0

Step 1: Develop the initial tableau.

1. Set up the variables:x = number of gallons of Alaska milk to be purchased per dayy = number of gallons of Nestle milk to be purchased per day= slack 1 or excess cases of constraint A(cheese)= slack 2 or excess cases of constraint B(butter)= slack 3 or excess cases of constraint C(cream)= artificial variable 1 or initial positive solution for constraint A= artificial variable 2 or initial positive solution for constraint B= artificial variable 3 or initial positive solution for constraint C

2. Set up the objective function where a relatively high cost is assigned per unit of an artificial variable (100 dollars for this case):

Minimize: E = 50x + 55y ++++++

Copy the coefficients to the Cj row. Assign the artificial variables as the initial solution and copy the coefficient to the Cj column.

3. Convert the constraints into equalities.

Constraint A(cheese)5x + 11y 1105x + 11y +++++Constraint B(butter)7x + 8y 1127x + 8y +++++ = 112Constraint C(cream)9x + 4y 729x + 4y +++++ = 72

Copy the coefficients to the , , and rows.

4. Compute the values.

= sumproduct(basic column, variable column)2,100 = 100(5)+100(7)+100(9)2,300 = 100(11)+100(8)+100(4)- 100 = 100(-1)+100(0)+100(0)- 100 = 100(0)+100(-1)+100(0)- 100 = 100(0)+100(0)+100(-1)100 = 100(1)+100(0)+100(0)100 = 100(0)+100(1)+100(0)100 = 100(0)+100(0)+100(1)29,400 = 100(110)+100(112)+100(72) = 29,400 dollars of total cost for this solution

5. Compute the ( - ) values.( - ) = ( row) – ( row)- 2050 = 50 - 2100- 2245 = 55 - 2300100 = 0 – (-100)100 = 0 – (-100)100 = 0 – (-100)0 = 100 – 1000 = 100 – 100 0 = 100 – 100

6. Determine the minimum negative ( - ) value.

-2245 = - minimum(-2050, -2245, 100, 100, 100, 0, 0, 0) = pivot column is y.

Table 1 Initial Tableau

Basic

50 55 0 0 0 100

100

100 Quantity

Soln x y

100 5 11 -1 0 0 1 0 0 110

100 7 8 0 -1 0 0 1 0 112

100 9 4 0 0 -1 0 0 1 72

Gross

2,100

2,300 -100

-100

-100

100 100 100 29,400

Net - -2,05

0

-2,245

100 100 100 0 0 0 TotalCost

Min-? Yes/No No Yes No No No No No No

Step 2: Determine the pivot row

1. Compute the quantity ratio ():

= Q/pivot column10 = 110/1114 = 112/818 = 72/4

2. Compute the minimum positive quantity ratio ():

10 = +minimum (10, 14, 18) = pivot row is

3. Determine the pivot number ): 11 = intersection of pivot column and pivot row

Step 3: Develop the second tableau.

1. Replace the pivot row with the pivot column in the solution mix: Exit = 100 Enter = 55y

2. Replace the row of the initial tableau with the y row in the second tableau:

y row = row of initial tableau/pivot number0.4545 = 5/111 = 11/11-0.091 = -1/110 = 0/110 = 0/110.0909 = 1/110 = 0/110 = 0/1110 = 110/11

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)3.3636 = 7 - 8(0.4545)0 = 8 – 8(1)0.7273 = 0 – 8(-0.091)-1 = -1 – 8(0)0 = 0 – 8(0)-0.727 = 0 – 8(0.0909)1 = 1 – 8(0)0 = 0 – 8(0)32 = 112 – 8(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)7.1818 = 9 – 4(0.4545)0 = 4 – 4(1)0.3636 = 0 – 4(-0.091)0 = 0 – 4(0)-1 = -1 – 4(0)-0.364 = 0 – 4(0.909)0 = 0 – 4(0)1 = 1 – 4(0)32 = 72 – 4(10)


= sumproduct(basic column, variable column)1079.5 = 55(0.4545)+100(3.3636)+100(7.1818)55 = 55(1)+100(0)+100(0)104.09 = 55(-0.091)+100(0.7273)+100(0.3636)- 100 = 55(0)+100(-1)+100(0)- 100 = 55(0)+100(0)+100(-1)-104.1 = 55(0.0909)+100(-0.727)+100(-0.364)100 = 55(0)+100(1)+100(0)100 = 55(0)+100(0)+100(1)6,950 = 55(10)+100(32)+100(32) = 6,960 dollars of total cost for this solution

6. Compute the ( - ) values.

( - ) = ( row) – ( row)- 1030 = 50 – 1079.50 = 55 - 55-104.1 = 0 – 104.09100 = 0 – (-100)100 = 0 – (-100)204.09 = 100 – (-104.1)0 = 100 – 100 0 = 100 – 100


-1030 =-minimum(-1030, 0, -104.1, 100, 100, 204.09,0, 0) = pivot column is x.

Table 2 Second Tableau

Basic

50 55 0 0 0 100 100

100

Quantity

Soln x y

55 y 0.4545

1 -0.091

0 0 0.0909

0 0 10

100 3.3636

0 0.7273

-1 0 -0.727

1 0 32

100 7.1818

0 0.3636

0 -1 -0.364

0 1 32

Gross

1079.5

55 104.09

-100

-100

-104.1

100

100

6,950

Net Basic -1030

0 -104.1

100

100

204.09

0 0 TotalCost

Min-?

Yes/No

Yes No No No No No No No



= Q/pivot column22 = 10/0.45459.5135 = 32/3.36364.4557 = 32/7.1818


4.4557 = +minimum (22, 9.5135, 4.4557)

= pivot row is

3. Determine the pivot number ): 7.1818 = intersection of pivot column and pivot row

Step 3: Develop the third tableau.

1. Replace the pivot row with the pivot column in the solution mix: Exit = 100 Enter = 50x

2. Replace the row of the second tableau with the x row in the third tableau:

x row = row of second tableau/pivot number1 = 7.1818/7.18180 = 0/7.18180.0506 = 0.3636/7.18180 = 0/7.1818-0.139 = -1/7.1818-0.051 = -0.364/7.18180 = 0/7.181801392 = 1/7.18184.4557 = 32/7.1818

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 = 0.4545 – 0.4545(1)1 = 1 – 0.4545(0)-0.114 = -0.091 – 0.4545(0.0506)0 = 0 – 0.4545(0)0.0633 = 0 – 0.4545(-0.139)0.1139 = 0.0909 – 0.4545(-0.051)0 = 0 – 0.4545(0)-0.063 = 0 – 0.4545(0.1392)7.9747 = 10 – 0.4545(4.4557)

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 = 3.3636 – 3.3636(1)0 = 0 – 3.3636(0)0.557 = 0.7273 – 3.3636(0.0506)-1 = -1 – 3.3636(0)0.4684 = 0 – 3.3636(-0.139)-0.557 = -0.727 – 3.3636(-0.051)1 = 1 – 3.3636(0)-0.468 = 0 – 3.3636(0.1392)17.013 = 32 – 3.3636(4.4557)


= sumproduct(basic column, variable column)50 = 55(0)+100(0)+50(1)55 = 55(1)+100(0)+50(0)51.962 = 55(-0.114)+100(0.557)+50(0.0506)- 100 = 55(0)+100(-1)+50(0)43.354 = 55(0.0633)+100(0.4684)+50(-0.139)-51.96 = 55(0.1139)+100(-0.557)+50(-0.051)100 = 55(0)+100(1)+50(0)-43.35 = 55(-0.063)+100(-0.468)+50(0.1392)2362.7 = 55(7.9747)+100(17.013)+50(4.4557) = 2362.7 dollars of total cost for this solution


( - ) = ( row) – ( row)0 = 50 – 500 = 55 - 55-51.96 = 0 – 51.962100 = 0 – (-100)-43.35 = 0 – 43.354151.96 = 100 – (-51.96)0 = 100 – 100 143.35 = 100 – (-43.35)


-51.96 =-minimum(0, 0, -51.96, 100, -43.35, 151.96,0, 143.35) = pivot column is .

Table 3 Third Tableau

Basic

50 55 0 0 0 100 100

100 Quantity

Soln x y

55 y 0 1 -0.114

0 0.0633

0.1139

0 -0.063

7.9747

100 0 0 0.557 -1 0.4684

-0.557

1 -0.468

17.013

50 x 1 0 0.0506

0 -0.139

-0.051

0 0.1392

4.4557

Gross

50 55 51.962

-100

43.354

-51.96

100

-43.35

2362.7

Net - 0 0 -51.96

100

-43.35

151.96

0 143.35

TotalCost

Min-? Yes/No

No No Yes No No No No No



= Q/pivot column-70 = 7.9747/-0.11430.545 = 17.013/0.55788 = 4.4557/0.0506


30.545 = +minimum (-70, 30.545, 88) = pivot row is

3. Determine the pivot number ): 0.557 = intersection of pivot column and pivot row

Step 3: Develop the fourth tableau.

1. Replace the pivot row with the pivot column in the solution mix: Exit = 100 Enter =

2. Replace the row of the third tableau with the row in the fourth tableau:

row = row of third tableau/pivot number0 = 0/0.5570 = 0/0.557 1 = 0.557/0.557-1.795 = -1/0.5570.8409 = 0.4684/0.557-1 = - 0.557/0.557 1.7955 = 1/0.557-0.841 = -0.468/0.557 30.545 = 17.013/0.557

3. Replace the y row of the third tableau with new values in the fourth tableau:

New y row = old y row – (number in old y row and pivot column)(x row)0 = 0 – (- 0.114)(0)1 = 1 – (- 0.114)(0)0 = - 0.114 – (- 0.114)(1)-0.205 = 0 – (- 0.114)(-1.795)0.1591 = 0.0633 – (- 0.114)(0.8409)0 = 0.1139 – (- 0.114)(-1)0.2045 = 0 – (- 0.114)(1.7955)-0.159 = - 0.063 – (- 0.114)(-0.841)11.455 = 7.9747 – (- 0.114)(30.545)

4. Replace the x row of the third tableau with new values in the fourth tableau:

New x row = old x row – (number in old x row and pivot column)(row)1 = 1 – 0.0506(0)0 = 0 – 0.0506(0)0 = 0.0506 – 0.0506(1)0.0909 = 0 – 0.0506(-1.795)-0.182 = -0.139 – 0.0506(0.8409)0 = - 0.051 – 0.0506(-1)-0.091 = 0 – 0.0506(1.7955)0.1818 = 0.1392 – 0.0506(-0.841)2.9091 = 4.4557 – 0.0506(30.545)


= sumproduct(basic column, variable column)50 = 55(0)+0(0)+50(1)55 = 55(1)+0(0)+50(0)0 = 55(0)+0(1)+50(0)- 6.705 = 55(-0.205)+0(-1.795)+50(0.0909)-0.341 = 55(0.1591)+0(0.8409)+50(-0.182)0 = 55(0)+0(-1)+50(0)6.7045 = 55(0.2045)+0(1.7955)+50(-0.091)0.3409 = 55(-0.159)+0(-0.841)+50(0.1818)775.45 = 55(11.455)+0(30.545)+50(2.9091) = 775.45 dollars of total cost for this solution


( - ) = ( row) – ( row)0 = 50 – 500 = 55 - 550 = 0 – 06.7045 = 0 – (-6.705)0.341 = 0 – (-0.341)100 = 100 – 093.295 = 100 – 6.704599.659 = 100 – 0.3409


NONE = - minimum(0, 0, 6.7045, 0.341, 100, 93.295, 99.659) = FINAL tableau is reached.

8. Determine the final solution:

y = 11.455 (11 gallons of Nestle milk should be purchased per day) = 30.545 (31 cases of excess cheese are produced per day)x = 2.9091 (3 gallons of Alaska milk should be purchased per day) = 775.45 (775.45 is the total cost per day)

Table 4 Fourth Tableau

Thus, Jacob should purchase 3 gallons of Alaska milk and 11 gallons of Nestle milk per day at a total cost of 775.45 with 31 cases of excess cheese made.

Basic

50 55 0 0 0 100

100 100 Quantity

Soln x y

55 y 0 1 0 -0.205

0.1591

0 0.2045

-0.159

11.455

0 0 0 1 -1.795

0.8409

-1 1.7955

-0.841

30.545

50 x 1 0 0 0.0909

-0.182

0 -0.091

0.1818

2.9091

Gross

50 55 0 -6.705

-0.341

0 6.7045

0.3409

775.45

Net - 0 0 0 6.7045

0.341 100

93.295

99.659

Total Cost

Min-? Yes/No None

Example 2: Coca-ColaKaren is the Head Buyer of Coca-Cola and she

wants to determine the supply mix that will result on minimum cost. She is able to determine the data necessary for her to make a decision. A kilogram of Equal sweetener can produce 4 liters of Coke Lite, 6 liters of Sprite Lite, and 10 liters of Coke Zero. A kilogram of Nutra sweetener can produce 12 liters of Coke Lite, 8 liters of Sprite Lite, and 5 liters of Coke Zero. Karen must produce at least 96 liters of Coke Lite, 96 liters of Sprite Lite, and 100 liters of Coke Zero per day. Equal sweetener costs 27 dollars per kilogram while Nutra sweetener costs 30 dollars per kilogram. How many kilograms of Equal sweetener and Nutra sweetener should she purchase per day to minimize costs? How much is the total cost?

Example 3:Lester is the Production Analyst of Dole Pineapple and he wants to determine the supply mix that will result to minimum cost. He is able to determine the data necessary for him to make a decision. A barrel of Absolute water can produce 9 cases of sliced pineapple, 8 cases of pineapple chunks, and 4 cases of crushed pineapple. A barrel of Wilkins water can produce 5 cases of sliced pineapple, 8 cases of pineapple chunks, and 11 cases of crushed pineapple. Lester must produce at least 90 cases of sliced pineapple, 128 cases of pineapple chunks, and 88 cases of crushed pineapple per day. Absolute water costs 20 per barrel while Wilkins water costs 22 per barrel. How many barrels of Absolute water and Wilkins water should he purchase per day to minimize costs? How much is the total cost?

Table: Coca- Cola

Product/Supplier

Cases/Barrel Sign Cases/Day

Absolute

Wilkins

Sliced 9 5 90

Chunks 8 8 128

Crushed 4 11 88

Cost/Barrel 20 22 = Min

Minimization model by simplex method

Business

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