LINEAR PROGRAMMING: MINIMIZATION

MODEL

The minimization model starts with an objective function with the purpose of minimizing a goal, which can be in the form of expenses or cost, travel time, distance, energy, or any variable where in less is desired.

Cost is used as the objective function in this chapter:

Minimize E = x + y

where:E = Total cost = cost per unit of x = cost per unit of yx = no. of units of resource 1 to be usedy = no. of units of resource 2 to be used

The constraints are then written in the form of inequalities.x + y Ax + y Bx + y Cx,y 0 (non negativity constraint)

Where = no. of units of product 1 made per unit of x= no. of units of product 1 made per unit of y

A = minimum units of product 1 to be made= no. of units of product 2 made per unit x= no. of units of product 2 made per unit y

B = Minimum units of product 2 to be made= no. of units of products 3 made per unit of x= no. of units of products 3 made per unit or

yC = Minimum units of products 3 to be made

The non negativity constraint will be assumed in the entire chapter.

Three methods in solving a minimization problem.

Section 8.1 Graphical methodSection 8.2 Simplex methodSection 8.3 Solver method

Section 8.1 Graphical Methoduses the corner- points technique drawn on a chart. This is, however, possible if there are only two variables due to the complexity of drawing a graph with three or more variables. In this method, one variable can represent the x-axis and the other the y-axis.

example 8.1 kraft (1)

jacob 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. he must produce atleast 110 cases of cheese, 112 cases of butter , and 72 cases of cream per day . alaska milk costs $ 50 per gallon while nestle milk costs $55 per gallon. the figures are summarized in the table 7.1.

how many gallons of alaska milk and nestle milk should he purchase per day to minimize costs ? how much is the total cost ? jacob wants to solve the problem using the graphical method.

TABLE 8.1 KRAFT

Product/Supplier

Cheese

Butter

Cream

Cost/gal($)

Cases per gal

Alaska Nestle

5 11

7 8

9 4

50 55

Sign Cases/Day

110

112

72

= Min

Step 1 Determine The Intercepts Of Each Equation 1. Compute The Intercepts Of The Cheese Line (Constraint A).

X - Intercept5x + 11y 110Inequality Of Cheese Product5x + 11y = 110 Convert To Equality5x + 11(0) = 110 Let Y= 05x = 110X = 110/5X = 22

Thus, The X – Intercept Is Point (22,0)

Y –Intercept5x + 11y = 110 Convert To Equality5(0) + 11y = 110 Let X = 011y = 11011y = 11o/11Y = 110/11Y = 10

Thus, The Y – Intercept Is Point (O,10)

2. Compute The Intercepts Of The Butter Line (Constraint B)

X – Intercept:7x + 8y 112 Inequality Of Butter Product7x + 8y = 112 Convert To Equality7x + 8(0)= 112 Let Y = 07x = 112X = 112/7X = 16

Thus, The X – Intercept Is Point (16,0)

Y – Intercept:7x + 8y = 112 Convert To Equality7(0) + 8y= 112 Let X = 08y = 112Y = 112/8Y = 14

Thus, The Y – Intercept Is Point (0, 14)

3. Compute The Intercept Of The Cream Line (Constraint C).

X – Intercept9x + 4y 72 Inequality Of Cream Product9x + 4y = 72 Convert To Equality9x + 4(0)= 72 Let Y = 09x = 72X = 72/9X = 8

Thus, The X – Intercept Is Point (8,0)

Y – Intercept:9x + 4y = 72 Convert To Equality9(0) + 4y= 72 Let X = 04y = 72Y = 72/4Y = 18

Thus, The Y- Intercept Is Point (0, 18)

Step 2. Draw The Graph Of The Feasible Or Common Region (Figure 8.1 )

1. Plot The X-intercept And Y-intercept Of Each Equation And Connect The Two To Form A Line .

2. Draw An Arrow From The Line Pointing Away From The Origin ( 0,0 ) To Determine The Region Covered By The Line And The ≥ Sign .

3. Determine The Corner Points Of The Lines Bordering The Feasible Region Or The Area Common To All The Arrows.Point A: Intersection Of The X-axis And The Cheese Line Point B: Intersection Of The Y-axis And The Cream Line Point C: Intersection Of The Cheese Line And The Butter LinePoint D: Intersection Of The Butter Line And The Cream Line

Step 3. Determine The Total Cost Of Each Corner Point In (Table 8.3)

1. Compute The Total Cost (E) At Point A (22,0)E = 50x + 55y1100 = 50(22) + 55(0)

= $ 1100 Of Total Cost At Corner Point A

2. Compute The Total Cost (E) At Point B (0, 18):E = 50x + 55y990 = 50(0) + 55(18)

= $ 990 Of Total Cost At Corner Point B

3. COMPUTE THE TOTAL COST (E) AT POINT C:

Point C Is The Intersection Of Cheese And Butter Lines:

5x + 11y = 110 Equation A (Cheese)7x + 8y = 112 Equation B (Butter)7(5x+11y=110)-5(7x+8y=112)35x+77y=770-35x-40y=-56037y=210y=5.6756

5x + 11y = 110 Equation A5X + 11(5.6756)= 110 Substitute For The Value Of Y5X= 110 -62.43325X=47.5676X= 9.5135 Point C Is (9.5135,5.657)

E = 50X + 55Y787.84 = 50(9.5135) + 55(5.6757)

= $787.84 OF TOTAL COST AT CORNER POINT C

4. Compute The Total Cost (E) At Point D:

Point D Is The Intersection Of Butter And Cream Lines:

7x + 8y = 112 Equation B(butter)9x + 4y = 72 Equation C (Cream)9(7x+8y=112)-7(9x+4y=72)63x+72y=1008-63x-28y=-50444y=504y=11.4545

7x + 8y = 112 Equation A7X + 8(11.4545)= 112 Substitute For The Value Of y7x+91.6363=1127x=112-91.63637x=20.3636x=2.9091

E = 50X + 55Y775.45 = 50(2.9091) + 55(11.4542)

= $775.45 OF TOTAL COST AT CORNER POINT C

5. Compute The Minimum Total Cost:

755.45 = Minimum (1,100,990, 787.84, 775.45)= $ 775.45 Of Minimum Cost Per Day At Point D

Table 8.3 Total Cost

Cornerpoins

alaska nestle Total cost Min?y/n

A 22 0 1,100

B 0 18 990

c 9.5135 5.6757 787.84

d 2.9091 11.455 775.45

No

no

no

yes

Thus, Jacob Should Buy 2,9091 Gallons Of Alaska Milk And 11,455 Gallon Of Nestle Milk A Day At A Total Cost Of $775.45.

Section 8.2 Simplex Method

Another Method To Solve Linear Programming Problems Is The Simplex Method. It May Be Necessary To Use This Method If There Are More Than Two Variables Involved .

Example 8.2 Kraft (Ii)Solve Example 8.1 Using The Simplex Method. The Figures Are Summarized In Table 8.4

Table 8.4 Kraft

Product/supplier Cases per gal

alaska nestle

sign Cases/day

Cheese

Butter

Cream

Cost/gal

5 11

7 8

9 4

50 55

110

112

72

= min

Step 1. Develop The Initial Tableau ( Table 8.5)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. Setup The Objective Function Where A Relatively High Cost Is Assigned Per Unit Of An Artificial Variable ( $100 For This Case ):

Minimize E = 50x + 55y + 0 + 0 + 0 + 100 + 100 + 100

Copy The Coefficients To The Cj Row . Assign The Artificial Variables As The Initial Solution And Copy The Coefficients To The Basic Cj Column.

3. Convert The Constraints Into Equalities

Constraint A (Cheese):5x + 11y 1105x + 11y – 1 + 0 + 0 + 1 + 0 + 0Convert To Equality Copy The Coefficients To The ROW.

Constraint B (Butter):7x + 8y 1127x + 8y + 0 - 1 + 0 + 0 +1 + 0 = 112Convert To Equality Copy The Coefficients To The Row.

Constrain C ( CREAM )

9X + 4Y9X + 4Y + O - 1 + O + O + O + 1 = 72 Convert To Equality

Copy The Coefficients To The Row.

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)

5. Compute The - ) Values.

- ) = Row) - ( Row)-2050 = 50 – 2100-2245 = 55 – 2300100 = 0 – (–100) 100 = 0 – (–100) 100 = 0 – (–100) 0 = 100 – 1000 = 100 – 1000 = 100 – 100

6. Determine The Minimum Negative - ) Values

–2245= –Minimum (–2050, –2245, 100, 100, 100, 0, 0, 0,

= Pivot Column Is Y

Table 8.5 Initial TableauBasic

100

100

100

Gross

Net

50 55 0 0 0 100 100 100

x y

5 11 -1 0 0 1 0 0

7 8 0 -1 -0 0 1 0

9 4 0 0 -1 0 0 1

2,100 2,300 -100 -100 -100 100 100 100

-2,050 -2,245 100 100 100 0 0 0

no yes no no no no no No

Soln

-

quantitiy

110

112

72

29,400

Total cost

Min-? Yes/no

Step 2. Determine The Pivot Row ( Table 8.6 ).

1.COMPUTE THE QUANTITY RATIO (): =Q/Pivot Column

10 =110/1114 = 112/818 = 72/4

2. Compute The Minimum Positive Quantity Ratio ( ) :

10 = +Minimum ( 10, 14, 18 )= Pivot Row Is

Determine The Pivot Number ( ):11 = Intersection Of Pivot Column And

Pivot Row

Table 8.6 First Pivot Row

SOLN MIX PivCol / Y QUANTITY Quant/PivCol Min + ? Y/N

11 110 10 Yes

8 112 14 No

4 72 18 No

Step 3. Develop The Second Tableau (Table 8.7)1. Replace The Pivot Row With The Pivot Column In The Solution Mix:

Exit = 100ENTER = 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 = 01/110 = 0/110.0909 = 1/110 = 0/110 = 0/1110 = 110/11

3. Replace T he Row Of The Initi al Tableau W ith New Values In The Sec ond Tabl eau:

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(-0.0909) -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(0.0909)0 = 0 – 4(0)1 = 1 – 4(0)32 = 72 – 4(10)

5. Compute The Values:

= Sum Product ( 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(7.1818)-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)6950 = 55(10) + 100(32) + 100(32)

= $6,960of 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 – 1000 = 100 – 100

7. Determine The Minimum Negative ( - ) Value:-1030 = -Minimum (-1030,0, -104,1,100,100,204.09,0,0)= Pivot Column Is X

TABLE 8.7 SECOND TABLEAU

Basic

55

100

100

Gross

Net

50 55 0 0 0 100 100 100

x y

0.454 1 -0.091 0 0 0.0909 0 0

3.3636 0 0.7273 -1 0 -0.727 1 0

7.1818 0 0.3636 0 -1 -0.364 0 1

1079.5 55 104.09 -100 -100 100 100 100

-1030 0 -104.1 100 100 204.09 0 0

yes no no no no no no No

Soln

-

quantitiy

10

32

32

6,950

Total cost

Min-? Yes/no

Step 4. Determine The Pivot Row (Table 8.8)1. COMPUTE For The Quantity Ration ():= Q/Pivot Column22 = 10/0.45459.5135 = 32/3.36364.4557 = 32/7.1818

2. Compute For The Minimum Positive Quantity Ration ():

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

Table 8.8 Second Pivot Row

SolnMix

Y

PivCol

X

0.4545

3.3636

7.1818

Quantity Quant/ Pivcol Min + ? Y/N

10 22 No

32 9.5135 No

32 4.4557 Yes

Step 5. Develop The Third Tableau (Table 8.9)1. Replace The Pivot Row With The Pivot Column In The Solution Mix:

Exit = 100Enter = 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 = 0/7.18180.1392 = 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(o.0506)0 = 0 – 0.4545(0)0.0633 = 0- 0.4545 (-0.139)01139 = 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)

TABLE 8.9 THIRD TABLEAU

Basic

55

100

100

Gross

Net

50 55 0 0 0 100 100 100

x y

0 1 -0.114 0 0.0633 0.1139 0 -0.063

0 0 0.557 -1 0.4684 -0.557 1 -0.468

1 0 0.0506 0 -0.139 -0.051 0 0.1392

50 55 51.962 -100 43.354 -51.96 100 -43.35

0 0 -51.96 100 -43.35 151.96 0 143.35

no no Yes no no No no No

Soln

-

quantity

7.9747

17.013

4.4557

2362.7

Total cost

Min-? Yes/no

Step 6. Determine The Pivot (Table 8.10)1. Compute The Quantity Ratio ():

= Q/Pivot Column-70 = 7.9747/-0.11430.545 = 17.013/0.55788 = 4.4557/0.0506

2. Compute The Minimum Positive Quantity Ratio ):

30.545 = +Minimum (-70,30.545,88)= Pivot Row Is

3. Determine The Pivot Number ( )0.557 = Intersection Of Pivot Column And Pivot Row

TABLE 8.10 THIRD PIVOT ROW

SolnMix

Y

PivCol

-0.114

0.557

0.0506

Quantity Quant/ Pivcol Min + ? Y/N

7.9747 -70 No

17.013 30.545 Yes

4.4557 88 No

Step 7. Develop The Fourth Tableau (Table 8.11)1. Replace The Pivot Row With The Pivot Column In The Solution Mix:

Exit = Enter = 0

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 =0/0.5571 = 0.557/0.557-1.795 = -1/0.5570.8409 = 0.4684/0.557-1 = -0.557/0.5571.7955 =1/0.557-0.841 = -0.468/0.55730.545 = 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 Row And Pivot Column) (0 = 0 – (-0.114)(0)1 = 1- (-0.114)(0)0 = -0.114- (-0.114)(1)-0.205 = 0- (-0.114)(0.8409)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 Third Tableau With New Values In The Fourth Tableau:New R Row = Old X Row – (Number In Old X Row And Pivot Column)(

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)

5. Compute The VALUES:= SUM PRODUCT(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 Of Cost For This Solution

6. Compute The ) Values:) = Row) – 0 = 50 – 500 = 55 – 550 = 0 – 06.7045 = 0 – (-6.705)0.3409 = 0 – (-0.341)100 = 100 – (0)93.295 = 100 – 6.704599. 695 = 100 – 0.3409

7. DETERMINE THE MINIMUM NEGATIVE - ) Value:

None = - Minimum (0, 0, 0, 6.7045, 0.3409, 100, 93.295, 99.659)

= Final Tableau Is Reached8. Determine The Final SolutionY = 11.455 (11 Gallons Of

Nestle Milk Should Be Purchased Per Day)

= 30.545 (31 Cases Of Excess Cheese Are Product 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 8.11 FOURTH TABLEAU

Basic

55

100

100

Gross

Net

Soln

Y

X

-

quantitiy

11.455

30.545

2.9091

775.45

Total cost

Min-? Yes/no

50 55 0 0 0 100 100 100

X Y

0 1 0 -0.205 0.1591 0 0.2045 -0.159

0 0 1 -1.795 0.8409 -1 1.7955 -0.841

1 0 0 0.0909 -0.182 0 -0.091 0.1818

50 55 0 -6.705 -0.341 0 6.7045 0.3409

0 0 0 6.7045 0.3409 100 93.295 99.659

NONE

Thus, Jacob Should Purchase 3 Gallons Of Alaska Milk 11 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. The Result Is The Same With The Answer Of Example 8.1

Section 8.3 Solver Method

Another Method To Solve Linear Programming Problems Is The Solver Method.

This Computerized Method Is Used With A Spreadsheet:

* Microsoft Excel’s Solver

* Open Office Calc’s Solver

Ro/Col B C D E F G

139 Product and Supplier Cases per Gail Sign Cases /Day

140 Alaska Nestle

141 Cheese 5 11 ≥ 110

142 Butter 7 8 ≥ 112

143 Cream 9 4 ≥ 72

144 Cost($) 50 55 = Min

Example 8.3 Kraft ( III )Table 8.12 Kraf Solve Example 8.1 using the Solver Method. The figures are summarized in Table 8.12. TABLE 8.12 KRAFT

STEP 1. SET UP ON A SPREADSHEET AS A SOLVER PROBLEM( TABLE 8.13 ). 1. DETERMINE THE VARIABLES ( ROW 150 ) : 

D150 = NUMBER OF GALLONS OF ALASKA MILK TO BE PURCHASED PER DAY

E150 = NUMBER OF GALLONS OF NESTLE MILK TO BE PURCHASED PER DAY

 2. DETERMINE THE FORMULA FOE THE TOTAL COLUMN ( COLUMN F ):

 F151 = SUMPRODUCT ( D150:E150,D151:E151 )F152 = SUMPRODUCT ( D150:E150,D152:E152 )F153 = SUMPRODUCT ( D150:E150,D153:E153 )F154 = SUMPRODUCT ( D150:E150,D154:E154 )

3.DETERMINE THE FORMULA FOR THE SLACK COLUMN ( COLUMN I ) :1151 = F151 - H1511152 = F152 - H1521153 = F153 – H153

Ro/Col B C D E F G H I149 Supplier Alaska Nestle Total Sign Cases/

DaySlack

150 Gallons

151 Cheese 5 11 0 ≥ 110 110

152 Butter 7 8 0 ≥ 112 112

153 Cream 9 4 0 ≥ 72 -72

154 Cost ($) 50 55 0 = Total Cost($)

Table 8.13 Solver Setup

STEP 2. SOLVE TABLE 8.13 USING THE SOLVER ADD-IN (TABLE 8.14). 

/TOOLS/SOLVER/*SET TARGET SELL : F154*EQUAL TO : MIN*BY CHANGING CELLS: D150:E150CONSTRAINTS : F151:F153≥ H151:H153/OPTIONS *ASSUME LINEAR MODEL*ASSUME NON-NEGATIVE OK/SOLVE/ANSWER REPORT

Ro/Col B C D E F G H I

149 Supplier Alaska Nestle Total Sign Cases/Day

Slack

150 Gallons 2.9091 11.455

151 Cheese 5 11 140.55 ≥ 110 30.545

152 Butter 7 8 112 ≥ 112 0

153 Cream 9 4 72 ≥ 72 0

154 Cost ($) 50 55 775.45 = Total Cost($)

Table 8.14 Solver Solution

STEP 3. ANALYZE THE ANSWER REPORT ( TABLE 8.15 ).F154 = 775.45 ( $774.45 IS THE TOTAL COST

PER DAY )D150 =2.9091 ( 3 GALLONS OF ALASKA MILK

SHOULD BE PURCHASED PER DAY)

E150 =11.455 ( 11GALLONS OF NESTLE MILK SHOULD BE PURCHASED PER DAY)

F151 SLACK= 30.545 ( 31 CASES EXCESS CHEESE ARE MADE PER DAY )

F152 SLACK= 0( 0 CASES EXCESS CHEESE ARE MADE PER DAY )

F153 SLACK= 0( 0 CASES EXCESS CHEESE ARE MADE PER DAY )

Microsoft Excel 9.0 Answer Report /Tools/SolverWorksheet: [Quameth.xls] Chapter 8 *Set Target Cell: F154

Target Cell (Min) *Equal To: MinCell Name Original Final *By Changing Cells: D150:E150

$F$154 Cost($)Total 0 775.45 Constrains: F151:F153>=H151:H153

Adjustable Cells /OptionsCell Name Original Final *Assume Linear Model

$D$150 Gallons Alaska 0 2.9091 *Assume Non-negative OK

$E$150 Gallons Nestle 0 11.455 /Solve/Answer Report

ConstraintsCell Name Value Formula Status Slack

$F$151 Cheese Total 140.55 $F$151≥$H$151 Not Binding 30.545

$F$152 Butter Total 112 $F$152≥$H$152 Binding 0

$F$153 Cream Total 72 $F$153≥$H$153 Binding 0

Thus, Jacob Should Buy 3 Gallons Of Alaska Milk 11 Gallons Of Nestle Milk Per Day At A Total Cost Of $775.45, With 31 Cases Of Excess Cheese Made. The Result Is The Same With The Answer Of Example 8.1

ProblemsSolve The Problems Using The Examples Discussed In The Chapter As A Guide.

Problem 8.1 Coca Cola Karen 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 Coke Lite, 96 Liters Of Sprite Lite, And 100 Liters Of Coke Zero Per Day. Equal Sweetener Costs $27 Per Kilogram While Nutra Sweetener Costs $30 Per Kilogram. The Figures Are Summarized In Table 8.16

How Many Kilograms Of Equal Sweetener And Nutra Sweetener Should She Purchase Per Day To Minimize Costs? How Much Is The Total Cost? Karen Wants To Solve The Problem Using Graphical Method.

TABLE 8.16 COCA- COLA

Product/supplier

Coke lite

Sprite lite

Coke zero

Cost/kg ($)

equal nutra

4 12

6 8

10 5

27 30 =

96

96

100

min

Liters per kg sign Liters/ day

Problem 8.2 Dole PineappleLester 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 Cases Of Crushed Pineapple Per Day. Absolute Water Costs $20 Per Barrel While Wilkins Water Costs $22 Per Barrel. The Figure Are Summarized In Table 8.19

How Many Barrels Of Absolute Water And Wilkins Water Should He Purchase Per Day To Minimize Costs? How Much Is The Total Cost? Lester Wants To Solve The Problem Using The Graphical Method.

TABLE 8.17 DOLE PINEAPPLE

Product/supplier

sliced

Chunks

Crushed

Cost/barrel ($)

absolute wilkins

9 5

8 8

4 11

20 22 =

90

128

88

min

Cases/barrel sign Liters/ day

Problem 8.5 Dole Pineapple (Ii)Solve Problem 8.4 Using The Simplex Method. The Figures Are Summarized In Table 8.20

Product/ Supplier

Sliced

Chunks

Crushed

Cases/Barrel

Absolute Wilkins

9 5

8 8

4 11

20 22

Sign Cases/ Day

90

128

88

= MinCost/ Barrel ($)

Prepared by:Honrade, Hanna Grace

C.Bellen, Michelle Ann T.