linear programming
DESCRIPTION
solving minimization problemTRANSCRIPT
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LINEAR PROGRAMMING: MINIMIZATION
MODEL
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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:
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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
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The constraints are then written in the form of inequalities.x + y Ax + y Bx + y Cx,y 0 (non negativity constraint)
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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
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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
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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.
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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 $55 per gallon. the figures are summarized in the table 7.1.
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how many gallons of alaska milk and nestle milk should he purchase per day to minimize costs ? how much is the total cost ? jacob wants to solve the problem using the graphical method.
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TABLE 8.1 KRAFT
Product/Supplier
Cheese
Butter
Cream
Cost/gal($)
Cases per gal
Alaska Nestle
5 11
7 8
9 4
50 55
Sign Cases/Day
110
112
72
= Min
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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)
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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)
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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)
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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)
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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)
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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)
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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 .
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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 .
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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
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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)
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= 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
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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.
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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.
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Constrain C ( CREAM )
9X + 4Y9X + 4Y + O - 1 + O + O + O + 1 = 72 Convert To Equality
Copy The Coefficients To The Row.
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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)
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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
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6. Determine The Minimum Negative - ) Values
–2245= –Minimum (–2050, –2245, 100, 100, 100, 0, 0, 0,
= Pivot Column Is Y
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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
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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
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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
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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
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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)
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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)
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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
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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
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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
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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
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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
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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:
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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
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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)
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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
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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
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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
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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
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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)
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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)
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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
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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
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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)
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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
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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 $775.45 With 31 Cases Of Excess Cheese Made. The Result Is The Same With The Answer Of Example 8.1
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Section 8.3 Solver Method
Another Method To Solve Linear Programming Problems Is The Solver Method.
This Computerized Method Is Used With A Spreadsheet:
* Microsoft Excel’s Solver
* Open Office Calc’s Solver
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Ro/Col B C D E F G
139 Product and Supplier Cases per Gail Sign Cases /Day
140 Alaska Nestle
141 Cheese 5 11 ≥ 110
142 Butter 7 8 ≥ 112
143 Cream 9 4 ≥ 72
144 Cost($) 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
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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 )
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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
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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
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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
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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 )
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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
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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.
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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 $30 Per Kilogram. The Figures Are Summarized In Table 8.16
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How Many Kilograms Of Equal Sweetener And Nutra Sweetener Should She Purchase Per Day To Minimize Costs? How Much Is The Total Cost? Karen Wants To Solve The Problem Using Graphical Method.
TABLE 8.16 COCA- COLA
Product/supplier
Coke lite
Sprite lite
Coke zero
Cost/kg ($)
equal nutra
4 12
6 8
10 5
27 30 =
96
96
100
min
Liters per kg sign Liters/ day
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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 $22 Per Barrel. The Figure Are Summarized In Table 8.19
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How Many Barrels Of Absolute Water And Wilkins Water Should He Purchase Per Day To Minimize Costs? How Much Is The Total Cost? Lester Wants To Solve The Problem Using The Graphical Method.
TABLE 8.17 DOLE PINEAPPLE
Product/supplier
sliced
Chunks
Crushed
Cost/barrel ($)
absolute wilkins
9 5
8 8
4 11
20 22 =
90
128
88
min
Cases/barrel sign Liters/ day
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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 ($)
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Prepared by:Honrade, Hanna Grace
C.Bellen, Michelle Ann T.