Chapter 8Chapter 8

Strategic Allocation of ResourcesStrategic Allocation of Resources

(Linear Programming)(Linear Programming)

A company makes 3 products: A, B and C.A company makes 3 products: A, B and C.

A B C AvailableA B C Available

ProfitProfit 35 45 25 35 45 25

Labor HrsLabor Hrs 5 7 3 5 7 3 2000 hrs 2000 hrs

FiberglassFiberglass 18 25 12 18 25 12 7000 lbs 7000 lbs

At least 100 units each must be made of A, B, CAt least 100 units each must be made of A, B, C

How many A’s, B’s, and C’s should be produced in How many A’s, B’s, and C’s should be produced in order to maximize total profits?order to maximize total profits?

Incorrect StrategyIncorrect Strategy: make as much as possible of the most : make as much as possible of the most profitable product (B), so make as little as possible of the profitable product (B), so make as little as possible of the other products (100 A’s and 100 C’s)other products (100 A’s and 100 C’s)

available: 2000 7000available: 2000 7000

Labor Fiberglass ProfitLabor Fiberglass Profit

make 100 A’smake 100 A’s

make 100 C’smake 100 C’s

remaining:remaining:

How many B’s?How many B’s?

500 1800 3500

300 1200 2500

1200 4000

1200/7= 171

4000/25= 160

We run out offiberglass 1st

make 160 B’s 1120 4000 7200

remaining: 80 0 $13,200 Total Profit

Optimal solution is $13,625 using LP (100 A, 100 B, 225 C) Difference of $425

Linear Programming FormulationLinear Programming Formulation

Max Z = 35A + 45B + 25CST

A = # of units of product A to produceB = # of units of product B to produceC = # of units of product C to produce

5A + 7B + 3C ≤ 2000 labor hours

18A + 25B + 12C ≤ 7000 fiberglass

A ≥ 100 minimum A

B ≥ 100 minimum BC ≥ 100 minimum C

Linear Programming using Lindo softwareLinear Programming using Lindo software

MaxMax 35 A + 45 B + 25 C35 A + 45 B + 25 C

Subject toSubject to2)2) 5 A + 7 B + 3 C <= 20005 A + 7 B + 3 C <= 20003)3) 18 A + 25 B + 12 C <= 18 A + 25 B + 12 C <=

700070004)4) A >= 100A >= 1005)5) B >= 100B >= 1006)6) C >= 100C >= 100

EndEndLP Optimum found at step 4LP Optimum found at step 4

Objective Function ValueObjective Function Value

1)1) 13625.000 13625.000

VariableVariable Value Value Reduced Cost Reduced Cost AA 100.000000100.000000 .000000.000000 BB 100.000000100.000000 .000000.000000 CC 225.000000225.000000 .000000.000000

RowRow Slack or Surplus Slack or Surplus Dual Prices Dual Prices2)2) 125.000000125.000000 .000000.0000003)3) .000000 .000000 2.083333 2.0833334)4) .000000 .000000 -2.500000 -2.5000005)5) .000000 .000000 -7.083333 -7.0833336)6) 125.000000125.000000 .000000.000000

No. Iterations = 4 No. Iterations = 4

Ranges in which the basis is unchanged:Ranges in which the basis is unchanged:

Obj Coefficient RangesObj Coefficient RangesVariableVariable CurrentCurrent Allowable Allowable AllowableAllowable

CoefCoef Increase Increase DecreaseDecrease AA 35.00000035.000000 2.500000 2.500000 InfinityInfinity BB 45.00000045.000000 7.083333 7.083333 InfinityInfinity CC 25.00000025.000000 Infinity Infinity 1.6666671.666667

Righthand Side RangesRighthand Side RangesRowRow CurrentCurrent Allowable Allowable AllowableAllowable

RHSRHS Increase Increase DecreaseDecrease2 2000.0000002 2000.000000 Infinity Infinity 125.000000 125.0000003 7000.0000003 7000.000000 500.000000 500.000000 1500.000000 1500.0000004 100.0000004 100.000000 83.333340 83.333340 100.000000 100.0000005 100.0000005 100.000000 60.000000 60.000000 100.000000 100.0000006 100.0000006 100.000000 125.000000 125.000000 Inifinity Inifinity

Example Using Excel SolverExample Using Excel Solver

10. A local brewery produces three types of beer: premium, regular, 10. A local brewery produces three types of beer: premium, regular, and light. The brewery has enough vat capacity to produce 27,000 and light. The brewery has enough vat capacity to produce 27,000 gallons of beer per month. A gallon of premium beer requires 3.5 gallons of beer per month. A gallon of premium beer requires 3.5 pounds of barley and 1.1 pounds of hops, a gallon of regular pounds of barley and 1.1 pounds of hops, a gallon of regular requires 2.9 pounds of barley and .8 pounds of hops, and a gallon of requires 2.9 pounds of barley and .8 pounds of hops, and a gallon of light requires 2.6 pounds of barley and .6 pounds of hops. The light requires 2.6 pounds of barley and .6 pounds of hops. The brewery is able to acquire only 55,000 pounds of barley and 20,000 brewery is able to acquire only 55,000 pounds of barley and 20,000 pounds of hops next month. The brewery’s largest seller is regular pounds of hops next month. The brewery’s largest seller is regular beer, so it wants to produce at least twice as much regular beer as it beer, so it wants to produce at least twice as much regular beer as it does light beer. It also wants to have a competitive market mix of does light beer. It also wants to have a competitive market mix of beer. Thus, the brewery wishes to produce at least 4000 gallons beer. Thus, the brewery wishes to produce at least 4000 gallons each of light beer and premium beer, but not more than 12,000 each of light beer and premium beer, but not more than 12,000 gallons of these two beers combined. The brewery makes a profit gallons of these two beers combined. The brewery makes a profit of $3.00 per gallon on premium beer, $2.70 per gallon on regular of $3.00 per gallon on premium beer, $2.70 per gallon on regular beer, and $2.80 per gallon on light beer. The brewery manager beer, and $2.80 per gallon on light beer. The brewery manager wants to know how much of each type of beer to produce next wants to know how much of each type of beer to produce next month in order to maximize profit.month in order to maximize profit.

Example Using Excel SolverExample Using Excel Solver

LP FormulationLP Formulation::

Max Z = 3P + 2.7R + 2.8LMax Z = 3P + 2.7R + 2.8L

STST

P + R + L < 27000P + R + L < 27000 capacitycapacity

3.5P + 2.9R + 2.6L < 550003.5P + 2.9R + 2.6L < 55000 barleybarley

1.1P + .8R + .6L < 200001.1P + .8R + .6L < 20000 hopshops

R – 2L > 0R – 2L > 0 2:1 ratio2:1 ratio

P > 4000P > 4000 minimum P requirementminimum P requirement

L > 4000L > 4000 minimum L requirementminimum L requirement

P + L < 12000P + L < 12000 maximum requirementmaximum requirement

02LR

2LR1

2

L

R

Instructions for Using Excel to Solve LP ModelsInstructions for Using Excel to Solve LP Models

1.1. Set up spreadsheet like example in packet. Set up spreadsheet like example in packet. (Z-value and LHS column should be formulas)(Z-value and LHS column should be formulas)

2.2. Select “Tools” on menu bar. Then select “Solver…”.Select “Tools” on menu bar. Then select “Solver…”.3.3. ““Set Target Cell:” should be the cell of your Z-value Set Target Cell:” should be the cell of your Z-value

formula.formula.4.4. Select “Min” or “Max”.Select “Min” or “Max”.5.5. ““By Changing Cells:” should be the range of cells for By Changing Cells:” should be the range of cells for

your decision variables values.your decision variables values.6.6. Select “Options…”Select “Options…”7.7. Check 2 boxes: “Assume Linear Model” and “Assume Check 2 boxes: “Assume Linear Model” and “Assume

Non-Negative”. Then click “OK”.Non-Negative”. Then click “OK”.8.8. Select “Add” to add constraints.Select “Add” to add constraints.

9. In “Cell Reference:” box point to LHS formula of first 9. In “Cell Reference:” box point to LHS formula of first constraint. Select <, =, or >. Click on “Constraint:” box constraint. Select <, =, or >. Click on “Constraint:” box and point to RHS value of first constraint. Click “Add” and point to RHS value of first constraint. Click “Add” for next constraint or “OK” if finished.for next constraint or “OK” if finished.

10. Repeat Step 9 for each other constraint.10. Repeat Step 9 for each other constraint.11. Select “Solve”.11. Select “Solve”.12. If it worked okay you should get the message “Solver 12. If it worked okay you should get the message “Solver

found a solution. All constraints and optimality found a solution. All constraints and optimality conditions are satisfied.” If you do not get this conditions are satisfied.” If you do not get this message you should modify your formulation or check message you should modify your formulation or check for mistakes.for mistakes.

13. In the Solver Results window under “Reports” click on 13. In the Solver Results window under “Reports” click on “Answer”. Then hold down the ‘Ctrl’ button while you “Answer”. Then hold down the ‘Ctrl’ button while you click on “Sensitivity”. Then click “OK”.click on “Sensitivity”. Then click “OK”.

14. Print your final worksheet showing the new values, 14. Print your final worksheet showing the new values, print the Answer Report and print the Sensitivity print the Answer Report and print the Sensitivity Report.Report.

AA BB CC DD EE FF GG

11 PP RR LL ObjectiveObjective

22 Dec VarsDec Vars 00 00 00 Value (Z)Value (Z)

33 Obj CoefObj Coef 33 2.72.7 2.82.8 =sumproduct(B3:D3,B$2:D$2)=sumproduct(B3:D3,B$2:D$2)

44

55 ConstraintsConstraints LHSLHS <,=,><,=,> RHSRHS

66 capacitycapacity 11 11 11 =sumproduct(B6:D6,B$2:D$2)=sumproduct(B6:D6,B$2:D$2) << 2700027000

77 barleybarley 3.53.5 2.92.9 2.62.6 =sumproduct(B7:D7,B$2:D$2)=sumproduct(B7:D7,B$2:D$2) << 5500055000

88 hopshops 1.11.1 0.80.8 0.60.6 =sumproduct(B8:D8,B$2:D$2)=sumproduct(B8:D8,B$2:D$2) << 2000020000

99 2:1 ratio2:1 ratio 11 -2-2 =sumproduct(B9:D9,B$2:D$2)=sumproduct(B9:D9,B$2:D$2) >> 00

1010 min req. Pmin req. P 11 =sumproduct(B10:D10,B$2:D$2)=sumproduct(B10:D10,B$2:D$2) >> 40004000

1111 min req. Lmin req. L 11 =sumproduct(B11:D11,B$2:D$2)=sumproduct(B11:D11,B$2:D$2) >> 40004000

1212 max req.max req. 11 11 =sumproduct(B12:D12,B$2:D$2)=sumproduct(B12:D12,B$2:D$2) << 1200012000

=sumproduct(B3:D3,B2:D2) is equivalent to =B3*B2 + C3*C2 + D3*D2

AA BB CC DD EE FF GG

11 PP RR LL ObjectiveObjective

22 Dec VarsDec Vars 40004000 9761.9059761.905 4880.9524880.952 Value (Z)Value (Z)

33 Obj CoefObj Coef 33 2.72.7 2.82.8 52023.8152023.81

44

55 ConstraintsConstraints LHSLHS <,=,><,=,> RHSRHS

66 capacitycapacity 11 11 11 18642.8618642.86 << 2700027000

77 barleybarley 3.53.5 2.92.9 2.62.6 5500055000 << 5500055000

88 hopshops 1.11.1 0.80.8 0.60.6 15138.115138.1 << 2000020000

99 2:1 ratio2:1 ratio 11 -2-2 00 >> 00

1010 min req. Pmin req. P 11 40004000 >> 40004000

1111 min req. Lmin req. L 11 4880.9524880.952 >> 40004000

1212 max req.max req. 11 11 8880.9528880.952 << 1200012000

Microsoft Excel 10.0 Answer ReportMicrosoft Excel 10.0 Answer Report

Worksheet: [Book1]Sheet1Worksheet: [Book1]Sheet1

Report Created: 1/15/2003 9:35:20 AMReport Created: 1/15/2003 9:35:20 AM

Target Cell (Max)Target Cell (Max)

CellCell NameName Original ValueOriginal Value Final ValueFinal Value

$E$3$E$3 Obj Coef Value (Z)Obj Coef Value (Z) 00 52023.8095252023.80952

Adjustable CellsAdjustable Cells

CellCell NameName Original ValueOriginal Value Final ValueFinal Value

$B$2$B$2 Dec Vars PDec Vars P 00 40004000

$C$2$C$2 Dec Vars RDec Vars R 00 9761.9047629761.904762

$D$2$D$2 Dec Vars LDec Vars L 00 4880.9523814880.952381

ConstraintsConstraints

CellCell NameName Cell ValueCell Value FormulaFormula StatusStatus SlackSlack

$E$6$E$6 capacity LHScapacity LHS 18642.8571418642.85714 $E$6<=$G$6$E$6<=$G$6 Not BindingNot Binding 8357.1428578357.142857

$E$7$E$7 barley LHSbarley LHS 5500055000 $E$7<=$G$7$E$7<=$G$7 BindingBinding 00

$E$8$E$8 hops LHShops LHS 15138.0952415138.09524 $E$8<=$G$8$E$8<=$G$8 Not BindingNot Binding 4861.9047624861.904762

$E$9$E$9 2:1 ratio LHS2:1 ratio LHS 00 $E$9>=$G$9$E$9>=$G$9 BindingBinding 00

$E$10$E$10 min req. P LHSmin req. P LHS 40004000 $E$10>=$G$10$E$10>=$G$10 BindingBinding 00

$E$11$E$11 min req. L LHSmin req. L LHS 4880.9523814880.952381 $E$11>=$G$11$E$11>=$G$11 Not BindingNot Binding 880.952381880.952381

$E$12$E$12 max req. LHSmax req. LHS 8880.9523818880.952381 $E$12<=$G$12$E$12<=$G$12 Not BindingNot Binding 3119.0476193119.047619

Microsoft Excel 10.0 Sensitivity ReportMicrosoft Excel 10.0 Sensitivity Report

Worksheet: [Book1]Sheet1Worksheet: [Book1]Sheet1

Report Created: 1/15/2003 9:35:20 AMReport Created: 1/15/2003 9:35:20 AM

Adjustable CellsAdjustable Cells

      FinalFinal ReducedReduced ObjectiveObjective AllowableAllowable AllowableAllowable

CellCell NameName ValueValue CostCost CoefficientCoefficient IncreaseIncrease DecreaseDecrease

$B$2$B$2 Dec Vars PDec Vars P 40004000 00 33 0.4166666670.416666667 1E+301E+30

$C$2$C$2 Dec Vars RDec Vars R 9761.9047629761.904762 00 2.72.7 0.4230769230.423076923 0.50.5

$D$2$D$2 Dec Vars LDec Vars L 4880.9523814880.952381 00 2.82.8 1E+301E+30 0.3793103450.379310345

ConstraintsConstraints

      FinalFinal ShadowShadow ConstraintConstraint AllowableAllowable AllowableAllowable

CellCell NameName ValueValue PricePrice R.H. SideR.H. Side IncreaseIncrease DecreaseDecrease

$E$6$E$6 capacity LHScapacity LHS 18642.8571418642.85714 00 2700027000 1E+301E+30 8357.1428578357.142857

$E$7$E$7 barley LHSbarley LHS 5500055000 0.9761904760.976190476 5500055000 18563.6363618563.63636 74007400

$E$8$E$8 hops LHShops LHS 15138.0952415138.09524 00 2000020000 1E+301E+30 4861.9047624861.904762

$E$9$E$9 2:1 ratio LHS2:1 ratio LHS 00 -0.130952381-0.130952381 00 2551.7241382551.724138 9034.4827599034.482759

$E$10$E$10 min req. P LHSmin req. P LHS 40004000 -0.416666667-0.416666667 40004000 2114.2857142114.285714 40004000

$E$11$E$11 min req. L LHSmin req. L LHS 4880.9523814880.952381 00 40004000 880.952381880.952381 1E+301E+30

$E$12$E$12 max req. LHSmax req. LHS 8880.9523818880.952381 00 1200012000 1E+301E+30 3119.0476193119.047619

1. The Ohio Creek Ice Cream Company is planning 1. The Ohio Creek Ice Cream Company is planning production for next week. Demand for Ohio Creek production for next week. Demand for Ohio Creek premium and light ice cream continue to outpace the premium and light ice cream continue to outpace the company’s production capacities. Ohio Creek earns a company’s production capacities. Ohio Creek earns a profit of $100 per hundred gallons of premium and $100 profit of $100 per hundred gallons of premium and $100 per hundred gallons of light ice cream. Two resources per hundred gallons of light ice cream. Two resources used in ice cream production are in short supply for next used in ice cream production are in short supply for next week: the capacity of the mixing machine and the week: the capacity of the mixing machine and the amount of high-grade milk. After accounting for required amount of high-grade milk. After accounting for required maintenance time, the mixing machine will be available maintenance time, the mixing machine will be available 140 hours next week. A hundred gallons of premium ice 140 hours next week. A hundred gallons of premium ice cream requires .3 hours of mixing and a hundred gallons cream requires .3 hours of mixing and a hundred gallons of light ice cream requires .5 hours of mixing. Only of light ice cream requires .5 hours of mixing. Only 28,000 gallons of high-grade milk will be available for 28,000 gallons of high-grade milk will be available for next week. A hundred gallons of premium ice cream next week. A hundred gallons of premium ice cream requires 90 gallons of milk and a hundred gallons of light requires 90 gallons of milk and a hundred gallons of light ice cream requires 70 gallons of milk.ice cream requires 70 gallons of milk.

P = # of gallons of Premium ice cream to makeP = # of gallons of Premium ice cream to make

L = # of gallons of Light ice cream to makeL = # of gallons of Light ice cream to make

Max Z = 100P + 100LMax Z = 100P + 100L

STST

.3P + .5L .3P + .5L ≤ 140≤ 140 capacity of mixing machinecapacity of mixing machine

90P + 70L ≤ 2800090P + 70L ≤ 28000 max milk availablemax milk available

Solution: P = 175; L = 175; Z = 35,000Solution: P = 175; L = 175; Z = 35,000

2. The Sureset Concrete Company produces 2. The Sureset Concrete Company produces concrete in a continuous process. Two concrete in a continuous process. Two ingredients in the concrete are sand, which ingredients in the concrete are sand, which Sureset purchases for $6 per ton, and gravel, Sureset purchases for $6 per ton, and gravel, which costs $8 per ton. Sand and gravel which costs $8 per ton. Sand and gravel together must make up exactly 75% of the together must make up exactly 75% of the weight of the concrete. Furthermore, no more weight of the concrete. Furthermore, no more than 40% of the concrete can be sand, and at than 40% of the concrete can be sand, and at least 30% of the concrete must be gravel. Each least 30% of the concrete must be gravel. Each day 2,000 tons of concrete are produced.day 2,000 tons of concrete are produced.

S = # tons of sand to add to mixtureS = # tons of sand to add to mixture

G = # tons of gravel to add to mixtureG = # tons of gravel to add to mixture

Min Z = 6S + 8GMin Z = 6S + 8G

STST

S + G = 1500S + G = 1500 sand & gravel are 75% of sand & gravel are 75% of 20002000

S S ≤ 800≤ 800 sand no more than 40% of 2000sand no more than 40% of 2000

G ≥ 600G ≥ 600 gravel at least 30% of 2000gravel at least 30% of 2000

Solution: S = 800; G = 700; Z = 10,400Solution: S = 800; G = 700; Z = 10,400

3. A ship has two cargo holds, one fore and one aft. The 3. A ship has two cargo holds, one fore and one aft. The fore cargo hold has a weight capacity of 70,000 pounds fore cargo hold has a weight capacity of 70,000 pounds and a volume capacity of 30,000 cubic feet. The aft hold and a volume capacity of 30,000 cubic feet. The aft hold has a weight capacity of 90,000 pounds and a volume has a weight capacity of 90,000 pounds and a volume capacity of 40,000 cubic feet. The shipowner has capacity of 40,000 cubic feet. The shipowner has contracted to carry loads of packaged beef and grain. contracted to carry loads of packaged beef and grain. The total weight of the available beef is 85,000 pounds; The total weight of the available beef is 85,000 pounds; the total weight of the available grain is 100,000 pounds. the total weight of the available grain is 100,000 pounds. The volume per mass of the beef is 0.2 cubic foot per The volume per mass of the beef is 0.2 cubic foot per pound, and the volume per mass of the grain is 0.4 cubic pound, and the volume per mass of the grain is 0.4 cubic foot per pound. The profit for shipping beef is $0.35 per foot per pound. The profit for shipping beef is $0.35 per pound, and the profit for shipping grain is $0.12 per pound, and the profit for shipping grain is $0.12 per pound. The shipowner is free to accept all or part of the pound. The shipowner is free to accept all or part of the available cargo; he wants to know how much meat and available cargo; he wants to know how much meat and grain to accept in order to maximize profit.grain to accept in order to maximize profit.

BF = # lbs beef to load in fore cargo holdBF = # lbs beef to load in fore cargo hold

BA = # lbs beef to load in aft cargo holdBA = # lbs beef to load in aft cargo hold

GF = # lbs grain to load in fore cargo holdGF = # lbs grain to load in fore cargo hold

GA = # lbs grain to load in aft cargo holdGA = # lbs grain to load in aft cargo hold

Max Z = .35 BF + .35BA + .12GF + .12 GAMax Z = .35 BF + .35BA + .12GF + .12 GA

STST

BF + GF BF + GF ≤ 70000≤ 70000 fore weight capacity – lbsfore weight capacity – lbs

BA + GA ≤ 90000BA + GA ≤ 90000 aft weight capacity – lbsaft weight capacity – lbs

.2BF + .4GF ≤ 30000.2BF + .4GF ≤ 30000 for volume capacity – cubic feetfor volume capacity – cubic feet

.2BA + .4GA ≤ 40000.2BA + .4GA ≤ 40000 for volume capacity – cubic feetfor volume capacity – cubic feet

BF + BA ≤ 85000BF + BA ≤ 85000 max beef availablemax beef available

GF + GA ≤ 100000GF + GA ≤ 100000 max grain availablemax grain available

4. The White Horse Apple Products Company purchases 4. The White Horse Apple Products Company purchases apples from local growers and makes applesauce and apples from local growers and makes applesauce and apple juice. It costs $0.60 to produce a jar of apple juice. It costs $0.60 to produce a jar of applesauce and $0.85 to produce a bottle of apple juice. applesauce and $0.85 to produce a bottle of apple juice. The company has a policy that at least 30% but not more The company has a policy that at least 30% but not more than 60% of its output must be applesauce.than 60% of its output must be applesauce.

The company wants to meet but not exceed the The company wants to meet but not exceed the demand for each product. The marketing manager demand for each product. The marketing manager estimates that the demand for applesauce is a maximum estimates that the demand for applesauce is a maximum of 5,000 jars, plus an additional 3 jars for each $1 spent of 5,000 jars, plus an additional 3 jars for each $1 spent on advertising. The maximum demand for apple juice is on advertising. The maximum demand for apple juice is estimated to be 4,000 bottles, plus an additional 5 bottles estimated to be 4,000 bottles, plus an additional 5 bottles for every $1 spent to promote apple juice. The company for every $1 spent to promote apple juice. The company has $16,000 to spend on producing and advertising has $16,000 to spend on producing and advertising applesauce and apple juice. Applesauce sells for $1.45 applesauce and apple juice. Applesauce sells for $1.45 per jar; apple juice sells for $1.75 per bottle. The per jar; apple juice sells for $1.75 per bottle. The company wants to know how many units of each to company wants to know how many units of each to produce and how much advertising to spend on each in produce and how much advertising to spend on each in order to maximize profit.order to maximize profit.

S = # jars apple Sauce to makeS = # jars apple Sauce to make

J = # bottles apple Juice to makeJ = # bottles apple Juice to make

SA = $ for apple Sauce AdvertisingSA = $ for apple Sauce Advertising

JA = $ for apple Juice AdvertisingJA = $ for apple Juice Advertising

Max Z = 1.45S + 1.75J - .6S - .85J – SA – JAMax Z = 1.45S + 1.75J - .6S - .85J – SA – JA

STST

S S ≥ .3(S + J)≥ .3(S + J) at least 30% apple sauceat least 30% apple sauce

S ≤ .6(S + J)S ≤ .6(S + J) no more than 60% apple sauceno more than 60% apple sauce

S ≤ 5000 + 3SAS ≤ 5000 + 3SA don’t exceed demand for apple saucedon’t exceed demand for apple sauce

J ≤ 4000 + 5JAJ ≤ 4000 + 5JA don’t exceed demand for apple juicedon’t exceed demand for apple juice

.6S + .85J + SA + JA ≤ 16000.6S + .85J + SA + JA ≤ 16000 budgetbudget

5. Dr. Maureen Becker, the head administrator at Jefferson County 5. Dr. Maureen Becker, the head administrator at Jefferson County Regional Hospital, must determine a schedule for nurses to make Regional Hospital, must determine a schedule for nurses to make sure there are enough nurses on duty throughout the day. During sure there are enough nurses on duty throughout the day. During the day, the demand for nurses varies. Maureen has broken the the day, the demand for nurses varies. Maureen has broken the day into 12 two-hour periods. The slowest time of the day day into 12 two-hour periods. The slowest time of the day encompasses the three periods from 12:00 A.M. to 6:00 A.M., encompasses the three periods from 12:00 A.M. to 6:00 A.M., which, beginning at midnight, require a minimum of 30, 20, and 40 which, beginning at midnight, require a minimum of 30, 20, and 40 nurses, respectively. The demand for nurses steadily increases nurses, respectively. The demand for nurses steadily increases during the next four daytime periods. Beginning with the 6:00 A.M. during the next four daytime periods. Beginning with the 6:00 A.M. – 8:00 A.M. period, a minimum of 50, 60, 80, and 80 nurses are – 8:00 A.M. period, a minimum of 50, 60, 80, and 80 nurses are required for these four periods, respectively. After 2:00 P.M. the required for these four periods, respectively. After 2:00 P.M. the demand for nurses decreases during the afternoon and evening demand for nurses decreases during the afternoon and evening hours. For the five two-hour periods beginning at 2:00 P.M. and hours. For the five two-hour periods beginning at 2:00 P.M. and ending at midnight, 70, 70, 60, 50, and 50 nurses are required, ending at midnight, 70, 70, 60, 50, and 50 nurses are required, respectively. A nurse reports for duty at the beginning of one of the respectively. A nurse reports for duty at the beginning of one of the two-hour periods and works eight consecutive hours (which is two-hour periods and works eight consecutive hours (which is required in the nurses’ contract). Dr. Becker wants to determine a required in the nurses’ contract). Dr. Becker wants to determine a nursing schedule that will meet the hospital’s minimum requirements nursing schedule that will meet the hospital’s minimum requirements throughout the day while using the minimum number of nurses.throughout the day while using the minimum number of nurses.

12 variables12 variables (one for each time block)(one for each time block)X1 = # of nurses starting at Midnight & working 8 hoursX1 = # of nurses starting at Midnight & working 8 hoursX2 = X2 = ““ 2am 2am ““X3 = X3 = ““ 4am 4am ““X4 = X4 = ““ 6am 6am ““X5 = X5 = ““ 8am 8am ““X6 = X6 = ““ 10am 10am ““X7 = X7 = ““ Noon Noon ““X8 = X8 = ““ 2pm 2pm ““X9 = X9 = ““ 4pm 4pm ““X10 = X10 = ““ 6pm 6pm ““X11 = X11 = ““ 8pm 8pm ““X12 =X12 = ““ 10pm 10pm ““

Min Z = X1 + X2 + X3 + X4 + X5 + X6 + ……. + X11 + X12Min Z = X1 + X2 + X3 + X4 + X5 + X6 + ……. + X11 + X12

STST

X1X1 + X10 + X11 + X12 + X10 + X11 + X12 ≥ 30≥ 30 midn – 2ammidn – 2am

X1 + X2X1 + X2 + X11 + X12 + X11 + X12 ≥ 20≥ 20 2am – 4am2am – 4am

X1 + X2 + X3 X1 + X2 + X3 + X12+ X12 ≥ 40≥ 40 4am – 6am4am – 6am

X1 + X2 + X3 + X4X1 + X2 + X3 + X4 ≥ 50≥ 50 6am – 8am6am – 8am

X2 + X3 + X4 + X5X2 + X3 + X4 + X5 ≥ 60≥ 60 8am – 10am8am – 10am

X3 + X4 + X5 + X6X3 + X4 + X5 + X6 ≥ 80≥ 80 10am–Noon10am–Noon

X4 + X5 + X6 + X7X4 + X5 + X6 + X7 ≥ 80≥ 80 Noon – 2pmNoon – 2pm

X5 + X6 + X7 + X8X5 + X6 + X7 + X8 ≥ 70≥ 70 2pm – 4pm2pm – 4pm

X6 + X7 + X8 + X9X6 + X7 + X8 + X9 ≥ 70≥ 70 4pm – 6pm4pm – 6pm

X7 + X8 + X9 + X10X7 + X8 + X9 + X10 ≥ 60≥ 60 6pm – 8pm6pm – 8pm

X8 + X9 + X10 + X11X8 + X9 + X10 + X11 ≥ 50≥ 50 8pm – 10pm8pm – 10pm

X9 + X10 + X11 + X12X9 + X10 + X11 + X12 ≥ 50≥ 50 10pm – midn10pm – midn

6. The Donnor meat processing firm produces wieners from four ingredients: chicken, beef, 6. The Donnor meat processing firm produces wieners from four ingredients: chicken, beef, pork, and a cereal additive. The firm produces three types of wieners: regular, beef, pork, and a cereal additive. The firm produces three types of wieners: regular, beef, and all-meat. The company has the following amounts of each ingredient available on a and all-meat. The company has the following amounts of each ingredient available on a daily basis.daily basis.

__________________________________________________________________________________________ lb/Daylb/Day Cost/lb($) Cost/lb($)

ChickenChicken 200200 .20.20BeefBeef 300300 .30.30PorkPork 150150 .50.50Cereal AdditiveCereal Additive 400400 .05.05

Each type of wiener has certain ingredient specifications, as follows.Each type of wiener has certain ingredient specifications, as follows.________________________________________________________________________________________________________________________________________________________________

SpecificationsSpecifications Selling Selling Price/lb($)Price/lb($)

RegularRegular Not more than 10% beef and pork combinedNot more than 10% beef and pork combinedNot less than 20% chickenNot less than 20% chicken $0.90$0.90

BeefBeef Not less than 75% beefNot less than 75% beef 1.25 1.25All-MeatAll-Meat No cereal additiveNo cereal additive

Not more than Not more than 50% beef and pork combined50% beef and pork combined 1.75 1.75

The firm wants to know the amount of wieners of each type to produce.The firm wants to know the amount of wieners of each type to produce.

14 variables14 variables (you could also formulate it with 11 (you could also formulate it with 11 variables)variables)

CR = # lbs Chicken ingredient in Regular wiener per dayCR = # lbs Chicken ingredient in Regular wiener per day

CB = # lbs Chicken ingredient in Beef wiener per dayCB = # lbs Chicken ingredient in Beef wiener per day

CM = # lbs Chicken ingredient in all-Meat wiener per dayCM = # lbs Chicken ingredient in all-Meat wiener per day

BR = # lbs Beef ingredient in Regular wiener per dayBR = # lbs Beef ingredient in Regular wiener per day

BB = # lbs Beef ingredient in Beef wiener per dayBB = # lbs Beef ingredient in Beef wiener per day

BM = # lbs Beef ingredient in all-Meat wiener per dayBM = # lbs Beef ingredient in all-Meat wiener per day

PR = # lbs Pork ingredient in Regular wiener per dayPR = # lbs Pork ingredient in Regular wiener per day

PB = # lbs Pork ingredient in Beef wiener per dayPB = # lbs Pork ingredient in Beef wiener per day

PM = # lbs Pork ingredient in all-Meat wiener per dayPM = # lbs Pork ingredient in all-Meat wiener per day

AR = # lbs Additive ingredient in Regular wiener per dayAR = # lbs Additive ingredient in Regular wiener per day

AB = # lbs Additive ingredient in Beef wiener per dayAB = # lbs Additive ingredient in Beef wiener per day

R = total lbs of Regular wienerR = total lbs of Regular wiener

B = total lbs of Beef wienerB = total lbs of Beef wiener

M = total lbs of all-Meat wienerM = total lbs of all-Meat wiener

Max Z = 0.90R + 1.25 B + 1.75 M - .2CR - .2CB - .2CM - .3BR - .3BB Max Z = 0.90R + 1.25 B + 1.75 M - .2CR - .2CB - .2CM - .3BR - .3BB

- .3BM - .5PR - .5PB - .5PM - .05AR - .05AB- .3BM - .5PR - .5PB - .5PM - .05AR - .05AB

STST

CR + BR + PR + AR = RCR + BR + PR + AR = R R is sum of all ingredients in RegularR is sum of all ingredients in Regular

CB + BB + PB + AB = BCB + BB + PB + AB = B B is sum of all ingredients in BeefB is sum of all ingredients in Beef

CM + BM + PM = MCM + BM + PM = M M is sum of all ingredients in MeatM is sum of all ingredients in Meat

CR + CB + CM CR + CB + CM ≤ 200≤ 200 max Chicken ingredient availablemax Chicken ingredient available

BR + BB + BM ≤ 300BR + BB + BM ≤ 300 max Beef ingredient availablemax Beef ingredient available

PR + PB + PM ≤ 150PR + PB + PM ≤ 150 max Pork ingredient availablemax Pork ingredient available

AR + AB ≤ 400AR + AB ≤ 400 max Additive ingredient availablemax Additive ingredient available

BR + PR ≤ .1RBR + PR ≤ .1R not more than 10% BR+PR combinednot more than 10% BR+PR combined

CR ≥ .2RCR ≥ .2R not less than 20% CR in Regularnot less than 20% CR in Regular

BB ≥ .75BBB ≥ .75B not less than 75% BB in Beefnot less than 75% BB in Beef

BM + PM ≤ .5MBM + PM ≤ .5M not more than 50% BM+PM not more than 50% BM+PM combinedcombined

7.7. The Jane Deere Company manufactures tractors in Provo, Utah. The Jane Deere Company manufactures tractors in Provo, Utah. Jeremiah Goldstein, the production planner, is scheduling tractor Jeremiah Goldstein, the production planner, is scheduling tractor production for the next three months. Factors that Mr. Goldstein production for the next three months. Factors that Mr. Goldstein must consider include sales forecasts, straight-time and overtime must consider include sales forecasts, straight-time and overtime labor hours available, labor cost, storage capacity, and carrying labor hours available, labor cost, storage capacity, and carrying cost. The marketing department has forecasted that the number of cost. The marketing department has forecasted that the number of tractors shipped during the next three months will be 250, 305, and tractors shipped during the next three months will be 250, 305, and 350. Each tractor requires 100 labor hours to produce. In each 350. Each tractor requires 100 labor hours to produce. In each month 29,000 straight-time labor hours will be available, and month 29,000 straight-time labor hours will be available, and company policy prohibits overtime hours from exceeding 10% of company policy prohibits overtime hours from exceeding 10% of straight-time hours. Straight-time labor cost rate is $20 per hour, straight-time hours. Straight-time labor cost rate is $20 per hour, including benefits. The overtime labor cost rate is 150% (time-and-including benefits. The overtime labor cost rate is 150% (time-and-a-half) of the straight-time rate. Excess production capacity during a a-half) of the straight-time rate. Excess production capacity during a month may be used to produce tractors that will be stored and sold month may be used to produce tractors that will be stored and sold during a later month. However, the amount of storage space can during a later month. However, the amount of storage space can accommodate only 40 tractors. A carrying cost of $600 is charged accommodate only 40 tractors. A carrying cost of $600 is charged for each month a tractor is stored (if not shipped during the month it for each month a tractor is stored (if not shipped during the month it was produced). Currently, no tractors are in storage.was produced). Currently, no tractors are in storage.

How many tractors should be produced in each month using How many tractors should be produced in each month using straight-time and using overtime in order to minimize total labor cost straight-time and using overtime in order to minimize total labor cost and carrying cost? Sales forecasts, straight-time and overtime labor and carrying cost? Sales forecasts, straight-time and overtime labor capacities, and storage capacity must be adhered to. (Tip: During capacities, and storage capacity must be adhered to. (Tip: During each month, all “sources” of tractors must exactly equal “uses” of each month, all “sources” of tractors must exactly equal “uses” of tractors.)tractors.)

9 variables9 variables

S1 = # tractors produced in month 1 using straight-timeS1 = # tractors produced in month 1 using straight-time

S2 = # tractors produced in month 2 using straight-timeS2 = # tractors produced in month 2 using straight-time

S3 = # tractors produced in month 3 using straight-timeS3 = # tractors produced in month 3 using straight-time

V1 = # tractors produced in month 1 using overtimeV1 = # tractors produced in month 1 using overtime

V2 = # tractors produced in month 2 using overtimeV2 = # tractors produced in month 2 using overtime

V3 = # tractors produced in month 3 using overtimeV3 = # tractors produced in month 3 using overtime

C1 = # tractors carried in warehouse at end of month 1C1 = # tractors carried in warehouse at end of month 1

C2 = # tractors carried in warehouse at end of month 2C2 = # tractors carried in warehouse at end of month 2

C3 = # tractors carried in warehouse at end of month 3C3 = # tractors carried in warehouse at end of month 3

sources of tractors = uses of tractorssources of tractors = uses of tractors (for each month)(for each month)

production + beg.inv. = sales + end.inv.production + beg.inv. = sales + end.inv.

Min Z = 2000S1 + 2000S2 + 2000S3 + 3000V1 + 3000V2 Min Z = 2000S1 + 2000S2 + 2000S3 + 3000V1 + 3000V2

+ 3000V3 + 600C1 + 600C2 + 600C3+ 3000V3 + 600C1 + 600C2 + 600C3

STST

S1 + V1 + 0 = 250 + C1S1 + V1 + 0 = 250 + C1 month 1: sources = usesmonth 1: sources = uses

S2 + V2 + C1 = 305 + C2S2 + V2 + C1 = 305 + C2 month 2: sources = usesmonth 2: sources = uses

S3 + V3 + C2 = 350 + C3S3 + V3 + C2 = 350 + C3 month 3: sources = usesmonth 3: sources = uses

100S1 100S1 ≤ 29000≤ 29000 straight-time capacity month 1straight-time capacity month 1

100S2 ≤ 29000100S2 ≤ 29000 straight-time capacity month 2straight-time capacity month 2

100S3 ≤ 29000100S3 ≤ 29000 straight-time capacity month 3straight-time capacity month 3

100V1 ≤ 2900100V1 ≤ 2900 overtime capacity month 1overtime capacity month 1

100V2 ≤ 2900100V2 ≤ 2900 overtime capacity month 2overtime capacity month 2

100V3 ≤ 2900100V3 ≤ 2900 overtime capacity month 3overtime capacity month 3

C1 ≤ 40C1 ≤ 40 storage capacity month 1storage capacity month 1

C2 ≤ 40C2 ≤ 40 storage capacity month 2storage capacity month 2

C3 ≤ 40C3 ≤ 40 storage capacity month 3storage capacity month 3

8.8. MadeRite, a manufacturer of paper stock for copiers and printers, MadeRite, a manufacturer of paper stock for copiers and printers, produces cases of finished paper stock at Mills 1, 2, and 3. The produces cases of finished paper stock at Mills 1, 2, and 3. The paper is shipped to Warehouses A, B, C, and D. The shipping cost paper is shipped to Warehouses A, B, C, and D. The shipping cost per case, the monthly warehouse requirements, and the monthly per case, the monthly warehouse requirements, and the monthly mill production levels are:mill production levels are:

Monthly MillMonthly Mill

DestinationDestination Production Production

AA B B C C D D (cases) (cases)

Mill 1Mill 1 $5.40$5.40 $6.20 $6.20 $4.10 $4.90 $4.10 $4.90 15,000 15,000

Mill 2Mill 2 4.00 4.00 7.10 7.10 5.60 3.90 5.60 3.90 10,000 10,000

Mill 3Mill 3 4.50 4.50 5.20 5.20 5.50 6.10 5.50 6.10 15,000 15,000Monthly WarehouseMonthly Warehouse

Requirement (cases)Requirement (cases) 9,0009,000 9,000 12,000 10,000 9,000 12,000 10,000

How many cases of paper should be shipped per month from each mill How many cases of paper should be shipped per month from each mill to each warehouse to minimize monthly shipping costs?to each warehouse to minimize monthly shipping costs?

A1 = # of units shipped from Mill 1 to Destination AA1 = # of units shipped from Mill 1 to Destination A

C3 = # of units shipped from Mill 3 to Destination CC3 = # of units shipped from Mill 3 to Destination C

(12 variables)(12 variables)

Min Z = 5.4A1 + 6.2B1 + 4.1C1 + 4.9D1 + 4.0A2 + 7.1B2Min Z = 5.4A1 + 6.2B1 + 4.1C1 + 4.9D1 + 4.0A2 + 7.1B2

+ 5.6C2 + 3.9D2 + 4.5A3 + 5.2B3 + 5.5C3 + 6.1D3+ 5.6C2 + 3.9D2 + 4.5A3 + 5.2B3 + 5.5C3 + 6.1D3

STST

A1 + B1 + C1 + D1 A1 + B1 + C1 + D1 ≤ 15000≤ 15000 Mill 1 capacityMill 1 capacity

A2 + B2 + C2 + D2 ≤ 10000A2 + B2 + C2 + D2 ≤ 10000 Mill 2 capacityMill 2 capacity

A3 + B3 + C3 + D3 ≤ 15000A3 + B3 + C3 + D3 ≤ 15000 Mill 3 capacityMill 3 capacity

A1 + A2 + A3 = 9000A1 + A2 + A3 = 9000 Destination A demandDestination A demand

B1 + B2 + B3 = 9000B1 + B2 + B3 = 9000 Destination B demandDestination B demand

C1 + C2 + C3 = 12000C1 + C2 + C3 = 12000 Destination C demandDestination C demand

D1 + D2 + D3 = 10000D1 + D2 + D3 = 10000 Destination D demandDestination D demand

9.9. A company has three research projects that it wants to A company has three research projects that it wants to do, and has three research teams that can do the do, and has three research teams that can do the projects. Any team could do any project but can only do projects. Any team could do any project but can only do one project. Some teams are better skilled at certain one project. Some teams are better skilled at certain projects and could do them at lower costs. The projects and could do them at lower costs. The estimated cost of each team doing each project (in estimated cost of each team doing each project (in $,000s) is shown below. Which team should do which $,000s) is shown below. Which team should do which project?project?

ProjectProject

11 22 33

AA 8787 6262 7676

TeamTeam BB 8181 7676 6464

CC 7777 5454 7070

A1 = 1 if team A does project 1A1 = 1 if team A does project 1

= 0 if not= 0 if not (9 variables)(9 variables)

Min Z = 87A1 + 62A2 + 76A3 + 81B1 + 76B2 + 64B3Min Z = 87A1 + 62A2 + 76A3 + 81B1 + 76B2 + 64B3

+ 77C1 + 54C2 + 70C3+ 77C1 + 54C2 + 70C3

STST

A1 + A2 + A3 = 1A1 + A2 + A3 = 1

B1 + B2 + B3 = 1B1 + B2 + B3 = 1 Each team does exactly one projectEach team does exactly one project

C1 + C2 + C3 = 1C1 + C2 + C3 = 1

A1 + B1 + C1 = 1A1 + B1 + C1 = 1

A2 + B2 + C2 = 1A2 + B2 + C2 = 1 Each project is done exactly onceEach project is done exactly once

A3 + B3 + C3 = 1A3 + B3 + C3 = 1