- 162 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Chapter 8' - thuyet

Download Now**An Image/Link below is provided (as is) to download presentation**

Download Now

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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

A B C Available

Profit 35 45 25

Labor Hrs 5 7 3 2000 hrs

Fiberglass 18 25 12 7000 lbs

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

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

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

available: 2000 7000

Labor Fiberglass Profit

make 100 A’s

make 100 C’s

remaining:

How many B’s?

500 1800 3500

300 1200 2500

1200 4000

We run out of

fiberglass 1st

1200/7

= 171

4000/25

= 160

make 160 B’s 1120 4000 7200

$13,200 Total Profit

remaining: 80 0

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

Difference of $425

Linear Programming Formulation

A = # of units of product A to produce

B = # of units of product B to produce

C = # of units of product C to produce

Max Z = 35A + 45B + 25C

ST

5A + 7B + 3C ≤ 2000 labor hours

18A + 25B + 12C ≤ 7000 fiberglass

A ≥ 100 minimum A

B ≥ 100 minimum B

C ≥ 100 minimum C

Linear Programming using Lindo software

Max 35 A + 45 B + 25 C

Subject to

2) 5 A + 7 B + 3 C <= 2000

3) 18 A + 25 B + 12 C <= 7000

4) A >= 100

5) B >= 100

6) C >= 100

End

LP Optimum found at step 4

Objective Function Value

1) 13625.000

Variable Value Reduced Cost

A 100.000000 .000000

B 100.000000 .000000

C 225.000000 .000000

Row Slack or Surplus Dual Prices

2) 125.000000 .000000

3) .000000 2.083333

4) .000000 -2.500000

5) .000000 -7.083333

6) 125.000000 .000000

No. Iterations = 4

Ranges in which the basis is unchanged:

Obj Coefficient Ranges

Variable Current Allowable Allowable

Coef Increase Decrease

A 35.000000 2.500000 Infinity

B 45.000000 7.083333 Infinity

C 25.000000 Infinity 1.666667

Righthand Side Ranges

Row Current Allowable Allowable

RHS Increase Decrease

2 2000.000000 Infinity 125.000000

3 7000.000000 500.000000 1500.000000

4 100.000000 83.333340 100.000000

5 100.000000 60.000000 100.000000

6 100.000000 125.000000 Inifinity

Example Using Excel Solver

10. A local brewery produces three types of beer: premium, regular, 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 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 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 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 does light beer. It also wants to have a competitive market mix of beer. Thus, the brewery wishes to produce at least 4000 gallons each of light beer and premium beer, but not more than 12,000 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 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 month in order to maximize profit.

Example Using Excel Solver

LP Formulation:

Max Z = 3P + 2.7R + 2.8L

ST

P + R + L < 27000 capacity

3.5P + 2.9R + 2.6L < 55000 barley

1.1P + .8R + .6L < 20000 hops

R – 2L > 0 2:1 ratio

P > 4000 minimum P requirement

L > 4000 minimum L requirement

P + L < 12000 maximum requirement

Instructions for Using Excel to Solve LP Models

- Set up spreadsheet like example in packet. (Z-value and LHS column should be formulas)
- Select “Tools” on menu bar. Then select “Solver…”.
- “Set Target Cell:” should be the cell of your Z-value formula.
- Select “Min” or “Max”.
- “By Changing Cells:” should be the range of cells for your decision variables values.
- Select “Options…”
- Check 2 boxes: “Assume Linear Model” and “Assume Non-Negative”. Then click “OK”.
- Select “Add” to add constraints.

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

10. Repeat Step 9 for each other constraint.

11. Select “Solve”.

12. If it worked okay you should get the message “Solver found a solution. All constraints and optimality conditions are satisfied.” If you do not get this message you should modify your formulation or check for mistakes.

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

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

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

P = # of gallons of Premium ice cream to make

L = # of gallons of Light ice cream to make

Max Z = 100P + 100L

ST

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

90P + 70L ≤ 28000 max milk available

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

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

S = # tons of sand to add to mixture

G = # tons of gravel to add to mixture

Min Z = 6S + 8G

ST

S + G = 1500 sand & gravel are 75% of 2000

S ≤ 800 sand no more than 40% of 2000

G ≥ 600 gravel at least 30% of 2000

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

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

BF = # lbs beef to load in fore cargo hold

BA = # lbs beef to load in aft cargo hold

GF = # lbs grain to load in fore cargo hold

GA = # lbs grain to load in aft cargo hold

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

ST

BF + GF ≤ 70000 fore weight capacity – lbs

BA + GA ≤ 90000 aft weight capacity – lbs

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

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

BF + BA ≤ 85000 max beef available

GF + GA ≤ 100000 max grain available

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

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

S = # jars apple Sauce to make

J = # bottles apple Juice to make

SA = $ for apple Sauce Advertising

JA = $ for apple Juice Advertising

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

ST

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

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

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

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

.6S + .85J + SA + JA ≤ 16000 budget

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

12 variables (one for each time block)

X1 = # of nurses starting at Midnight & working 8 hours

X2 = “ 2am “

X3 = “ 4am “

X4 = “ 6am “

X5 = “ 8am “

X6 = “ 10am “

X7 = “ Noon “

X8 = “ 2pm “

X9 = “ 4pm “

X10 = “ 6pm “

X11 = “ 8pm “

X12 = “ 10pm “

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

ST

X1 + X10 + X11 + X12 ≥ 30 midn – 2am

X1 + X2 + X11 + X12 ≥ 20 2am – 4am

X1 + X2 + X3 + X12 ≥ 40 4am – 6am

X1 + X2 + X3 + X4 ≥ 50 6am – 8am

X2 + X3 + X4 + X5 ≥ 60 8am – 10am

X3 + X4 + X5 + X6 ≥ 80 10am–Noon

X4 + X5 + X6 + X7 ≥ 80 Noon – 2pm

X5 + X6 + X7 + X8 ≥ 70 2pm – 4pm

X6 + X7 + X8 + X9 ≥ 70 4pm – 6pm

X7 + X8 + X9 + X10 ≥ 60 6pm – 8pm

X8 + X9 + X10 + X11 ≥ 50 8pm – 10pm

X9 + X10 + X11 + X12 ≥ 50 10pm – midn

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, and all-meat. The company has the following amounts of each ingredient available on a daily basis.

_____________________________________________

lb/Day Cost/lb($)

Chicken 200 .20

Beef 300 .30

Pork 150 .50

Cereal Additive 400 .05

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

________________________________________________________________________________

Specifications Selling Price/lb($)

Regular Not more than 10% beef and pork combined

Not less than 20% chicken $0.90

Beef Not less than 75% beef 1.25

All-Meat No cereal additive

Not more than

50% beef and pork combined 1.75

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

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

CR = # lbs Chicken ingredient in Regular wiener per day

CB = # lbs Chicken ingredient in Beef wiener per day

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

BR = # lbs Beef ingredient in Regular wiener per day

BB = # lbs Beef ingredient in Beef wiener per day

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

PR = # lbs Pork ingredient in Regular wiener per day

PB = # lbs Pork ingredient in Beef wiener per day

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

AR = # lbs Additive ingredient in Regular wiener per day

AB = # lbs Additive ingredient in Beef wiener per day

R = total lbs of Regular wiener

B = total lbs of Beef wiener

M = total lbs of all-Meat wiener

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

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

ST

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

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

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

CR + CB + CM ≤ 200 max Chicken ingredient available

BR + BB + BM ≤ 300 max Beef ingredient available

PR + PB + PM ≤ 150 max Pork ingredient available

AR + AB ≤ 400 max Additive ingredient available

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

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

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

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

7. The Jane Deere Company manufactures tractors in Provo, Utah. Jeremiah Goldstein, the production planner, is scheduling tractor production for the next three months. Factors that Mr. Goldstein must consider include sales forecasts, straight-time and overtime labor hours available, labor cost, storage capacity, and carrying cost. The marketing department has forecasted that the number of tractors shipped during the next three months will be 250, 305, and 350. Each tractor requires 100 labor hours to produce. In each month 29,000 straight-time labor hours will be available, and company policy prohibits overtime hours from exceeding 10% of straight-time hours. Straight-time labor cost rate is $20 per hour, including benefits. The overtime labor cost rate is 150% (time-and-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 during a later month. However, the amount of storage space can 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 was produced). Currently, no tractors are in storage.

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

9 variables

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

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

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

V1 = # tractors produced in month 1 using overtime

V2 = # tractors produced in month 2 using overtime

V3 = # tractors produced in month 3 using overtime

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

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

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

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

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

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

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

ST

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

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

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

100S1 ≤ 29000 straight-time capacity month 1

100S2 ≤ 29000 straight-time capacity month 2

100S3 ≤ 29000 straight-time capacity month 3

100V1 ≤ 2900 overtime capacity month 1

100V2 ≤ 2900 overtime capacity month 2

100V3 ≤ 2900 overtime capacity month 3

C1 ≤ 40 storage capacity month 1

C2 ≤ 40 storage capacity month 2

C3 ≤ 40 storage capacity month 3

8. MadeRite, a manufacturer of paper stock for copiers and printers, 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 per case, the monthly warehouse requirements, and the monthly mill production levels are:

Monthly Mill

Destination Production

A B C D (cases)

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

Mill 2 4.00 7.10 5.60 3.90 10,000

Mill 3 4.50 5.20 5.50 6.10 15,000

Monthly Warehouse

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

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

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

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

(12 variables)

Min 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

ST

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

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

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

A1 + A2 + A3 = 9000 Destination A demand

B1 + B2 + B3 = 9000 Destination B demand

C1 + C2 + C3 = 12000 Destination C demand

D1 + D2 + D3 = 10000 Destination D demand

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

Project

1 2 3

A 87 62 76

Team B 81 76 64

C 77 54 70

A1 = 1 if team A does project 1

= 0 if not (9 variables)

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

+ 77C1 + 54C2 + 70C3

ST

A1 + A2 + A3 = 1

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

C1 + C2 + C3 = 1

A1 + B1 + C1 = 1

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

A3 + B3 + C3 = 1

Download Presentation

Connecting to Server..