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Chapter 8 Linear Programming: Modeling Examples

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Introduction to Management Science

8th Edition

by

Bernard W. Taylor III

Chapter 8

Linear Programming: Modeling

Examples

Chapter 8 - Linear Programming: Modeling Examples

Chapter Topics

- A Product Mix Example
- A Diet Example
- An Investment Example
- A Marketing Example
- A Transportation Example
- A Blend Example
- A Multi-Period Scheduling Example
- A Data Envelopment Analysis Example

Chapter 8 - Linear Programming: Modeling Examples

Problem Definition (1 of 7)

- Four-product T-shirt/sweatshirt manufacturing company.
- Must complete production within 72 hours
- Truck capacity = 1,200 standard sized boxes.
- Standard size box holds12 T-shirts.
- One-dozen sweatshirts box is three times size of standard box.
- $25,000 available for a production run.
- 500 dozen blank T-shirts and sweatshirts in stock.
- How many dozens (boxes) of each type of shirt to produce?

Chapter 8 - Linear Programming: Modeling Examples

Model Construction (3 of 7)

Decision Variables:

x1 = sweatshirts, front printing

x2 = sweatshirts, back and front printing

x3 = T-shirts, front printing

x4 = T-shirts, back and front printing

Objective Function:

Maximize Z = $90x1 + $125x2 + $45x3 + $65x4

Model Constraints:

0.10x1 + 0.25x2+ 0.08x3 + 0.21x4 72 hr

3x1 + 3x2 + x3 + x4 1,200 boxes

$36x1 + $48x2 + $25x3 + $35x4 $25,000

x1 + x2 500 doz. sweatshirts

x3 + x4 500 doz. T-shirts

Chapter 8 - Linear Programming: Modeling Examples

Computer Solution with Excel (4 of 7)

Exhibit 8.1

Chapter 8 - Linear Programming: Modeling Examples

Solution with Excel Solver Window (5 of 7)

Exhibit 8.2

Chapter 8 - Linear Programming: Modeling Examples

Solution with QM for Windows (6 of 7)

Exhibit 8.3

Chapter 8 - Linear Programming: Modeling Examples

Solution with QM for Windows (7 of 7)

Exhibit 8.4

Chapter 8 - Linear Programming: Modeling Examples

Data and Problem Definition (1 of 5)

Breakfast to include at least 420 calories, 5 milligrams of iron, 400 milligrams of calcium, 20 grams of protein, 12 grams of fiber, and must have no more than 20 grams of fat and 30 milligrams of cholesterol.

Chapter 8 - Linear Programming: Modeling Examples

Model Construction – Decision Variables (2 of 5)

x1 = cups of bran cereal

x2 = cups of dry cereal

x3 = cups of oatmeal

x4 = cups of oat bran

x5 = eggs

x6 = slices of bacon

x7 = oranges

x8 = cups of milk

x9 = cups of orange juice

x10 = slices of wheat toast

Chapter 8 - Linear Programming: Modeling Examples

Model Summary (3 of 5)

Minimize Z = 0.18x1 + 0.22x2 + 0.10x3 + 0.12x4 + 0.10x5 + 0.09x6 + 0.40x7 + 0.16x8 + 0.50x9 + 0.07x10

subject to:

90x1 + 110x2 + 100x3 + 90x4 + 75x5 + 35x6 + 65x7 + 100x8 + 120x9 + 65x10 420

2x2 + 2x3 + 2x4 + 5x5 + 3x6 + 4x8 + x10 20

270x5 + 8x6 + 12x8 30

6x1 + 4x2 + 2x3 + 3x4+ x5 + x7 + x10 5

20x1 + 48x2 + 12x3 + 8x4+ 30x5 + 52x7 + 250x8 + 3x9 + 26x10 400

3x1 + 4x2 + 5x3 + 6x4 + 7x5 + 2x6 + x7+ 9x8+ x9 + 3x10 20

5x1 + 2x2 + 3x3 + 4x4+ x7 + 3x10 12 xi 0, i=1,…,10.

Chapter 8 - Linear Programming: Modeling Examples

Computer Solution with Excel (4 of 5)

Exhibit 8.5

Chapter 8 - Linear Programming: Modeling Examples

Solution with Excel Solver Window (5 of 5)

Exhibit 8.6

Chapter 8 - Linear Programming: Modeling Examples

Model Summary (1 of 4)

Maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4

subject to:

x1 14,000

– x1 + x2 – x3 – x4 0

x2 + x3 21,000

–1.2x1 + x2 + x3 – 1.2 x4 0

x1 + x2 + x3 + x4 = 70,000

x1, x2, x3, x4 0

where

x1 = amount invested in municipal bonds ($)

x2 = amount invested in certificates of deposit ($)

x3 = amount invested in treasury bills ($)

x4 = amount invested in growth stock fund($)

Chapter 8 - Linear Programming: Modeling Examples

Computer Solution with Excel (2 of 4)

Exhibit 8.7

Chapter 8 - Linear Programming: Modeling Examples

Solution with Excel Solver Window (3 of 4)

Exhibit 8.8

Chapter 8 - Linear Programming: Modeling Examples

Sensitivity Report (4 of 4)

Exhibit 8.9

Chapter 8 - Linear Programming: Modeling Examples

Data and Problem Definition (1 of 6)

- Budget limit $100,000
- Television time for four commercials
- Radio time for 10 commercials
- Newspaper space for 7 ads
- Resources for no more than 15 commercials and/or ads

Chapter 8 - Linear Programming: Modeling Examples

Model Summary (2 of 6)

Maximize Z = 20,000x1 + 12,000x2 + 9,000x3

subject to:

15,000x1 + 6,000x 2+ 4,000x3 100,000

x1 4

x2 10

x3 7

x1 + x2 + x3 15

x1, x2, x3 0

where

x1 = Exposure from Television Commercial (people)

x2 = Exposure from Radio Commercial (people)

x3 = Exposure from Newspaper Ad (people)

Chapter 8 - Linear Programming: Modeling Examples

Solution with Excel (3 of 6)

Exhibit 8.10

Chapter 8 - Linear Programming: Modeling Examples

Solution with Excel Solver Window (4 of 6)

Exhibit 8.11

Chapter 8 - Linear Programming: Modeling Examples

Integer Solution with Excel (5 of 6)

Exhibit 8.12

Exhibit 8.13

Chapter 8 - Linear Programming: Modeling Examples

IntegerSolution with Excel (6 of 6)

Exhibit 8.14

Chapter 8 - Linear Programming: Modeling Examples

Problem Definition and Data (1 of 3)

Warehouse supply of Retail store demand Television Sets: for television sets:

1 - Cincinnati 300 A - New York 150

2 - Atlanta 200 B - Dallas 250

3 - Pittsburgh 200 C - Detroit 200

Total 700 Total 600

Chapter 8 - Linear Programming: Modeling Examples

Model Summary (2 of 4)

Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C

subject to:

x1A + x1B+ x1C 300

x2A+ x2B + x2C 200

x3A+ x3B + x3C 200

x1A + x2A + x3A = 150

x1B + x2B + x3B = 250

x1C + x2C + x3C= 200

xij 0, i=1,2,3 and j= A,B,C.

Chapter 8 - Linear Programming: Modeling Examples

Solution with Excel (3 of 4)

Exhibit 8.15

Chapter 8 - Linear Programming: Modeling Examples

Solution with Solver Window (4 of 4)

Exhibit 8.16

Chapter 8 - Linear Programming: Modeling Examples

Problem Definition and Data (1 of 6)

Chapter 8 - Linear Programming: Modeling Examples

Problem Statement and Variables (2 of 6)

- Determine the optimal mix of the three components in each grade of motor oil that will maximize profit. Company wants to produce at least 3,000 barrels of each grade of motor oil.
- Decision variables: The quantity of each of the three components used in each grade of motor oil (9 decision variables); xij = barrels of component i used in motor oil grade j per day, where i = 1, 2, 3 and j = s (super), p (premium), and e (extra).

Chapter 8 - Linear Programming: Modeling Examples

component availability

blend specification; e.g.:

x1s/(x1s+x2s+x3s) ≥ 0.50

x1s ≥ 0.50·(x1s+x2s+x3s)

0.50·(x1s–x2s–x3s) ≥ 0

A Blend Example

Model Summary (3 of 6)

Maximize Z = 11x1s + 13x2s + 9x3s + 8x1p + 10x2p + 6x3p + 6x1e + 8x2e + 4x3e

subject to:

x1s + x1p + x1e 4,500

x2s + x2p + x2e 2,700

x3s + x3p + x3e 3,500

0.50·(x1s – x2s – x3s) 0

0.70x2s – 0.30·(x1s + x3s) 0

0.60x1p – 0.40·(x2p + x3p) 0

0.25·(3·x3p – x1p – x2p) 0

0.20·(2·x1e – 3·x2e- – 3·x3e) 0

0.10·(9·x2e – x1e – x3e) 0

x1s + x2s + x3s 3,000

x1p+ x2p + x3p 3,000

x1e+ x2e + x3e 3,000

xij 0, i=1,2,3, and j=s,p,e.

Chapter 8 - Linear Programming: Modeling Examples

Solution with Excel (4 of 6)

Exhibit 8.17

Chapter 8 - Linear Programming: Modeling Examples

Solution with Solver Window (5 of 6)

Exhibit 8.18

Chapter 8 - Linear Programming: Modeling Examples

Sensitivity Report (6 of 6)

Exhibit 8.19

Chapter 8 - Linear Programming: Modeling Examples

A Multi-Period Scheduling Example

Problem Definition and Data (1 of 5)

Production Capacity: 160 computers per week

50 more computers with overtime

Assembly Costs: $190 per computer regular time; $260 per computer overtime

Inventory Cost: $10/comp. per week

Order schedule: Week (j) Computer Orders (dj)

1 105

2 170

3 230

4 180

5 150

6 250

Chapter 8 - Linear Programming: Modeling Examples

A Multi-Period Scheduling Example

Decision Variables (2 of 5)

Decision Variables:

rj = regular production of computers per week j

(j = 1,…,6) -- decision variable

oj = overtime production of computers per week j

(j = 1,…,6) -- decision variable

ij = extra computers carried over as inventory in week j

(j = 1,…,5) — computed as ij = rj + oj + ij–1 – dj, where

dj is the demand (orders) as tabulated on p.35.

Note: the ij variables are being calculated, so that they are not independent from the rj and oj decision variables: the ij’s are functions of rj and oj.

Chapter 8 - Linear Programming: Modeling Examples

Individual constraints

Nontrivial for the ij!

A Multi-Period Scheduling Example

Model Summary (3 of 5)

Model summary:

Minimize Z = $190(r1 + r2 + r3 + r4 + r5 + r6) + $260(o1 + o2 + o3 + o4 + o5 +o6) + $10(i1 + i2 + i3 + i4 + i5)

subject to:

rj 160 (j = 1, …, 6)

oj 150 (j = 1, …, 6)

r1 + o1 - i1 105

r2 + o2 + i1 - i2 170

r3 + o3 + i2 - i3 230

r4 + o4 + i3 - i4 180

r5 + o5 + i4 - i5 150

r6 + o6 + i5 250

rj, oj, ij 0

Chapter 8 - Linear Programming: Modeling Examples

A Multi-Period Scheduling Example

Solution with Excel (4 of 5)

Exhibit 8.20

Chapter 8 - Linear Programming: Modeling Examples

A Multi-Period Scheduling Example

Solution with Solver Window (5 of 5)

Exhibit 8.21

Chapter 8 - Linear Programming: Modeling Examples

A Data Envelopment Analysis (DEA) Example

Problem Definition (1 of 5)

DEA compares a number of service units of the same type based on their inputs (resources) and outputs. The result indicates if a particular unit is less productive, or efficient, than other units.

Elementary school comparison:

Input 1 = teacher to student ratio

Output 1 = average reading SOL score

Input 2 = supplementary $/student

Output 2 = average math SOL score

Input 3 = parent education level

Output 3 = average history SOL score

Chapter 8 - Linear Programming: Modeling Examples

A Data Envelopment Analysis (DEA) Example

Problem Data Summary (2 of 5)

Chapter 8 - Linear Programming: Modeling Examples

A Data Envelopment Analysis (DEA) Example

Decision Variables and Model Summary (3 of 5)

Decision Variables:

xi = a price per unit of each output where i = 1, 2, 3

yi = a price per unit of each input where i = 1, 2, 3

Model Summary:

Maximize Z = 81·x1 + 73·x2 + 69·x3

subject to:

.06y1 + 460y2 + 13.1y3 = 1

86x1 + 75x2 + 71x3.06y1 + 260y2 + 11.3y3

82x1 + 72x2 + 67x3 .05y1 + 320y2 + 10.5y3

81x1 + 79x2 + 80x3 .08y1 + 340y2 + 12.0y3

81x1 + 73x2 + 69x3 .06y1 + 460y2 + 13.1y3

xi, yi 0, i=1,2,3.

Chapter 8 - Linear Programming: Modeling Examples

A Data Envelopment Analysis (DEA) Example

Solution with Excel (4 of 5)

Exhibit 8.22

Chapter 8 - Linear Programming: Modeling Examples

A Data Envelopment Analysis (DEA) Example

Solution with Solver Window (5 of 5)

Exhibit 8.23

Chapter 8 - Linear Programming: Modeling Examples

Problem Statement and Data (1 of 5)

- Canned cat food, Meow Chow; dog food, Bow Chow.
- Ingredients/week: 600 lb horse meat; 800 lb fish; 1000 lb cereal.
- Recipe requirement: Meow Chow at least half (1/2) fish;Bow Chow at least half (1/2) horse meat.
- 2,250 sixteen-ounce cans available each week.
- Profit /can: Meow Chow $0.80; Bow Chow $0.96.
- How many cans of Bow Chow and Meow Chow should be produced each week in order to maximize profit?

Chapter 8 - Linear Programming: Modeling Examples

Model Formulation (2 of 5)

Step 1: Define the Decision Variables

xij = ounces of ingredient i in pet food j per week, where i = h (horse meat), f (fish) and c (cereal), and j = m (Meow chow) and b (Bow Chow).

Step 2: Formulate the Objective Function

Maximize Z = $0.05(xhm + xfm + xcm) + 0.06(xhb + xfb + xcb)

Chapter 8 - Linear Programming: Modeling Examples

Model Formulation (3 of 5)

Step 3: Formulate the Model Constraints

Amount of each ingredient available each week:

xhm + xhb 9,600 ounces of horse meat

xfm + xfb 12,800 ounces of fish

xcm + xcb 16,000 ounces of cereal additive

Recipe requirements:

Meow Chow

xfm/(xhm + xfm + xcm) 1/2 or - xhm + xfm- xcm 0

Bow Chow

xhb/(xhb + xfb + xcb) 1/2 or xhb- xfb - xcb 0

Can Content Constraint

xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces

Chapter 8 - Linear Programming: Modeling Examples

Model Summary (4 of 5)

Step 4: Model Summary

Maximize Z = $0.05xhm + $0.05xfm + $0.05xcm + $0.06xhb + 0.06xfb + 0.06xcb

subject to:

xhm + xhb 9,600 ounces of horse meat

xfm + xfb 12,800 ounces of fish

xcm + xcb 16,000 ounces of cereal additive

- xhm + xfm- xcm 0

xhb- xfb - xcb 0

xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces xij 0, i=h,f,m, and j=m,b.

Chapter 8 - Linear Programming: Modeling Examples

Solution with QM for Windows (5 of 5)

Exhibit 8.24

Chapter 8 - Linear Programming: Modeling Examples

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