by bernard w. taylor iiicgungor/modellemeyegiris/acrobats/...modellemeye giriş-yrd.doç.dr.ceyda...

Modellemeye Giriş-Yrd.Doç.Dr.Ceyda ŞEN 1

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ÇALIŞMA SORULARI - Çözümlü

Breakfast Food Cal

Fat (g)

Cholesterol (mg)

Iron (mg)

Calcium (mg)

Protein (g)

Fiber (g)

Cost ($)

1. Bran cereal (cup) 2. Dry cereal (cup) 3. Oatmeal (cup) 4. Oat bran (cup) 5. Egg 6. Bacon (slice) 7. Orange 8. Milk-2% (cup) 9. Orange juice (cup)

10. Wheat toast (slice)

90 110 100 90 75 35 65

100 120 65

0 2 2 2 5 3 0 4 0 1

0 0 0 0

270 8 0 12 0 0

6 4 2 3 1 0 1 0 0 1

20 48 12 8

30 0

52 250

3 26

3 4 5 6 7 2 1 9 1 3

5 2 3 4 0 0 1 0 0 3

0.18 0.22 0.10 0.12 0.10 0.09 0.40 0.16 0.50 0.07

A Diet Example

Data and Problem Definition (1 of 5)

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Introduction to Management Science 8th Edition

By Bernard W. Taylor III

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.


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

A Diet Example

Model Construction – Decision Variables (2 of 5)

3



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

A Diet Example

Model Summary (3 of 5)

4




A Diet Example

Computer Solution with Excel (4 of 5)

5Introduction to Management Science 8th Edition


A Diet Example

Solution with Excel Solver Window (5 of 5)

6



diet.xls


Make vs. Buy Decision Example

The Electro-Poly Corporation (1 of 4)

Electro-Poly is a leading maker of slip-rings.

A $750,000 order has just been received.

The company has 10,000 hours of wiring capacity and 5,000

hours of harnessing capacity.

Model 1 Model 2 Model 3

Number ordered 3,000 2,000 900

Hours of wiring/unit 2 1.5 3

Hours of harnessing/unit 1 2 1

Cost to Make $50 $83 $130

Cost to Buy $61 $97 $145

“Spreadsheet Modeling and Decision Analysis: A Practical

Introduction to Management Science” by Cliff T. Ragsdale7


Defining the Decision Variables (2 of 4)

M1 = Number of model 1 slip rings to make in-house



B1 = Number of model 1 slip rings to buy from competitor







Defining the Objective Function (3 of 4)

Minimize the total cost of filling the order.

MIN: 50M1+ 83M2+ 130M3+ 61B1+ 97B2+ 145B3




Defining the Constraints (4 of 4)

Demand Constraints

M1 + B1 = 3,000 } model 1

M2 + B2 = 2,000 } model 2

M3 + B3 = 900 } model 3

Resource Constraints

2M1 + 1.5M2 + 3M3 <= 10,000 } wiring

1M1 + 2.0M2 + 1M3 <= 5,000 } harnessing

Non-negativity Conditions

M1, M2, M3, B1, B2, B3 >= 0

Implementing the Model



make-or-buy.xls

make-or-buy.xls


A Transportation Problem-1: Tropicsun (1 of 4)

Mt. Dora

1

Eustis

2

Clermont

3

Ocala

4

Orlando

5

Leesburg

6

Distances (in miles)CapacitySupply

275,000

400,000

300,000 225,000

600,000

200,000

GrovesProcessing

Plants

21

50

40

35

30

22

55

25

20



A Transportation Problem


Xij = # of bushels shipped from node i to node j

Specifically, the nine decision variables are:

X14 = # of bushels shipped from Mt. Dora (node 1) to Ocala (node 4)

X15 = # of bushels shipped from Mt. Dora (node 1) to Orlando (node 5)

X16 = # of bushels shipped from Mt. Dora (node 1) to Leesburg (node 6)

X24 = # of bushels shipped from Eustis (node 2) to Ocala (node 4)

X25 = # of bushels shipped from Eustis (node 2) to Orlando (node 5)

X26 = # of bushels shipped from Eustis (node 2) to Leesburg (node 6)

X34 = # of bushels shipped from Clermont (node 3) to Ocala (node 4)

X35 = # of bushels shipped from Clermont (node 3) to Orlando (node 5)

X36 = # of bushels shipped from Clermont (node 3) to Leesburg (node 6)“Spreadsheet Modeling and Decision Analysis: A Practical





Minimize the total number of bushel-miles.

MIN: 21X14 + 50X15 + 40X16 +

35X24 + 30X25 + 22X26 +

55X34 + 20X35 + 25X36




Defining the Constraints (4 of 4)

Capacity constraints

X14 + X24 + X34 <= 200,000 } Ocala

X15 + X25 + X35 <= 600,000 } Orlando

X16 + X26 + X36 <= 225,000 } Leesburg

Supply constraints

X14 + X15 + X16 = 275,000 } Mt. Dora

X24 + X25 + X26 = 400,000 } Eustis

X34 + X35 + X36 = 300,000 } Clermont

Non-negativity conditionsXij >= 0 for all i and j




transportation.xls


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

Shipping Costs

From Warehouse To Store

A B C

1 $16 $18 $11

2 14 12 13

3 13 15 17

A Transportation Problem-2

Problem Definition and Data (1 of 3)

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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

A Transportation Example


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Solution with Excel (3 of 4)

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Solution with Solver Window (4 of 4)

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transportation-2.xls


A Production Planning Problem:

The Upton Corporation (1 of 6)

Upton is planning the production of their heavy-duty air

compressors for the next 6 months.

• Beginning inventory = 2,750 units

• Safety stock = 1,500 units

• Unit carrying cost = 1.5% of unit production cost

• Maximum warehouse capacity = 6,000 units

1 2 3 4 5 6

Unit Production Cost $240 $250 $265 $285 $280 $260

Units Demanded 1,000 4,500 6,000 5,500 3,500 4,000

Maximum Production 4,000 3,500 4,000 4,500 4,000 3,500

Minimum Production 2,000 1,750 2,000 2,250 2,000 1,750

Month

19“Spreadsheet Modeling and Decision Analysis: A Practical

Introduction to Management Science” by Cliff T. Ragsdale

Pi = number of units to produce in month i, i=1 to 6

Bi = beginning inventory month i, i=1 to 6

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A Production Planning Problem





Minimize the total cost production

& inventory costs.

MIN: 240P1+250P2+265P3+285P4+280P5+260P6

+ 3.6(B1+B2)/2 + 3.75(B2+B3)/2 + 3.98(B3+B4)/2

+ 4.28(B4+B5)/2 + 4.20(B5+ B6)/2 + 3.9(B6+B7)/2

Note: The beginning inventory in any month is the same as the ending inventory in the previous month.

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Production levels

2,000 <= P1 <= 4,000 } month 1

1,750 <= P2 <= 3,500 } month 2

2,000 <= P3 <= 4,000 } month 3

2,250 <= P4 <= 4,500 } month 4

2,000 <= P5 <= 4,000 } month 5

1,750 <= P6 <= 3,500 } month 6

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Defining the Constraints-I (4 of 6)




Ending Inventory (EI = BI + P - D)

1,500 < B1 + P1 - 1,000 < 6,000 } month 1

1,500 < B2 + P2 - 4,500 < 6,000 } month 2

1,500 < B3 + P3 - 6,000 < 6,000 } month 3

1,500 < B4 + P4 - 5,500 < 6,000 } month 4

1,500 < B5 + P5 - 3,500 < 6,000 } month 5

1,500 < B6 + P6 - 4,000 < 6,000 } month 6

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Defining the Constraints-II (5 of 6)



Beginning Balances

B1 = 2750

B2 = B1 + P1 - 1,000

B3 = B2 + P2 - 4,500

B4 = B3 + P3 - 6,000

B5 = B4 + P4 - 5,500

B6 = B5 + P5 - 3,500

B7 = B6 + P6 - 4,000

Notice that the Bi

can be computed

directly from the

Pi. Therefore,

only the Pi need to

be identified as

changing cells.

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Defining the Constraints-III (6 of 6)




productionplanning.xls


Component Maximum Barrels

Available/day Cost/barrel

1 4,500 $12

2 2,700 10

3 3,500 14

Grade Component Specifications Selling Price ($/bbl)

Super At least 50% of 1

Not more than 30% of 2 $23

Premium At least 40% of 1

Not more than 25% of 3

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Extra At least 60% of 1 At least 10% of 2

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A Blend Example

Problem Definition and Data (1 of 5)

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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 gasoline (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).

A Blend Example

Problem Statement and Variables (2 of 5)

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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

0.50x1s - 0.50x2s - 0.50x3s 0

0.70x2s - 0.30x1s - 0.30x3s 0

0.60x1p - 0.40x2p - 0.40x3p 0

0.75x3p - 0.25x1p - 0.25x2p 0

0.40x1e- 0.60x2e- - 0.60x3e 0

0.90x2e - 0.10x1e - 0.10x3e 0

x1s + x2s + x3s 3,000

x1p+ x2p + x3p 3,000

x1e+ x2e + x3e 3,000

xij 0

A Blend Example


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A Blend Example

Solution with Excel (4 of 5)

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blendexample.xls


A Blend Example

Solution with Solver Window (5 of 5)

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Example Problem Solution

Problem Statement and Data (1 of 4)

Canned cat food, Meow Chow; dog food, Bow Chow.

Ingredients/week: 600 pound horse meat; 800 pound fish;

1000 lb cereal.

Recipe requirement: Meow Chow at least half fish; Bow

Chow at least half 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?

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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)


Model Formulation (2 of 4)

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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


Model Formulation (3 of 4)

32




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



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by bernard w. taylor iiicgungor/modellemeyegiris/acrobats/...modellemeye giriş-yrd.doç.dr.ceyda...

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