Transportation, Transshipment, and Assignment...

Transportation, Transshipment, and Assignment Problems

Prof. Yongwon Seo

([email protected])

College of Business Administration, CAU

TRANSPORTATION PROBLEM


Transportation Model:Example Problem Definition and Data

• How many tons of wheat to transport from each grain elevator to each mill on a monthly basis in order to minimize the total cost of transportation?

3

Grain Elevator Supply Mill Demand

1. Kansas City 150 A. Chicago 200

2. Omaha 175 B. St. Louis 100

3. Des Moines 275 C. Cincinnati 300

Total 600 tons Total 600 tons

Transport Cost from Grain Elevator to Mill ($/ton)

Grain Elevator A. Chicago B. St. Louis C. Cincinnati

1. Kansas City 2. Omaha 3. Des Moines

$ 6 7 4

$ 8 11

5

$ 10 11 12

Transportation Model: Schematic Diagram

4

1

2

3

A

B

C

6

8

10

7

11

11

4

5

12

150

175

205

200

100

300

Transportation Model: Formulation

5

Minimize Z = $6x1A + 8x1B + 10x1C + 7x2A + 11x2B + 11x2C +

4x3A + 5x3B + 12x3Csubject to:

x1A + x1B + x1C = 150

x2A + x2B + x2C = 175

x3A + x3B + x3C = 275

x1A + x2A + x3A = 200

x1B + x2B + x3B = 100

x1C + x2C + x3C = 300

xij 0

xij = tons of wheat from each grain elevator, i, i = 1, 2, 3,

to each mill j, j = A,B,C

Transportation Table

6

From

To

A. Chicago B. St.Louis C.Cincinnati

Supply

1. Kansas 150

2. Omaha 175

3. DesMoines

275

Demand 200 100 300 600

6 8 10

7 11 11

4 5 12

Excel Solver

7

Objective function

=D5+D6+D7Decision variables in cells

C5:E7

=C7+D7+E7

Cost array in

cells K5:M7

Excel Solver

8

Supply constraints

Demand constraints

Solution

9

• Ex) Harley’s Sand and Gravel Pit supplies topsoil for three residential housing developments from three different “farms.”

Schematic Diagram of a Transportation Problem

10

Unit Transportation Cost

Farms Project

Transportation Table for Harley’s Sand and Gravel

11

LP formulation

12

x11

x21

x31

x12

x22

x32

x13

x23

x33

0 variablesAll

300 :3Project

150 :2Project

50 : 1Project

200 :C Farm

200 :B Farm

100 :A Farm

36795824

332313

322212

312111

333231

232221

131211

333231232221131211

xxx

xxx

xxx

xxx

xxx

xxx

tosubject

xxxxxxxxxZMinimize

LP formulation : General Form

13

jiallforx

mjDx

niSx

tosubject

xcMinimize

ij

n

i

jij

m

j

iij

n

i

m

j

ijij

,,0

,,1,

,,1,

1

1

1 1

Using Excel to solve transportation prob.

14

When the Shipping Route between Farm B and Project 1 Is Prohibited

15

TRANSSHIPMENT PROBLEM


Transshipment Problems

• A transportation problem in which some locations are used as intermediate shipping points, thereby serving both as origins and as destinations.

• Involve the distribution of goods from intermediate nodes in addition to multiple sources and multiple destinations.

17

Transshipment: Example

18

Transshipment: Formulation

19

Minimize Z = $16x13 + 10x14 + 12x15 + 15x23 + 14x24+ 17x25 + 6x36 + 8x37 + 10x38 + 7x46 + 11x47+ 11x48 + 4x56 + 5x57 + 12x58

subject to:

x13 + x14 + x15 = 300

x23 + x24 + x25 = 300

x36 + x46 + x56 = 200

x37 + x47 + x57 = 100

x38 + x48 + x58 = 300

x13 + x23 - x36 - x37 - x38 = 0

x14 + x24 - x46 - x47 - x48 = 0

x15 + x25 - x56 - x57 - x58 = 0

xij 0

Transshipment: Excel Solver

20

Objective function

=SUM(B6:D6)

=SUM(B6:B7)

=SUM(C13:C15) =SUM(C13:E13)

Cost arrays

Constraints for transshipment flows; i.e.,

shipments in = shipments out

Transshipment: Excel Solver

21

Transshipment

constraints in cells

C20:C22

Transshipment: Solution

22

Ex.

23

The manager of Harley’s Sand and Gravel Pit has decided to utilize two intermediate nodes as transshipment points for temporary storage of topsoil.

Farm A

Farm B

Farm C

Formulation

24

ji, allfor ,0

300 :8 Node

150 :7 Node

50 :6 Node

:5 Node

:4 Node

200 :3 Node

200 :2 Node

100 :1 Node

tosubject

5423

5848

5747

5646

585756352515

484746342414

3534

2524

1514

58481514

ijx

xx

xx

xx

xxxxxx

xxxxxx

xx

xx

xx

xxxxZMinimize

ASSIGNMENT PROBLEM


Assignment Problems

• Involve the matching or pairing of two sets of items such as jobs and machines, secretaries and reports, lawyers and cases, and so forth.

• Have different cost or time requirements for different pairings.

• Special form of linear programming model similar to the transportation model.– Supply at each source and demand at each destination limited

to one unit.

27

Assignment Model

• Problem: Assign four teams of officials to four games in a way that will minimize total distance traveled by the officials. Supply is always one team of officials, demand is for only one team of officials at each game.

28

Formulation

29

Minimize Z = 210xAR + 90xAA + 180xAD + 160xAC + 100xBR+70xBA+ 130xBD + 200xBC + 175xCR + 105xCA +140xCD+ 170xCC + 80xDR + 65xDA + 105xDD + 120xDC

subject to:

xAR + xAA + xAD + xAC = 1 xij 0

xBR + xBA + xBD + xBC = 1

xCR + xCA + xCD + xCC = 1

xDR + xDA + xDD + xDC = 1

xAR + xBR + xCR + xDR = 1

xAA + xBA + xCA + xDA = 1

xAD + xBD + xCD + xDD = 1

xAC + xBC + xCC + xDC = 1

Excel Solver

30

Objective function

Decision

variables,

C5:F8

=C5+D5+E5+F5

=D5+D6+D7+D8

Mileage array

Excel Solver

31

Simplex LP

Solution

32

Assignment Problem: Example

• A manager has prepared a table that shows the cost of performing each of five jobs by each of five employees (see Table 6-8). According to this table, job I will cost $15 if done by Al. $20 if it is done by Bill, and so on. The manager has stated that his goal is to develop a set of job assignments that will minimize the total cost of getting all four jobs done. It is further required that the jobs be performed simultaneously, thus requiring one job being assigned to each employee.

• In the past, to find the minimum-cost set of assignments, the manager has resorted to listing all of the different possible assignments (i.e., complete enumeration) for small problems such as this one. But for larger problems, the manager simply guesses because there are too many possibilities to try to list them. For example, with a 5X5 table, there are 5! = 120 different possibilities; but with, say, a 7X7 table, there are 7! = 5,040 possibilities.

33

Problem

34

jiallforx

xxxxx

xxxxx

xxxxx

xxxxx

xxxxx

xxxxx

toSubject

xxxxZMinimize

ij ,,0

1

1

1

1

1

1

16182015

5545352515

5242322212

5141312111

5554535251

2524232221

1514131211

55541211

Integer?

Multi-Criteria Decision-Making Models

Prof. Yong Won Seo

([email protected])

College of Business Administration, CAU

Multi-Criteria Decision-Making

■ Study of problems with several criteria, i.e., multiple criteria, instead of a single objective when making a decision.

■ Three techniques discussed: goal programming, the analytical hierarchy process and scoring models.

■ Goal programming is a variation of linear programming considering more than one objective (goals) in the objective function.

■ The analytical hierarchy process develops a score for each decision alternative based on comparisons of each under different criteria reflecting the decision makers’ preferences.

■ Scoring models are based on a relatively simple weighted scoring technique.

37

Applications of MCDA

• Some of the MCDM methods are:

– Analytic hierarchy process (AHP)

– Goal programming (GP)

– Data Envelopment Analysis (DEA)

– Inner Product of Vectors (IPV)

– Multi-attribute value theory (MAVT)

– Multi-attribute utility theory (MAUT)

– Multi-attribute global inference of quality (MAGIQ)

– ELECTRE (Outranking)

– PROMETHÉE (Outranking)

– The evidential reasoning approach

– Dominance-based Rough Set Approach (DRSA)

– Aggregated indices randomization method (AIRM)

– Nonstructural fuzzy decision support system (NSFDSS)

– Grey relational analysis (GRA)

– Superiority and inferiority ranking method (SIR method)… (source:: Wikipedia)

38

• Goal Programming (GP)

– A variation of linear programming that allows multiple objectives (goals)—soft (goal) constraints or a combination of soft and hard (nongoal) constraints

– There are priorities in the goals. (prioritized goal programming model)

– In order to obtain an acceptable solution when there are conflicts, it becomes necessary to make trade-offs: satisfying hard constraints and achieving higher levels of certain goals with sacrificing other goals.

Goal Programming

Standard LP Example

• Beaver Creek Pottery Company Example

Maximize Z = 40x1 + 50x2

subject to:

1x1 + 2x2 40 hours of labor (per day)

4x1 + 3x2 120 pounds of clay (per day)

x1, x2 0

Where: x1 = number of bowls produced

x2 = number of mugs produced

40

Modified Problem

• Labor : overtime allowed (but not desirable)

• Storage space can be added (but not desirable)

• The company has the following objectives, listed in order of importance:

1. To avoid layoffs, the company does not want to use fewer than 40 hours of labor per day

2. The company would like to achieve a satisfactory profit level of $1600 per day

3. Because the clay must be stored in a special place so that it does not dry out, the company prefers not to prepare more than 120 pounds on hand each day

4. Because of high overtime cost, the company would like to minimize the amount of overtime.

41

Goal Constraints

Labor goal:

x1 + 2x2 + d1- - d1

+ = 40 (hours/day)

Profit goal:

40x1 + 50 x2 + d2- - d2

+ = 1,600 ($/day)

Material goal:

4x1 + 3x2 + d3- - d3

+ = 120 (lbs of clay/day)

42

Objective Function

• Minimize P1d1-, P2d2

-, P3d3+, P4d1

+

• Add one by one based on priorities

1. Min labor constraint (priority 1 - less than 40 hours labor) Minimize P1d1

-

2. Add profit goal constraint (priority 2 - achieve profit of $1,600): Minimize P1d1

-, P2d2-

3. Add material goal constraint (priority 3 - avoid keeping more than 120 pounds of clay on hand) Minimize P1d1

-, P2d2-, P3d3

+

4. Add overtime constraint (priority 4 - minimum overtime): Minimize P1d1

-, P2d2-, P3d3

+, P4d1+

43

Graphical Interpretation

44

Minimize P1d1-, P2d2

-, P3d3+, P4d1

+

subject to:

x1 + 2x2 + d1- - d1

+ = 40

40x1 + 50 x2 + d2 - - d2

+ = 1,600

4x1 + 3x2 + d3 - - d3

+ = 120

x1, x2, d1 -, d1

+, d2 -, d2

+, d3 -, d3

+ 0


45


-, P3d3+, P4d1

+

subject to:

x1 + 2x2 + d1- - d1

+ = 40

40x1 + 50 x2 + d2 - - d2

+ = 1,600

4x1 + 3x2 + d3 - - d3

+ = 120

x1, x2, d1 -, d1

+, d2 -, d2

+, d3 -, d3

+ 0


46


-, P3d3+, P4d1

+

subject to:

x1 + 2x2 + d1- - d1

+ = 40

40x1 + 50 x2 + d2 - - d2

+ = 1,600

4x1 + 3x2 + d3 - - d3

+ = 120

x1, x2, d1 -, d1

+, d2 -, d2

+, d3 -, d3

+ 0


47


-, P3d3+, P4d1

+

subject to:

x1 + 2x2 + d1- - d1

+ = 40

40x1 + 50 x2 + d2 - - d2

+ = 1,600

4x1 + 3x2 + d3 - - d3

+ = 120

x1, x2, d1 -, d1

+, d2 -, d2

+, d3 -, d3

+ 0


48


-, P3d3+, P4d1

+

subject to:

x1 + 2x2 + d1- - d1

+ = 40

40x1 + 50 x2 + d2 - - d2

+ = 1,600

4x1 + 3x2 + d3 - - d3

+ = 120

x1, x2, d1 -, d1

+, d2 -, d2

+, d3 -, d3

+ 0

Transportation, Transshipment, and Assignment...

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