Linear Programming

Transportation Model

Transportation Model:General Ideas

A model deals with the determination of aminimum-cost plan for transporting asingle commodity from a number ofsources (e.g., factories) to a number ofdestinations (e.g., warehouses)

Other applications of the model: Inventory Control

Employment Scheduling

Personnel Assignment

Transportation Model:General Ideas

Transportation model: linear program thatcan be solved by regular simplex method

Definition and Applications

Transportation Model includes:Single Commodity

Number of sources

Number of destinations

Level of supply at each source

Amount of demand of each destination

Unit transportation cost of commodity from anumber of sources to a number ofdestinations

Sources

1a1

2a2

mam

.

.

.

1 b1

Destinations

2 b2

n bn

.

.

.

Unitsofsupply

Units ofdemands

c11: x11

cmn: xmn

ai : the amount of supply at source-i

bj : the amount of demand at destination-j

cij : unit transportation cost between

source-i and destination-j

xij : amount transported from source-i to

destination-j

LP Model for TransportationProblem

Objective Function:

Minimize

Subject to:

m

i

n

jijij xcz

1 1

i

n

jij ax

1

j

m

jij bx

1

i = 1,2,3,…, m

j = 1,2,3,…, n

0ijx for all i and j

Balanced Tansportation Model

i

n

jij ax

1

j

m

jij bx

1

i = 1,2,3,…, m

j = 1,2,3,…, n

Example

Three Plants(Factories): LA, DT,NO

Two Warehouses:DV, MI

Table of cij

68102(3)NO

108100(2)DT

21580(1)LA

(2)(1)

MIDV

Example

Three Plants (Factories): LA, DT, NO

Two Warehouses: DV, MI

Table of cij

68102(3)NO

108100(2)DT

21580(1)LA

(2)(1)

MIDV

Example

Sources and Supply

Destinations and Demand

SupplySource

1200NO

1500DT

1000LA

14002300

Destination

Demand

MIDV

370014002300Demand

1200NO (3)

1500DT (2)

1000LA (1)

SupplyMI (2)DV (1)

x11

x21

x12

x22

x31 x32

80 215

100 108

102 68

Sources

Destinations

Balanced Transportation Model

14002300

1200NO (3)

1300DT (2)

1000LA (1)

MI (2)DV (1)

80 215

100 108

102 68

0 0DummyPlant

200

Unbalanced Transportation Model

40014001900

1200NO (3)

1500DT (2)

1000LA (1)

MI (2)DV (1)

80 215

100 108

102 68

0

0

Dummy

DistributionCentre

0

Unbalanced Transportation Model

Transportation – Production

5. Production and Inventory

cost from period-i to period-j

5. Transportation cost from

source-i to destination-j

4. Demand per period-j4. Demand at destination-j

3. Production capacity of

period-i

3. Supply at source-i

2. Demand period-j2. Destination-j

1. Production period-i1. Source-i

Production SystemTransportation System

The Transportation Technique

Step 1: Determine a starting feasiblesolution

Step 2: Determine an entering variablefrom the non-basic variable

If all such variables satisfy the optimalitycondition (of simplex method): STOP;

Else, GOTO Step 3

The Transportation Technique

Step 3: Determine a leaving variable(using the feasibility condition) fromamong the variables of the current basicsolution;

Find the new basic solution

RETURN TO Step 2

Determination of the StartingSolution

Northwest-corner (NWC) Rule

Least Cost Method

Vogel’s Approximation Method (VAM)

Northwest-corner

Allocate the maximum amount allowable by thesupply and demand to the variable x11 (at thenorthwest corner of the tableau)

The satisfied column (row) is then crossed out

Adjust the amounts of supply and demand for alluncrossed-out rows and columns

Do iteration until exactly one row or one columnremains uncrossed-out

Northwest-corner

Allocate the maximum amount allowable by thesupply and demand to the variable x11 (at thenorthwest corner of the tableau)

The satisfied column (row) is then crossed out

Adjust the amounts of supply and demand for alluncrossed-out rows and columns

Do iteration until exactly one row or one columnremains uncrossed-out

NWC: Example

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4321source Supply

Destination

10 0 20 11

x12x11 x13 x14

x21

x31

x24x23x22

x32 x33 x34

12

0 14

7 9

16

20

18

NWC: Example

1015155Demand

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252

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

NWC: Example

1015155Demand

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252

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

NWC: Example

1015155Demand

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5

NWC: Example

1015155Demand

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5 15

NWC: Example

1015155Demand

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5 15 5

NWC: Example

1015155Demand

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5 15 5

5

1015155Demand

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5 15 5

5

Starting Solution: 5 10 + 10 0 + 5 7 +15 9 + 5 20 + 5 18 = 410

Basic Variables: x11, x12, x22, x23, x24, x34

Non-basic Variables: x13, x14, x21, x31, x32,x33

Determination of Entering VariableMethod of Multiplier

For each basic variable xij

ui : multiplier of row-i

vj : multiplier of column-j

ui + vj = cij

number of equations = m + n – 1

u1 = 0 (usually)

Determination of Entering VariableMethod of Multiplier

For each non-basic variable xpq

up : multiplier of row-p

vq : multiplier of column-q

cpq = up + vq – cpq

Determination of Entering VariableMethod of Multiplier

For each non-basic variable xpq

up : multiplier of row-p

vq : multiplier of column-q

cpq = up + vq – cpq

Maximization: xpq with most negative cpq

Minimization: xpq with the most positive cpq

1015155Demand

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5 15 5

5

Iteration #1

Basic Variable

u3 = 5

u2 = 7

u2 = 7

u2 = 7

u1 = 0

u1 = 0

v4 = 13u3 + v4 = c34 = 18x34 :

v4 = 13u2 + v4 = c24 = 20x24 :

v3 = 2u2 + v3 = c23 = 9x23 :

v2 = 0u2 + v2 = c22 = 7x22 :

v2 = 0u1 + v2 = c12 = 0x12 :

v1 = 10u1 + v1 = c11 = 10x11 :

Non-basic Variable

c33 = u3 + v3 – c33 = 5 + 2 – 16 = –9x33 :

c32 = u3 + v2– c32 = 5 + 0 – 14 = –9x32 :

c31 = u3 + v1 – c31 = 5 + 10 – 0 = 15x31 :

c21 = u2 + v1 – c21 = 7 + 10 – 12 = 5x21 :

c14 = u1 + v4 – c14 = 0 + 13 – 11 = 2x14 :

c13 = u1 + v3 – c13 = 0 + 2 – 20 = –18x13 :

x31 : entering variable

1015155Demand

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252

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5 15 5

5

-18 2

5

15 -9 -9

x31

Determination of Leaving Variable:Loop Construction Equivalent to applying feasibility condition in

simplex method

The loop starts and end at the designated non-basic variable

Every corner of the loop should be a cell withbasic variable (except the start and the end)

Change the value of every corner alternatively(+) or (–)

Leaving variable is the one with smallest value

A tie is arbitrarily broken

1015155Demand

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5 15 5

5

x31

1015155Demand

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4321source Supply

Destination

10 0 20 11

5

12

0 14

7 9

16

20

18

10

5 15 5

5

x31

+

– +

– +

–

x11, x22, x34 = 0 Leaving variable: x34

1015155Demand

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4321source Supply

Destination

10 0 20 11

0

12

0 14

7 9

16

20

18

15

0 15 10

x34

5

Iteration#1

Solution: 0 10 + 15 0 + 0 7 + 15 9+ 10 20 + 5 0 = 335

Basic Variables: x11, x12, x22, x23, x24, x31

Non-basic Variables: x13, x14, x21, x32, x33,x34

1015155Demand

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252

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4321source Supply

Destination

10 0 20 11

0

12

0 14

7 9

16

20

18

15

0 15 10

x34

5

Iteration #2

Basic Variable

u3 = –10

u2 = 7

u2 = 7

u2 = 7

u1 = 0

u1 = 0

v1 = 10u3 + v1 = c31 = 0x31 :

v4 = 13u2 + v4 = c24 = 20x24 :

v3 = 2u2 + v3 = c23 = 9x23 :

v2 = 0u2 + v2 = c22 = 7x22 :

v2 = 0u1 + v2 = c12 = 0x12 :

v1 = 10u1 + v1 = c11 = 10x11 :

Non-basic Variable

c34 = u3 + v4 – c34 = –10 + 13 – 18 = –15x34 :

c33 = u3 + v3 – c33 = –10 + 2 – 16 = –24x33 :

c32 = u3 + v2– c32 = –10 + 0 – 14 = –24x32 :

c21 = u2 + v1 – c21 = 7 + 10 – 12 = 5x21 :

c14 = u1 + v4 – c14 = 0 + 13 – 11 = 2x14 :

c13 = u1 + v3 – c13 = 0 + 2 – 20 = –18x13 :

x21 : entering variable

1015155Demand

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252

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4321source Supply

Destination

10 0 20 11

0

12

0 14

7 9

16

20

18

15

0 15 10x21

5

+

+

–

–

x11, x22=0 Leaving Variable = x11

1015155Demand

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4321source Supply

Destination

10 0 20 11

x11

12

0 14

7 9

16

20

18

15

0 15 100

5

Iteration #2

Iteration#2

Solution: 0 12 + 15 0 + 0 7 + 15 9+ 10 20 + 5 0 = 335

Basic Variables: x12, x21, x22, x23, x24, x31

Non-basic Variables: x11, x13, x14, x21, x32,x33, x34

1015155Demand

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252

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4321source Supply

Destination

10 0 20 11

x11

12

0 14

7 9

16

20

18

15

0 15 100

5

Iteration #3

Basic Variable

v1= 5u2 = 7u2 + v1 = c21 = 12x21 :

u3 = –5

u2 = 7

u2 = 7

u2 = 7

u1 = 0

v1 = 5u3 + v1 = c31 = 0x31 :

v4 = 13u2 + v4 = c24 = 20x24 :

v3 = 2u2 + v3 = c23 = 9x23 :

v2 = 0u2 + v2 = c22 = 7x22 :

v2 = 0u1 + v2 = c12 = 0x12 :

Non-basic Variable

c11 = u1 + v1 – c11 = 0 + 5 – 10 = –5x11 :

c34 = u3 + v4 – c34 = –5 + 13 – 18 = –10x34 :

c33 = u3 + v3 – c33 = –5 + 2 – 16 = –19x33 :

c32 = u3 + v2– c32 = –5 + 0 – 14 = –19x32 :

c14 = u1 + v4 – c14 = 0 + 13 – 11 = 2x14 :

c13 = u1 + v3 – c13 = 0 + 2 – 20 = –18x13 :

x14 : entering variable

1015155Demand

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4321source Supply

Destination

10 0 20 11

x14

12

0 14

7 9

16

20

18

15

0 15 100

5

x24 = 0

+

+

–

–

Leaving Variable = x24

1015155Demand

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252

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4321source Supply

Destination

10 0 20 11

10

12

0 14

7 9

16

20

18

5

10 15 x240

5

Iteration #3

Iteration#3

Solution: 5 0 + 10 11 + 0 12 + 10 7+ 15 9 + 5 0 = 315

Basic Variables: x12, x14, x21, x22, x23, x31

Non-basic Variables: x11, x13, x24, x32, x33,x34

1015155Demand

53

252

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4321source Supply

Destination

10 0 20 11

10

12

0 14

7 9

16

20

18

5

10 15 x240

5

Iteration #4

Basic Variable

v1 = 5u3 = –5u3 + v1 = c31 = 0x31 :

v1= 5u2 = 7u2 + v1 = c21 = 12x21 :

u1 = 0

u2 = 7

u2 = 7

u1 = 0

v4 = 11u1 + v4 = c14 = 11x14 :

v3 = 2u2 + v3 = c23 = 9x23 :

v2 = 0u2 + v2 = c22 = 7x22 :

v2 = 0u1 + v2 = c12 = 0x12 :

Non-basic Variable

c34 = u3 + v4 – c34 = –5 + 11 – 18 = –12x34 :

c33 = u3 + v3 – c33 = –5 + 2 – 16 = –19x33 :

c11 = u1 + v1 – c11 = 0 + 5 – 10 = –5x11 :

c24 = u2 + v4 – c24 = 7 + 11 – 20 = –2x24 :

c32 = u3 + v2– c32 = –5 + 0 – 14 = –19x32 :

c13 = u1 + v3 – c13 = 0 + 2 – 20 = –18x13 :

All non-basic variables are now negative = END

(Class) Assignment #3Improved Starting Solution

Elaborate the Least-Cost Method

Elaborate Vogel’s Approximation Method(VAM)

Present the elaborations

(Class) Assignment #4The Assignment Model

Elaborate the assignment model asspecific transportation model

Present the elaboration

The End

This is the end of Chapter 6A