a5f14transportation Problem Module

8/3/2019 a5f14transportation Problem Module

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AIBS

MBA-IB, MIB 201

OPERATIONS RESEARCH

REMICA AGGARWAL


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Transportation Problem


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DescriptionA transportation problem basically deals with theproblem, which aims to find the best way to fulfill thedemand of n demand points using the capacities of msupply points.

While trying to find the best way, generally a variablecost of shipping the product from one supply point to ademand point or a similar constraint should be taken intoconsideration.


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Formulating TransportationProblems

Example 1: Powerco has three electric power plantsthat supply the electric needs of four cities.

The associated supply of each plant and demand ofeach city is given in the table 1.

The cost of sending 1 million kwh of electricity froma plant to a city depends on the distance theelectricity must travel.


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Transportation tableau

A transportation problem is specified by the

supply, the demand, and the shipping costs. So the relevant data can be summarized in atransportation tableau.

The transportation tableau implicitly expresses

the supply and demand constraints and theshipping cost between each demand and supplypoint.


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Table 1. Shipping costs, Supply, and

Demand for Powerco Example

From To

City 1 City 2 City 3 City 4 Supply

(Million kwh)Plant 1 $8 $6 $10 $9 35

Plant 2 $9 $12 $13 $7 50

Plant 3 $14 $9 $16 $5 40

Demand(Million kwh)

45 20 30 30

Transportation Tableau


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Solution1. Decision Variable:

Since we have to determine how much electricity is

sent from each plant to each city;

Xij = Amount of electricity produced at plant i andsent to city j

X14 = Amount of electricity produced at plant 1 andsent to city 4


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2. Objective function

Since we want to minimize the total cost of shippingfrom plants to cities;

Minimize Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24+14X31+9X32+16X33+5X34


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3. Supply Constraints

Since each supply point has a limited production capacity;

X11+X12+X13+X14


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4. Demand Constraints

Each demand point has a demand ;

X11+X21+X31 >= 45

X12

+X22

+X32

>= 20

X13+X23+X33 >= 30

X14+X24+X34 >= 30


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5. Sign Constraints

Since a negative amount of electricity can not beshipped all Xijs must be non negative;

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)


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LP Formulation of Powercos Problem

Min Z =8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24+14X31+9X32+16X33+5X34

S.T.: X11+X12+X13+X14 = 30

X14+X24+X34 >= 30

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)


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General Description of a Transportation

Problem

1. A set of m supply pointsfrom which a goodis shipped. Supply point ican supply atmost siunits.

2. A set of n demand pointsto which the goodis shipped. Demand pointjmust receive at

least diunits of the shipped good.3. Each unit produced at supply point iand

shipped to demand pointjincurs a variablecostof cij.


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Xij = number of units shipped from supply point itodemand point j

),...,2,1;,...,2,1(0

),...,2,1(

),...,2,1(..

min

1

1

1 1

njmiX

njdX

misXts

Xc

ij

mi

i

jij

nj

j

iij

mi

i

nj

j

ijij


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Balanced Transportation Problem

If Total supply equals to total demand,the problem is said to be a balanced

transportation problem:

nj

j

j

mi

i

i ds11


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Balancing a TP if total supply exceedstotal demand

If total supply exceeds total demand, wecan balance the problem by addingdummy demand point. Since shipments

to the dummy demand point are not real,they are assigned a cost of zero.


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Balancing a transportation problem if totalsupply is less than total demand

If a transportation problem has a total supply that isstrictly less than total demand the problem has nofeasible solution. There is no doubt that in such a caseone or more of the demand will be left unmet. we canbalance the problem by adding dummy supply point.

Since shipments to the dummy supply point are not real,they are assigned a cost of zero.


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MG Auto has three plants in Los Angeles,Detroit and New Orleans, and two major

distribution centers in Denver and Miami.The capacities of the three plants during thenext quarter are 1000, 1500, 1200 cars.The

quarterly demand at the two distributioncenters are 2300 and 1400 cars.

ANOTHER EXAMPLE


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The transportation cost per car on the different routes, rounded to the

closest dollar are calculated as given below:

Denver Miami

Los Angeles $80 $215

Detroit $100 $108

New Orleans $102 $68

Formulate the LP model.

Let xij be defined as the no. of cars shipped from ith source to jth destination.


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Min Z = 80x11 + 215x12 + 100x21 + 108x22 + 102x31 + 68x32

st x11 +x12


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Transportation Table

O1 c11 c12 c1n a1

a2

am

O2 c21 c22 c2n

cm1 cm2 cmn

Om b1 b2 bnRequirement

x11 x12 x1n

x21 x22 x2n

xm1 xm2 xmn

Avai

lability

D1 D2 Dn


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There are (m+n) constraints and mn variables. Because of neccand suff conditions of feasibility, any feasible solution satisfying

m + n1 of m + n constraints equations will automatically satisfy the last

constraint. A feasible solution to a TP can have at most m + n1 strictly positive

components.

Algorithm of TP

Formulate the problem and set up in the matrix form.

Obtain an initial basic feasible solution.

Test the initial solution for optimality.

Update the solution.

Finding Basic Feasible Solution for TP


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Methods to find the BFS for a balancedTP

There are three basic methods:

1. Northwest Corner Method

2. Minimum Cost Method

3. Vogels Method


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1. Northwest Corner MethodTo find the bfs by the NWC method:

First of all check whether there exists a feasible solution by

checking whether total demand = total supply. Select the north-west (upper left ) corner of the transportationtableau & allocate as much as possible so that either thecapacity of the first row is exhausted or the requirement ofthe first column is satisfied. i.e x11= min ( a1, b1)

Now if b1>a1 move down & make second allocation of the ofmagnitude x21= min(a2 , b1-x11)

If b1


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According to the explanations in the previousslide we can set x11=3 (meaning demand of

demand point 1 is satisfied by supply point 1).5

6

2

3 5 2 3

3 2

6

2

X 5 2 3


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After we check the east and south cells, we saw that wecan go east (meaning supply point 1 still has capacity to

fulfill some demand).

3 2 X

6

2

X 3 2 3

3 2 X

3 3

2

X X 2 3


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After applying the same procedure, we saw that we can

go south this time (meaning demand point 2 needs moresupply by supply point 2).

3 2 X

3 2 1

2

X X X 3

3 2 X

3 2 1 X

2

X X X 2


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Finally, we will have the following bfs, which is:x11=3, x12=2, x22=3, x23=2, x24=1, x34=2

3 2 X

3 2 1 X

2 X

X X X X


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The Northwest Corner Method does notutilize shipping costs.

It can yield an initial bfs easily but the totalshipping cost may be very high.

The minimum cost method uses shippingcosts in order come up with a bfs that hasa lower cost.

SHORTCOMINGS OF NWC METHOD


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To begin the minimum cost method, first we find thedecision variable with the smallest shipping cost (Xij). Then

assign Xijits largest possible value, which is the minimumof siand dj.After that, as in the Northwest Corner Method we shouldcross out row i and column j and reduce the supply ordemand of the noncrossed-out row or column by the valueof Xij.

Then we will choose the cell with the minimum cost ofshipping from the cells that do not lie in a crossed-out rowor column and we will repeat the procedure.

2. Minimum Cost Method


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An example for Minimum Cost MethodStep 1: Select the cell with minimum cost.

2 3 5 6

2 1 3 5

3 8 4 6

5

10

15

12 8 4 6


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Step 2: Cross-out column 2

2 3 5 6

2 1 3 5

8

3 8 4 6

12 X 4 6

5

2

15


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Step 3: Find the new cell with minimum shippingcost and cross-out row 2

2 3 5 6

2 1 3 5

2 8

3 8 4 6

5

X

15

10 X 4 6


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Step 4: Find the new cell with minimum shippingcost and cross-out row 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

X

X

15

5 X 4 6


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Step 5: Find the new cell with minimum shippingcost and cross-out column 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5

X

X

10

X X 4 6


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Step 6: Find the new cell with minimum shippingcost and cross-out column 3

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4

X

X

6

X X X 6


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Step 7: Finally assign 6 to last cell. The bfs is

found as: X11=5, X21=2, X22=8, X31=5, X33=4 andX34=6

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4 6

X

X

X

X X X X


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3. Vogels Method1. Begin with computing each row and column a penalty.

2. The penalty will be equal to the difference between the two smallest

shipping costs in the row or column.

3. Identify the row or column with the largest penalty.

4. Find the first basic variable which has the smallest shipping cost in thatrow or column.

5. Then assign the highest possible value to that variable, and cross-outthe row or column as in the previous methods.

6. Compute new penalties and use the same procedure.


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An example for Vogels MethodStep 1: Compute the penalties.

Supply Row Penalty

6 7 8

15 80 78

Demand

Column Penalty 15-6=9 80-7=73 78-8=70

7-6=1

78-15=63

15 5 5

10

15


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Step 2: Identify the largest penalty and assign thehighest possible value to the variable.

Supply Row Penalty

6 7 8

5

15 80 78

Demand

Column Penalty 15-6=9 _ 78-8=70

8-6=2

78-15=63

15 X 5

5

15


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Supply Row Penalty

6 7 8

5 5

15 80 78

Demand

Column Penalty 15-6=9 _ _

_

_

15 X X

0

15


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Supply Row Penalty

6 7 8

0 5 5

15 80 78

Demand

Column Penalty _ _ _

_

_

15 X X

X

15


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Step 5: Finally the bfs is found as X11=0, X12=5,X13=5, and X21=15

Supply Row Penalty6 7 8

0 5 5

15 80 78

15

Demand

Column Penalty _ _ _

_

_

X X X

X

X


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EXAMPLE

TRANSPORTATION PROBLEM

S1 10 2 20 11 15

25

10

S2 12 7 9 20

S3 4 14 16 18

5 15 15 15Demand


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Other examples

North West corner Rule

S1 10 2 20 11 15

2510

S2 12 7 9 20

S3 4 14 16 18

5 15 15 15Demand

Supply

The starting basic solution is x11 = 5, x12 = 10, x22 = 5, x23 = 15, x24 = 5, x34 = 10.

The associated cost of the schedule is Z = 10 5 + 2 10 + 7 5 + 9 15 +

20 5 + 18 10 = 520.

5 10

155

10

D1 D2 D3 D4

5


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Least Cost Method

S1 10 2 20 11 15

25

10

S2 12 7 9 20

S3 4 14 16 18

5 15 15 15Demand

Supply

15

15 10

5

0

5


The associated cost of the schedule is Z = 2 15 + 11 0 + 9 15 + 20 10 +

4 5 + 18 5 = 475.

D1 D2 D3 D4

0


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Test the initial solution for optimality

Step 1 :Construct the transportation table entering the origin capacities ai

,the destination requirement bj and the costs cij.

Step 2 :Determine the initial BFS by using any of the discussed methods.

Step 3 : For all the occupied (basic) variables xij solve the system of

equations ui + vj = cij, starting with some ui = 0 and entering successivelythe values ui and vj on the transportation table.

Step 4 : Compute the net evaluations cij zij = cij (ui + vj) for all the non-basic cells and enter them in the corresponding cells.

Step 5 : Examine the sign of each cij zij . If all cij zij >=0 then the currentbfs is optimal . Otherwise select the unoccupied cell having the smallestnegative net evaluation to enters the basic. Allocate an unknown quantity to this cell .


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1 10 2 20 11 15

25

10

2 12 7 9 20

3 4 14 16 18

5 15 15 15

5 10

155

10

5

Supply

u1 = 0

u3 = 3

u2

= 5

v1=10 v2=2 v3=4 v4=15

Basic variables satisfy the equations ui + vj = cij

u1 + v1 = 10 u1 + v2 = 2 u2 + v2 = 7 u2 + v3 = 9u2 + v4 = 20

u3 + v4 = 18

(16) (-4)

(-3)

(9) (9)(-9) 10-

5+5-

10+5-

D1 D2 D3 D4


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Step 6 : Allocate a quantity to that cell and alternately subtract andadd to and from the transition cells of the loop in such a way thatthe rim requirements remain satisfied.

Step 7 : Assign a max value to in such a way that the value of onebasic variable becomes zero and other basic variable remains nonnegative. The basic cell whose allocation has been reduced to zero ,leaves the basis.

The new values of the variable will remain nonnegative if

x11 = 5 - 0

x22 = 5 - 0

x34 = 10 - 0The maximum value of can be 5.

Step 8: return to step 3 and repeat the procedure until an optimum basicfeasible solution is obtained.


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1 10 2 20 11 15

25

10

2 12 7 9 20

3 4 14 16 18

5 15 15 15Demand

5

15

150

5

1 2 3 4

10

Supply

u1 = 0

u3 = 3

u2

= 5

v1=1 v2=2 v3=4 v4=15

(9) (16) (-4)

(6)

(9) (9)

Because each unit shipped through route (3,1) reduces the shipping cost

by 9(=c13 - u3+ v1), the total cost associated with the schedule is 9* 5 =45

less than the previous schedule. Thus the new cost is Z = 475.

15-

0+ 10-


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1 10 2 20 11 15

25

10

2 12 7 9 20

3 4 14 16 18

5 15 15 15Demand

5

5

1510

5

1 2 3 4

10

Supply

u1 = 0

u3 = 7

u2

= 5

v1=-3 v2=2 v3=4 v4=11

(13) (16)

(10) (4)

(5) (5)


The associated cost of the schedule is Z = 435


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Unbalanced TransportationProblem

When Supply(availability) Requirement(demand)

n

1j

j

m

1i

i ba

Two Possibilities:

(1) Supply is less than the demand.

Add a fictitious source with availability

m

1ii

n

1jj ab

Assign zero cost of transportation from this source to any destination.

Solve as usual TP.


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(2) Availability(Supply) is more than the demand.

Add a fictitious destination with requirement

n

1j

j

m

1i

i ba

Assign zero cost of transportation from each source to that destination.

Solve as usual TP.

Prohibited Routes.

If there is a restriction on the routes available for transportation , we

assign a very large cost element M to each of such routes which are

not available.


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Consider the following TP

D1 D2 D3 D4 D5

800

500

900

S1 5 8 6 6 3

S2 4 7 7 6 5

S3 8 4 6 6 4400 400 500 400 800

2200a3

1i

i

2500b5

1j

j


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D1 D2 D3 D4 D5

S1 5 8 6 6 3 800

S2 4 7 7 6 5 500

S3 8 4 6 6 4 900

SF 0 0 0 0 0 300400 400 500 400 800

800

300

400

400

0

100

300200

u1 = 0

u3 = 1

u2 = 1

v1=3 v2=3 v3=5 v4=5

u4 = -5

v5=3


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Maximization Problem

To solve a transportation problem where the objective is tomaximize the total value or profit.

All the values of the profit matrix are subtracted from the

highest profit value in the matrix. After that the optimal solution is obtained as for the

minimization problems.

Finally the value of the objective function is determined with

reference to the original profit matrix. If a maximization type of TP is unbalanced, then a dummy

source or destination is introduced first and then it isconverted into a minimization TP.


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Solve the following TP for maximizing profit.

Market

A B C DX 12 18 6 25

Y 8 7 10 18

Z 14 3 11 20

Per Unit Profit(Rs)

Availability at warehouses:X : 200

Y :500

Z :300Demand in the markets(in units):

A B C D

180 320 100 400


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Unique vs Multiple Solutions:

A transportation problem is optimal when all cij zij aregreater than , or equal to zero.

If all nonbasic cells are strictly positive, then it is unique.

If some cell has cij zij = 0 for a nonbasic cell, then multiplesolutions exist i.e there exist a transportation pattern otherthan the one obtained which can satisfy all the rimrequirements for the same cost.

Trace a closed loop beginning with the cell having cij zij =0 and get the revised solution in the same way as a solutionis improved.

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a5f14transportation Problem Module