Transportation Models Lect 2

8/8/2019 Transportation Models Lect 2

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Transportation Models


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Consider a commodity which is produced at

various centers called SOURCES and is

demanded at various other DESTINATIONS.


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The production capacity of each source(availability) and the requirement of each

destination are known and fixed.


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The cost of transporting one unit of thecommodity from each source to each destination

is also known.


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The commodity is to be transported from various

sources to different destinations in such a way that

the requirement of each destination is satisfied andat the same time the total cost of transportation in

minimized.


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This optimum allocation of the commodity from

various sources to different destinations is calledTRANSPORTATION PROBLEM.


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Different methods of obtaining initial basic

feasible solution to a balanced minimizationtransportation problem are

NORTH WEST CORNER RULE

LEAST COST METHOD

VOGELS APPROXIMATION METHOD


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A transportation problem can be statedmathematically as follows:

Let there be m SOURCES and nDESTINATIONS

Let ai : the availability at the ith sourcebj : the requirement of the j

th destination.

Cij : the cost of transporting one unit of

commodity from the ith source to the jth

destination

xij : the quantity of the commodity

transported from ith source to the jth

destination (i=1, 2, m; j=1,2, ..n)


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Source D1 D2 D3 D4 Availability

S1 C11 C12 C13 C14 a1

S2 C21 C22 C23 C24 a2

S3 C31 C32 C33 C34 a3

Requirement b1 b2 b3 b4 ai = bj

Destination


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Th

e problem is to determine th

e values ofxij such that total cost of transportation is

minimized.

We assume that the total quantity available

is the same as the total requirement.

i.e. ai = bj


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A solution where the row total of

allocations is equal to theavailabilities and the column total is

equal to the requirements is called a

Feasible Solution.

The solution with m+n-1 allocationsis called a Basic Solution.


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1. Destination

Source D1 D2 D3 D4 D5 vailabilityS1 5 3 8 6 6 1100

S2 4 5 7 6 7 900

S3 8 4 4 6 6 700

Requirement 800 400 500 400 600


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2. DestinationSource

D1 D2 D3 D4 D5 Availability

S1 6 4 4 7 5 100S2 5 6 7 4 8 125

S3 3 4 6 3 4 175

Requirement 60 80 85 105 70


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Algorithm of NORTH-WEST CORNER RULE

Step I:Given a balanced transportation problem; make the

first allocation in the north-west cell (i.e. topmost

left cell) of the table. This allocation will be x11 =minimum (a1, b1). This allocation will exhaust

either the availability at source 1 or the

requirement at destination 1.


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Step II:

If the availability of source 1 gets exhausted,strike out the remaining cells of row 1 and if the

requirement of destination 1 gets exhausted,

strike out the remaining cells of column 1.

At this stage the table gets reduced to the size

either m x (n-1) or (m-1) x n.

Step III:

Repeat steps I, and II till all the availabilities get

exhausted and all the requirements get fulfilled.


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Algorithm of LEAST-COST METHOD or

MATRIX MINIMA METHODStep I:

Given a balanced transportation problem the first

allocation is made in the cell with the minimumcost.

This allocation will be xij = minimum (ai, bj).


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Step II:

If xij = bj (i.e. the requirement of the jthdestination is fulfilled),

strike out the remaining cells of the jth column.

If xij = ai (i.e. the availability of the ith source is

exhausted),

strike out the remaining cells of the ith row.


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At this stage the table gets reduced to the sizeeither m x (n-1) or (m-1) x n.

Step III:Repeat steps I, and II till all the availabilities

get exhausted and all the requirements get

fulfilled.


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Algorithm of VOGELS APPROXIMATION

METHOD (VAM)

Step I:

Given a balanced transportation problem,

calculate the penalty for each row and for each

column by taking the difference between the lowestand the next lowest cost in that row or column.

Step II:

Select the row or column having largest penalty. Ifthere is a tie the selection can be made randomly.

Suppose the largest penalty corresponds to the ithrow and Cij be the lowest cost in that row.


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Step III:

Allocate maximum possible quantity, xij =minimum (ai, bj), in the cell (i, j) and strike out the

remaining cell of the ith row or the jth column

depending upon whether minimum is at ai or bj.

Step IV:

Repeat steps I, II and III till the availabilities at all

the sources are exhausted and the requirements of

all the destinations are fulfilled.


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Algorithm of MODIFIED DISTRIBUTION(MODI) METHOD

Step I: For an initial basic feasible solutionwith (m+n-1) occupied (basic) cells,calculate ui and vj values for rows and

columns respectively using the relationshipCij = ui + vj for all allocated cells only. Tostart with assume any one of the ui or vj to bezero.

Step II: For the unoccupied (non-basic)cells, calculate the cell evaluations or thenet evaluations as ij = Cij (ui + vj).


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Step III:

a) If all ij > 0, the current solution is optimaland unique.

b) If any ij

= 0, the current solution isoptimal, but an alternate solution exists.

c) If any ij < 0, then an improved solutioncan be obtained; by converting one of the

basic cells to a non basic cells and one of thenon basic cells to a basic cell. Go to step IV.


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Step IV: Select the cell corresponding to

most negative cell evaluation. This cell iscalled the entering cell. Identify a closed

path or a loop which starts and ends at the

entering cell and connects some basic cells atevery corner.

Step V: Put a + sign in the entering cell andmark the remaining corners of the loop

alternately with and + signs.


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Step VI: From the cells marked with sign,

select the smallest quantity (say ). Add to each quantity of the cell marked with +

sign and subtract from each quantity of

the cell marked with sign. In case of a tie,make zero allocation to any one of the cells.

This will make one non-basic cells as basic

and vice-versa.

Step VII: Return to step I.


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Unbalanced transportation problem

When the total availability is equal to the

total requirement the problem (i.e. ai = bj)

is said to be a balanced transportation

problem. If the total availability at different

sources is not equal to the total requirement

at different destinations, (i.e. ai bj), the

problem is said to be an unbalanced

transportation problem.

Steps to convert an unbalanced problem to a

balanced one are


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If ai > bj i.e. the total availability is greater

than the total requirement, a dummy

destination is introduced in the transportation

problem with requirement = ai - bj. Theunit cost of transportation from each source

to this destination is assumed to be zero.


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If ai < bj i.e. the total availability is less

than the total requirement, a dummy source

is introduced in the transportation problem

with requirement = bj - ai. The unit cost of

transportation from each destination to this

source is assumed to be zero.

After making the necessary modifications in thegiven problem to convert it to a balanced problem,

it can be solved using any of the methods.

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Transportation Models Lect 2