8Transportation Model

TRANSPORTATION MODEL

EXAMPLE: CONSIDER A MANUFACTURER WHO OPERATES THREE FACTORIES AND DISPATCHES HIS PRODUCTS TO FIVE DIFFERENT RETAIL SHOPS. THE TABLE BELOW INDICATES THE CAPACITIES OF THE THREE FACTORIES, THE QUANTITY OF PRODUCTS REQUIRED AT THE VARIOUS RETAIL SHOPS AND THE COST OF SHIPPING ONE UNIT OF THE PRODUCT FROM EACH OF THREE FACTORIES TO EACH OF THE FIVE RETAIL SHOPS.

Retail Shops R1R2R3R4R5CapacitiesFactories F11913365150F2241216201100F3143312336150Requirements 10070504040300

NORTHWEST CORNER METHOD


The North-west Corner To be filled up first


CapacityRequirement


CapacityRequirementRequirement > Capacity


CapacityRequirementRequirement > Capacity


50

NEW CapacityNEW RequirementFactory Capacity over

50

Other values to be ignored.The NEW North-west Corner

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Other values to be ignored.RequirementCapacity

50

Other values to be ignored.RequirementCapacityRequirement < Capacity

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Other values to be ignored.RequirementCapacityRequirement < Capacity

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Other values to be ignored.NEW CapacityShop Requirement overIgnore Values NEW Requirement

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Other values to be ignored.Ignore ValuesThe NEW North-west Corner

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Other values to be ignored.Ignore ValuesRequirementCapacity

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Other values to be ignored.Ignore ValuesRequirementCapacityRequirement > Capacity

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Other values to be ignored.Ignore ValuesRequirementCapacityRequirement > Capacity

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Other values to be ignored.Ignore ValuesRequirementCapacityRequirement > CapacityOther values to be ignored.

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Other values to be ignored.Ignore ValuesRequirementCapacityFactory Capacity OverOther values to be ignored.

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Other values to be ignored.Ignore ValuesOther values to be ignored.The NEW North-west Corner

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Other values to be ignored.Ignore ValuesOther values to be ignored.RequirementCapacity

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Other values to be ignored.Ignore ValuesOther values to be ignored.RequirementCapacityRequirement < Capacity

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Thus the assignment with respective to the values stand as under after the northwest rule.

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LEAST COST METHODLeast cost cells


ASSIGN MAXIMUM POSSIBLE/REQUIRED TO LEAST COST CELLS504050


CAPACITY OF F1 FACTORY IS OVER AND REQUIREMENT OF R3 AND R5 ARE FULFILLED, SO WE CROSS THEM.504050


NOW AMONG REMAINING CELLS --Next least cost cell505040


ASSIGN MAXIMUM POSSIBLE/REQUIRED TO LEAST COST CELL60504050


CAPACITY OF FACTORY F2 IS OVER SO WE CROSS IT ALSO.50605040


NOW THE ONLY OPTION LEFT FOR FULFILLING REMAINING REQUIREMENTS IS ALLOCATING FROM FACTORY 350604050501040


THUS, ALL REQUIREMENTS HAVE BEEN FULFILLED BY THE GIVEN CAPACITIES.50604050104050


VOGEL APPROXIMATION METHOD (VAM)Vogel Approximation method for finding a basic feasible solution involves the following steps.

i)From the transportation Table we determine the penalty for each row and column. The penalties are calculated for each row (column) by subtracting the lowest cost element in that row (column) from the next lowest cost element in the same row (column).

ii)We identify the row or column with the largest penalty among all the rows and columns. If the penalties corresponding to two or more rows or columns are equal we select the topmost row and the extreme left column.iii)We select Xij as a basic variable if Cij is the minimum cost in the row or column with largest penalty. We choose the numerical value of xij as high as possible subject to the row and the column constraints. Depending upon whether ai or bj is the smaller of the two, ith row or jth column is eliminated.iv)The step (ii) is now performed on the reduced matrix until all the basic variables have been identified.

STEP 1: DETERMINATION OF PENALTY

Least cost element in rowSecond least cost element in row Penalty = 17-4 = 13

Des.:Ori.:1234C:Penalty (P)120221741201322437977033237201550Req.:604030110240Penalty (P)

STEP 1: DETERMINATION OF PENALTY

AS SHOWN IN THE PREVIOUS SLIDE, WE CALCULATE THE PENALTY FOR ALL ROWS AND ALL COLUMNS.

TABLE : COMPUTATION OF PENALTY FOR VAM:

Des.:Ori.:1234C:Penalty (P)12022174120132243797702332372015505Req.:604030110240Penalty (P)41583

STEP 2: IDENTIFYING THE LARGEST PENALTY Largest Penalty


STEP 3: IDENTIFYING THE LEAST COST IN THE ROW OR COLUMN WITH LARGEST PENALTYLeast cost element in that columnLargest Penalty


CapacityRequirement


CapacityRequirementRequirement < Capacity


CapacityRequirementRequirement< Capacity


CapacityRequirementRequirement < Capacity40


New CapacityNew Requirement40Retail Shop 2 Requirement exhausted

40Retail Shop 2 Requirement exhaustedNew Highest Penalty

STEP 3: IDENTIFYING THE LEAST COST IN THE ROW OR COLUMN WITH LARGEST PENALTYOther values to be ignoredLeast cost element in that rowLargest Penalty40

CapacityRequirement40


CapacityRequirementRequirement > Capacity40


CapacityRequirementRequirement > Capacity8040


New CapacityNew Requirement8040Factory 1 Capacity exhausted

8040New Largest Penalty

AS SHOWN IN THE PREVIOUS SLIDES, THE ITERATIVE PROCEDURE IS FOLLOWED UNTIL ALL THE BASIC VARIABLES HAVE BEEN IDENTIFIED

TABLE : INITIAL BASIC FEASIBLE SOLUTION

408030301050


THE TRANSPORTATION COST CORRESPONDING TO THIS CHOICE OF BASIC VARIABLES IS: 22 X 40 + 4 X 80 + 9 X 30 + 7 X 30 + 24 X 10 + 32 X 50 = 3520.

THE VAM PROVIDES A BASIC FEASIBLE SOLUTION WHOSE COST IS QUITE CLOSE TO THE MINIMUM TRANSPORTATION COST AND HENCE IS A POPULAR TECHNIQUE.

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8Transportation Model

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