assignment problem

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1 Assignment Problem Assignment problem is also known as a special case of LP problem or transportation problem; with which unit of demand and supply is “1” Its LP formulation Our objective here is to determine its solution using heuristic algorithm – similar to what we did in the transportation lecture. (to p2) (to p3)

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Assignment Problem. Assignment problem is also known as a special case of LP problem or transportation problem; with which unit of demand and supply is “1” Its LP formulation - PowerPoint PPT Presentation

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Page 1: Assignment Problem

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Assignment Problem

• Assignment problem is also known as a special case of LP problem or transportation problem; with which unit of demand and supply is “1”

• Its LP formulation

• Our objective here is to determine its solution using heuristic algorithm – similar to what we did in the transportation lecture.

(to p2)

(to p3)

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LP formulation

Total 1 1 1 1

Total

1111

LP: Min 210Xar + 90Xaa + 180Xad + ……..+ 120 Xdcs.t. Xar+Xaa+Xad+Xac = 1 ; Xar+Xbr+Xcr+Xdr = 1 Xbr+Xba+Xbd+Xbc = 1 ; Xaa+Xba+Xca+Xda = 1

Xcr+Xca+Xcd+Xcc = 1 ; Xad+Xbd+Xcd+Xdd = 1Xdr+Xda+Xdd+Xdc = 1 ; Xac+Xbc+Xcc+Xdc = 1

all Xij = 0 or 1 for i=a,b,c,d & j=r,a,d,c

(to p1)

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Heuristic algorithm

• Its logical flow:– We make use of the “opportunity cost” concept– It is defined as follows:

How it works? (to p4)

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StepsStep 1: For each column/row, find its minimum cost and subtract

from its respective column/row

Step 2: Determine its feasible solution by crossing

rows/columns with most “0” values

Step 3: Solution is obtained if

total crossed lines = total numbers of rows/column

Otherwise,

select min cost of uncrossed cells and subtracting it

from all uncrossed and add it to double crossed cells

Step 4: Repeat step 4 until solution is obtained.

Example (to p5)

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Example

Consider the following example:

Step 1 : For row, select its min and subtract from them(to p6)

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Step 1

Step 1: for column, select min cost and subtract from them

Step 2: Determine its feasible solution (to p7)

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Step 2

Step 3: Only 3 lines. No good since we need four linesThus, we select the min cost for uncrossed = 15

We subtract them from uncrossed cells and add to it double crossedWhich resulting as ………. (to p8)

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Steps 3 & 4

Step 4: We have four line above, Stop. Optimal solution is obtainedSolution is:

or

Important notes (to p9)

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Important Note

Note 1: It is a (nxn) matrix

i.e. total supply= total demand

If not, we add row/column to them

Note 2: We assign a big value M to

a route that is not feasible oneHow computer package works?

Tutorial (to p10)

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Tutorial

• Appendix B– 37, 38, 40, 46

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