In the least cost method, we allocate the maximum in the lowest cost cell and then keep moving to the next lowest cost cells. So now let’s start solving by using least cost method.

Distribution centers

AnX Y Supply

Rs. 2000 Rs. 5380 1000

X_{11} – 1000 X_{12 }– 0

Factories B Rs. 2500 Rs. 2700 1500

X_{21 }– 1300 X_{22} – 200

C Rs. 2550 Rs. 1700 1200

X_{21 }= 0 X_{32 }= 1200

Demand 2300 1400

So here we see that the lowest cost cell CY which is 1700 had the maximum possible allocation that meets the entire supply and demand requirement which is made 1200. This, as a result,complies with thesupply of row 3 and thus cancelled.

Then coming to the next least cost cell AX (2000). Maximum possible allocation i.e. 1000 is done and then crossed. Next, comes the next lowest cost cell BX 2500, maximum allocation done here is 1300. So the total demand of column X is 2300, and we did allocate 1000 in cell AX. Now coming to the next lowest cost cell CX where it is 2500. Here only 0 can be allocated to meet the demand (2300) and supply (1200). Then in the next lowest Supply BY the allocation possible is 200. AY is the next to last lowest cost cell where 0 allocations could be done.

So, therefore

Z = Rs. (2000 x 1000 + 2500 x 1300 + 2700 x 200 + 1700 x 1200)

= Rs. 78, 30, 000.

**Links of Previous Main Topic:-**

- Concept of capital expenditure
- Learning objectives and chapter outline
- Limitations of operations research
- Linear programming learning objectives and outline of chapter
- Introduction learning objectives
- Duality in linear programming
- Learning objectives

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- Performing optimality test
- The stepping stone method
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