QUESTION 1 (80 marks) An importer of European vehicles has Mercedes available at

Question

QUESTION 1 (80 marks) An importer of European vehicles has Mercedes available at three locations and orders for Mercedes to be delivered to four outlets as indicated below Location A Mercedes 25 available Mercedes required 30 20 15 20 The cost in dollars of delivering a Mercedes to each outlet is as follows: From A 260 500 240 420 360 460 280 180 460 300 240 320 What is the optimum distribution plan to minimize the delivery costs and what is the minimum total cost to the importer of delivering Mercedes to the four outlets? (Evidence of appropriate analysis must be provided in order to obtain full marks) (80 marks)

Explanation / Answer

In the given problem total supply = 75 cars

Total demand = 65 cars

toatal supply > total demand so it is an unbalanced transportation problem

The Balanced transportation problem is given below

where His a dummy destination and costs in the corresponding row are equal to 0

Inital basic feasible solution by lowest cost entry method

1.select loest cost cell (H,C) and transport maximum amount which is equal to min (20,10) = 10 to that cell

2. select next lowest cost cell (G,B) and transport maximum possible amount which is equal to min (30,15 )= 15 to that cell

3. select next lowest cost cell (F,A) and transport maximum amount which is equal to min (25,20) =20 to that cell

4 select next min cost cell (D,A) and transport maximum amount which is equal to min ( 5,15)= 5 to that cell

5.select next lowest cost cell (E,C) and transport maximum amount which is equal to (10,15) =10 to that cell

6. select next min cost cell (D,B) and transport min ( 15,10)=10 to that cell

7. last allocation to the cell (C,B) and the amount is 5 units

The solution is

In the above solution the no.of occupied cells = 7

The value of m+n-1 = no.of rows + no.of columns -1= 5+3-1 =7

the no.of occupid cells = m+n-1 =7 hence the given solution is initial basic feasible solution to the given transportation problem

the transportation cost = 260x5+360x10+460x5+300x10+240x20+180x15+0x10=17700 USD

now to test whether the above initial basic feasible solution is optimum or not can be test by the following procedure

optimum test procedure UV- method

Matrix cij for occupied cells in the above I.B.F.S

Matrix Ui, Vj for occupied cells

construct the matrix Cij, Ui+ Vj for emty cells

matrix Cij - (Ui+Vj) for emty cells

in the above matrix some of the cells having -ve values it means the current I.B.F.S is not an optimum solution

New initial basic feasible solution

find most negative cell in the above matrix and concider it as new entering cell and consider one of the cells as leaving cell

now the new initial basic feasible solution is given as

the transportation cost is = 16900 USD

Again apply the above optimum test procedue to test its optimality

Repeat this procedure until to get optimum solution

The optimum solution is given asfollows

Minimum transportation cost = 16400USD

To|From A B C total D 260 360 460 15 E 500 460 300 15 F 240 280 240 20 G 420 180 320 15 H 0 0 0 10 Total 25 30 20 75

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