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Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E and F whose requirements are 60, 50 and 40 quintals respectively.

6. Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of  transportation per quintal from the godowns to the shops are given in the following table:

Transportation cost per quintal (in Rs)
From/To A B
D 6 4
E 3 2
F 2.50 3

How should the supplies be transported in order that the transportation cost is minimum? What is the                   minimum cost?

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Let godown A supply  x and y quintals of grain to shops D and E respectively. Then , (100-x-y) will be  supplied to shop F. Requirements at shop D is  60 since godown A supply  x .Therefore remaining (60-x) quintals of grain will be transported from godown B.

Similarly, (50-y) quintals and 40-(100-x-y)=(x+y-60) will be transported from godown B to shop E and F respectively. The problem can be represented diagrammatically as follows:

   

          x,y\geq 0  and  100-x-y\geq 0  

          x,y\geq 0  and  x+y\leq 100

60-x\geq 0,50-y\geq 0\, \, \, and\, \, x+y-60\geq 0

 \Rightarrow \, \, \, \, x\leq 60,y\leq 50,x+y\geq 60

Total transportation cost z is given by , 

z=6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)

z=2.5x+1.5y+410

  Mathematical formulation of given problem is as  follows:

Minimize : z=2.5x+1.5y+410

Subject to constraint ,

                           x+y\leq 100

                           x\leq 60

                        y\leq 50

                         x+y\geq 60

                            x,y\geq 0

The feasible  region determined by constraints is as follows:

       

The corner points of feasible region are A(60,0),B(60,40),C(50,50),D(10,50)

The value of Z at corner points is as shown :  

 corner points 

z=2.5x+1.5y+410

 
   A(60,0)             560  

B(60,40)

            620  
  C(50,50)             610  
   D(10,50)             510 minimum

therefore  510  may or may not be minimum value of Z .

Hence , Z has miniimum value  510  at  point  D(10,50)

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