Browse Questions

# 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:

 From/To A B D 6 4 E 3 2 F 2.5 3

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

Toolbox:
• First formulate the objective function and identify the constraints from the problem statement, To solve a Linear Programming problem graphically, first plot the constraints for the problem. This is done by plotting the boundary lines of the constraints and identifying the points that will satisfy all the constraints.
• Let $R$ be the feasible region for a linear programming problem and let $Z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
• If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Let $x$ quintals of grains be transported from godown A to ration shop D and y quintals of grains be transported from godown A to ration shop E.
Therefore the grains transported from godown A to shop F will be $100-(x+y)$ quintals.
Because capacity of godown A is 100 quintals.
Therefore we have $x\geq 0,y\geq 0$ and $100-(x+y)\geq 0$ (i.e)$x+y\leq 100$
As requirements of grains at shop D is 60 quintals,$(60-x)$ quintals of grain will be transported from godown B to D
Step 2:
Similarly $(50-y)$ quintals and $40-[100-(x+y)]$ should be transported from godown B to shops E and F respectively.
Hence $(60-x)\geq 0$
$\Rightarrow x\leq 60$
Similarly $(50-y)\geq 0$
$\Rightarrow y \leq 50$
$40-[100-(x+y)]\geq 0$
$-60+x+y\geq 0$
$x+y\geq 60$
Step 3:
The cost of transportation is
$Z=6x+3y+250(100-x-y)+4(60-x)+2(50-y)+3(40-100-x+y]$
Simplifying this we get,
$\Rightarrow 6x+3y+250-2.5x-2.5y+240-4x+100-2y+120-300+3x+3y$
(i.e) $Z=2.5x+1.5y+410$
Subject to constraints :
$x\leq 60,y\leq 50,x+y \geq 60,x+y \leq 100,y\geq 0$
Step 4:
Let us plot the graph of equations
$x=60,y=50,x+y=60$ and $x+y=100$
The feasible region ABCD shows the shaded area in the figure satisfies the inequalities $x\leq 60,y\leq 50,x+y\geq 60,x+y\leq 100$ and $x,y\geq 0$
Step 5:
Let us calculate the values of $Z$ at the corner points $A(10,50),B(50,50)C(60,40)$ and $D(60,0)$
At the points (x,y) the value of the objective function $Z=2.5x+1.5y+410$
At $A(10,50)$ the value of the objective function $Z=2.5\times 10+1.5\times 50+410=25+75+410=510$
At $B(50,50)$ the value of the objective function $Z=2.5\times 50+1.5\times 50+410=125+75+410=610$
At $C(60,40)$ the value of the objective function $Z=2.5\times 60+1.5\times 40+410=150+60+410=620$
At $D(60,0)$ the value of the objective function $Z=2.5\times 60+1.5\times 0+410=150+0+410=560$
This implies the minimum cost 510 at $A(10,50)$