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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost? The distances (in km) between the depots and the petrol pumps is given in the following table: \[\] $\begin{matrix} & \underline{\text{Distance in Km}} & \\ \underline{\text{From} \setminus\text{To}}& \text{A}&& \text{B}\\ \text{D}& 7 &&3 \\ \text{E}& 6 &&4 \\ \text{F} & 3 && 2 \end{matrix}$

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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 $x$ litres of oil be supplied from depot $A$ to petrol pump $D$ and $y$ litres of oil be supplied from depot $A$ to petrol pump $E$,then $7000-(x+y)\geq 0$ (i.e) $x+y\leq 7000$
The requirements of oil at petrol pumps $D,E$ and $F$ ar $4500l,3000l$ and $3500l$
The quantity of oil transported from depot $B$ to petrol pump $D,E$ and $F$ are $(4500-x),(3000-y)$ and $3500-[7000-(x+y)]$litres respectively.
(i.e)$4500-x\geq 0$
$\Rightarrow x\leq 4500$
$3000-y\geq 0$
$\Rightarrow y\leq 3000$
$3500-[7000-(x+y)]\geq 0$
$\Rightarrow x+y\geq 3500$
Step 2:
The cost of transportation per km for 10litre oil is Rs1.
Hence the cost of transportation/km/litre=Rs$\large\frac{1}{10}=$Rs 0.1
$\Rightarrow Z=0.7x+0.6y+0.3[7000-(x+y)]+0.3(4500-x)+0.4(3000-y)+0.2[(x+y)-3500]$
On simplifying we get,
$Z=0.3x+0.1y+3950$
Step 3:
The graph can be plotted according to the inequalities above and the feasible area is the shaded portion ABECD
Now the corner points are $A(500,3000),B(3500,0),E(4500,0),C(4500,2500)$ and $D(4000,3000)$
Step 4:
Now let us calculate the values of the objective function as follows :
At the points $(x,y)$ the value of the objective function subjected to $Z=0.3x+0.1y+3950$
At $A(500,3000)$ the value of the objective function subjected to $Z=0.3\times 500+0.1\times 3000+3950=4400$
At $B(3500,0)$ the value of the objective function subjected to $Z=0.3\times 3500+0+3950=5000$
At $E(4500,0)$ the value of the objective function subjected to $Z=0.3\times 4500+0+3950=5300$
At $C(4500,2500)$ the value of the objective function subjected to $Z=0.3\times 4500+0.1\times 2500+3950=5550$
At $C(4000,3000)$ the value of the objective function subjected to $Z=0.3\times 4000+0.1\times 3000+3950=5450$
Step 5:
It is clear that the minimum transport charges are Rs4400 at $A(500,3000)$.
When $x=500$ and $y=3000$,
500litres,3000 litres and 3500 litres of oil should be transported from depot A to petrol pumps $D,E$ and $F$ and 4000litres,0 litre of oil be transported from depot B to petrol pumps D,E and F with minimum cost of transportation of Rs4400.
answered Aug 29, 2013 by sreemathi.v
edited Aug 29, 2013 by sreemathi.v
 

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