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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. The cost of transportation per quintal from the godowns to the shops are given in the following table:
How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Solution:
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Verified Answer
Let $x$ quintals of grains are transported from godown A to ration shop D and Y quintals of grains are transported from godown A to ration shop E.
The cost of transportation.
$\mathrm{Z}=6 \mathrm{x}+3 \mathrm{y}+2 \cdot 50(100-\mathrm{x}-\mathrm{y})+4(60-\mathrm{x})+2$
$(50-y)+3[40-(100-(x+y))]$
$=6 x+3 y+2 \cdot 50-2 \cdot 50 x-2 \cdot 50 y+240-4 x+100-2 y+$
$120-300+3 x+3 y$
$=2 \cdot 50+1 \cdot 5 \mathrm{y}+410$
So, L.P.P is to minimize $(Z)=2 \cdot 5 x+1 \cdot 5 y+410$.
Subject to constraints are $x \leq 60, y \leq 50$,
$x+y \geq 60, x+y \leq 100, x, y \geq 0$
$$
\begin{array}{ll}
\text { AtA }(10,50), & \mathrm{Z}=2 \cdot 5 \mathrm{x}+1 \cdot 5 \mathrm{y}+410 \\
& =2 \cdot 5 \times 10+1 \cdot 5 \times 50+410 \\
& =510 \mathrm{~min}^{\mathrm{m}} \\
\operatorname{AtB}(50,50), & \mathrm{Z}=2 \cdot 5 \times 50+1.5 \times 50+410 \\
& =125+75+410=610 \\
\operatorname{AtC}(60,40), & \mathrm{Z}=2.5 \times 60+1.5 \times 40+410 \\
& =150+60+410=620 \\
\text { At D }(60,0), & \mathrm{Z}=2.5 \times 60+1.5 \times 0+410 \\
& =150+410=660
\end{array}
$$
Thus $\mathrm{Z}$ is minimum i.e., ₹ 510 at $\mathrm{A}$.
i.e., when $x=10$ and $y=50$
The cost of transportation.
$\mathrm{Z}=6 \mathrm{x}+3 \mathrm{y}+2 \cdot 50(100-\mathrm{x}-\mathrm{y})+4(60-\mathrm{x})+2$
$(50-y)+3[40-(100-(x+y))]$
$=6 x+3 y+2 \cdot 50-2 \cdot 50 x-2 \cdot 50 y+240-4 x+100-2 y+$
$120-300+3 x+3 y$
$=2 \cdot 50+1 \cdot 5 \mathrm{y}+410$
So, L.P.P is to minimize $(Z)=2 \cdot 5 x+1 \cdot 5 y+410$.
Subject to constraints are $x \leq 60, y \leq 50$,
$x+y \geq 60, x+y \leq 100, x, y \geq 0$
$$
\begin{array}{ll}
\text { AtA }(10,50), & \mathrm{Z}=2 \cdot 5 \mathrm{x}+1 \cdot 5 \mathrm{y}+410 \\
& =2 \cdot 5 \times 10+1 \cdot 5 \times 50+410 \\
& =510 \mathrm{~min}^{\mathrm{m}} \\
\operatorname{AtB}(50,50), & \mathrm{Z}=2 \cdot 5 \times 50+1.5 \times 50+410 \\
& =125+75+410=610 \\
\operatorname{AtC}(60,40), & \mathrm{Z}=2.5 \times 60+1.5 \times 40+410 \\
& =150+60+410=620 \\
\text { At D }(60,0), & \mathrm{Z}=2.5 \times 60+1.5 \times 0+410 \\
& =150+410=660
\end{array}
$$
Thus $\mathrm{Z}$ is minimum i.e., ₹ 510 at $\mathrm{A}$.
i.e., when $x=10$ and $y=50$
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