28. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L, respectively. The company has to supply oil to three petrol pumps, D, E and F, whose requirements are 4500L, 3000L and 3500L, respectively. The distances (in km) between the depots and the petrol pumps are given in the following table:

Distance in (km)
From/To	A	B
D	7	3
E	6	4
F	3	2

Assuming that the transportation cost of 10 litres of oil is ?. 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, $\left(7000–x–y\right)$ will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.

Similarly $,\left(3000–y\right)Land3500–\left(7000–x–y\right)=\left(x+y–3500\right)$ L will be transported from depot B to petrol pumps E and F, respectively.

The given problem can be represented diagrammatically as given below:

$\begin{array}{}\end{array}x\ge 0,y\ge 0,and\left(7000–x–y\right)\ge 0$

$Then,x\ge 0,y\ge 0,andx+y\le 7000$

$4500–x\ge 0,3000–y\ge 0,andx+y–3500\ge 0$

$Then,x\le 4500,y\le 3000,andx+y\ge 3500$

Cost of transporting 10 L of petrol = Rs. 1

Cost of transporting 1 L of petrol = Rs. 1/10

Hence, the total transportation cost is given by,

$\begin{array}{l}z = \left(7/10\right) x + \left(6/10\right) y + 3 / 10 \left(7000–x–y\right) + 3 / 10 \left(4500–x\right) + 4 / 10 \left(3000–y\right) + 2 / 10 \left(x+y–3500\right)\hfill \end{array}$

$\begin{array}{l}= 0.3x + 0.1y + 3950\hfill \end{array}$

The problem can be formulated as given below:

$Minimisez=0.3x+0.1y+3950\dots \dots \dots .\left(i\right)$

Subject to

...more

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, $\left(7000–x–y\right)$ will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.

Similarly $,\left(3000–y\right)Land3500–\left(7000–x–y\right)=\left(x+y–3500\right)$ L will be transported from depot B to petrol pumps E and F, respectively.

The given problem can be represented diagrammatically as given below:

$\begin{array}{}\end{array}x\ge 0,y\ge 0,and\left(7000–x–y\right)\ge 0$

$Then,x\ge 0,y\ge 0,andx+y\le 7000$

$4500–x\ge 0,3000–y\ge 0,andx+y–3500\ge 0$

$Then,x\le 4500,y\le 3000,andx+y\ge 3500$

Cost of transporting 10 L of petrol = Rs. 1

Cost of transporting 1 L of petrol = Rs. 1/10

Hence, the total transportation cost is given by,

$\begin{array}{l}z = \left(7/10\right) x + \left(6/10\right) y + 3 / 10 \left(7000–x–y\right) + 3 / 10 \left(4500–x\right) + 4 / 10 \left(3000–y\right) + 2 / 10 \left(x+y–3500\right)\hfill \end{array}$

$\begin{array}{l}= 0.3x + 0.1y + 3950\hfill \end{array}$

The problem can be formulated as given below:

$Minimisez=0.3x+0.1y+3950\dots \dots \dots .\left(i\right)$

Subject to constraints,

$\begin{array}{}\end{array}x+y\le 7000\dots \dots \dots .\left(ii\right)$

$x\le 4500\dots \dots \dots ..\left(iii\right)$

$y\le 3000\dots \dots .\left(iv\right)$

$x+y\ge 3500\dots \dots \dots \left(v\right)$

$x,y\ge 0\dots \dots \dots \dots \left(vi\right)$

The feasible region determined by the constraints is given below:

A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000) and E (500, 3000) are the corner points of the feasible region.

The values of z at these corner points are given below:

Corner Point	z = 0.3x + 0.1y + 3950
A (3500, 0)	5000
B (4500, 0)	5300
C (4500, 2500)	5550
D (4000, 3000)	5450
E (500, 3000)	4400	Minimum

The minimum value of z is 4400 at (500, 3000).

Hence, the oil supplied from depot A is 500 L, 3000 L and 3500 L, and from depot B is 4000 L, 0 L and 0 L to petrol pumps D, E and F, respectively.

Therefore, the minimum transportation cost is ?. 4400.

less

<p>Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So,&nbsp;<span title="Click to copy mathml"><math><mrow><mrow><mo>(</mo><mrow><mn>7</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>x</mi><mo>–</mo><mi>y</mi></mrow><mo>)</mo></mrow></mrow></math></span>&nbsp;will be supplied from A to petrol pump F.</p><p>The requirement at petrol pump D is 4500 L. Since x L are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.</p><p>Similarly&nbsp;<span title="Click to copy mathml"><math><mrow><mo>,</mo><mrow><mo>(</mo><mrow><mn>3</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>y</mi></mrow><mo>)</mo></mrow><mi>L</mi><mi>a</mi><mi>n</mi><mi>d</mi><mn>3</mn><mn>5</mn><mn>0</mn><mn>0</mn><mo>–</mo><mrow><mo>(</mo><mrow><mn>7</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>x</mi><mo>–</mo><mi>y</mi></mrow><mo>)</mo></mrow><mo>=</mo><mrow><mo>(</mo><mrow><mi>x</mi><mo>+</mo><mi>y</mi><mo>–</mo><mn>3</mn><mn>5</mn><mn>0</mn><mn>0</mn></mrow><mo>)</mo></mrow></mrow></math></span>&nbsp;L will be transported from depot B to petrol pumps E and F, respectively.</p><p>The given problem can be represented diagrammatically as given below:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1732774193phpFjki7A_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732774193phpFjki7A.jpeg" alt="" width="221" height="188"></picture></div></div><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"></mtable><mrow><mi>x</mi><mo>≥</mo><mn>0</mn><mo>,</mo><mi>y</mi><mo>≥</mo><mn>0</mn><mo>,</mo><mi>a</mi><mi>n</mi><mi>d</mi><mrow><mo>(</mo><mrow><mn>7</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>x</mi><mo>–</mo><mi>y</mi></mrow><mo>)</mo></mrow><mo>≥</mo><mn>0</mn></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>T</mi><mi>h</mi><mi>e</mi><mi>n</mi><mo>,</mo><mi>x</mi><mo>≥</mo><mn>0</mn><mo>,</mo><mi>y</mi><mo>≥</mo><mn>0</mn><mo>,</mo><mi>a</mi><mi>n</mi><mi>d</mi><mi>x</mi><mo>+</mo><mi>y</mi><mo>≤</mo><mn>7</mn><mn>0</mn><mn>0</mn><mn>0</mn></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mn>4</mn><mn>5</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>x</mi><mo>≥</mo><mn>0</mn><mo>,</mo><mn>3</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>y</mi><mo>≥</mo><mn>0</mn><mo>,</mo><mi>a</mi><mi>n</mi><mi>d</mi><mi>x</mi><mo>+</mo><mi>y</mi><mo>–</mo><mn>3</mn><mn>5</mn><mn>0</mn><mn>0</mn><mo>≥</mo><mn>0</mn></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>T</mi><mi>h</mi><mi>e</mi><mi>n</mi><mo>,</mo><mi>x</mi><mo>≤</mo><mn>4</mn><mn>5</mn><mn>0</mn><mn>0</mn><mo>,</mo><mi>y</mi><mo>≤</mo><mn>3</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>,</mo><mi>a</mi><mi>n</mi><mi>d</mi><mi>x</mi><mo>+</mo><mi>y</mi><mo>≥</mo><mn>3</mn><mn>5</mn><mn>0</mn><mn>0</mn></mrow></mrow></math></span></p><p>Cost of transporting 10 L of petrol = Rs. 1</p><p>Cost of transporting 1 L of petrol = Rs. 1/10</p><p>Hence, the total transportation cost is given by,</p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mi>z</mi>&nbsp;<mo>=</mo>&nbsp;<mrow><mo>(</mo><mrow><mn>7</mn><mo>/</mo><mn>1</mn><mn>0</mn></mrow><mo>)</mo></mrow>&nbsp;<mi>x</mi>&nbsp;<mo>+</mo>&nbsp;<mrow><mo>(</mo><mrow><mn>6</mn><mo>/</mo><mn>1</mn><mn>0</mn></mrow><mo>)</mo></mrow>&nbsp;<mi>y</mi>&nbsp;<mo>+</mo>&nbsp;<mn>3</mn>&nbsp;<mo>/</mo>&nbsp;<mn>1</mn><mn>0</mn>&nbsp;<mrow><mo>(</mo><mrow><mn>7</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>x</mi><mo>–</mo><mi>y</mi></mrow><mo>)</mo></mrow>&nbsp;<mo>+</mo>&nbsp;<mn>3</mn>&nbsp;<mo>/</mo>&nbsp;<mn>1</mn><mn>0</mn>&nbsp;<mrow><mo>(</mo><mrow><mn>4</mn><mn>5</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>x</mi></mrow><mo>)</mo></mrow>&nbsp;<mo>+</mo>&nbsp;<mn>4</mn>&nbsp;<mo>/</mo>&nbsp;<mn>1</mn><mn>0</mn>&nbsp;<mrow><mo>(</mo><mrow><mn>3</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>–</mo><mi>y</mi></mrow><mo>)</mo></mrow>&nbsp;<mo>+</mo>&nbsp;<mn>2</mn>&nbsp;<mo>/</mo>&nbsp;<mn>1</mn><mn>0</mn>&nbsp;<mrow><mo>(</mo><mrow><mi>x</mi><mo>+</mo><mi>y</mi><mo>–</mo><mn>3</mn><mn>5</mn><mn>0</mn><mn>0</mn></mrow><mo>)</mo></mrow></mtd></mtr></mtable></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mo>=</mo>&nbsp;<mn>0</mn><mo>.</mo><mn>3</mn><mi>x</mi>&nbsp;<mo>+</mo>&nbsp;<mn>0</mn><mo>.</mo><mn>1</mn><mi>y</mi>&nbsp;<mo>+</mo>&nbsp;<mn>3</mn><mn>9</mn><mn>5</mn><mn>0</mn></mtd></mtr></mtable></mrow></math></span></p><p>The problem can be formulated as given below:</p><p><span title="Click to copy mathml"><math><mrow><mi>M</mi><mi>i</mi><mi>n</mi><mi>i</mi><mi>m</mi><mi>i</mi><mi>s</mi><mi>e</mi><mi>z</mi><mo>=</mo><mn>0</mn><mo>.</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>0</mn><mo>.</mo><mn>1</mn><mi>y</mi><mo>+</mo><mn>3</mn><mn>9</mn><mn>5</mn><mn>0</mn><mo>…</mo><mo>…</mo><mo>…</mo><mo>.</mo><mrow><mo>(</mo><mrow><mi>i</mi></mrow><mo>)</mo></mrow></mrow></math></span></p><p>Subject to constraints,</p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"></mtable><mrow><mi>x</mi><mo>+</mo><mi>y</mi><mo>≤</mo><mn>7</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>…</mo><mo>…</mo><mo>…</mo><mo>.</mo><mrow><mo>(</mo><mrow><mi>i</mi><mi>i</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>x</mi><mo>≤</mo><mn>4</mn><mn>5</mn><mn>0</mn><mn>0</mn><mo>…</mo><mo>…</mo><mo>…</mo><mo>.</mo><mo>.</mo><mrow><mo>(</mo><mrow><mi>i</mi><mi>i</mi><mi>i</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>y</mi><mo>≤</mo><mn>3</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>…</mo><mo>…</mo><mo>.</mo><mrow><mo>(</mo><mrow><mi>i</mi><mi>v</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>x</mi><mo>+</mo><mi>y</mi><mo>≥</mo><mn>3</mn><mn>5</mn><mn>0</mn><mn>0</mn><mo>…</mo><mo>…</mo><mo>…</mo><mrow><mo>(</mo><mrow><mi>v</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>x</mi><mo>,</mo><mi>y</mi><mo>≥</mo><mn>0</mn><mo>…</mo><mo>…</mo><mo>…</mo><mo>…</mo><mrow><mo>(</mo><mrow><mi>v</mi><mi>i</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p>The feasible region determined by the constraints is given below:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1732774210phpuLEEaT_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732774210phpuLEEaT.jpeg" alt="" width="267" height="256"></picture></div></div><p>A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000) and E (500, 3000) are the corner points of the feasible region.</p><p>The values of z at these corner points are given below:</p><table><tbody><tr><td><p>Corner Point</p></td><td><p>z = 0.3x + 0.1y + 3950</p></td><td>&nbsp;</td></tr><tr><td><p>A (3500, 0)</p></td><td><p>5000</p></td><td>&nbsp;</td></tr><tr><td><p>B (4500, 0)</p></td><td><p>5300</p></td><td>&nbsp;</td></tr><tr><td><p>C (4500, 2500)</p></td><td><p>5550</p></td><td>&nbsp;</td></tr><tr><td><p>D (4000, 3000)</p></td><td><p>5450</p></td><td>&nbsp;</td></tr><tr><td><p>E (500, 3000)</p></td><td><p>4400</p></td><td><p>Minimum</p></td></tr></tbody></table><p>The minimum value of z is 4400 at (500, 3000).</p><p>Hence, the oil supplied from depot A is 500 L, 3000 L and 3500 L, and from depot B is 4000 L, 0 L and 0 L to petrol pumps D, E and F, respectively.</p><p>Therefore, the minimum transportation cost is ?. 4400.</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment