Advertisements
Advertisements
Question
An oil company has two depots, A and B, with capacities of 7000 litres and 4000 litres respectively. The company is to supply oil to three petrol pumps, D, E, F whose requirements are 4500, 3000 and 3500 litres respectively. The distance (in km) between the depots and petrol pumps is given in the following table:
Figure
Assuming that the transportation cost per km is Rs 1.00 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
Advertisements
Solution
Let x and y litres of oil be supplied from A to the petrol pumps D and E. Then, (7000 − x − y) L 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, (3000 − y) L and [3500 − (7000 − x − y)] L i.e. (x + y − 3500) L will be transported from depot B to petrol pump E and F. respectively.
The given problem can be represented diagrammatically as follows. 
Since, quantity of oil are non-negative quantities.Therefore,
x ≥0 , y ≥ 0, and (7000 - x - y ) ≥ 0
⇒ x ≥ 0 , y ≥ 0, and x + y ≥ 7000
4500 - x ≥ 0 , 3000 - y ≥ 0 , and x + y -3500 ≥ 0
⇒ x ≤ 4500, y ≤ 3000, and x + y ≥ 3500
Cost of transporting 10 L of petrol = Re 1
Cost of transporting 1 L of petrol = \[Rs \frac{1}{10}\]
Therefore, total transportation cost is given by,
`z = 7/10 xx x+6/10 y + 3/10 (7000 - x -y ) + 3/10 (4500 - x) + 4/10 ( 3000 - y ) + 2/10( x + y - 3500)`
= 0.3x + 0.1y + 3950
The problem can be formulated as follows.
Minimize Z = 0.3x + 0.1y + 3950
subject to the constraints,
\[x + y \leq 7000\]
\[x \leq 4500\]
\[y \leq 3000\]
\[x + y \geq 3500\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows:
x + y = 7000, x = 4500, y = 3000, x + y = 3500, x = 0 and y = 0
Region represented by x + y ≤ 7000:
The line x + y = 7000 meets the coordinate axes at A1(7000, 0) and \[B_1 \left( 0, 7000 \right)\] respectively. By joining these points we obtain the line x + y = 7000 . Clearly (0,0) satisfies the x + y = 7000 . So, the region which contains the origin represents the solution set of the inequation x + y ≤ 7000.
Region represented by x ≤ 4500:
The line x = 4500 is the line passes through C1(4500, 0) and is parallel to Y axis. The region to the left of the line x = 4500 will satisfy the inequation
x ≤ 4500.
Region represented by y ≤ 3000:
The line y = 3000 is the line passes through D1(0, 3000) and is parallel to X axis. The region below the the line y = 3000 will satisfy the inequation
y ≤ 3000.
Region represented by x + y ≥ 3500:
The line x + y = 7000 meets the coordinate axes at E1(3500, 0) and \[F_1 \left( 0, 3500 \right)\] respectively. By joining these points we obtain the line
x + y = 3500 . Clearly (0,0) satisfies the x + y = 3500. So, the region which contains the origin represents the solution set of the inequation x + y ≥ 3500.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 7000, x ≤ 4500, y ≤ 3000, x + y ≥ 3500, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E1(3500, 0), C1(4500, 0), I1(4500, 2500), H1(4000, 3000), and G1(500, 3000).
The values of Z at these corner points are as follows.
| Corner point | Z = 0.3x + 0.1y + 3950 |
| E1(3500, 0) | 5000 |
| C1(4500, 0) | 5300 |
| I1(4500, 2500) | 5550 |
| H1(4000, 3000) | 5450 |
| G1(500, 3000) | 4400 |
The minimum value of Z is 4400 at G1(500, 3000).
Thus, 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.
The minimum transportation cost is Rs 4400.
APPEARS IN
RELATED QUESTIONS
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at Rs 7 profit and B at a profit of Rs 4. Find the production level per day for maximum profit graphically.
Solve the following L.P.P. graphically:
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0 and x, y ≥ 0
Solve the following L.P.P. graphically Maximise Z = 4x + y
Subject to following constraints x + y ≤ 50
3x + y ≤ 90,
x ≥ 10
x, y ≥ 0
In order to supplement daily diet, a person wishes to take X and Y tablets. The contents (in milligrams per tablet) of iron, calcium and vitamins in X and Y are given as below :
| Tablets | Iron | Calcium | Vitamin |
| x | 6 | 3 | 2 |
| y | 2 | 3 | 4 |
The person needs to supplement at least 18 milligrams of iron, 21 milligrams of calcium and 16 milligrams of vitamins. The price of each tablet of X and Y is Rs 2 and Rs 1 respectively. How many tablets of each type should the person take in order to satisfy the above requirement at the minimum cost? Make an LPP and solve graphically.
Maximize Z = 5x + 3y
Subject to
\[3x + 5y \leq 15\]
\[5x + 2y \leq 10\]
\[ x, y \geq 0\]
Maximize Z = 50x + 30y
Subject to
\[2x + y \leq 18\]
\[3x + 2y \leq 34\]
\[ x, y \geq 0\]
Maximize Z = 4x + 3y
subject to
\[3x + 4y \leq 24\]
\[8x + 6y \leq 48\]
\[ x \leq 5\]
\[ y \leq 6\]
\[ x, y \geq 0\]
Maximize Z = 3x1 + 4x2, if possible,
Subject to the constraints
\[x_1 - x_2 \leq - 1\]
\[ - x_1 + x_2 \leq 0\]
\[ x_1 , x_2 \geq 0\]
Solved the following linear programming problem graphically:
Maximize Z = 60x + 15y
Subject to constraints
\[x + y \leq 50\]
\[3x + y \leq 90\]
\[ x, y \geq 0\]
Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, y ≥ 0
To maintain one's health, a person must fulfil certain minimum daily requirements for the following three nutrients: calcium, protein and calories. The diet consists of only items I and II whose prices and nutrient contents are shown below:
| Food I | Food II | Minimum daily requirement | |
| Calcium Protein Calories |
10 5 2 |
4 6 6 |
20 20 12 |
| Price | Rs 0.60 per unit | Rs 1.00 per unit |
Find the combination of food items so that the cost may be minimum.
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
| Food | Vitamin A | Vitamin B | Vitamin C |
| X | 1 | 2 | 3 |
| Y | 2 | 2 | 1 |
One kg of food X costs ₹16 and one kg of food Y costs ₹20. Find the least cost of the mixture which will produce the required diet?
A manufacturer makes two types A and B of tea-cups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below:
| Machines | |||
| I | II | III | |
| A B |
12 6 |
18 0 |
6 9 |
Each machine is available for a maximum of 6 hours per day. If the profit on each cup A is 75 paise and that on each cup B is 50 paise, show that 15 tea-cups of type A and 30 of type B should be manufactured in a day to get the maximum profit.
A firm manufactures headache pills in two sizes A and B. Size A contains 2 grains of aspirin, 5 grains of bicarbonate and 1 grain of codeine; size B contains 1 grain of aspirin, 8 grains of bicarbonate and 66 grains of codeine. It has been found by users that it requires at least 12 grains of aspirin, 7.4 grains of bicarbonate and 24 grains of codeine for providing immediate effects. Determine graphically the least number of pills a patient should have to get immediate relief. Determine also the quantity of codeine consumed by patient.
A chemical company produces two compounds, A and B. The following table gives the units of ingredients, C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.
| Compound | Minimum requirement | ||
| A | B | ||
| Ingredient C Ingredient D |
1 3 |
2 1 |
80 75 |
| Cost (in Rs) per kg | 4 | 6 | - |
A manufacturer makes two products A and B. Product A sells at Rs 200 each and takes 1/2 hour to make. Product B sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consists of 40 hours of production and weekly turnover must not be less than Rs 10000. If the profit on each of product A is Rs 20 and on product B is Rs 30, then how many of each should be produced so that the profit is maximum. Also, find the maximum profit.
A man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 sq.m of ground per tree and costs Rs 20 per tree and type B requires 20 sq.m of ground per tree and costs Rs 25 per tree. When fully grown, type A produces an average of 20 kg of fruit which can be sold at a profit of Rs 2.00 per kg and type B produces an average of 40 kg of fruit which can be sold at a profit of Rs. 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit?
A company sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of unit of A that can be sold is 70 and that for B is 125. If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B, how many units of each product should be sold to maximize profit?
A manufacturing company makes two models A and B of a product. Each piece of model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of ₹8000 on each piece of model A and ₹12000 on each piece of model B. How many pieces of model A and model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week?
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:
| From \ To | Cost (in ₹) | ||
| A | B | C | |
| P | 160 | 100 | 150 |
| Q | 100 | 120 | 100 |
How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?
There are two types of fertilisers 'A' and 'B' . 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If 'A' costs ₹10 per kg and 'B' cost ₹8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requiremnets are met at a minimum cost
By graphical method, the solution of linear programming problem
\[\text{ Subject } to \text{ 3 } x_1 + 2 x_2 \leq 18\]
\[ x_1 \leq 4\]
\[ x_2 \leq 6\]
\[ x_1 \geq 0, x_2 \geq 0, \text{ is } \]
The point at which the maximum value of x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is obtained, is ______.
The graph of the inequality 3X − 4Y ≤ 12, X ≤ 1, X ≥ 0, Y ≥ 0 lies in fully in
Find the feasible solution of linear inequation 2x + 3y ≤ 12, 2x + y ≤ 8, x ≥ 0, y ≥ 0 by graphically
The minimum value of z = 2x + 9y subject to constraints x + y ≥ 1, 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 is ______.
The maximum value of z = 3x + 10y subjected to the conditions 5x + 2y ≤ 10, 3x + 5y ≤ 15, x, y ≥ 0 is ______.
If 4x + 5y ≤ 20, x + y ≥ 3, x ≥ 0, y ≥ 0, maximum 2x + 3y is ______.
The minimum value of z = 7x + 9y subject to 3x + y ≤ 6, 5x + 8y ≤ 40, x ≥ 0, y ≥ 2 is ______.
Maximise and Minimise Z = 3x – 4y subject to x – 2y ≤ 0, – 3x + y ≤ 4, x – y ≤ 6, x, y ≥ 0
A manufacturer wishes to produce two commodities A and B. The number of units of material, labour and equipment needed to produce one unit of each commodity is shown in the table given below. Also shown is the available number of units of each item, material, labour, and equipment.
| Items | Commodity A | Commodity B | Available no. of Units |
| Material | 1 | 2 | 8 |
| Labour | 3 | 2 | 12 |
| Equipment | 1 | 1 | 10 |
Find the maximum profit if each unit of commodity A earns a profit of ₹ 2 and each unit of B earns a profit of ₹ 3.
Any point in the feasible region that gives the optional value (maximum or minimum) of the objective function is called:-
The constraints –x1 + x2 ≤ 1, –x1 + 3x2 ≤ 9, x1x2 ≥ 0 define on ______.
Solve the following Linear Programming Problem graphically:
Maximize: z = – x + 2y,
Subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
What is an optimal solution?
For a bounded feasible region, where do the maximum and minimum values occur?
