Advertisements
Advertisements
Question
Tow godowns, A and B, have grain storage 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:
| Transportation cost per quintal(in Rs.) | ||
| From-> | A | B |
| To | ||
| D | 6.00 | 4.00 |
| E | 3.00 | 2.00 |
| F | 2.50 | 3.00 |
How should the supplies be transported in order that the transportation cost is minimum?
Advertisements
Solution
Let godown A supply x quintals and y quintals of grain to the shops D and E respectively.
Then, (100 − x − y) will be supplied to shop F.
The requirement at shop D is 60 quintals since, x quintals are transported from godown A.
Therefore, the remaining (60 − x) quintals will be transported from godown B.
Similarly, (50 − y) quintals and 40 − (100 − x − y) i.e. (x + y − 60) quintals will be transported from godown B to shop E and F respectively.
The given problem can be represented diagrammatically as follows.
Quantity of the grain cannot be negative.Therefore,
x ≥ 0 , y ≥ 0 and 100 - x - y ≥ 0
⇒ x ≥ 0 , y ≥ 0 , and x + y ≤ 100
60 - x ≥ 0 , 50 - y ≥ 0 , and x + y - 60 ≥ 0
⇒ x ≤ 60 , y ≤ 50 , and x + y ≥ 60
Total transportation cost Z is given by,
\[Z = 6x + 3y + 2 . 5\left( 100 - x - y \right) + 4\left( 60 - x \right) + 2\left( 50 - y \right) + 3\left( x + y - 60 \right)\]
\[ = 6x + 3y + 250 - 2 . 5x - 2 . 5y + 240 - 4x + 100 - 2y + 3x + 3y - 180\]
\[ = 2 . 5x + 1 . 5y + 410\]
The given problem can be formulated as:
Minimize Z = 2.5x + 1.5y + 410
subject to the constraints,
\[x + y \leq 100\]
\[x \leq 60\]
\[y \leq 50\]
\[x + y \geq 60\]
\[x, y \geq 0\]
First we will convert inequations into equations as follows:
x + y = 100, x = 60, y = 50, x + y =60, x = 0 and y = 0
Region represented by x + y ≤ 100:
The line x + y = 100 meets the coordinate axes at A1(100, 0) and B1(0, 100) respectively. By joining these points we obtain the line x + y = 100. Clearly (0,0) satisfies the x + y = 100. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 100.
Region represented by x ≤ 60:
x = 60 is the line that passes (60, 0) and is parallel to the Y axis.The region to the left of the line x = 60 will satisfy the inequation x ≤ 60.
Region represented by y ≤ 50:
y = 50 is the line that passes (0, 50) and is parallel to the X axis.The region below the line y = 50 will satisfy the inequation y ≤ 50.
Region represented by x + y ≥ 60:
The line x + y = 60 meets the coordinate axes at C1(60, 0) and \[D_1 \left( 0, 60 \right)\] respectively. By joining these points we obtain the line x + y = 60. Clearly (0,0) does not satisfies the inequation x + y ≥ 60. So,the region which does not contain the origin represents the solution set of the inequation x + y ≥ 60.
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 ≤ 100, x ≤ 60, y ≤ 50, x + y ≥ 60, x ≥ 0 and y ≥ 0 are as follows.
The corner points are C1(60, 0), G1(60, 40), F1(50, 50), and E1(10, 50).
The values of Z at these corner points are as follows.
| Corner point | Z = 2.5x + 1.5y + 410 |
| C1(60, 0) | 560 |
| G1(60, 40) | 620 |
| F1(50, 50) | 610 |
| E1(10, 50) | 510 |
The minimum value of Z is 510 at E1(10, 50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.
The minimum cost is Rs 510.
APPEARS IN
RELATED QUESTIONS
Minimize :Z=6x+4y
Subject to : 3x+2y ≥12
x+y ≥5
0 ≤x ≤4
0 ≤ y ≤ 4
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 Rs 10 per kg and 'B' cost Rs 8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost
Solve the following LPP by graphical method:
Maximize: z = 3x + 5y
Subject to: x + 4y ≤ 24
3x + y ≤ 21
x + y ≤ 9
x ≥ 0, y ≥ 0
Also find the maximum value of z.
Solve the following linear programming problem graphically :
Maximise Z = 7x + 10y subject to the constraints
4x + 6y ≤ 240
6x + 3y ≤ 240
x ≥ 10
x ≥ 0, y ≥ 0
Minimize Z = 2x + 4y
Subject to
\[x + y \geq 8\]
\[x + 4y \geq 12\]
\[x \geq 3, y \geq 2\]
Maximize Z = 3x1 + 4x2, if possible,
Subject to the constraints
\[x_1 - x_2 \leq - 1\]
\[ - x_1 + x_2 \leq 0\]
\[ x_1 , x_2 \geq 0\]
Find the minimum value of 3x + 5y subject to the constraints
− 2x + y ≤ 4, x + y ≥ 3, x − 2y ≤ 2, x, y ≥ 0.
A hospital dietician wishes to find the cheapest combination of two foods, A and B, that contains at least 0.5 milligram of thiamin and at least 600 calories. Each unit of Acontains 0.12 milligram of thiamin and 100 calories, while each unit of B contains 0.10 milligram of thiamin and 150 calories. If each food costs 10 paise per unit, how many units of each should be combined at a minimum cost?
A dietician mixes together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below:
| Vitamin A |
Vitamin B |
Vitamin |
Vitamin D |
|
| Food X Food Y |
1 2 |
1 1 |
1 3 |
2 1 |
One kg food X costs Rs 5, whereas one kg of food Y costs Rs 8. Find the least cost of the mixture which will produce the desired diet.
A fruit grower can use two types of fertilizer in his garden, brand P and Q. The amounts (in kg) of nirogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicates that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
| kg per bag | ||
| Brand P | Brand P | |
| Nitrogen | 3 | 3.5 |
| Phosphoric acid | 1 | 2 |
| Potash | 3 | 1.5 |
| Chlorine | 1.5 | 2 |
If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
A manufacturer has three machines installed in his factory. machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:
| Item | Number of hours required by the machine | ||
A B |
I | II | III |
| 1 2 |
2 1 |
1 5/4 |
|
He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B. Assuming that he can sell all that he produces, how many of each item should he produces so as to maximize his profit? Determine his maximum profit. Formulate this LPP mathematically and then solve it.
A factory manufactures two types of screws, A and B, each type requiring the use of two machines - an automatic and a hand-operated. It takes 4 minute on the automatic and 6 minutes on the hand-operated machines to manufacture a package of screws 'A', while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws 'A' at a profit of 70 P and screws 'B' at a profit of Rs 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
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 manufacturer of Furniture makes two products : chairs and tables. processing of these products is done on two machines A and B. A chair requires 2 hrs on machine A and 6 hrs on machine B. A table requires 4 hrs on machine A and 2 hrs on machine B. There are 16 hrs of time per day available on machine A and 30 hrs on machine B. Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively. Find with the help of graph what should be the daily production of each of the two products so as to maximize his profit.
A factory uses three different resources for the manufacture of two different products, 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.
A company manufactures two types of novelty Souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 paise each for type A and 60 paise each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp while it takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at most 20 hours and the grinding/cutting machine for at most 12 hours. The profit from the sale of a lamp is ₹5.00 and a shade is ₹3.00. Assuming that the manufacturer sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
A producer has 30 and 17 units of labour and capital respectively which he can use to produce two type of goods x and y. To produce one unit of x, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit of capital is required to produce one unit of y. If x and y are priced at Rs 100 and Rs 120 per unit respectively, how should be producer use his resources to maximize the total revenue? Solve the problem graphically.
A company manufactures two types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
If a young man drives his vehicle at 25 km/hr, he has to spend ₹2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to ₹5 per km. He has ₹100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
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?
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and that on a chain is Rs 190, find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an LPP and solve it graphically.
A cooperative society of farmers has 50 hectares of land to grow two crops X and Y. The profits from crops X and Y per hectare are estimated as ₹10,500 and ₹9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at the rate of 20 litres and 10 litres per hectare, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society?
A manufacturer has three machine I, II, III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit each of M and N on the three machines are given in the following table:
| Items | Number of hours required on machines | ||
| I | II | III | |
| M | 1 | 2 | 1 |
| N | 2 | 1 | 1.25 |
She makes a profit of ₹600 and ₹400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?
A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at ₹100 and ₹120 per unit respectively, how should he use his resources to maximise the total revenue? Form the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?
A carpenter has 90, 80 and 50 running feet respectively of teak wood, plywood and rosewood which is used to product A and product B. Each unit of product A requires 2, 1 and 1 running feet and each unit of product B requires 1, 2 and 1 running feet of teak wood, plywood and rosewood respectively. If product A is sold for Rs. 48 per unit and product B is sold for Rs. 40 per unit, how many units of product A and product B should be produced and sold by the carpenter, in order to obtain the maximum gross income? Formulate the above as a Linear Programming Problem and solve it, indicating clearly the feasible region in the graph.
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A
require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours and 20 minutes available for cutting and 4 hours available for assembling. The profit is Rs. 50 each for type A and Rs. 60 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize profit? Formulate the above LPP and solve it graphically and also find the maximum profit.
A company manufactures two types of products A and B. Each unit of A requires 3 grams of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 grams of chromium. The firm can produce 9 grams of nickel and 8 grams of chromium. The profit is ₹ 40 on each unit of the product of type A and ₹ 50 on each unit of type B. How many units of each type should the company manufacture so as to earn a maximum profit? Use linear programming to find the solution.
The graph of the inequality 3X − 4Y ≤ 12, X ≤ 1, X ≥ 0, Y ≥ 0 lies in fully in
Draw the graph of inequalities x ≤ 6, y −2 ≤ 0, x ≥ 0, y ≥ 0 and indicate the feasible region
The maximum value of z = 6x + 8y subject to x - y ≥ 0, x + 3y ≤ 12, x ≥ 0, y ≥ 0 is ______.
Area of the region bounded by y = cos x, x = 0, x = π and X-axis is ______ sq.units.
Maximise and Minimise Z = 3x – 4y subject to x – 2y ≤ 0, – 3x + y ≤ 4, x – y ≤ 6, x, y ≥ 0
Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and ____________.
In Corner point method for solving a linear programming problem the first step is to ____________.
The objective function Z = ax + by of an LPP has maximum vaiue 42 at (4, 6) and minimum value 19 at (3, 2). Which of the following is true?
Solve the following Linear Programming Problem graphically.
Maximise Z = 5x + 2y subject to:
x – 2y ≤ 2,
3x + 2y ≤ 12,
– 3x + 2y ≤ 3,
x ≥ 0, y ≥ 0
