Advertisements
Advertisements
Question
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?
Advertisements
Solution
Here, demand of the commodity (5 + 5 + 4 = 14 units) is equal the supply of the commodity (8 + 6 = 14 units). So, no commodity would be left at the two factories.
Let x units and y units of the commodity be transported from the factory P to the depots at A and B, respectively.
Then, (8 − x − y) units of the commodity will be transported from the factory P to the depot C.
Now, the weekly requirement of depot A is 5 units of the commodity. Now, x units of the commodity are transported from factory P, so the remaining (5 − x) units of the commodity are transported from the factory Q to the depot A.
The weekly requirement of depot B is 5 units of the commodity. Now, y units of the commodity are transported from factory P, so the remaining (5 − y) units of the commodity are transported from the factory Q to the depot B.
Similarly, 6 − (5 − x) − (5 − y) = (x + y − 4) units of the commodity will be transported from the factory Q to the depot C.
Since the number of units of commodity transported are from the factories to the depots are non-negative, therefore,
x ≥ 0, y ≥ 0, 8 − x − y ≥ 0, 5 − x ≥ 0, 5 − y ≥ 0, x + y − 4 ≥ 0
Or x ≥ 0, y ≥ 0, x + y ≤ 8, x ≤ 5, y ≤ 5, x + y ≥ 4
Total transportation cost = 160x + 100y + 150(8 − x − y) + 100(5 − x) + 120(5 − y) + 100(x + y − 4) = 10x − 70y + 1900
Thus, the given linear programming problem is
Minimise Z = 10x − 70y + 1900
subject to the constraints
x + y ≤ 8
x ≤ 5
y ≤ 5
x + y ≥ 4
x ≥ 0, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(4, 0), B(5, 0), C(5, 3), D(3, 5), E(0, 5) and F(0, 4).
The value of the objective function at these points are given in the following table.
| Corner Point | Z = 10x − 70y + 1900 |
| (4, 0) | 10 × 4 − 70 × 0 + 1900 = 1940 |
| (5, 0) | 10 × 5 − 70 × 0 + 1900 = 1950 |
| (5, 3) | 10 × 5 − 70 × 3 + 1900 = 1740 |
| (3, 5) | 10 × 3 − 70 × 5 + 1900 = 1580 |
| (0, 5) | 10 × 0 − 70 × 5 + 1900 = 1550 → Minimum |
| (0, 4) | 10 × 0 − 70 × 4 + 1900 = 1620 |
The minimum value of Z is 1550 at x = 0, y = 5.
Hence, for minimum transportation cost, factory P should supply 0, 5, 3 units of commodity to depots A, B, C respectively and factory Q should supply 5, 0, 1 units of commodity to depots A, B, C respectively.
The minimum transportation cost is ₹1,550.
APPEARS IN
RELATED QUESTIONS
Minimize: Z = 6x + 4y
Subject to the conditions:
3x + 2y ≥ 12,
x + y ≥ 5,
0 ≤ x ≤ 4,
0 ≤ y ≤ 4
Solve the following L.P.P graphically:
Maximize: Z = 10x + 25y
Subject to: x ≤ 3, y ≤ 3, x + y ≤ 5, x ≥ 0, y ≥ 0
A company manufactures bicycles and tricycles each of which must be processed through machines A and B. Machine A has maximum of 120 hours available and machine B has maximum of 180 hours available. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B. Manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B.
If profits are Rs. 180 for a bicycle and Rs. 220 for a tricycle, formulate and solve the L.P.P. to determine the number of bicycles and tricycles that should be manufactured in order to maximize the profit.
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 L. P. P. graphically:Linear Programming
Minimize Z = 6x + 2y
Subject to
5x + 9y ≤ 90
x + y ≥ 4
y ≤ 8
x ≥ 0, 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
Solve the following LPP graphically :
Maximise Z = 105x + 90y
subject to the constraints
x + y ≤ 50
2x + y ≤ 80
x ≥ 0, 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 = 9x + 3y
Subject to
2x + 3y ≤ 13
3x + y ≤ 5
x, y ≥ 0
Maximize Z = 10x + 6y
Subject to
\[3x + y \leq 12\]
\[2x + 5y \leq 34\]
\[ x, y \geq 0\]
Maximize Z = 3x + 4y
Subject to
\[2x + 2y \leq 80\]
\[2x + 4y \leq 120\]
Maximize Z = 7x + 10y
Subject to
\[x + y \leq 30000\]
\[ y \leq 12000\]
\[ x \geq 6000\]
\[ x \geq y\]
\[ x, y \geq 0\]
Maximize Z = 2x + 3y
Subject to
\[x + y \geq 1\]
\[10x + y \geq 5\]
\[x + 10y \geq 1\]
\[ x, y \geq 0\]
One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no storage of the other ingredients used in making the cakes.
If a young man drives his vehicle at 25 km/hr, he has to spend Rs 2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs 5/per km. He has Rs 100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
A publisher sells a hard cover edition of a text book for Rs 72.00 and paperback edition of the same ext for Rs 40.00. Costs to the publisher are Rs 56.00 and Rs 28.00 per book respectively in addition to weekly costs of Rs 9600.00. Both types require 5 minutes of printing time, although hardcover requires 10 minutes binding time and the paperback requires only 2 minutes. Both the printing and binding operations have 4,800 minutes available each week. How many of each type of book should be produced in order to maximize profit?
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 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 small firm manufacturers items A and B. The total number of items A and B that it can manufacture in a day is at the most 24. Item A takes one hour to make while item B takes only half an hour. The maximum time available per day is 16 hours. If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160, how many of each type of item be produced to maximize the profit? Solve the problem graphically.
A firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically.
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 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 makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
| Types of Toys | Machines | ||
| I | II | III | |
| A | 12 | 18 | 6 |
| B | 6 | 0 | 9 |
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.
From the details given below, calculate the five-year moving averages of the number of students who have studied in a school. Also, plot these and original data on the same graph paper.
| Year | 1993 | 1994 | 1995 | 1996 | 1997 | 1998 | 1999 | 2000 | 2001 | 2002 |
| Number of Students | 332 | 317 | 357 | 392 | 402 | 405 | 410 | 427 | 405 | 438 |
Find the feasible solution of linear inequation 2x + 3y ≤ 12, 2x + y ≤ 8, x ≥ 0, y ≥ 0 by graphically
The minimum value of z = 10x + 25y subject to 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≥ 5 is ______.
The maximum value of Z = 5x + 4y, Subject to y ≤ 2x, x ≤ 2y, x + y ≤ 3, x ≥ 0, y ≥ 0 is ______.
The feasible region of an LPP is shown in the figure. If z = 3x + 9y, then the minimum value of z occurs at ______.
The point which provides the solution to the linear programming problem: Max P = 2x + 3y subject to constraints: x ≥ 0, y ≥ 0, 2x + 2y ≤ 9, 2x + y ≤ 7, x + 2y ≤ 8, is ______
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 ____________.
Which of the statements describe the solution set for `-2(x + 8) = - 2x + 20`?
If x – y ≥ 8, x ≥ 3, y ≥ 3, x ≥ 0, y ≥ 0 then find the coordinates of the corner points of the feasible region.
Find feasible solution for the following system of linear inequation graphically.
3x + 4y ≥ 12, 4x + 7y ≤ 28, x ≥ 0, y ≥ 0
A linear programming problem is given by Z = px + qy, where p, q > 0 subject to the constraints x + y ≤ 60, 5x + y ≤ 100, x ≥ 0 and y ≥ 0.
- Solve graphically to find the corner points of the feasible region.
- If Z = px + qy is maximum at (0, 60) and (10, 50), find the relation of p and q. Also mention the number of optimal solution(s) in this case.
