Formulation of Linear Programming Problem(LPP) as a mathematical model consist of the following basic steps:
- Find the key-decision to be made from the study of the solution.
- Identify the variables.
- Express the possible alternatives mathematically in terms of variables.
- Mention the objective quantitatively & express it as a linear function of variables.
- Express the constraints as linear equations/ in-equations in terms of variables.
Example 01
Problem:
Consider, a factory manufactures article(product) A and B. A certain machine requires 1.5 hours and in addition a craftsman requires 2 hours to manufacture article A. Similarly, the machine requires 2.5 hours and in addition the craftsman requires 1.5 hours to manufacture article B.In a week the factory can avail 80hours of machine and 70 hours for craftsman's time. The profit on each article A is Rs. 5/- and on each article B is Rs. 4/-. If all the articles produced can be sold away, find how many of each kind of articles should be produced to earn maximum profit per week.
Solution:
The given data can be summarized in the following table:
Decision Article Hours on Profit per
Variable Machine Craftsman unit
-----------------------------------------------------------X A 1.5 2.0 Rs. 5/-
Y B 2.5 1.5 Rs. 4/-
-----------------------------------------------------------
Hours available 80 70
(per week) (Maximum) (Maximum)
Let, X and Y be the number of articles A and B to be manufactured, respectively.
Objective Function:
Total profit can be earned from article A is 5X.
Total profit can be earned form article B is 4Y.
Let, Z be the total profit of the factory(per week).
It can be expressed algebraically,
maximize Z= 5X + 4Y
This is called Objective function.
Constraints:
- Total Machine time <= 80, can be expressed as 1.5X + 2.5Y <= 80
- Total craftsman time <= 70, can be expressed as 2X + 1.5Y <= 70
- Non-negativity constraints X>=0 & Y>=0.
So, the LP problem becomes as,
Maximize Z = 5X + 4Y
Subject to,
1.5X + 2.5Y <= 80
2X + 1.5Y <= 70
X >= 0 and Y >= 0
Example 02
Problem:
Consider,
vitamin A and B are found in two different foods F1 and F2. One unit of F1 contains 2 units of vitamin A and 3 units of vitamin B. One unit of F2 contains 4 units of vitamin A and 2 units of vitamin B.One unit of food F1 and F2 cost Rs. 5 and Rs. 2.5 respectively. Minimum daily requirement of vitamin A and B for a person is 40 and 50 units respectively. Find the optimal mix of food F1 & F2 that satisfy the requirement with minimum cost.
Solution:
The given data can be summarized in the following table:
Decision Food Vitamin Cost per
Variable A B unit
-----------------------------------------------------------X F1 2 3 Rs. 5/-
Y F2 4 2 Rs. 2.5/-
-----------------------------------------------------------
Daily Requirement 40 50
(per day) (Minimum) (Minimum)
Let, X and Y be the amount of food F1 and F2, respectively.
Objective Function:
Total cost can be for food F1 is 5X.
Total cost can be for food F2 is 2.5Y.
Let, Z be the total cost for food F1 and F2 together(per day).
It can be expressed algebraically,
minimized Z= 5X + 2.5Y
This is called Objective function.
Constraints:
- Requirement of vitamin A >= 40, can be expressed as 2X + 4Y >= 40
- Requirement of vitamin B >= 50, can be expressed as 3X + 2Y >= 50
- Non-negativity constraints X>=0 & Y>=0.
So, the LP problem becomes as,
Minimize Z = 5X + 2.5Y
Subject to,
2X + 4Y >= 40
3X + 2Y >= 50
X >= 0 and Y >= 0
Try Your Own
- There are three products P, Q, R from three raw material A, B, C. One unit of P requires 2 units of A and 3 units of B. One unit of Q requires 2 units of B and 5 units of C. One unit of R requires 3 units of A, 2 units of B and 4 units of C.Profit per unit for P, Q, R are Rs. 3, Rs. 5 and Rs. 4, respectively. Availability of the raw material are 8 units of A, 10 units of B and 15 units of C. Formulate the problem mathematically to maximize the profit.
- There are two products A and B.Cost of production(per unit) of A and B are Rs. 60 and Rs. 80, respectively. The company has to supply at least 200 units of B. One unit of product A requires 1 machine hour whereas B has a machine hours available abundantly within the company. Total machine hour available for product A is 400 hours. One unit of each product requires 1 labour hour. Total labour hour available are 500. Formulate the LP problem to minimize production cost.
- Consider the following data
----------------------------------------------------------------
Food1 Food2 Minimum Daily Need
(per gm) (per gm)
----------------------------------------------------------------
Cost(per gm) Rs. 0.60 Rs. 1.00
Calcium 10 4 20
Protein 5 5 20
Calories 2 6 12
----------------------------------------------------------------
Formulate the LP problem which will satisfy the daily need in minimum cost.
Last 3 questions is the first Home Assignment(HA).
ReplyDeleteHA No: 01
Date of Assignment: 19-08-2020
Last Date of Submission: 21-08-2020 (Time: 5pm)
Submit .pdf file Named as HA_01_YourName.pdf through e-mail.
Write the document in your own wording.
Avoid Copy-Paste tendency.
After reading, please ask if you have any question/query regarding HA here....
In example 2,
ReplyDeleteThere is a minimum requirement of 40units of vitamin A and 50units of vitamin B.
And as per the solution, X and Y are the amount of food F1 and F2 respectively.
If X becomes 0 or Y becomes zero then how can we get 40units of vitamin A and 50units of vitamin B.
How can X>=0 & Y>=0?
Not understood this portion Sir.
Objective Function must be minimized, for that, if X or Y becomes 0, it's ok.....
DeleteSo, Finishing the Class....
ReplyDeleteOkay sir.
DeleteDONE Sir
DeleteOK sir. Beriya gelam.
DeleteThis comment has been removed by the author.
DeleteOk sir.
DeleteThank you sir, I clearly understood.
DeleteOk sir.Thank you sir.
DeleteSir, it's a request from my side. I think I need another class with some more examples like this. It would help me a lot.
ReplyDeleteYes sir, i need it too.
DeleteYes sir.I also agree to it,if possible.
DeleteI'm agreed
DeleteYes Sir, I need an extra class if possible
DeleteSir I also need a class of examples.
Delete