Friday, July 17, 2020

Formulation of Problem(Introduction)


Formulation of Linear Programming Problem(LPP) as a mathematical model consist of the following basic steps: 
  1. Find the key-decision to be made from the study of the solution.
  2. Identify the variables.
  3. Express the possible alternatives mathematically in terms of variables.
  4. Mention the objective quantitatively & express it as a linear function of variables.
  5. 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:
  1. Total Machine time <= 80, can be expressed as 1.5X + 2.5Y <= 80  
  2. Total craftsman time <= 70, can be expressed as 2X + 1.5Y <= 70 
  3. 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:
  1. Requirement of vitamin A >= 40, can be expressed as 2X + 4Y >= 40 
  2. Requirement of vitamin B >= 50, can be expressed as 3X + 2Y >= 50 
  3. 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
  1. 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.
  2. 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.
  3. 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.


17 comments:

  1. Last 3 questions is the first Home Assignment(HA).

    HA 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....

    ReplyDelete
  2. In example 2,
    There 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.

    ReplyDelete
    Replies
    1. Objective Function must be minimized, for that, if X or Y becomes 0, it's ok.....

      Delete
  3. Sir, 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.

    ReplyDelete