Linear Programming Question 135
Question: A brick manufacture has two depots A and B, with stocks of 30000 and 20000 bricks respectively. He receive orders form three builders P, Q and R for 15000, 20,000 and 15000 bricks respectively. The cost (in) of transporting 1000 bricks to the builders form the deposits as given in the table.
To From Transportation cost per 1000 bricks (in Rs.)
| P | Q | R | |
| A | 40 | 20 | 20 |
| B | 20 | 60 | 40 |
| The manufacturer wished to find how to fulfill the order so that transportation cost is minimum. Formulation of the L.P.P., is given as |
Options:
A) Minimize $ Z=40x-20y $ Subject to, $ x+y\ge 15,x+y\le 30,x\ge 15,y\le 20, $
$ x\ge 0,y\ge 0 $
B) Minimize $ Z=40x-20y $ Subject to, $ x+y\ge 15,x+y\le 30,x\le 15,y\ge 20, $
$ x\ge 0,y\ge 0 $
C) Minimize $ Z=40x-20y $ Subject to, $ x+y\ge 15,x+y\le 30,x\le 15,y\le 20, $
$ x\ge 0,y\ge 0 $
D) Minimize $ Z=40x-20y $ Subject to, $ x\ge 0,y\ge 0 $
Show Answer
Answer:
Correct Answer: C
Solution:
The given information can be expressed as given in the diagram:
In order to simply, we assume that 1 unit = 1000 bricks
Suppose that depot A supplies x units to P and y units to Q so that depot A supplies (30-x-y)
Bricks to builder R.
Now, as P requires a total of 15000 bricks, it requires (15-x) units from depot B.
Similarly, Q requires (20-y) units from B and R requires $ 15-(30-x-y)=x+y-15 $ units from B. Using the transportation cost given in table, total transportation cost.
$ Z=40x+20y+20(30-x-y)+20(15-x)+60 $
$ (20-y)+40(x+y-15) $
$ =40x-20y+1500 $
Obviously the constraints are that all quantities of bricks supplied form A and B to P, Q, R are non-negative.
$ \therefore x\ge 0,y\ge 0,30-x-y\ge 0,15-x\ge 0,20-y\ge 0, $
$ x+y-15\ge 0. $
Since, 1500 is a constant, hence instead of minimizing $ Z=40x-20y+1500 $ , we can minimize $ Z=40x-20y. $
Hence, mathematical formulation of the given
LPP is minimize $ Z=40x-20y, $
Subject to the constraints:
$ x+y\ge 15,x+y\le 30, $
$ x\le 15,y\le 20,x\ge 0,y\ge 0 $
BETA
