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Question

Mathematics Question on Linear Programming Problem

A company has two factories located at PP and QQ and has three depots situated at A,BA, B and CC. The weekly requirement of the depots at A,BA, B and CC is respectively 5,55, 5 and 44 units, while the production capacity of the factories at PP and QQ are respectively 88 and 66 units. The cost of transportation per unit is given below: How many units should be transported from each factory to each depots A,B,CA, B, C respectively in order that the transportation cost is minimum?

A

a

B

b

C

c

D

d

Answer

c

Explanation

Solution

The given data can be represented as given in the following diagram : Total weekly production of PP and Q=8+6=14Q = 8 + 6 = 14 units, and total weekly requirement at depots A,B,C=5+5+4=14A, B, C = 5 + 5 + 4 = 14 units There is no mismatch between supply and demand. Let factory PP supply xx units per week to depot AA and yy units to depot BB, then it supplies 8xy8 - x - y units to depot CC. Clearly, 0x5,0y5,08xy4 0 \le x \le 5, 0\le y \le 5, 0\le 8 - x -y \le 4. As depot AA' s requirement is 55 units and it receives xx units from factory PP, it must receive (5x)(5 - x) units from factory QQ. Similarly, depot BB receives (5y)(5 - y) units from factory QQ and depot CC receives (x+y4)(x + y - 4) units from factory QQ. Thus total transportation cost (in ??) =16x+10y+15(8xy)+10(5x)+12(5y)= 16x + 10y + 15(8 - x - y) + 10(5 - x) + 12(5 - y) +10(x+y4)+ 10(x+ y-4) =x7y+190= x - 7y+ 190 Mathematical formulation of LPPLPP is: Minimize z=x7y+190z = x- 7y + 190 subject to the constraints: x+y4,x+y8,x0,x5,y0,y5 x + y\ge 4 , x + y \le 8, x\ge 0, x \le 5, y \ge 0, y \le 5 Now, we draw the graphs of the lines x+y=4,x+y=8,x=5,y=5,x=0,y=0x + y = 4, x + y = 8, x = 5, y = 5, x = 0, y = 0 The feasibler egion corresponding to these constraints is shown in the figure. It is the bounded (convex) region ABCDEFABCDEF. Vertices of the feasible region are A(4,0),B(5,0),C(5,3),D(3,5),E(0,5)A(4, 0), B(5, 0), C(5, 3), D(3, 5), E(0, 5) and F(0,4)F(0, 4) Z=x7y+190\because Z= x- 7y+ 190 \therefore At A(4,0),Z=194A(4, 0), Z = 194 At B(5,0),Z=195B(5, 0), Z = 195 At C(5,3),Z=174C(5, 3), Z = 174 At D(3,5),Z=158D(3, 5), Z = 158 At E(0,5),Z=155E(0, 5), Z = 155 At F(0,4),Z=162F(0, 4), Z = 162 Thus ZZ is minimum at E(0,5)E(0, 5) i.e., when x=0,y=5x = 0, y = 5. Thus, for minimum transportation cost, factory PP should supply 0,5,30, 5, 3 units to depots A,B,CA, B, C respectively and factory QQ should supply 5,0,15, 0, 1 units respectively to depots A,B,CA, B, C.