Question
Mathematics Question on Linear Programming Problem
A company has two factories located at P and Q and has three depots situated at A,B and C. The weekly requirement of the depots at A,B and C is respectively 5,5 and 4 units, 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: How many units should be transported from each factory to each depots A,B,C respectively in order that the transportation cost is minimum?
a
b
c
d
c
Solution
The given data can be represented as given in the following diagram : Total weekly production of P and Q=8+6=14 units, and total weekly requirement at depots A,B,C=5+5+4=14 units There is no mismatch between supply and demand. Let factory P supply x units per week to depot A and y units to depot B, then it supplies 8−x−y units to depot C. Clearly, 0≤x≤5,0≤y≤5,0≤8−x−y≤4. As depot A' s requirement is 5 units and it receives x units from factory P, it must receive (5−x) units from factory Q. Similarly, depot B receives (5−y) units from factory Q and depot C receives (x+y−4) units from factory Q. Thus total transportation cost (in ?) =16x+10y+15(8−x−y)+10(5−x)+12(5−y) +10(x+y−4) =x−7y+190 Mathematical formulation of LPP is: Minimize z=x−7y+190 subject to the constraints: x+y≥4,x+y≤8,x≥0,x≤5,y≥0,y≤5 Now, we draw the graphs of the lines x+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 ABCDEF.
Vertices of the feasible region are A(4,0),B(5,0),C(5,3),D(3,5),E(0,5) and F(0,4) ∵Z=x−7y+190 ∴ At A(4,0),Z=194 At B(5,0),Z=195 At C(5,3),Z=174 At D(3,5),Z=158 At E(0,5),Z=155 At F(0,4),Z=162 Thus Z is minimum at E(0,5) i.e., when x=0,y=5. Thus, for minimum transportation cost, factory P should supply 0,5,3 units to depots A,B,C respectively and factory Q should supply 5,0,1 units respectively to depots A,B,C.