question_answer

There are two factories located at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements   of the   depots   are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are 8 and 6 units respectively.                       The cost of transportation per unit is given below

To From	Cost (in Rs.)
	A	B	C
P	160	100	150
Q	100	120	100

How many units should be transported from each factory to each depot in order that the transportation cost is minimum? What is the minimum transportation cost?

Answer:

Let x unit and y units of commodity be transported from the factory at P to the depots at A and B, respectively/ Then, \[(8-x-y)\] units will be transported to depot C and the units transported from Q to A is \[5-x,\] to B is \[5-y\] and to C is \[4-(8-x-y),\] i.e. \[x+y-4.\]             \[\therefore \]The objective function is to minimise              \[Z=160x+100y+150(8-x-y)+100(5-x)\]                                           \[+120(5-y)+100(x+y-4)\] \[=160x+100y+1200-150x-150y+500\]                       \[-100x+600-120y+100x+100y-400\] \[=10x-70y+1900\] \[=10(x-7y+190)\] Also, we have             \[5-x\ge 0\]         \[\Rightarrow \]   \[x\le 5\]                         \[5-y\ge 0\]         \[\Rightarrow \]   \[y\le 5\]                         \[8-x-y\ge 0\]      \[\Rightarrow \]   \[x+y\le 8\]                         \[x+y-4\ge 0\]    \[\Rightarrow \]   \[x+y\ge 4\]             and       \[x\ge 0,\] \[y\ge 0\] Therefore, the required linear programming problem (LPP) is Minimise \[Z=10(x\,-7y+190)\] Subject to the constraints                \[x\le 5,\] \[y\le 5,\]              \[x+y\le 8,\] \[x+y\ge 4,\]             \[x\ge 0\] and \[y\ge 0\] Consider the inequality as equations, we get             x = 5                                        ?(i)             y = 5                                        ?(ii)             \[x+y=8\]                                 ?(iii)             \[x+y=4\]                                 ?(iv) Table for \[x+y=8\] is

x	0	8
y	8	0

  So, line \[x+y=8\] passes through (0, 8) and (8, 0). On putting (0, 0) in the inequality              \[x+y\le 8,\] we get \[0+0\le 8\] \[\therefore \] The shaded region is towards the origin. Table for \[x+y=4\] is

x	0	4
y	4	0

So, line \[x+y=4\] passes through (4, 0) and (0, 4). On putting (0, 0) in the inequality \[x+y\ge 4,\] we get \[0+0\ge 4\]                                              [false] \[\therefore \] The shaded region is away from the origin. Now, the lines x = 5 and y = 5 are perpendicular to X-axis and Y-axis, respectively. So, the shaded region is towards the Y-axis and X-axis, respectively. On solving Eqs. (i) and (iii), we get C (5, 3). On solving Eqs. (ii) and (iii), we get D(3, 5).   Now, plotting the graph, the shaded portion ABCDEF represents the feasible region which is bounded.             The comer points of feasible region are (4, 0), B(5, 0), C(5, 3), D(3, 5), E(0, 5) and F(0, 4), respectively. Now, the values of Z at each corner point are given below

Corner points	\[\mathbf{Z=10(x-7y+190)}\]
A(4, 0)	\[Z=10(4-7\times 0+190)\] \[=10\times 194=1940\]
B(5, 0)	\[\begin{align}   & Z=10(5-7\times 0+190) \\  & =10\times 195=1950 \\ \end{align}\]
C(5, 3)	\[Z=10(5-3\times 7+190)\] \[=10\times 174=1740\]
D(3, 5)	\[Z=10(3-5\times 7+190)\] \[=10\times 158=1580\]
E(0, 5)	\[Z=10(0-5\times 7+190)\] \[=10\times 155=1550\] (Minimum)
F(0, 4)	\[Z=10(0-4\times 7+190\] \[=10\times 162=1620\]

\[\therefore \] Minimum value of Z is 1550 at the point E(0, 5). Hence, the minimum transportation cost will be Rs. 1550, when 0, 5 and 3 units from the factory P and 5, 0 and 1 units from the factory Q will be delivered to the depots A B and C, respectively.