Question

l(a) Consider the Linear Programming (LP) problem below, Max Z = 5x + 4y s.t 6x + 4y = 24 6x + 3y < 22.5 x + y = 5 x + 2y 56 -x+ys1 y <2. x, y 2 0. Solve the problem. AN[10] (b) Out of a stock of three engineering components A,B and C given as 200, 120 and 150 respectively, it is required to produce two types of circuits. Circuit type I requires 20 resistors, 10 transistors and 10 capacitors, while printed circuit type II requires 10 resistors, 20 transistors and 30 capacitors. If a type 1 circuit sells at Gh.5 cedis each and type Il circuit at Gh.12 cedis each. (1) Formulate the linear Programming problem to get the maximum Financial returns. AP[5] (ii)How many of each type should be made to get the maximum financial returns algebraically. AP[10] 10:20 AM

please answer 1b only with clear steps

please I need you to answer only the sub question (b) for me.

Answer 1

1)

(b)

Solution:

(I)

given three components are Resistor,Transistor,Capacitor.

let x1,x2,x3 are the variables to this linear programing problem.

Formulation of linear progamming problem is,

Max Z = 200 x1 + 120 x2 + 150 x3

subject to

20 x1 + 10 x2 + 10 x3 = 5 10 x1 + 20 x2 + 30 x3 = 12

and x1,x2,x3≥0;

(II)

Max Z = 200 x1 + 120 x2 + 150 x3

subject to

20 x1 + 10 x2 + 10 x3 = 5 10 x1 + 20 x2 + 30 x3 = 12

and x1,x2,x3≥0;

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '=' we should add artificial variable A1

2. As the constraint-2 is of type '=' we should add artificial variable A2

After introducing artificial variables

Max Z = 200 x1 + 120 x2 + 150 x3 - M A1 - M A2

subject to

20 x1 + 10 x2 + 10 x3 + A1 = 5 10 x1 + 20 x2 + 30 x3 + A2 = 12

and x1,x2,x3,A1,A2≥0

Iteration-1		Cj	200	120	150	-M	-M
B	CB	XB	x1	x2	x3	A1	A2	MinRatio XBx3
A1	-M	5	20	10	10	1	0	510=0.5
A2	-M	12	10	20	(30)	0	1	1230=0.4→
Z=-17M		Zj	-30M	-30M	-40M	-M	-M
		Zj-Cj	-30M-200	-30M-120	-40M-150↑	0	0

Negative minimum Zj-Cj is -40M-150 and its column index is 3. So, the entering variable is x3

Minimum ratio is 0.4 and its row index is 2. So,the leaving basis variable is A2

Iteration-2		Cj	200	120	150	-M
B	CB	XB	x1	x2	x3	A1	MinRatio XBx1
A1	-M	1	(16.6667)	3.3333	0	1	116.6667=0.06→
x3	150	0.4	0.3333	0.6667	1	0	0.40.3333=1.2
Z=-M+60		Zj	-16.6667M+50	-3.3333M+100	150	-M
		Zj-Cj	-16.6667M-150↑	-3.3333M-20	0	0

Negative minimum Zj-Cj is -16.6667M-150 and its column index is 1. So,the entering variable is x1

Minimum ratio is 0.06 and its row index is 1. So,the leaving basis variable is A1

Iteration-3		Cj	200	120	150
B	CB	XB	x1	x2	x3	MinRatio
x1	200	0.06	1	0.2	0
x3	150	0.38	0	0.6	1
Z=69		Zj	200	130	150
		Zj-Cj	0	10	0

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=0.06,x2=0,x3=0.38

Max Z=69 should be made for each to get maximum financial returns algebrically.

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(Thank You).