Question

Problem 4 company manufactures three types of heaters, Hi, Ha, and Hs, using two ra, and ts, respectively, that can be manufactured per hour in order to machines, M and M2. The price charged for each of the three heaters is and $88, respectively. Determine the number of each type of heater, maximize the company's hourly revenue. Each heater Hi takes 2 minutes on machine M, and 5 minutes on machine M2. Each heater H2 takes 8 minutes on machine Mi and 3 minutes on machine M2. Each heater Hs takes 4 min- utes on machine Mi and 6 minutes on machine M2. In order to maintain full employment in the factory, the total number of heaters manufactured each hour must be at least 100. ) Formulate the linear programming problem to be solved for the production mounts zi, 2, and s to maximize hourly revenue for the three heaters. n the simplez tableau w many variables are in the basic and nonbasic sets for the optimal Convert the linear programming problem into its standard form and write

Answer 1

M2 1M1 H1 2 min 3 50 42 H 3 oin atee to be manuased In a houss to o maximle ofal o Subie toonlacnfu

Solution:
Problem is

Max Z = 40 x1 + 63 x2 + 88 x3

subject to

2 x1 + 8 x2 + 4 x3 ≤ 60 5 x1 + 3 x2 + 6 x3 ≤ 60 x1 + x2 + x3 ≤ 100

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 slack variable S1

2. As the constraint-2 is of type '≤' we should add slack variable S2

3. As the constraint-3 is of type '≤' we should add slack variable S3

After introducing slack variables

Max Z = 40 x1 + 63 x2 + 88 x3 + 0 S1 + 0 S2 + 0 S3

subject to

2 x1 + 8 x2 + 4 x3 + S1 = 60 5 x1 + 3 x2 + 6 x3 + S2 = 60 x1 + x2 + x3 + S3 = 100

and x1,x2,x3,S1,S2,S3≥0

Iteration-1		Cj	40	63	88	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	MinRatio XB/x3
S1	0	60	2	8	4	1	0	0	60/4=15
S2	0	60	5	3	(6)	0	1	0	60/6=10→
S3	0	100	1	1	1	0	0	1	100/1=100
Z=0		Zj	0	0	0	0	0	0
		Zj-Cj	-40	-63	-88↑	0	0	0

Negative minimum Zj-Cj is -88 and its column index is 3. So, the entering variable is x3.

Minimum ratio is 10 and its row index is 2. So, the leaving basis variable is S2.

∴ The pivot element is 6.

Entering =x3, Departing =S2, Key Element =6

R2(new)=R2(old)÷6

R1(new)=R1(old) - 4R2(new)

R3(new)=R3(old) - R2(new)

Iteration-2		Cj	40	63	88	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	MinRatio XB/x2
S1	0	20	-4/3	(6)	0	1	-2/3	0	20/6=3.3333→
x3	88	10	5/6	1/2	1	0	1/6	0	10/(1/2)=20
S3	0	90	1/6	1/2	0	0	-1/6	1	90/(1/2)=180
Z=880		Zj	220/3	44	88	0	44/3	0
		Zj-Cj	100/3	-19↑	0	0	44/3	0

Negative minimum Zj-Cj is -19 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 3.3333 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 6.

Entering =x2, Departing =S1, Key Element =6

R1(new)=R1(old)÷6

R2(new)=R2(old) - 1/2 R1(new)

R3(new)=R3(old) - 1/2 R1(new)

Iteration-3		Cj	40	63	88	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	MinRatio
x2	63	10/3	-2/9	1	0	1/6	-1/9	0
x3	88	25/3	17/18	0	1	-1/12	2/9	0
S3	0	265/3	5/18	0	0	-1/12	-1/9	1
Z=2830/3		Zj	622/9	63	88	19/6	113/9	0
		Zj-Cj	262/9	0	0	19/6	113/9	0

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=0, x2=10/3, x3=25/3
So zero H1 heaters, approximately 4 H2 heaters and 9 H3 heaters
Max Z=2830/3

Basic variables : 2 they are x2 and x3

Non basic variables: 1 and it is x1