Question

Clapper Electronics produces two models of telephone answering devices, model 102 and model H23. Production process consists of two stages, manufacturing and inspection. Manufacturing standard times are 2 hours per unit for model 102 and 1 hour per unit for model H23. Inspection standard times are 1 hour per unit for model 102 and 3 hours per unit for model H23. During next production period, a total of 400 hours are available at manufacturing department and 300 hours are available at inspection department. Each unit of model 102 represent a profit of $9.00 per unit. Each unit of model H23 represent a profit of $7.00 per unit. Management wants to determine the quantity of each product to be produced in order to maximize profit and what will be the amount of that profit.

For the above situation: a) Present the complete Linear Programming Formulation of the problem. b) Solve problem using the Simplex Method in tableau form. All tableaus and calculations must be shown. c) Provide clear answer to management request

Answer 1

Let x1 is the number of model 102 to be produced

x2 is the number of model H32 to be produced

Objective function:

Maximize Z= 9x1+7x2

Constraints:

Manufacturing hour constraint: 2x1+1x2<=400

Inspection Hour constraint: 1x1+3x2<=300

x1,x2>=0 (non negativity constraint)

b) Simplex

Solution:
Problem is

Max Z = 9 x1 + 7 x2

subject to

2 x1 + x2 ≤ 400 x1 + 3 x2 ≤ 300

and x1,x2≥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

After introducing slack variables

Max Z = 9 x1 + 7 x2 + 0 S1 + 0 S2

subject to

2 x1 + x2 + S1 = 400 x1 + 3 x2 + S2 = 300

and x1,x2,S1,S2≥0

Iteration-1		Cj	9	7	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XBx1
S1	0	400	(2)	1	1	0	4002=200→
S2	0	300	1	3	0	1	3001=300
Z=0		Zj	0	0	0	0
		Zj-Cj	-9↑	-7	0	0

Negative minimum Zj-Cj is -9 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 200 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 2.

Entering =x1, Departing =S1, Key Element =2

R1(old) =	400	2	1	1	0
R1(new)=R1(old)÷2	200	1	0.5	0.5	0

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

R2(old) =	300	1	3	0	1
R1(new) =	200	1	0.5	0.5	0
R2(new)=R2(old) - R1(new)	100	0	2.5	-0.5	1

Iteration-2		Cj	9	7	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XBx2
x1	9	200	1	0.5	0.5	0	2000.5=400
S2	0	100	0	(2.5)	-0.5	1	1002.5=40→
Z=1800		Zj	9	4.5	4.5	0
		Zj-Cj	0	-2.5↑	4.5	0

Negative minimum Zj-Cj is -2.5 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 40 and its row index is 2. So, the leaving basis variable is S2.

∴ The pivot element is 2.5.

Entering =x2, Departing =S2, Key Element =2.5

R2(new)=R2(old)÷2.5

R2(old) =	100	0	2.5	-0.5	1
R2(new)=R2(old)÷2.5	40	0	1	-0.2	0.4

R1(new)=R1(old) - 0.5R2(new)

R1(old) =	200	1	0.5	0.5	0
R2(new) =	40	0	1	-0.2	0.4
0.5×R2(new) =	20	0	0.5	-0.1	0.2
R1(new)=R1(old) - 0.5R2(new)	180	1	0	0.6	-0.2

Iteration-3		Cj	9	7	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio
x1	9	180	1	0	0.6	-0.2
x2	7	40	0	1	-0.2	0.4
Z=1900		Zj	9	7	4	1
		Zj-Cj	0	0	4	1

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=180,x2=40

Max Z=1900

c) Produce Model 102 180 units and produce H32 40 units which meet the constraints and gives the highest profit of 1900.

Please rate me

Thanks