Question

A furniture maker has 4,200 wood units and 8,000 hours available, during which it will manufacture decorative screens. Previously, 2 models have been sold well, so it will be limited to producing these 2 types of furniture. He estimates that model one requires 5 units of wood and 7 hours of available time, while model 2 requires 2 units of wood and 8 hours. The utility that each model provides is 120 dollars. and 80 dollars, respectively. What is the maximum utility that this manufacturer can achieve? USING SIMPLEX METHOD

Answer 1

LPP Formulation:

Let us say that Model 1 = x1 and Model 2 = x2

Now the objective function to maximize the utility

Max Z = 120x1 + 80x2

Constraints:

5x1 + 2x2 <= 4200 (Wood Constraint)
7x1 + 8x2 <= 8000 (Hours constraint)

and x1,x2 >= 0 (non-negativity constraint)

Below is the simplex method solution:

Solution:
Problem is

Max Z = 120 x1 + 80 x2

subject to

5 x1 + 2 x2 ≤ 4200 7 x1 + 8 x2 ≤ 8000

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 = 120 x1 + 80 x2 + 0 S1 + 0 S2

subject to

5 x1 + 2 x2 + S1 = 4200 7 x1 + 8 x2 + S2 = 8000

and x1,x2,S1,S2≥0

Iteration-1		Cj	120	80	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XBx1
S1	0	4200	(5)	2	1	0	42005=840→
S2	0	8000	7	8	0	1	80007=1142.8571
Z=0		Zj	0	0	0	0
		Zj-Cj	-120↑	-80	0	0

Negative minimum Zj-Cj is -120 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 840 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 5.

Entering =x1, Departing =S1, Key Element =5

R1(new)=R1(old)÷5

R1(old) =	4200	5	2	1	0
R1(new)=R1(old)÷5	840	1	0.4	0.2	0

R2(new)=R2(old) - 7R1(new)

R2(old) =	8000	7	8	0	1
R1(new) =	840	1	0.4	0.2	0
7×R1(new) =	5880	7	2.8	1.4	0
R2(new)=R2(old) - 7R1(new)	2120	0	5.2	-1.4	1

Iteration-2		Cj	120	80	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XBx2
x1	120	840	1	0.4	0.2	0	8400.4=2100
S2	0	2120	0	(5.2)	-1.4	1	21205.2=407.6923→
Z=100800		Zj	120	48	24	0
		Zj-Cj	0	-32↑	24	0

Negative minimum Zj-Cj is -32 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 407.6923 and its row index is 2. So, the leaving basis variable is S2.

∴ The pivot element is 5.2.

Entering =x2, Departing =S2, Key Element =5.2

R2(new)=R2(old)÷5.2

R2(old) =	2120	0	5.2	-1.4	1
R2(new)=R2(old)÷5.2	407.6923	0	1	-0.2692	0.1923

R1(new)=R1(old) - 0.4R2(new)

R1(old) =	840	1	0.4	0.2	0
R2(new) =	407.6923	0	1	-0.2692	0.1923
0.4×R2(new) =	163.0769	0	0.4	-0.1077	0.0769
R1(new)=R1(old) - 0.4R2(new)	676.9231	1	0	0.3077	-0.0769

Iteration-3		Cj	120	80	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio
x1	120	676.9231	1	0	0.3077	-0.0769
x2	80	407.6923	0	1	-0.2692	0.1923
Z=113846.1538		Zj	120	80	15.3846	6.1538
		Zj-Cj	0	0	15.3846	6.1538

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=676.9231,x2=407.6923

Max Z=113846.1538

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