Question

An appliance store sells three brands of TV sets, brands A B and C. The profit per sot is $30 for brand A 500 for brand B and $80 for brand C. The total warehouse space fitted to all brands is sutient for 700 sets, and the inventory is delivered only once per month At least 150 customers per month will demand brand A, at least 100 will demand brand Band at least 175 will demand either brand Bor brand How can the appliance store at all of these constraints and can maximum probit? The appliance store will maximize its prodits by soling brand television brand television, and brand Clovisions

Answer 1

x₁ = number of brand A TV

x₂ = number of brand B TV

x₃ = number of brand C TV

.

The profit per set is $30 for brand A S00 for brand B and $80 for brand C

Mar Z = 30.01 + 60.02 +80.13

.

The total warehouse space allotted to all brands is sufficient for 700 sets

21 + 12 + 13 < 700

Al least 150 customers per month will demand brand A,

Τι 2 150

at least 100 will demand brand B,

C2 > 100

and at least 175 will demand either brand B or brand C

22 +.13 > 175

.

our system is

Mar Z = 30.01 + 60.02 +80.13

subject to

21 + 12 + 13 < 700

22 +.13 > 175

Τι 2 150

C2 > 100

After introducing slack, surplus, artificial variables

.

Mar Z = 30.61 +60:c2+80.63 +0S1+052+0.53 +0S4-MA-MA2- MAz

subject to

21 +.22 +13 + Si = 700

22 +33 - S2 + 41 = 175

21 - S3 + A2 = 150

C2 - 54 + 43 = 100

Iteration-1		Cj	30	60	80	0	0	0	0	-M	-M	-M
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	A1	A2	A3	MinRatio XB/x2
S1	0	700	1	1	1	1	0	0	0	0	0	0	700/1=700
A1	-M	175	0	1	1	0	-1	0	0	1	0	0	175/1=175
A2	-M	150	1	0	0	0	0	-1	0	0	1	0	---
A3	-M	100	0	(1)	0	0	0	0	-1	0	0	1	100/1=100→
Z=-425M		Zj	-M	-2M	-M	0	M	M	M	-M	-M	-M
		Zj-Cj	-M-30	-2M-60↑	-M-80	0	M	M	M	0	0	0

Negative minimum Zj-Cj is -2M-60 and its column index is 2.

Minimum ratio is 100 and its row index is 4.

The pivot element is 1.

Entering =x₂, Departing =A₃,

R1 + R1 - R4

R2 + R2 - R4

Iteration-2		Cj	30	60	80	0	0	0	0	-M	-M
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	A1	A2	MinRatio XB/x3
S1	0	600	1	0	1	1	0	0	1	0	0	600/1=600
A1	-M	75	0	0	(1)	0	-1	0	1	1	0	75/1=75→
A2	-M	150	1	0	0	0	0	-1	0	0	1	---
x2	60	100	0	1	0	0	0	0	-1	0	0	---
Z=-225M+6000		Zj	-M	60	-M	0	M	M	-M-60	-M	-M
		Zj-Cj	-M-30	0	-M-80↑	0	M	M	-M-60	0	0

Negative minimum Zj-Cj is -M-80 and its column index is 3.

Minimum ratio is 75 and its row index is 2.

The pivot element is 1.

Entering =x₃, Departing =A₁

R1 + R1 - R

Iteration-3		Cj	30	60	80	0	0	0	0	-M
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	A2	MinRatio XB/x1
S1	0	525	1	0	0	1	1	0	0	0	525/1=525
x3	80	75	0	0	1	0	-1	0	1	0	---
A2	-M	150	(1)	0	0	0	0	-1	0	1	150/1=150→
x2	60	100	0	1	0	0	0	0	-1	0	---
Z=-150M+12000		Zj	-M	60	80	0	-80	M	20	-M
		Zj-Cj	-M-30↑	0	0	0	-80	M	20	0

Negative minimum Zj-Cj is -M-30 and its column index is 1.

Minimum ratio is 150 and its row index is 3.

The pivot element is 1.

Entering =x₁, Departing =A₂

R1 + R1 - R3

Iteration-4		Cj	30	60	80	0	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	MinRatio XB/S2
S1	0	375	0	0	0	1	(1)	1	0	375/1=375→
x3	80	75	0	0	1	0	-1	0	1	---
x1	30	150	1	0	0	0	0	-1	0	---
x2	60	100	0	1	0	0	0	0	-1	---
Z=16500		Zj	30	60	80	0	-80	-30	20
		Zj-Cj	0	0	0	0	-80↑	-30	20

Negative minimum Zj-Cj is -80 and its column index is 5.

Minimum ratio is 375 and its row index is 1

The pivot element is 1.

Entering =S₂, Departing =S₁

R2 + R2 + R1

Iteration-5		Cj	30	60	80	0	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	MinRatio
S2	0	375	0	0	0	1	1	1	0
x3	80	450	0	0	1	1	0	1	1
x1	30	150	1	0	0	0	0	-1	0
x2	60	100	0	1	0	0	0	0	-1
Z=46500		Zj	30	60	80	80	0	50	20
		Zj-Cj	0	0	0	80	0	50	20

Since all  

Hence, the optimal solution is arrived

Zj-Cio

21 = 150, C2 = 100, 13 = 450

Mar Ζ = 46500

.

150 brand A TV

100 brand B TV

450 brand C TV

maximum profit is $46500