Question

A certain company makes two products: P₁ and P₂ . Each P₁ requires 5 kg of material M and 3 kg of matetrial N. Each P₂ requires 3 kg of M and 3 kg of N. In the warehouse, there are 350 kg of M and 270 kg of N available. The profit is $50 for each P₁ and $40 for each P₂. What mis of P₁ and P₂ products should the company make and sell in order to maximize its total profit?

Answer 1

The given problem can be converted into a LPP as below.

Maximize Z = 50P1 + 40P2 [Profit]

subject to

5P1 + 3P2 <= 350 [Constraint 1 - Material M]
3P1 + 3P2 <= 270 [Constraint 2 - Material N]
P1, P2 >=0 [Non-negativity constraint]

We solve the LPP using the simplex method as below.

After introducing slack variables, the standard form of the LPP is

Maximize Z = 50P1 + 40P2 + 0S1 + 0S2

subject to

5P1 + 3P2 + S1 = 350
3P1 + 3P2 + S2 = 270
P1, P2, S1, S2 >=0

Iteration-1		Cj	50	40	0	0
B	CB	XB	P1	P2	S1	S2	MinRatio XB/P1
S1	0	350	(5)	3	1	0	350/5=70→
S2	0	270	3	3	0	1	270/3=90
Z=0		Zj	0	0	0	0
		Zj-Cj	-50↑	-40	0	0

Negative minimum Zj-Cj is -50 and its column index is 1. So, the entering variable is P1.

Minimum ratio is 70 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 5.

Entering =P1, Departing =S1, Key Element =5

R1(new)=R1(old)÷5

R1(old) =	350	5	3	1	0
R1(new)=R1(old)÷5	70	1	0.6	0.2	0

R2(new)=R2(old) - 3R1(new)

R2(old) =	270	3	3	0	1
R1(new) =	70	1	0.6	0.2	0
3×R1(new) =	210	3	1.8	0.6	0
R2(new)=R2(old) - 3R1(new)	60	0	1.2	-0.6	1

Iteration-2		Cj	50	40	0	0
B	CB	XB	P1	P2	S1	S2	MinRatio XB/P2
P1	50	70	1	0.6	0.2	0	70/0.6=116.6667
S2	0	60	0	(1.2)	-0.6	1	60/1.2=50→
Z=3500		Zj	50	30	10	0
		Zj-Cj	0	-10↑	10	0

Negative minimum Zj-Cj is -10 and its column index is 2. So, the entering variable is P2.

Minimum ratio is 50 and its row index is 2. So, the leaving basis variable is S2.

∴ The pivot element is 1.2.

Entering =P2, Departing =S2, Key Element =1.2

R2(new)=R2(old)÷1.2

R2(old) =	60	0	1.2	-0.6	1
R2(new)=R2(old)÷1.2	50	0	1	-0.5	0.8333

R1(new)=R1(old) - 0.6R2(new)

R1(old) =	70	1	0.6	0.2	0
R2(new) =	50	0	1	-0.5	0.8333
0.6×R2(new) =	30	0	0.6	-0.3	0.5
R1(new)=R1(old) - 0.6R2(new)	40	1	0	0.5	-0.5

Iteration-3		Cj	50	40	0	0
B	CB	XB	P1	P2	S1	S2	MinRatio
P1	50	40	1	0	0.5	-0.5
P2	40	50	0	1	-0.5	0.8333
Z=4000		Zj	50	40	5	8.3333
		Zj-Cj	0	0	5	8.3333

Since all Zj-Cj≥0

Hence, the optimal solution arrives with the value of variables as
P1=40,P2=50; Max Z=4000

Therefore, in order to maximize its total profit mix of P₁ and P₂ products should the company make and sell are

P1 = 40 kg
P2 = 50 kg

Maximum profit = $4000