Question

7) A furniture company manufactures four models o first constructed in the carpentry shop and is nex where it is varnished, w required in each shop is as shown in the display below axed, and polished. The number of man hours of labor Desk 1 |Desk 2 | Desk 3 Desk 4|Available Carpentry Shop| 2 Finishing Shop 2 1 6000 4000 10 40 ecause of limitations in capacity of the plant, no more than can be expected in the carpentry shop and 4,000 in the finis next six months. The profit (revenue minus labor costs) fron in the item is as follows: abor costs) from the sale of ench Desk 1 | Desk 2 Desk 3 Desk 4 Profit 12 18 40 20 a) Determine the quantities to make of each type product which profit. b) The Company is plamning to expand one of ts resources (oither the Carp try Shop or Finishing Shop), so that the capacity of that resource will ncr by 1,000 man hours. Which of the two resources will give the largest incr in profit? How much will be the increment? ither the Carpen-

Answer 1

a) let x1, x, x3, x4 be the number of desk1, 2 , 3 , 4 respectively

Max Z = 12 x1 + 20 x2 + 18 x3 + 40 x4

subject to

2 x1 + 9 x2 + 7 x3 + 10 x4 ≤ 6000 2 x1 + x2 + 3 x3 + 40 x4 ≤ 4000

and x1,x2,x3,x4≥0;

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate

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 = 12 x1 + 20 x2 + 18 x3 + 40 x4 + 0 S1 + 0 S2

subject to

2 x1 + 9 x2 + 7 x3 + 10 x4 + S1 = 6000 2 x1 + x2 + 3 x3 + 40 x4 + S2 = 4000

and x1,x2,x3,x4,S1,S2≥0

Iteration-1		Cj	12	20	18	40	0	0
B	CB	XB	x1	x2	x3	x4	S1	S2	MinRatio XB/x4
S1	0	6000	2	9	7	10	1	0	6000/10=600
S2	0	4000	2	1	3	(40)	0	1	4000/40=100→
Z=0		Zj	0	0	0	0	0	0
		Zj-Cj	-12	-20	-18	-40↑	0	0

Negative minimum Zj-Cj is -40 and its column index is 4. So, the entering variable is x4.

Minimum ratio is 100 and its row index is 2. So, the leaving basis variable is S2.

∴ The pivot element is 40.

Entering =x4, Departing =S2, Key Element =40

R2(new)=R2(old)÷40

R1(new)=R1(old) - 10R2(new)

Iteration-2		Cj	12	20	18	40	0	0
B	CB	XB	x1	x2	x3	x4	S1	S2	MinRatio XBx2
S1	0	5000	3/2	(35/4)	25/4	0	1	-1/4	5000 / 35/4=4000 / 7=571.4286→
x4	40	100	1/20	1/40	3/40	1	0	1/40	100 / 1/40=4000
Z=4000		Zj	2	1	3	40	0	1
		Zj-Cj	-10	-19↑	-15	0	0	1

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

Minimum ratio is 571.4286 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 35/4.

Entering =x2, Departing =S1, Key Element =35/4

R1(new)=R1(old) ×4/35

R2(new)=R2(old) - 1/40R1(new)

Iteration-3		Cj	12	20	18	40	0	0
B	CB	XB	x1	x2	x3	x4	S1	S2	MinRatio XBx1
x2	20	4000/7	6/35	1	5/7	0	4/35	-1/35	4000/7 / 6/35=10000/3=3333.3333
x4	40	600/7	(8/175)	0	2/35	1	-1/350	9/350	600/7 / 8/175=1875→
Z=104000/7		Zj	184/35	20	116/7	40	76/35	16/35
		Zj-Cj	-236/35↑	0	-10/7	0	76/35	16/35

Negative minimum Zj-Cj is -236/35 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 1875 and its row index is 2. So, the leaving basis variable is x4.

∴ The pivot element is 8/175.

Entering =x1, Departing =x4, Key Element =8/175

R2(new)=R2(old) ×175/8

R1(new)=R1(old) - 6/35R2(new)

Iteration-4		Cj	12	20	18	40	0	0
B	CB	XB	x1	x2	x3	x4	S1	S2	MinRatio
x2	20	250	0	1	1/2	-15/4	1/8	-1/8
x1	12	1875	1	0	5/4	175/8	-1/16	9/16
Z=27500		Zj	12	20	25	375/2	7/4	17/4
		Zj-Cj	0	0	7	295/2	7/4	17/4

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=1875,x2=250,x3=0,x4=0

Max Z=27500

so max profit is 27500

and 1875 desk 1, 250 desk 2 should be made

b) if 1000 man hrs is added to carpentry shop than

Max Z = 12 x1 + 20 x2 + 18 x3 + 40 x4 + 0 S1 + 0 S2

subject to

2 x1 + 9 x2 + 7 x3 + 10 x4 + S1 = 7000 2 x1 + x2 + 3 x3 + 40 x4 + S2 = 4000

and x1,x2,x3,x4,S1,S2≥0

optimal solution is arrived with value of variables as :
x1=3625/2

x2=375

x3=0

x4=0

Max Z=29250

if 1000 man hrs is added to finishing shop than

Max Z = 12 x1 + 20 x2 + 18 x3 + 40 x4

subject to

2 x1 + 9 x2 + 7 x3 + 10 x4 ≤ 6000 2 x1 + x2 + 3 x3 + 40 x4 ≤ 5000

and x1,x2,x3,x4≥0;

Since all Zj-Cj≥0

optimal solution is arrived with value of variables as :
x1=4875/2

x2=125

x3=0

x4=0

Max Z=31750

as we can see the profit value is more in case of finishing shop

so finishing shop will give the largest increment in profit

the increment will be

31750 - 27500 = 4250