Derana Furnitures manufactures three products: Single Bed, The Dining Table, and Dining Chair. Derana Furnitures wants to

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Derana Furnitures manufactures three products: Single Bed, The Dining Table, and Dining Chair. Derana Furnitures wants to maximize their profits by producing their products effectively. Each product contributes Rs. 40, Rs. 500 and Rs. 300 to the company's profits. However, the company says it does not need to manufacture all three products. They say they do not produce non-profit products. However, the company hopes to find an optimal solution to this problem. The company uses three special machines for production. They are Machine A, Machine B and Machine C. It takes machine A 6 hours, 4 hours and 2 hours to produce one unit of each product. Machine A is available in 98 hours. Also, it will take 4 hours, 2 hours and 1 hour for machine B respectively. The existing machinery in machine B is 200 hours. Also, for machine C it takes 2 hours, 3 hours and 1 hour respectively. The existing machine hours on machine C are 82 hours. In addition, there are raw materials used in the manufacture of all products. Resource availability is 80. Therefore, this ingredient must be compatible with all 3 products. Not only that, it takes 9 hours, 6 hours and 1 hour of labor to produce one unit from one product. Labor has 180 hours. That is the maximum capacity. So, these are the limits of this company's problem.

Further information is given below.

Resources
Requirement
for one unit

Machines

Material
(Units)

Labours
(Hours)

Profit
Per
Unit
(LKR)

Machine A

Machine B

Machine C

Single Bed

The Dining Table

500

Dining Chair

300

Available resources

200

180

Model Development

Objective - Maximizing Profit

Define the decision variable

X1: - Number of Single Bed produce by the company per week.

X2: - Number of The Dining Table produce by the company per week.

X3: - Number of Dining Chair produce by the company per week.

Objective function

Maximize Z = 40X1 + 500X2 + 300X3

Subject to,

6X₁ + 4X₂ + 2X₃ ≤ 98 (Machine A hours constraint)

4X₁ + 2X₂ + X₃ ≤ 200 (Machine B hours constraint)

2X₁+ 3X₂ + X₃ ≥ 82 (Machine C hours constraint)

X₁ + X₂ + 2X₃ ═ 80 (Material units constraint)

9X₁ + 6X₂ + X₃ ≤ 180 (Labour hours constraint)

X₁, X₂, X₃ ≥ 0 ( Non-negativity constraint)

Developing a Solution Using the Simplex Method of Linear Programming

Converting LP model into standard model

Objective function

Maximize Z = 40X1 + 500X2 + 300X3

Converting…

Maximize Z = 40X1 + 500X2 + 300X3

Subject to,

6X₁ + 4X₂ + 2X₃ ≤ 98 6X₁ + 4X₂ + 2X₃+ S₁ ═ 98

4X₁ + 2X₂ + X₃ ≤ 200 4X₁ + 2X₂ + X₃ + S₂ ═ 200

2X₁+ 3X₂ + X₃ ≥ 82 2X₁+ 3X₂ + X₃-S₃+ A₁ ═ 82

X₁ + X₂ + 2X₃ ═ 80 X₁ + X₂ + 2X₃ + A₂ ═ 80

9X₁ + 6X₂ + X₃ ≤ 180 9X₁ + 6X₂ + X₃+S₄ ═ 180

X₁, X₂, X₃ ≥ 0 X₁, X₂, X₃, S₁, S₂, S₃, S₄, A₁, A₂ ≥ 0

Initial Tableau

C_j C_b	Basic	Solution	40	500	300	0	0	0	0	-M	-M
C_j C_b	Basic	Solution	X₁	X₂	X₃	S₁	S₂	S₃	S₄	A₁	A₂
0	S₁	98	6	4	2	1	0	0	0	0	0
0	S₂	200	4	2	1	0	1	0	0	0	0
-M	A₁	82	2	3	1	0	0	-1	0	1	0
-M	A₂	80	1	1	2	0	0	0	0	0	1
0	S₄	180	9	6	1	0	0	0	1	0	0
	Z_J	-162M	-3M	-4M	-3M	0	0	M	0	-M	-M
	C_J- Z_J	--	40+3M	500+4M	300+3M	0	0	-M	0	0	0

X₂ - Entering Variable

( Assume M = 100 )

4 - Pivot Element

S₁ – Leaving Variable

Value of Entering Variable

New values for entering variable (X2)

S2	A1	A2	S4
200 - (2×24.5) = 151	82 - (3×24.5) = 8.5	80 - (1×24.5) = 55.5	180 - (6×24.5) = 33
4 - (2×1.5) = 1	2 - (3×1.5) = -2.5	1- (1×1.5) = -0.5	9 - (6×1.5) = 0
2 – (2×1) = 0	3 – (3×1) = 0	1 – (1×1) = 0	6 – (6×1) = 0
1 – (2×0.5) = 0	1 – (3×0.5) = -0.5	2 – (1×0.5) = 1.5	1 – (6×0.5) = -2
0 – (2×0.25) = -0.5	0 – (3×0.25) = -0.75	0 – (1×0.25) = -0.25	0 – (6×0.25) = -1.5
1 – (2×0) = 1	0 – (3×0) = 0	0 – (1×0) = 0	0 – (6×0) = 0
0 – (2×0) = 0	-1 – (3×0) = -1	0 – (1×0) = 0	0 – (6×0) = 0
0 – (2×0) = 0	0 – (3×0) = 0	0 – (1×0) = 0	1 – (6×0) = 1
0 – (2×0) = 0	1 – (3×0) = 1	0 – (1×0) = 0	0 – (6×0) = 0
0 – (2×0) = 0	0 - (3×0) = 0	1 - (1×0) = 1	0 - (6×0) = 0

2nd Tableau

C_j C_b	basic	solution	40	500	300	0	0	0	0	-M	-M
C_j C_b	basic	solution	X₁	X₂	X₃	S₁	S₂	S₃	S₄	A₁	A₂
500	X₂	24.5	1.5	1	0.5	0.25	0	0	0	0	0
0	S₂	151	1	0	0	-0.5	1	0	0	0	0
-M	A₁	8.5	-2.5	0	-0.5	-0.75	0	-1	0	1	0
-M	A₂	55.5	-0.5	0	1.5	-0.25	0	0	0	0	1
0	S₄	33	0	0	-2	-1.5	0	0	1	0	0
	Z_j	12,250+64M	750+3M	500	250-M	125+M	0	M	0	-M	-M
	C_{j -}Z_j	--	-710-3M	0	50+M	-125-M	0	-M	0	0	0

X₃ - Entering Variable

( Assume M = 100 )

1.5 - Pivot Element

A₂ – Leaving Variable

Value of Entering Variable

New values for entering variable (X3)

X2	S2	A1	S4
24.5 - (0.5×37) = 6	151 - (0×37) = 151	8.5 - (-0.5×37) = 55.5	33 - (-2×37) = 107
1.5- (0.5× -0.33) = 1.67	1 - (0×-0.33) = 1	-2.5 - (-0.5×-0.33) = -2.67	0 - (-2×-0.33) = -0.66
1 – (0.5×0) = 1	0 – (0×0) = 0	0 – (-0.5×0) = 0	0 – (-2×0) = 0
0.5 – (0.5×1) = 0	0– (0×1) = 0	-0.5 – (-0.5×1) = 1.5	-2 – (-2×1) = 0
0 .25– (0.5× -0.17) = 0.34	-0.5 – (0×-0.17) = -0.5	-0.75 – (-0.5×-0.17) = -0.25	-1.5 – (-2×-0.17) = -1.84
0 – (0.5×0) = 0	1 – (0×0) = 1	0 – (-0.5×0) = 0	0 – (-2×0) = 0
0 – (0.5×0) = 0	0– (0×0) = 0	-1 – (-0.5×0) = 0	0 – (-2×0) = 0
0 – (0.5×0) = 0	0 – (0×0) = 0	0 – (-0.5×0) = 0	1 – (-2×0) = 1
0 – (0.5×0) = 0	0– (0×0) = 0	1– (-0.5×0) = 0	0 – (-2×0) = 0
0 – (0.5×0.67) = -0.37	0 - (0×0.67) = 0	0 - (-0.5×0.67) = 0.34	0 - (-2×0.67) = 1.34

3rd Tableau

Cj Cb	Basic	Solution	40	500	300	0	0	0	0	-M	-M
Cj Cb	Basic	Solution	X1	X2	X3	S1	S2	S3	S4	A1	A2
300	X3	37	-0.33	0	1	-0.17	0	0	0	0	0.67
500	X2	6	1.67	1	0	0.34	0	0	0	0	-0.37
0	S2	151	1	0	0	-0.5	1	0	0	0	0
-M	A1	27	-2.67	0	0	-0.84	0	-1	0	1	0.34
0	S4	107	-0.66	0	0	-1.84	0	0	1	0	1.34
	Zj	14100-27M	736+2.67M	500	300	119+0.84M	0	M	0	-M	06-0.34M
	Cj-Zj	--	-696-2.67M	0	0	-119-0.84M	0	-M	0	0	-16-0.66M