Question

3. a manufacturing company produces 2 producrs popular wian nome renovators: 10 to ceiling fans. Both the chandeliers and pans require * Moducts popular win nome renovators: old fashioned candeliers ond 2 step production process more It takes about anrs wo wire each chandelier and 3 hours to wire a ceiling FOO. It takes doo assemble chandeliers and Ihr to assemble electric Roo. The producrion Copaditity is suon mot only 105 OF Wiring time and Sons of assembly time are available. toon onondeliers produced ners che im Prand each Fan PO.Hanagement wants to genieve several goals, each equal in priority os Pollows: • To produce a profit Of 30 if possible • To fully utilize the available wiring dept. turs • To Ovoid overtime in one ossembly dept • To meer a contracruol requirement to produce or lost ceiling fons

a. formulate the goal programming
b. solve the problem using graphical method
c. which is underachieved?
d. which is overachieved?

Answer 1

b.

Max 7C + 6F

Subject to,

2C + 3F <= 12
6C + 5F <= 30
C, F >=0 and Intergers

Graphical method

When the iso-profit line moves towards the feasible region, it touches the integer point (5,0) first. So, the optimal integer solution is C=5, F=0

Simplex Tableau

Initial Simplex Tableau
CBi	Cj	7	6	0	0	Solution	Ratio
Basic variables	C	F	s1	s2
0	s1	2	3	1	0	12	6
0	s2	6	5	0	1	30	5
Zj	0	0	0	0
Cj - Zj	7	6	0	0
First iteration
CBi	Cj	7	6	0	0	Solution	Ratio
Basic variables	C	F	s1	s2
0	s1	0	1.33333	1	-0.3333	2	1.5
7	C	1	0.83333	0	0.16667	5	6
Zj	7	5.83	0	1.17
Cj - Zj	0	0.17	0	-1.17
Second iteration
CBi	Cj	7	6	0	0	Solution
Basic variables	C	F	s1	s2
6	F	0.00	1.00	0.75	-0.25	1.50
7	C	1.00	0.00	-0.63	0.38	3.75
Zj	7	6	0.125	1.125	35.25
Cj - Zj	0	0.00	-0.125	-1.13

Non-integer optimal solution is thus C=3.75 and F=1.50

Gomory's Cut method

C = 3+3/4; F=1+1/2

We take the equation corresponding to C i.e.

1C + 0F - 0.63s1 + 0.38s2 = 3.75

or, (1+0)C + (0+0)F - 0.63s1 + 0.38s2 = 3+0.75

Considering only the fractional part,

-0.63s1 + 0.38s2 >= 0.75

Gomory's cuequation, therefore, is 0.63s1 - 0.38s2 + G1 = -0.75

Including this into the original inital tableau are re-calculating the optimal value gives us the following result

Initial Simplex Tableau
CBi	Cj	7	6	0	0	0	Solution	Ratio
Basic variables	C	F	s1	s2	G1
0	s1	2	3	1	0	0	12	6
0	s2	6	5	0	1	0	30	5
0	G1	0	0	0.63	-0.38	1	-0.75	-
Zj	0	0	0	0	0
Cj - Zj	7	6	0	0	0
First iteration
CBi	Cj	7	6	0	0	0	Solution
Basic variables	C	F	s1	s2	G1
0	s1	0	1.333333	1	-0.33333	0	2
7	C	1	0.833333	0	0.166667	0	5
0	G1	0	0	0.63	-0.38	1	-0.75	-
Zj	7	5.833333	0	1.166667	0	35
Cj - Zj	0	0.166667	0	-1.16667	0

So, the final integer optimal solution is C=5, F=0, and Z=35