Question

A chemical manufacturer produces three different chemical products. The profit per unit, the labor and raw material required per unit, and the pollution emitted per unit are given in the file. This company’s two objectives are to maximize profit and minimize pollution produced. The company wants to identify a set of points that constitute a trade-off curve for these two objectives. Formulate and solve an appropriate optimization model in the given shell file to help the chemical manufacturer find this trade-off curve.

Product1 4 Product 2 Product 3 Labor hours/unit Raw material/unit Pollution/unit Profit/unit 10 $10 $9 $8 Production plan Product Units produced Constraints on resources Labor hours Used 1300 1000 Raw material Actual Bound Total pollution Profit

Answer 1

Calculations: For maximizing profit Initial Iteration 10 0 0 1300 4. 325 1000 333.3333 |<-min -10 min Here x1 is incoming & xs is outgoing vector First Iteration: 10 0 0 1296 |с-min 10 0.333333 500 333.3333 0.666667 2.33 min 3333.33 0.00 2.00 0.00 3.33 Here x2 is incoming & x4 is outgoing vector Final Iteration: 10 0 0 4. 2.33333 12.67 10 0.33333 1197.333 7.00 309.33 0.00 0.00 6.67 Here all the values under zs are positive for each y, so this gives the optimum solution as x, -1197.33 1197, x2-1296 & x3-0 & Maximum Z $309.33

Calculations:
For maximizing profit:
Initial Iteration:
			10	9	8	0	0
yB	cB	xB	y1	y2	y3	y4	y5	xB/y1
x4	0	1300	4	3	2	1	0	325
x5	0	1000	3	2	3	0	1	333.3333	<- min
	zj	0	-10	-9	-8	0	0
			min
Here x1 is incoming & x5 is outgoing vector
First Iteration:
			10	9	8	0	0
yB	cB	xB	y1	y2	y3	y4	y5	xB/y2
x4	0	1296	0	-1	-2	-3	-4	-1296	<- min
x1	10	333.3333	1	0.666667	1	0	0.333333	500
	zj	3333.33	0.00	-2.33	2.00	0.00	3.33
				min
Here x2 is incoming & x4 is outgoing vector
Final Iteration:
			10	9	8	0	0
yB	cB	xB	y1	y2	y3	y4	y5
x2	9	-1296	0	1	2	3	4
x1	10	1197.333	1	0	-0.33333	-2	-2.33333
	zj	309.33	0.00	0.00	6.67	7.00	12.67
Here all the values under zj's are positive for each yj, so this gives the optimum solution as x1 = 1197.33 ≈ 1197, x2 = 1296 & x3 = 0 & Maximum Z = $309.33