Linear Programming Model

The two products that case chemical makes- cs01 and cs02 yield excessive amounts of three different pollutants: a,b,and c. the state government has ordered the company to install and to employ antipollution devices. the following table provides the current daily emissions in kg/1000liters and the maximum of each pollutant allowed in kg.

pollutant======cs01========cs02=====max allowed

A ============ 25 ========40 ====== 43
B ============ 10 ======== 15 ======20
C ============ 80 ======== 60 ======50

the manager of the production department has approved the installation of two antipollution devices. the emissions from each product can be handled by either device in any proportion. ( the emissions are sent through a device only once, that is, the output of one device cannot be the input to the other or back to itself). The following table shows the percentage of each pollutant from each product that is removed by each device.

============device 1============device2
pollutant cs01 cs02 cs01 cs02
A ===========40 === 40 =========== 30==== 20
B ===========60==== 60 =========== 0 ==== 0
C ===========55==== 55 =========== 65==== 80

For example, if the emission from cs01 is sent through device 1, 40% of pollutant A, 60% of pollutant B, and 55 % of pollutant C are removed. Manufacturing considerations dictate that cs01 and cs02 be produced in the ratio of 2 to 1.
Formulate a LP model to determine a plan that maximizes the total daily production ( amount of cs01 plus amount of CS02) while meeting gov requirements.

Solution. Assume that the daily production for CS01 and CS02 are x(1000litres) and y(1000litres) respectively. Then the total production is f(x,y)=x+y.

We consider the following four cases. Now we form the constraints.

(1) The emissions from CS01 and CS02 are sent through device 1
The pollutant A is at most 43 from Table 1, so we have

(25*0.6)x+(40*0.6)y<=43
Since the pollutant B is at most 20 from Table 1, we have,

(10*0.4)x+(15*0.4)y<=20

Since the pollutant C is at most 50 from Table 1, we have,

(80*0.45)x+(60*0.45)y<=50
In addition, since the CS01 and CS02 be produced in the ratio 2 to 1,

x-2y=0.

So we formulate the LP as follows at this case.

Maximize f(x,y)=x+y
Subject to: 15x+24y<=43
4x+6y<=20
34x+27y<=50
x-2y=0

(2) The emissions from CS01 and CS02 are sent through device 2
The pollutant A is at most 43 from Table 1, so we have

(25*0.7)x+(40*0.8)y<=43
Since the pollutant B is at most 20 from Table 1, we have,

10x+15y<=20

Since the pollutant C is at most 50 from Table 1, we have,

(80*0.35)x+(60*0.20)y<=50
In addition, since the CS01 and CS02 be produced in the ratio 2 to 1,

x-2y=0.

So we formulate the LP as follows at this case.

Maximize f(x,y)=x+y
Subject to: 17.5x+32y<=43
10x+15y<=20
28x+12y<=50
x-2y=0

(3) The emissions from CS01 and CS02 are sent through device 1 and 2 , respectively
The pollutant A is at most 43 from Table 1, so we have

(25*0.6)x+(40*0.8)y<=43
Since the pollutant B is at most 20 from Table 1, we have,

(10*0.4)x+15y<=20

Since the pollutant C is at most 50 from Table 1, we have,

(80*0.45)x+(60*0.20)y<=50
In addition, since the CS01 and CS02 be produced in the ratio 2 to 1,

x-2y=0.

So we formulate the LP as follows at this case.

Maximize f(x,y)=x+y
Subject to: 15x+32y<=43
4x+15y<=20
34x+12y<=50
x-2y=0

(4) The emissions from CS01 and CS02 are sent through device 2 and 1, respectively.
The pollutant A is at most 43 from Table 1, so we have

(25*0.7)x+(40*0.6)y<=43
Since the pollutant B is at most 20 from Table 1, we have,

10x+(15*0.4)y<=20

Since the pollutant C is at most 50 from Table 1, we have,

(80*0.35)x+(60*0.45)y<=50
In addition, since the CS01 and CS02 be produced in the ratio 2 to 1,

x-2y=0.

So we formulate the LP as follows at this case.

Maximize f(x,y)=x+y
Subject to: 17.5x+24y<=43
10x+6y<=20
28x+27y<=50
x-2y=0

We can solve the above four linear programming and find which has the biggest production. Then the corresponding plan is optimal. In fact, with the constraint x-2y=0, the above LP can be easily solved.

For (1) we can get x=20/19 and y=5/19&#61672; x+y=25/19 (1000litres)=1.3158(KL);
For (2) we can get x=8/7 and y=4/7 &#61672; x+y=12/7(1000litres)=1.7143(KL);
For (3) we can get x=10/8 and y=5/8&#61672;x+y=15/8 (1000litres)=1.875(KL);
For (4) we can get x=100/83 and y=50/83&#61672;x+y=150/83(1000litres)=1.8072(KL).

Comparing the above four solution, we know the third is optimal. So we should choose the emission CS01 is sent to device 1 and the emission from CS02 is sent through device 2. The biggest production is 1.875 (1000litres).