Is there any way to solve it in a SINGLE LP model? The two liquid products that

Question

Is there any way to solve it in a SINGLE LP model?

The two liquid products that Knox-Shays Chemical makes - KS-01 and KS-02 - 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 of each of the three pollutants, in kg/1000 liters of product (kilograms per 1000 liters), and the maximum of each pollutant allowed in kg.

Product

Pollutant: KS-01, KS-02, Max Allowed

A: 20, 35, 40

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.

Device1 Device2

POLLUTANT: Device 1: KS-01 KS-02 Device 2: KS-01 KS-02

A: Device 1: 40, 30 Device 2: 40, 30

B: Device 1: 60, 70 Device 2: 0, 0

C: Device 1: 50, 40 Device 2: 50, 75

For example, if the emission from KS-01 is sent through Device 1, 40% of pollutant A, 60% of pollutant B, and 50% of pollutant C are removed. Manufacturing considerations dictate that KS- 01 and KS-02 be produced in the ratio of 1 to 3.Formulate a single LP model to determine a plan that maximizes the total daily production (amount of KS-01 plus amount of KS-02) while meeting governmental requirements.

Is there any way to solve it in a SINGLE LP model?

Explanation / Answer

From the provided information:

Assume that the daily production for KS01 and KS02 are x(1000 litres)y(1000 litres) respectively.

Then the tota production is f(x,y)=x+y

We consider the following four cases. Let us form the constraints

1)

The emissions from KSO1 and KS02 are sent through device 1.

The pollutant A highest at 40, so we have

(20*0.6)x+(35*0.6)y40

Since the pollutant B is at most 20 we have

(10*0.4)x+(15*0.4)y20

Since the pollutant c is at most 50 we have

(80*0.5x)+(60*0.5)y 50

In addition, since KS01 and KS02 are produced in the ratio of 1:3

3x-y=0

So we formulate the LP as following

MAX Z , f(x,y)=x+y

S.T:

1.12x+21y40

2.4x+6y20

3.40x+30y50

From the above equations, we can get

x=20/19 and y=5/19

therefore,

x+y=25/19

=1.3158(KL)

2)

The emission from KS01 and KS02 are sent through device 2

The pollutant A is highest at 40 so

(20*0.7)x+(35*0.8)y40

Since the pollutant B is highest at 20 so we have

10x+15y20

Since the pollutant c is highest at 50 we have

(80*0.35)x+(60*0.20)y50

The ratio of KS01 and KS02 are 1:3

3x-y=0

The LP is as follows

MAX Z f(x,y)=x+y

S.T:

1.14x+28y40

2.10x+15y20

3.28x+12y50

From the above equations we get

X=8/7 and y+4/7

Therefore

x+y=12/7(1000 litres)

=1.7143(KL)

3)

the emissions from KS01 and KS02 are sent through device 1 and 2 respectively.

The polluatant A is highest at 40 so we have

(20*0.6)x+(35*0.8)y40

Since the pollutant B is highest at 20 so we have

(10*0.4)x+15y20

Since the pollutant c is highest at 50 we have

(80*0.5)x+(60*0.2)50

The ratio of KS01 and KS02 are 1:3

3x-y=0

MaxZ,f(x,y)=x+y

S.T

1.12x+28y40

2.4x+15y20

3.40x+12y50

Solving the above equation we get

X=10/8 and y=5/8

Therefore,

x+y=15/8

=1.875(KL)

4) The emission from K01 and KS02 are sent through device 2 and. respectively

The pollutant A is highest at 40 so

(20*0.7)x+(35*0.6)y40

Since the pollutant B is highest at 20 so we have

10x+(15*0.4)y20

Since the pollutant c is highest at 50 we have

(80*0.35)x+(60*0.5)y50

The ratio of KS01 and KS02 are 1:3

3x-y=0

So we form the LP as the following

MaxZ f(x,y)=x+y

S.T:

1.14x+21y40

2.10x+6y20

3.28x+30y50

Solving the equation we get

X=100/83 and y=50/83

Therefore,

x+y=150/83

=1.8072(KL)

Comparing all the four solutions we know the third is the optimal solution

So the optimal solution is reached when we sent the emissions KS01 is send to device 1 and KS02 is sent to device 2.

The optimal production is 1.875 KL

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