EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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Chapter 16, Problem 5P

A chemical plant makes three major products on a weekly basis. Each of these products requires a certain quantity of raw chemical and different production times, and yields different profits. Thepertinentin formation is in Table P16.5. Note that there is sufficient warehouse space at the plant to store a total of 450 kg/week.

TABLE P16.5

Product 1 Product 2 Product 3 Resource Availability
Raw chemical 7 k g / k g 5 k g / k g 13kg/kg 3000 k g
Production time 0.005 h r / k g 0.1 h r / k g 0.2hr/kg 55 h r / w e e k
Product $ 30 / k g $ 30 / k g $ 35 / k g

(a) Set up a linear programming problem to maximize profit.

(b) Solve the linear programming problem with the simplex method.

(c) Solve the problem with a software package.

(d) Evaluate which of the following options will raise profits the most: increasing raw chemical, production time, or storage.

(a)

Expert Solution
Check Mark
To determine

A linear programming problem for a chemical plant that makes three products on a weekly basis. The data is given below:

Product 1Product 2Product 3Resource AvailabilityRaw Chemical7 kg/kg5 kg/kg13 kg/kg3000 kgProduction Time0.05 hr/kg0.1 hr/kg0.2 hr/kg55 hr/weekProfit$30/kg$30/kg$35/kg

Answer to Problem 5P

Solution:

The Linear Programming formulation is given as:

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

Explanation of Solution

Given Information:

Following data is given for a chemical plant that makes three products on a weekly basis:

Product 1Product 2Product 3Resource AvailabilityRaw Chemical7 kg/kg5 kg/kg13 kg/kg3000 kgProduction Time0.05 hr/kg0.1 hr/kg0.2 hr/kg55 hr/weekProfit$30/kg$30/kg$35/kg

Also, the warehouse can store a total of 450 kg/week.

Let x1, x2, x3 be respectively the weights needed of Product 1, Product 2 and Product 3 to maximize profit and satisfy the provided conditions.

The Linear Programming Model can be set up as follows:

As Profit is to be maximized, the objective function is Maximize C= 30x1+30x2+35x3.

The constraints are:

The raw chemical constraint is 7x1+5x2+13x33000.

As the total production time must be equal to or less than 55 hr. Thus, the time constraint is 0.05x1+0.1x2+0.2x355.

The storage available is 450 kg/week. Thus, the storage constraint is x1+x2+x3450.

Also, the weights can never be negative. Thus, positivity constraint is x10, x20, x30

Hence, the Linear Programming formulation is given as:

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

(b)

Expert Solution
Check Mark
To determine

To calculate: The solution of the linear programming problem given below:

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

Answer to Problem 5P

Solution:

The values of variables are x1=0, x2=356.25, x3=93.75. The maximum C=13968.75.

Explanation of Solution

Given Information:

A linear programming problem,

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

Calculation:

Consider the provided linear programming problem,

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

First convert the above problem to standard form by adding slack variables.

As the constraints are subjected to less than condition, non- negative slack variables are added to reach equality.

Let the slack variables be S10, S20 and S30.

Thus, the linear programming model would be:

Maximize C= 30x1+30x2+35x30S1+0S2+0S3

7x1+5x2+13x3+S1=30000.05x1+0.1x2+0.2x3+S2=55x1+x2+x3+S3=450

x10, x20, x30, S10, S20 and S30

The above linear programming models consist of three non-basic variables (x1, x2, x3) and three basic variables (S1, S2, S3).

Now the apply the Simplex method and solve the above problem as:

BasisCx1x2x3S1S2S3SolutionMinimum RatioP13030350000S1075131003000230.7692S200.050.10.201055275S30111001450450

The negative minimum, P is 35 and it corresponds to variable x3. So, the entering variable is x3

The minimum ratio is 230.7692 and it corresponds to basis variable S1. So, the leaving variable is S1.

Therefore, the pivot element is 13.

BasisCx1x2x3S1S2S3SolutionMinimum RatioP111.153816.538502.69231008076.923x300.538460.3846210.0769200230.7692600S200.057690.0230800.01538108.846154383.3333S300.461540.6153900.0769201219.2308356.25

The negative minimum, P is 16.5385 and it corresponds to variable x2. So, the entering variable is x2.

The minimum ratio is 356.25 and it corresponds to basis variable S3. So, the leaving variable is S3.

Therefore, the pivot element is 0.61539.

BasisCx1x2x3S1S2S3SolutionMinimum RatioP11.25000.625026.87513968.75x300.25010.12500.62593.75750S200.075000.012510.03750.62550x200.75100.12501.625356.252580

Since P0, optimal solution is obtained.

Hence, the values of variables are x1=0, x2=356.25, x3=93.75. The maximum C=13968.75.

(c)

Expert Solution
Check Mark
To determine

To calculate: The solution of the linear programming problem given below using software package:

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

Answer to Problem 5P

Solution:

The values of variables are x1=0, x2=356.25, x3=93.75. The maximum C=13968.75.

Explanation of Solution

Given Information:

A linear programming problem,

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

Calculation:

Consider the provided linear programming problem,

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

The solution can be obtained using Excel.

Set up the values and use the formula as shown below:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 16, Problem 5P , additional homework tip  1

The values obtained are:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 16, Problem 5P , additional homework tip  2

Now press Solver under Data tab and enter the constraints and objective as shown below:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 16, Problem 5P , additional homework tip  3

The resulting solution is:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 16, Problem 5P , additional homework tip  4

Hence, the values of variables are x1=0, x2=356.25, x3=93.75. The maximum C=13968.75.

(d)

Expert Solution
Check Mark
To determine

The factor among increasing raw material, production time or storage that will rise profits the most for the linear programming problem given below:

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

Answer to Problem 5P

Solution:

Increasing storage will result in the most profits.

Explanation of Solution

Given Information:

A linear programming problem,

Maximize C= 30x1+30x2+35x3

Subject to Constraints:

7x1+5x2+13x33000

0.05x1+0.1x2+0.2x355

x1+x2+x3450

x10, x20, x30

Calculation:

Open the Excel sheet of part (c), then Press Solver under Data and select Simplex LP as a solving method as shown below:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 16, Problem 5P , additional homework tip  5

Press solve then select Sensitivity in reports as shown below:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 16, Problem 5P , additional homework tip  6

The sensitivity report obtained is:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 16, Problem 5P , additional homework tip  7

The high shadow price for storage implies that increasing storage will result in the most profits.

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Chapter 16 Solutions

EBK NUMERICAL METHODS FOR ENGINEERS

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