logo for solving-math-problems.com
leftimage for solving-math-problems.com

Linear Programming Model











































A manufacturing firm produces two products. Each product must undergo an assembly process and a finishing process. It is then transferred to the warehouse, which has space for only a limited number of items. The firm has 144 hours available for assembly and 240 hours for finishing, and it can store a maximum of 15 units in the warehouse. Each unit of product 1 has a profit of $75 and requires 12 hours to assemble and 6 hours to finish. Each unit of product 2 has a profit of $95 and requires 8 hours to assemble and 20 hours to finish. The firm wants to determine the quantity of each product to produce in order to maximize profit.


A. Formulate a linear programming model for this problem.


B. Solve this model by using grapchical analysis.



Comments for Linear Programming Model

Click here to add your own comments

Sep 25, 2011
Linear Programming Model
by: Staff


---------------------------------------------------------------

Part III

B3. Circle the CORNER VALUES on the graph:


Open the following link to view the Corner Values:


(1) If your browser is Firefox, click the following link to VIEW ; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/graph-corner-values-2011-09-24.png



B4. Compute the values which will yield the maximum profit:

P = 75n₁ + 95n₂

The maximum and minimum profit can be found by computing the corner values circled along the edge of the “bounded feasible region”.

Computation of corner values:

Lower left:

The origin

n₁ = 0

n₂ = 0

Profit

P = 75n₁ + 95n₂

P = 75*0 + 95*0

P = 0 (zero profit)




Upper left:

The intersection of

n₁ = 0

&

6n₁ + 20n₂ = 240

At the intersection, n₁ = 0 and n₂ = 12

Profit

P = 75n₁ + 95n₂

P = 75*0 + 95*12

P = 0 + 1140

P = $1,140


Upper right:

The intersection of

n₁ + n₂ = 15

&

6n₁ + 20n₂ = 240

At the intersection, n₁ = 4.28571 and n₂ = 10.7143

Profit

P = 75n₁ + 95n₂

P = 75*4.28571 + 95*10.7143

P = $1339.29




Middle right:

The intersection of

12n₁ + 8n₂ ≤ 144

&

n₁ + n₂ = 15

At the intersection, n₁ = 6 and n₂ = 9

Profit

P = 75n₁ + 95n₂

P = 75*6 + 95*9

P = $1,305.00




Lower right:

The intersection of

n₂ = 0

&

12n₁ + 8n₂ = 144

At the intersection, n₁ = 12 and n₂ = 0

Profit

P = 75n₁ + 95n₂

P = 75*12 + 95*0

P = $900.00



The Maximum Profit occurs when


n₁ (product 1) = 4.28571 units and n₂ (product 2) = 10.7143 units


P = $1339.29


Since you cannot manufacture fractional units, the closest values are:

n₁ (product 1) = 4 units and n₂ (product 2) = 11 units

at these manufacturing levels, the profit is:

P = 75n₁ + 95n₂

P = 75*4 + 95*11

P = $1345.00


However, be aware that if n₁ (product 1) = 4 units and n₂ (product 2) = 11 units, the total number finishing hours will exceed boundary limitation of 240 hours by 4 hours:




6x + 20y ≤ 240

6*4+20*11 = 244 hours, 4 hours more than the boundary limitation





Thanks for writing.

Staff
www.solving-math-problems.com


Sep 25, 2011
Linear Programming Model
by: Staff

---------------------------------------------------------------

Part II

The equations for the manufacturing process are:


n₁ + n₂ ≤ 15

12n₁ + 8n₂ ≤ 144

6n₁ + 20n₂ ≤ 240

P = 75n₁ + 95n₂


A. Linear programming model

Maximize the profit

P = 75n₁ + 95n₂


Subject to:

n₁ + n₂ ≤ 15

12n₁ + 8n₂ ≤ 144

6n₁ + 20n₂ ≤ 240



B. Solve this model by using graphical analysis.

B1. Plot the BOUNDARY CONDITIONS identified by the three inequalities on the same two dimensional graph:

n₁ + n₂ ≤ 15

12n₁ + 8n₂ ≤ 144

6n₁ + 20n₂ ≤ 240


Open the following link to view the graph:


(1) If your browser is Firefox, click the following link to VIEW ; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/graph-3-inequalities-2011-09-24.png



B2. Highlight the BOUNDED FEASIBLE REGION on the graph:

All possible values of n₁ and n₂ (the number of units manufactured of Product 1 and Product 2) within the “bounded feasible region” satisfy all three boundary conditions.

n₁ + n₂ ≤ 15

12n₁ + 8n₂ ≤ 144

6n₁ + 20n₂ ≤ 240


Open the following link to view the Bounded Feasible Region:


(1) If your browser is Firefox, click the following link to VIEW ; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/graph-bounded-feasible-region-2011-09-24.png


---------------------------------------------------------------

Sep 25, 2011
Linear Programming Model
by: Staff


Part I


The question:

A manufacturing firm produces two products. Each product must undergo an assembly process and a finishing process. It is then transferred to the warehouse, which has space for only a limited number of items. The firm has 144 hours available for assembly and 240 hours for finishing, and it can store a maximum of 15 units in the warehouse. Each unit of product 1 has a profit of $75 and requires 12 hours to assemble and 6 hours to finish. Each unit of product 2 has a profit of $95 and requires 8 hours to assemble and 20 hours to finish. The firm wants to determine the quantity of each product to produce in order to maximize profit.

A. Formulate a linear programming model for this problem.

B. Solve this model by using graphical analysis.



The answer:


1. Storage

store a maximum of 15 units in the warehouse

n = total number of units stored ≤ 15

n₁ = number of Product 1 units produced and stored
n₂ = number of Product 2 units produced and stored

n = n₁ + n₂ ≤ 15


2. Hours for assembly

The firm has 144 hours available for assembly
product 1 requires 12 hours to assemble
product 2 requires 8 hours to assemble

h_a = total number of assembly hours ≤ 144

a₁ = number of assembly hours for Product 1 per unit = 12

a₂ = number of assembly hours for Product 2 per unit = 8

h_a = a₁n₁ + a₂n₂ ≤ 144

h_a = 12n₁ + 8n₂ ≤ 144



3. Hours for finishing

The firm has 240 hours available for finishing
product 1 requires 6 hours to finish
product 2 requires 20 hours to finish


h_f = total number of finishing hours ≤ 240

f₁ = number of finishing hours for Product 1 per unit = 6

f₂ = number of finishing hours for Product 2 per unit = 20

h_f = f₁n₁ + f₂n₂ ≤ 240

h_f = 6n₁ + 20n₂ ≤ 240


4. Profit



P = Total Profit

Each unit of product 1 has a profit of $75

Each unit of product 2 has a profit of $95

p₁ = Profit earned from Product 1 per unit = $75

p₂ = Profit earned from Product 2 per unit = $95

P = p₁n₁ + p₂n₂

P = 75n₁ + 95n₂

---------------------------------------------------------------------------

---------------------------------------------------------------

Click here to add your own comments

Join in and write your own page! It's easy to do. How? Simply click here to return to Math Questions & Comments - 01.



Copyright © 2008-2015. All rights reserved. Solving-Math-Problems.com