Question

3. Suzuki Robotics (SR) manufactures two lightweight robots designed for easier house-cleaning. The Alpha-ONE model is older, heavier, and is designed for carpet cleaning. The Alpha-TWO model is newer, lighter, and is designed primarily for wooden floor cleaning. SR produces the components for both models at its Indiana plant. Each Alpha-ONE model requires 5 hours of manufacturing time and each Alpha-TWO model requires 4 hours of manufacturing time. The Indiana plant has 2500 hours of robot manufacturing time available for the next production period. The frames for each model are obtained from a third-party supplier, who can supply as many Alpha-ONE frames as needed. However, Alpha-TWO frame is newer and lighter, and the supplier can only provide up to 350 Alpha-TWO frames for the next production period. Final assembly and testing require 2 hours for each Alpha-ONE model and 2.5 hours for each Alpha-TWO model. Maximum of 1200 hours of assembly and testing time are available for the next production period. The company’s accounting department projects a profit contribution of $300 for each Alpha-ONE model produced and $250 for each Alpha-TWO model produced.

a. Formulate a linear programming model that can be used to determine the number of units of each model that should be produced in order to maximize the total contribution to profit. b. Solve the problem graphically. What is the optimal solution?

c. Which constraints are binding?

Answer 1

Answer a)

Decision Variable:

Let x1 = Units of Alpha-One Model and x2 = Units of Alpha-Two Model

Objective Function:

MaxZ = 300x1 + 250x2 (Total Profit Maximization)

Subject to Constraints:

C1 = 5x1 + 4x2 < 2500 (Manufacturing Time)

C2 = x2 < 350 (Frames for Alpha Two)

C3 = 2x1 + 2.5x2 < 1200 (Assembly and Testing)

Non-Negativity Condition:

x1, x2 > 0

Answer b) We will solve the given LP using the following steps of Graphical Method:

Step 1:To draw constraint 5x1 + 4x2 ≤ 2500 (C1)
Treat it as 5x1+4x2=2500
When x1=0 then x2=?
⇒5(0)+4x2=2500
⇒4x2=2500
⇒x2=2500 / 4=625
When x2=0 then x1=?
⇒5x1+4(0)=2500
⇒5x1=2500
⇒x1=2500 / 5=500

Step 2 To draw constraint x2≤350→(2)
Treat it as x2=350
Here the line is parallel to X-axis

Step 3: To draw constraint 2x1+2.5x2≤1200→(3)
Treat it as 2x1+2.5x2=1200
When x1=0 then x2=?
⇒2(0)+2.5x2=1200
⇒2.5x2=1200
⇒x2=1200 / 2.5=480
When x2=0 then x1=?
⇒2x1+2.5(0)=1200
⇒2x1=1200
⇒x1=1200 / 2=600

Step 4: Draw the diagram and find the coordinate Points

Where x-axes = Values of x1, and y-axes = values of x2

C1 = Red Line, C2 = Green Line, and C3 = Blue Line, Feasible Region = Orange Lines

Coordinates:

O = (0,0) (Clearly observed from the graph)

A = (500,0) (Clearly observed from the graph)

For B, multiply C1 with 2.5 and C3 with 4 and substract C3 from C1:

by solving this equation, we get  x1 = 322.22, and x2 = 222.22 B = (322.22 , 222.22)

For C, replace x2=350 in C3

Hence, we get 2x1 + 2.5 (350) = 1200

Hence, x1 = (1200 - 875) / 2 = 162.5

Thus, C = (162.5 , 350)

D = (0,350) (Clearly observed from the graph)

Step 5: Determine the optimal Solution:

The highest value occus at the point B. Hence, the optimal solution to the given LP problem is : x1=322.22, x2=222.22 and maxZ =152222.22.

Answer c) The constraints on which lines the optimal solution exists, are called as 'Binding Constraints'.

The optimal soultion B (322.22, 222.22) exists on constraint lines of C1 and C2.

Hence, C1 = C1 = 5x1 + 4x2 < 2500, and C3 = 2x1 + 2.5x2 < 1200