Question

% Problem B.12 Brief Par, Inc.produces a standard golf bag anda deluxe golfbag on a weekly basis. Each golfbag requires time for outing and dyeing and time for sewing and finishing, as shown in the following table Product Standard Bag Deluxe Bag Hours Required per Bag Cutting and Sewing and Dyeing 1/2 The profts per bag and weekly hours available for cutting and dyeing and for sewing and finishing are as follows: Proper Unt (5) Product Sandrag Deluxe Bag Weekly Hours Available Cutting and Dyeing Sewing and Finishing Parine, wil so whatever quantities it produces of these two products The aim of the objective function for Parine, should be to Maximize the objective value The aim on Number of standard bageproduced per week nder your sponse as a whole number Enter your answer in the answer box and then click Check Answer 2. 2 remaining

Answer 1

Answer:

Step A: We will first formulate the given problem as an LP model:

Decision Variable:

Let x1 = No. of Standard Bags, and x2 = No. of Delux Bags

Objective Function:

As the objective is to maximize the total profit, the objective function =

MaxZ = 10x1 + 4x2

Subject to Constraints:

0.5 x1 + x2 ≤ 300 (Hours available for cutting and dying)

x1 + 0.67x2 ≤ 480 (Hours available for Sewing and Finishing)

x1,x2 ≥ 0

Step B: Solve the LP Model:

As no specific information is mentioned in the question, we will solve this LP using the Simplex Method:

Step 1: The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate

1. As the constraint-1 is of type '≤' we should add slack variable S1
2. As the constraint-2 is of type '≤' we should add slack variable S2

After introducing slack variables Max = 10x1 +4x2+0S1+0 S2 subject to 0.5x1 + x2+S1 =300 x1 +0.67x2 + S2=480 and x1x2,S1,S20

Step 2: Prepare the first Iteration table as mentioned below:

Iteration-1 Ci 10 4 0 0 1 Si S2 MinRatio XB /X1 300 0.5 480 (1) Zj0 300 /0.5=600 480 /1=480 0.67 0 0

We will find the iterations till we arrive Zj-Cj ≥ 0

Negative minimum Zj-Cj is -10 and its column index is 1. So, the entering variable is x1.

The minimum ratio is 480 and its row index is 2. So, the leaving basis variable is S2.

∴ The pivot element is 1.

Entering =x1, Departing =S2, Key Element =1

Step 3: Prepare the second Iteration table as mentioned below:

RI(old) = 300 05 1111 0 R2(new) = 480 1 0.67 0 1 0.5xR2(new) = 240 0.5 0.3350 0.5 RI(new)=R1(old) - 0.5R2(new) 60 00.6651 -0.5 R2(old) = 480 10.67011 R2(new)=R2(old) 4801 0.67 01

Iteration 2 L C 10 x1 MinRatio CB | 0 10 Si x1 =4800 0 XB 600 4801 Zj Zi- C o 4 0 12 S1 0.665 1 0.670 6. 7 0 2.7 0 0 S2 .5 1 10 10 10 |

Since all Zj-Cj ≥ 0

Hence, optimal solution is arrived with value of variables as :

x1= No. of Standard Bags 208.8722 = 480 (Whole Number),

x2 = No. of Delux Bags = 0 (Whole Number)

Max Z= 10 (480) + 8 (0) = 4800 $