Question

1. A manufacturing firm produces two machine parts, using lathes, milling machines and grinding machines.

For lathe operations, part 1 takes 10 minutes for each unit, and part II takes 5 minutes per unit. The firm has a total lathe time of 2500 minutes per week.

For milling operations, part 1 takes 4 minutes, and Part II takes 10 minutes per unit. There is a total of 2000 minutes available on the milling machines.

For grinding, part 1 takes 1 minute and part II takes 1.5 minutes per unit. A total of 450 minutes is available on grinding machines.

The firm makes $50 profit on each unit of part I and $100 on each unit of part II. The company would like to produce enough volumes of these two parts to maximize the total profit, subject to capacity constraints.

Assuming linear relationships, formulate this problem as a Linear Programming problem (denote by X1 and X2 the no. of units to be produced for parts I and part II respectively).

1. Define all the variables used in the formulation.
2. Write down the objective function,
3. Write down all the constraints, including the non-zero constraints.

            (Formulation only. You do not have to solve this problem)

(2) Let X be a random variable, having a standard Gaussian (Normal) distribution (with a mean of 0, and standard deviation 1).

1. Find z such that P(X<z)= 0.95
2. Find z such that P(X>z) = 0.14

(3) Demand for a luxury item at a retail store is forecast to be normally distributed, with a mean of 120 and a standard distribution of 20. If the retail store wants to keep an inventory of this item so as to satisfy the demand 90% of the time, find the amount to be stocked.

Answer 1

Ans no 1

Let the volume of part 1 be X and Part 2 be Y

The objective function is

Z= 50x + 100 y

The constraints are

10X + 5Y <= 2500

4X + 10Y <= 2000

X + 1.5Y <=450

X != 0

Y != 0

Ans no 2

a)Here mean =0, and std.dev = 1.

p(X<z)= 0.95

p ( Z < (z-0)/1) = 0.95

p ( Z < z) = 0.95

we know that for 95 percentile, Z= 1.645 ( use norm.inv function in excel)

so, z = 1.645

b)

p ( x>z ) = 0.14

so for p ( x< z) = 1- 0.14

                          = 0.86

p ( Z < (z-0)/1) = 0.86

p ( Z < z) = 0.86

we know that for 86 percentile, Z= 1.08 ( use norm.inv function in excel)

so, z = 1.08

Ans no 3

let the stock be X.

Here mean = 120 , and std.dev = 20

so

p ( Z < (x- 120)/20)

for 90 percent z = 1.28

so, (x-120)/20= 1.28

i.e. x = 25.6 + 120

       X = 145.6