# Speciality Toys.

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• Published : February 28, 2011

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The Synopsis of the problem:

Specialty Toys, a retailer of Children’s toys is planning to launch a new toy called “Weather Teddy”. Sales Managers at the stores are working relentlessly to forecast the most appropriate demand order quantity in such a way that profit could be maximized. The analysis of the problem calls for an ideal demand order quantity situation with lower probability of stock-out option.

Following is the statistical information given:

The cost of goods sold per unit = \$ 16
The cost of Sales price Per Unit = \$ 24
Surplus inventory sales price per unit = \$ 5
Cost of excess inventory per unit = \$ 16- \$ 5 = \$ 11

Expected Demand predicted by Sales Forecaster= 20,000 units Probability of demand between 10,000 units and 30,000 units = 0.95

Solution of the problem:

In order to make the most appropriate recommendation to Management, a Managerial report has been prepared that addresses following issues:

1. Normal Probability Distribution sketch used to approximate the demand distribution with the its mean and Standard Deviation

0.4750

0.4750

0.95

10,00020,00030,000
10,00020,00030,000

Figure 1: Normal Probability Distribution Curve for Expected demand

Expected Demand= 20,000

Hence, Mean ()= 20,000

The probability of demand units to be in between 10,000 and 20,000 is 0.95 as also given in the Figure 1.

In order to compute Standard Deviation, we will use following statistical formula:

Z=(x - )/ …………………………………………………………..(1.1)

Where,
Z= mean of normal distribution
X= The demand units
= The mean of demand order quantity
= Standard deviation of mean order quantity

In order to calculate z value from the probability given, we need to calculate the exact probability @ 30,000 units order quantity.

Since the 0.95 probability of demand being in between 10,000 units and 20,000 units represent the orange shaded region in Figure 1, the probability at 30,000 units has been calculated as below:

Probability at 30,000 units= 0.4750 + 0.50=0.9750
From the Standard Normal Distribution Table, we can derive the Z value for the probability at 0.9750, which will be 1.96

Now,
X=30,000
=20,000
z=1.96
=?

Using the formula 1.0, the standard deviation () = (30,000-20,000)/1.96

Standard Deviation ()= 5102

2. The probability of a stock-out for the order quantities suggested by the Management Team:

Order quantities suggested by Management team are as follows: a. 15,000
b. 18,000
c. 24,000
d. 28,000

a. Probability of stock-out @ 15,000 units of order
Again from using the formula 1.1,

Z= (15,000-20,000)/5102=-0.98

From the Normal Distribution Table, the Z value of -0.98 brings probability value of 0.1635
P(z) = 0.1635
P(z>15000)= 1- 0.6315 = 0.8365 = 83%

The probability of stock-out if the Management orders 15,000 units w ill be 83%.

b. Probability of stock-out @ 18,000 units of order
Using formula 1.1, z = (18,000-20,000)/5102= -0.932
From the Normal Distribution Table, the Z value of -0.932 has probability value of 0.3483
P(z) = 0.3483
P( z>18,000)= 1-0.3483 = 0.6517 = 65%
c. Probability of stock-out @ 24,000 units of order
Using formula 1.1, z = (24,000 – 20,000)/5102= 0.784
From the Normal Distribution Table, the Z value of 0.784 has probability value of 0.7823
P(z) = 0.7823
P(z>24,000) = 1- 0.7823 = 0.2177 = 21%

d. Probability of stock-out @ 28,000 units of order
Using formula 1.1, z = (28,000-20,000)/5102= 1.568
From the Normal Distribution Table, the Z value of 1.568has probability of 0.9406
P(z) = 0.9406
P(z>28,000) = 1-0.9406 = 0.05 = 5 %

3. The projected profit for the order quantities suggested by the management under three scenarios: a. The worst case, in which sales = 10,000
b. The Most likely case, in which sales = 20,000...