Fashionable Is a Franchisee

  SCM 403 Homework 1

(Jason J Young #0636477)

Problem 1

Fashionable is a franchisee of The Limited. Prior to the winter season, The Limited offers Fashionables the choice of five different colors of a particular sweater design. The sweaters are knit overseas by hand, and because of the lead-time involved, Fashionables will need to order its assortment in advance of the selling season (i.e., before knowing the actual demand). Per the contracting terms offered by The Limited, Fashionables will not be able to cancel, modify, return, or reorder during the selling season. Demand for each color during the season is normally distributed with a mean of 500 and a standard deviation of 200. Further, the reason of simplicity, you may assume that demand for each color is independent of the others. The Limited offers the sweaters to Fashionables at the wholesale price of $40 per sweater and Fashionables plan to sell each sweater at a retail price of $70 per unit.

1a) If Fashionable wishes to ensure a 97.5% in-stock probability, what should be the order quantity for each type (i.e., color) of sweater?

μ = 500      σ = 200     SL or In-Stock Probability = 97.5%     Q = unknown

Convert .975 into Z from the Standard Normal Distribution Table: z = 1.96

Calculate: Q = μ + zσ

               Q = 500 + (1.96)(200)

               Q = 892 sweaters for each color should be ordered to achieve a 97.5% SL

1b) What is the chance that the demand for a type of sweater falls in between 300 and 700 units?

μ = 500     σ = 200     D1 = 300     D2 = 700     Pr{300≤X≤700} = unknown

Need to find the probability (i.e. chance) that X (demand) will fall between 300 and 700 sweaters. Convert D1 (300) & D2 (700) into the equivalent z-statistics with standard normal distribution:

z = D1 – μ / σ                                       z = D2 - μ / σ                                      

z = 300 – 500 / 200                              z = 700 – 500 / 200

z = -1                                                  z = 1

Probability = .1587                               Probability = .8413

Thus, probability {-1 ≤ z ≤ 1}

= Pr{z≤1} – Pr{z≤-1}

= .8413 - .1587

= .6826

= 68.26% chance that demand for a type of sweater falls between 300 & 700 units

Problem 1 continued…

1c) Suppose Fashionables orders 725 of each color of sweater. What is the stock-out probability of each color?

μ = 500     σ = 200     Q = 725     Stock-Out Probability{D > 725} = unknown

Need to find the probability that X (demand) will be more than 725 sweaters for each color. Convert Q (725) into the equivalent z-statistic with standard normal distribution: