Case Study Specialty Toys

September 18, 2017 Management

1. Use the sales forecaster’s predication to describe a normal probability distribution that can be used to approximate the demand distribution. Sketch the distribution and show its mean and standard deviation. Let’s assume that the expected sales distribution is normally distributed, with a mean of 20,000, and 95% falling within 10,000 and 20,000. We know that +/- 1. 96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by: z = (x – mu)/sigma = 1. 96 sigma = (x – mu)/z Sigma = (30,000-20,000) / 1. 6 = 5,102 units. So, we have a distribution with a mean of 20,000 and a standard deviation of 5,102. 2. Compute the probability of a stock-out for the order quantities suggested by members of the management team. Using the normal distribution theory, we discover that as the ordered quantity increases the probability of stockout decreases. At 15,000 the probability of stockout will be 0. 8365 At 18,000 the probability of stockout will be 0. 6517 At 24,000 the probability of stockout will be 0. 2177 At 28,000 the probability of stockout will be 0. 0582 . Compute the projected profit for the order quantities suggested by the management team under three scenarios: worst case in which sales = 10,000 units, most likely case in which sales = 20,000 units and best case in which sales = 30,000 units: Order Quantity: 15,000 were cost price is \$16, selling price \$24 & after holiday selling price \$5 |Unit Sales |Profit | |10,000 |25,000 | |20,000 |120,000 | |30,000 |120,000 |

Order Quantity: 18,000 were cost price is \$16, selling price \$24 & after holiday selling price \$5 |Unit Sales |Profit | |10,000 |-8,000 | |20,000 |144,000 | |30,000 |144,000 | Order Quantity: 24,000 were cost price is \$16, selling price \$24 & after holiday selling price \$5 |Unit Sales |Profit | |10,000 |-74,000 | |20,000 |116,000 | 30,000 |192,000 | Order Quantity: 28,000 were cost price is \$16, selling price \$24 & after holiday selling price \$5 |Unit Sales |Profit | |10,000 |-118,000 | |20,000 |72,000 | |30,000 |224,000 | 4. One of specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs.

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What quantity would be ordered under this policy, and what is the projected profit under the three sales scenarios? The area under the curve to the left of the unknown quantity must be 0. 7 (70%). So, we must first find the z value that cuts off an area of 0. 7 in the left tail of standard normal distribution. Using the cumulative probability table, we see that z=0. 53. To find the quantity corresponding z=0. 53, we have Z=x-u/sigma. Were x is the required quantity. x = 20,000 + 0. 53(5102) = 22,704

The projected profits under the 3 scenarios are computed below for Order Quantity: 22,704. |Unit Sales |Profit | |10,000 |-58,968 | |20,000 |131,032 | |30,000 |181,224 | 5- Provide your own recommendation for an order quantity and note the associated profit projections. Provide a rationale for your recommendation. By observing the projected profit obtained under the 3 scenarios used in parts 3 and 4.

An order quantity in the range of 18,000 to 20,000 will be a good order quantity between the risk of a loss and generating good profits. So an order quantity of 19,000 units will be best order. The profit projections at this quantity are: |Unit Sales |Profit | |10,000 |-19,000 | |20,000 |152,000 | |30,000 |152,000 | ———————– .95 .025 00 30,0 00 10,0 .025 00 20,0

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