### Solution to Case Problem Specialty Toys

Solution to Case Problem Specialty Toys 10/24/2012 I. Introduction: The Specialty Toys Company faces a challenge of deciding how many units of a new toy should be purchased to meet anticipated sales demand. If too few are purchased, sales will be lost; if too many are purchased, profits will be reduced because of low prices realized in clearance sales. Here, I will help to analyze an appropriate order quantity for the company. II. Data Analysis: 1. 20,0 00 .025 10,0 00 30,0 00 .025 .95 20,0 00 .025 10,0 00 30,0 00 .025 .95 Since the expected demand is 2000, thus, the mean µ is 2000.

Through Excel, we get the z value given a 95% probability is 1. 96. Thus, we have: z= (x-µ)/ ? =(30000-20000)/ ? =1. 96, so we get the standard deviation ? =(30000-20000)/1. 96=5102. The sketch of distribution is above. 95. 4% of the values of a normal random variable are within plus or minus two standard deviations of its mean. 2. At order quantity of 15,000, z= (15000-20000)/5102=-0. 98, P(stockout) = 0. 3365 + 0. 5 = 0. 8365 At order quantity of 18,000, z= (18000-20000)/5102=-0. 39, P(stockout) = 0. 1517 + 0. 5= 0. 6517 At order quantity of 24,000, z= (24000-20000)/5102=0. 8, P (stockout) = 0. 5 – 0. 2823 = 0. 2177 At order quantity of 28,000, z= (28000-20000)/5102=1. 57, P (stockout) = 0. 5 – 0. 4418 = 0. 0582 3. Order Quantity = 15,000| Unit Sales| Total Cost| Sales at $24| Sales at $5| Profit| 10,000| 240,000| 240,000| 25,000| 25,000| 20,000| 240,000| 360,000| 0| 120,000| 30,000| 240,000| 360,000| 0| 120,000| Order Quantity = 18,000| Unit Sales| Total Cost| Sales at $24| Sales at $5| Profit| 10,000| 288,000| 240,000| 40,000| -8000| 20,000| 288,000| 432,000| 0| 144,000| 30,000| 288,000| 432,000| 0| 144,000|

Order Quantity = 24,000| Unit Sales| Total Cost| Sales at $24| Sales at $5| Profit| 10,000| 384,000| 240,000| 70,000| -74,000| 20,000| 384,000| 480,000| 20,000| 116,000| 30,000| 384,000| 576,000| 0| 192,000| Order Quantity =28,000| Unit Sales| Total Cost| Sales at $24| Sales at $5| Profit| 10,000| 448,000| 240,000| 90,000| -118,000| 20,000| 448,000| 480,000| 40,000| 72,000| 30,000| 448,000| 672,000| 0| 224,000| 4. According to the background information, we get the sketch of distribution above. Since z= (Q-20,000)/5102 =0. 52, so we get Q=20,000+0. 2*5102=22,653. Thus, the quantity would be ordered under this policy is 22,653. The projected profits under the three sales scenarios are below: Order Quantity =22,653| Unit Sales| Total Cost| Sales at $24| Sales at $5| Profit| 10,000| 362,488| 240,000| 63,265| -59,183| 20,000| 362,488| 480,000| 13,265| 130,817| 30,000| 362,488| 543,672| 0| 181,224| 5. From the information we get above, I would recommend an order quantity that can maximize the expected profit, and it can be calculated by the formula below: P(Demand<=Q) = C1/(C1+C2).

The probability here suggests the demand that is less than or equal to the recommended order quantity, and C1 is the cost of stock out loss per unit, and C2 is the cost of unsold inventory per unit. According to the background information, Specialty will sell Weather Teddy for $24 per unit and the cost is $16 per unit. So, we can see that C1= $24 – $16 = $8. If inventory remains after the holiday season, Specialty will sell all surplus inventory for $5 a unit. So, C2 = $16 – $5 = $11.

Therefore, we get P(Demand <=Q) = 8/(8+11) = 0. 4211. And the sketch of distribution is below: Since P= 0. 4211, we know z = (Q-20000)/5102 = -0. 2, then we get Q= 20000-0. 2*5102 =18980. And the projected profits obtained under the 3 scenarios are below: Order Quantity =18,980| Unit Sales| Total Cost| Sales at $24| Sales at $5| Profit| 10,000| 303,680| 240,000| 44,900| -18,780| 20,000| 303,680| 455,520| 0| 151,840| 30,000| 303,680| 455,520| 0| 151,840| So the Order Quantity I recommend is 18,980.