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).

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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.

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