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Question 3. (30 Points) The J&B; Card Shop sells calendars with different coral

ID: 351572 • Letter: Q

Question

Question 3. (30 Points) The J&B; Card Shop sells calendars with different coral reef pictures shown for each month. The once-a-year order for each year's calendar arrives in September From past experience, the September-to-July demand for these calendars can be approximated by a Normal distribution with mean of 500 and standard deviation of 120. The calendars cost $1.50 each, and J&B; sells them for $6 each. (a) (10 pts) If J&B; reduces the calendar price to $0.5 per card at the end of July, how many calendars should be ordered? (b) (10 pts) Now assume that the demand is Uniformly distributed between 100 and 900 and everything else is the same. How many calendars should J&B; order now? (c) (10 pts Suppose that J&B; is very concerned about loss of goodwill, so J&B; does not allow any unsatisfied demand. In order to achieve this, J&B; buys additional calendars from its competitors at their retail price of $5, when J&B; is in stock-out. Find the number of calendars J&B; should order. (Assume that demand follows a Normal distribution as in part (a))

Explanation / Answer

(a) Mean demand, m = 500

Std deviation of demand, s = 120

Shortage or Underage cost, Cu = selling price - cost = 6 - 1.5 = 4.5 (lost opportunity to earn profit due to shortage)

Excess or Overage cost, Co = Cost - Discounted selling price = 1.5 - 0.5 = 1

Critical ratio = Cu/(Cu+Co) = 4.5/(4.5+1) = 0.8182

Corresponding value of z = NORMSINV(0.8182) = 0.9085

Optimal Order quantity (Q) = m+z*s = 500+0.9085*120 = 609 calendars

(b) If demand is uniformly distributed, then the optimal number of calendars to be ordered = 100 + 0.8182*(900-100) = 755 calendars

(c) In this scenario, the cost of shortage is equal to the difference of cost of calendar and the retail price of calendar purchased from competitor. Therefore, Cu = 5 - 1.5 = 3.5

Overage cost remains unchanged as in part a

Critical ratio = 3.5/(3.5+1) = 0.7778

z value = NORMSINV(0.7778) = 0.7648

Optimal number of calendars to be ordered = 500+0.7648*120 = 592 calendars

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