The J&B Card Shop sells calendars featuring a different colonial picture for eac

Question

The J&B Card Shop sells calendars featuring a different colonial picture for each month. The once-a-year order for each year's calendar arrives in September. From past experience, the Septemberto-July demand for the calendars can be approximated by a normal distribution with = 300 and standard deviation = 20. The calendars cost $6.50 each, and J&B sells them for $15 each a. Suppose that J&B throws out all unsold calendars at the end of July. Using marginal economic analysis, how many calendars should be ordered? b. If J&B sells surplus calendars for $1 at the end of July and can sell all of them at this price, how many calendars should be ordered?

Explanation / Answer

For a normal distribution it is always optimal to have service level (here the optimum level of Quantity ordered for calendars) = µ + Z; where µ is Mean of normal distribution and standard deviation.

Z is Safety Stock/ order quantity and µ + Z is order up to level also called base order quantity.

In the problem it is given µ = 300 and = 20

Unit purchase cost(PC) = $6.50 and Unit Selling Price (SP) = $15

First calculate Marginal Benefit (MB) = SP – PC = 15 = 6.5 = $8.5

Then calculate Marginal Cost (MC)= PC – SV = 6.5 – 0 =$ 6.5 (Note :here all unused colanders are thrown away so salvage value is zero. )

Now Probability of Annual Demand (D) being less than ordered quantity (Q) is:

P( D <= Q*) = MB/ MB + MC = 8.5 / ( 8.5 + 6.5) = 0.57

Now Look for P(Z z) = 0.57 in Standard Normal table or for use MS Excel and apply this formula “=NORMSINV(0.57)” in excel you will get 0.176

P(Z<=(Q*- µ /) = 0.57 => (Q- µ)/ = 0.176

Q* = µ + 0.176 = 300 + 20 x 0.176 = 303.52 or 304

      

Calendar being physical object can’t be in fraction so quantity is rounded off and the J&B to optimize profit should order 304 calendars.

For part ‘b’ only difference is salvage value of $1,

Marginal Cost (MC) will change to $ 5.5 rest all values will remain same. Applying the above method we get,

P( D <= Q*) = MB/ MB + MC = 8.5 / ( 8.5 + 5.5) = 0.607

Now Look for P(Z z) = 0.607 in Standard Normal table or for use MS Excel and apply this formula “=NORMSINV(0.607)” in excel you will get 0.272

P(Z<=(Q*- µ /) = 0.607 => (Q- µ)/ = 0.272

Q* = µ + 0.272 = 300 + 20 x 0.272 = 305.44 or 306

J&B to optimize profit should order 306 calendars when salvage value is $1.

  

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