Question

cA Christmas tree provider sells trees on Dec. 20th. They sell trees for 5 full days, closing the lot at the end of the day on December 24th. Based on past sales data, they decided to cut down 220 trees for this year's selling season. Each cut tree has a 70% chance of being a healthy tree that a customer would purchase and a 30% chance of being an unhealthy tree that nobody wants to purchase. On each of the 5 selling days, the number of customers who show up prepared to purchase a tree is normally distributed with an average of 30 and a standard deviation of 5.

a) On average, How many healthy trees do they have?

b) What is the likelihood they end up with 168 or more healthy trees?

c) What is the likelihood that 168 customers or more visit to purchase a tree?

d) What is the likelihood that more customers visit over the 5 days than they can provide healthy trees for? (stockout)

e) How many trees do they need to cut down to ensure stocking out is less than or equal to 10%?

f) Suppose it costs $25 for each tree they cut down and healthy trees are sold for $100. Is it better to stick with the 220 tree strategy or change it to what you found in part e?

Answer 1

Solution

Let X = per day demand (number of trees)

Then, selling season (5 days) demand, Y = 5X.

Given X ~ N(30, 5²), Y ~ N(150, 25²) …………………………………………………………. (1)

Now to work out the solution,

Part (a)

On average, number of healthy trees per season = 220 x 0.7 = 154 ANSWER

Part (b)

Likelihood of ending up with 168 or more healthy trees

= P(S ≥ 168), where S ~ B(220, 0.7)

= 0.0149 [Using Excel Function: Statistical BINOMDIST] ANSWER

Part (c)

Likelihood that 168 customers or more visit to purchase a tree

= P(Y ≥ 168) [vide (1)]

= 0.2358 [Using Excel Function: Statistical NORMDIST][vide (1)] ANSWER

Part (d)

Likelihood that more customers visit over the 5 days than they can provide healthy trees for i.e., (stockout)

= P(Y ≥ 154) [vide (1)]

= 0.4364 [Using Excel Function: Statistical NORMDIST][vide (1)] ANSWER

Part (e)

Let t = trees they need to cut down to ensure stocking out is less than or equal to 10%

Then, we should have; P(Y > 0.7t) ≤ 0.1

=> 0.7t = 183 [Using Excel Function: Statistical NORMINV][vide (1)]  

=> t = 262 ANSWER

Part (f)

Suppose it costs $25 for each tree they cut down and healthy trees are sold for $100. Is it better to stick with the 220 tree strategy or change it to what you found in part e?

With 220 tree strategy, expected profit = {154(100 - 25) – (66 x 25)} = 11400

With 262 tree strategy (as obtained in ‘e’), expected profit = {183(100 - 25) – (79 x 25)} = 11750

Since 11750 > 11400, ‘e’ strategy is better. ANSWER

[Elaboration of above analysis: out of 220 trees cut, only 154 (i.e., 70%) are expected to be healthy. These yield a profit of 75 per tree. The other 66 unhealthy incur per tree, a loss of 25, being the cost per tree cut. Similarly, out of 262 trees cut, only 183 (i.e., 70%) are expected to be healthy. These yield a profit of 75 per tree. The other 79 unhealthy incur per tree, a loss of 25, being the cost per tree cut.]

DONE