Newsvendor Model — Christmas Tree Purchasing: 23-Ind-A1 Operations Research, December 2017, Q4

Question 4: Newsvendor Model — Christmas Tree Purchasing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Purchase (cost) price $c=\$10$/tree; selling price $r=\$25$/tree; demand $X\sim N(\mu=100,\sigma=30)$; unsold trees have no stated salvage value.

Find. The order quantity $Q^*$ that maximizes Joe's expected profit (the classical single-period "newsvendor" quantity).

Approach. Balance the cost of ordering one tree too many (overage) against the cost of ordering one tree too few (underage) via the newsvendor critical ratio, then read the corresponding fractile off the normal demand distribution. Because Joe's margin ($15) is 1.5× his cost exposure ($10), the critical ratio comes out above 0.5, so the profit-maximizing policy is deliberately to buy a few more trees than the average expected demand — a small, calculated risk of unsold trees is cheaper in expectation than turning away paying customers.

Overage and underage costs. Overage $C_o$ = loss on each unsold tree (no salvage stated, so the full purchase price is lost) = $\$10$. Underage $C_u$ = forgone profit on each tree demanded but not stocked = selling price − cost = $25-10=\$15$.
Critical ratio (probability the optimal $Q^*$ should not be exceeded by demand): $$CR=\frac{C_u}{C_u+C_o}=\frac{15}{15+10}=0.60$$
Convert the critical ratio to a standard-normal fractile $z=\Phi^{-1}(0.60)$: $$z=\Phi^{-1}(0.60)=0.2533$$
Optimal order quantity $Q^*=\mu+z\sigma$: $$Q^*=100+0.2533(30)=100+7.60$$ $$\boxed{Q^*\approx108\text{ trees}}$$ At this order quantity Joe stocks out (sells every tree, turning some customers away) with probability $1-CR=0.40$, and carries leftover, unsold trees the other 60% of the time — exactly the 60/40 split the critical ratio was built to produce, since a $15 lost sale is only 1.5× as painful as a $10 unsold tree. If instead trees were nearly free to dispose of (a low $C_o$) the same logic would push $Q^*$ even further above the mean.

Final results — Question 4
Item	Value
Overage cost $C_o$	$10/tree
Underage cost $C_u$	$15/tree
Critical ratio	0.60
$z$-value	0.2533
Optimal order quantity $Q^*$	≈108 trees

Topic: Newsvendor (single-period, perishable-product) inventory model.

Key relations: Critical ratio $CR=C_u/(C_u+C_o)$; for normally distributed demand, $Q^*=\mu+z\sigma$ where $z=\Phi^{-1}(CR)$ — a direct consequence of setting the derivative of expected profit w.r.t. $Q$ to zero.

Why this works: Every unit above demand costs $C_o$ and every unit below demand costs $C_u$ in forgone margin; the expected-profit-maximizing $Q^*$ is exactly the quantile where the marginal expected loss from stocking one more unit (its overage probability times $C_o$) equals the marginal expected gain (its service probability times $C_u$), which is precisely the $CR$ fractile.

Common pitfall: Using $C_o=$ full retail price or ignoring that a higher margin item (large $C_u/C_o$) pushes $Q^*$ above the mean, not to it — ordering exactly the mean (100 trees) is a common but sub-optimal shortcut whenever $CR\ne0.5$.

Source: Nahmias, Production and Operations Analysis (7th ed.), ch. 5 (the newsvendor problem).

Question 4 of 9: Newsvendor Model — Christmas Tree Purchasing

Question 4: Newsvendor Model — Christmas Tree Purchasing (20 marks)