Question 9 of 10: Markov Chain Brand Switching — Steady State and Price-Reduction Optimization

Question 9: Markov Chain Brand Switching — Steady State and Price-Reduction Optimization (15 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. Two-state Markov chain on monthly brand choice: $P(\text{Hearts}\to\text{Hearts})=0.80$, $P(\text{Corporal}\to\text{Corporal})=0.90$ (so $P(\text{H}\to\text{C})=0.20$, $P(\text{C}\to\text{H})=0.10$). Selling price $1.00/box, cost $0.80/box (margin $0.20/box); 40 million owners, one purchase/owner/month.

Find. (a) Steady-state market share and Hearts' annual expected profit; (b) the price cut $x$ (cents) maximizing Hearts' annual expected profit, given that a cut of $x$ cents raises the Hearts retention probability from 0.80 to $(0.80+0.01x)$.

Approach. Solve the two-state Markov chain for its long-run (steady-state) share of Hearts purchases, convert that share into an annual box count and profit, then re-derive the steady state as a function of the modified retention probability and optimize expected annual profit over $x$ by calculus.

1. Part (a) — steady state. Let $\pi_H,\pi_C$ be the long-run purchase-probabilities. Balance equations: $\pi_H=0.8\pi_H+0.1\pi_C$ and $\pi_H+\pi_C=1$. The first gives $0.2\pi_H=0.1\pi_C\Rightarrow\pi_C=2\pi_H$; substituting, $\pi_H+2\pi_H=1$: $$\boxed{\pi_H=1/3,\qquad \pi_C=2/3.}$$
2. Part (a) — annual profit. With 40 million owners buying 12 times/year, total purchases/year $=480$ million, of which the Hearts share $\pi_H=1/3$ are Hearts boxes: $480\text{M}\times\tfrac13=160$ million boxes/year. At a $0.20 margin/box: $$\boxed{\text{Annual profit}=160{,}000{,}000\times\$0.20=\$32{,}000{,}000.}$$
3. Part (b) — steady state as a function of the price cut. Let $x$ cents be the price cut, so the Hearts retention probability becomes $a=0.80+0.01x$ (Corporal's transition probabilities are unaffected). Re-solving the balance equations with $a$ in place of $0.8$: $\pi_H(1-a)=0.1(1-\pi_H)\Rightarrow\pi_H=\dfrac{0.1}{1.1-a}=\dfrac{0.1}{0.3-0.01x}$.
4. Part (b) — profit as a function of $x$. The margin per box falls to $(0.20-0.01x)$ dollars (price down $x$ cents, cost unchanged), so $$\Pi(x)=480{,}000{,}000\times\dfrac{0.1}{0.3-0.01x}\times(0.20-0.01x)=\dfrac{48{,}000{,}000\,(x-20)}{x-30}.$$ (Verified algebraically and numerically that this simplification is exact.)
5. Part (b) — optimize. $\dfrac{d\Pi}{dx}=48{,}000{,}000\cdot\dfrac{(x-30)-(x-20)}{(x-30)^2}=\dfrac{48{,}000{,}000\times(-10)}{(x-30)^2}$, which is strictly negative for every $x\ne30$ — so $\Pi(x)$ has no interior maximum; it is strictly decreasing over the whole feasible range $0\le x<20$ (feasible because the margin $0.20-0.01x$ must stay positive). The maximum is therefore at the boundary $x=0$: $$\boxed{x^*=0\text{ cents — Hearts should not cut its price at all; any cut only lowers expected annual profit below the }\$32\text{M found in part (a).}}$$ (Sanity check: $\Pi(10)=48\text{M}\times(-10)/(-20)=\$24$M $<\$32$M; $\Pi(0)=48\text{M}\times(-20)/(-30)=\$32$M, confirming the corner optimum.)