Markov Steady State — Textbook Buyback Cycle: 23-Ind-A1 Operations Research, May 2017, Q6

Question 6: Markov Steady State — Textbook Buyback Cycle (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. Total annual demand $D=5{,}000{,}000$ book-sales/yr (all conditions combined); resale rates 90% (new→once-used), 80% (once-used→twice-used), 60% (twice-used→thrice-used); a thrice-used copy's cover falls off after its 4th use, so it is never sold back (0% resale); profits $6/$3/$2/$1 for new/once/twice/thrice-used.

Find. (a) Steady-state annual sales of new copies $N$. (b) Steady-state average profit per book sold.

Steady-state cascade: each generation's sold copies split between resold-forward (continues the chain) and kept/discarded (leaves the system). A thrice-used copy always leaves after its 4th use.

Approach. In steady state, this year's crop of once-, twice-, and thrice-used copies are simply last year's new, once-, and twice-used copies scaled down by their respective resale rates; write every generation's volume as a fixed multiple of the new-copy volume $N$, then use "all four generations sum to the total demand $D$" to solve for $N$.

Express each generation's steady-state annual sales as a multiple of $N$. Once-used sales $=0.9N$ (90% of new copies get resold and reappear as once-used stock); twice-used sales $=0.8(0.9N)=0.72N$; thrice-used sales $=0.6(0.72N)=0.432N$.
Part (a) — total-demand balance (every book sold each fall is new, once-, twice-, or thrice-used): $$N+0.9N+0.72N+0.432N=3.052N=D=5{,}000{,}000$$ $$\boxed{N=5{,}000{,}000/3.052=1{,}638{,}270\text{ new copies/yr}}$$ Once-used $=1{,}474{,}443$, twice-used $=1{,}179{,}554$, thrice-used $=707{,}733$ (sum $=5{,}000{,}000$, check).
Part (b) — steady-state average profit per book, a demand-weighted average across the four generations: $$\bar\pi=\frac{6N+3(0.9N)+2(0.72N)+1(0.432N)}{D}=\frac{10.572N}{3.052N}=\frac{10.572}{3.052}$$ $$\boxed{\bar\pi=\$3.46\text{ per book}}$$

Final results — Question 6
Item	Value
(a) New copies sold/yr	1,638,270
(a) Once-used sold/yr	1,474,443
(a) Twice-used sold/yr	1,179,554
(a) Thrice-used sold/yr	707,733
(b) Average profit per book	$3.46

Topic: Steady-state analysis of a Markov-chain-like generational flow (textbook resale cycle).

Key relations: Steady-state generation volumes $g_k=N\prod_{j<k}\rho_j$ for resale rates $\rho_j$; total-flow balance $\sum_k g_k=D$; steady-state weighted average $\bar\pi=\sum_k\pi_k g_k/D$.

Why this works: Once the system has run long enough to reach steady state, this year's "once-used" supply is not an independent unknown — it is entirely determined by last year's new-copy volume and the fixed 90% resale rate, so the whole four-generation cascade reduces to one unknown, $N$, and one balance equation.

Common pitfall: Treating the four resale percentages as independent probabilities of a single random book's fate (a genuine Markov-chain transition analysis) instead of as the deterministic fraction of each generation's VOLUME that flows forward — both approaches converge to the same answer here because the population is large, but the volume-cascade approach is far faster and is what "steady state... how many... each year" is asking for.

Source: Hillier & Lieberman, Introduction to Operations Research (11th ed.), ch. 16 (Markov chains — steady-state probabilities).

Question 6 of 8: Markov Steady State — Textbook Buyback Cycle

Question 6: Markov Steady State — Textbook Buyback Cycle (20 marks)