Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 — 98-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 160 marks across 8 questions (each worth 20) and only 100 marks are required, so a candidate would normally answer 5 — all eight are solved below for completeness.
Reference texts: Hillier & Lieberman, Introduction to Operations Research (11th ed., McGraw-Hill) — linear programming and the simplex method & sensitivity analysis (ch. 3–4/6), network optimization & PERT/CPM (ch. 9–10), integer programming (ch. 12), Markov chains (ch. 16), decision analysis (ch. 15). Nahmias, Production and Operations Analysis — the single-period (newsvendor) inventory model.
Given. Five states: 0, 25, 50, 75, 100 MW. From every "producing" state (25, 50, 75, 100): 5% chance of total failure to state 0 next hour, 95% chance of continuing at the same output level. From state 0 (under repair): 50% stay at 0, 40% to 100, 6% to 75, 3% to 50, 1% to 25.
State-transition diagram, 5-state Markov chain of generator output (self-loops omitted for clarity).
Find. The long-term (steady-state) expected output of the unit, in MW.
Approach. Solve the steady-state balance equations $\pi P=\pi$, $\sum\pi_i=1$ for the stationary distribution, then take the probability-weighted average of the output levels.
Write the transition matrix over states $(0,25,50,75,100)$:
$$P=\begin{pmatrix}0.50&0.01&0.03&0.06&0.40\\0.05&0.95&0&0&0\\0.05&0&0.95&0&0\\0.05&0&0&0.95&0\\0.05&0&0&0&0.95\end{pmatrix}$$
Solve the steady-state equations$\pi P=\pi$ with $\sum_i\pi_i=1$. The balance equation at state 0, $\pi_0=0.50\pi_0+0.05(1-\pi_0)$, together with each producing state's own balance $\pi_i=0.95\pi_i+(\text{share of }0.05\pi_0\text{ flowing back in})$, solves exactly to
$$\pi_0=\tfrac{1}{11},\quad \pi_{25}=\tfrac{1}{55},\quad \pi_{50}=\tfrac{3}{55},\quad \pi_{75}=\tfrac{6}{55},\quad \pi_{100}=\tfrac{8}{11}$$
(check: $\tfrac{1}{11}+\tfrac{1}{55}+\tfrac{3}{55}+\tfrac{6}{55}+\tfrac{8}{11}=\tfrac{5+1+3+6+40}{55}=1$. ✓)
Compute the long-term expected output as the probability-weighted average:
$$E[\text{output}]=0\!\left(\tfrac1{11}\right)+25\!\left(\tfrac1{55}\right)+50\!\left(\tfrac3{55}\right)+75\!\left(\tfrac6{55}\right)+100\!\left(\tfrac8{11}\right)=\frac{925}{11}$$
$$\boxed{E[\text{output}]\approx 84.09\text{ MW}}$$
