Question 9 of 10: Non-Linear Inventory Model with Planned Backorders
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 98-Ind-A1 Operations Research. Three-hour, open-book exam (any non-communicating calculator permitted); the paper totals 200 marks across 10 questions and only 100 marks are required, so a candidate would normally answer a subset — all ten are solved below for completeness.
Reference texts: Hillier & Lieberman, Introduction to Operations Research (11th ed., McGraw-Hill) — linear/integer programming, network optimization, dynamic programming, decision analysis and queueing theory; Nahmias, Production and Operations Analysis — inventory models with planned backorders.
Question 9: Non-Linear Inventory Model with Planned Backorders (20 marks)
Given. Demand rate $d$ units/day at price $r$ $/unit; production run of $x$ units every $y$ days; set-up cost $k$ $/run, variable cost $c$ $/unit; holding cost $h$ $/unit/day while in stock; backorder penalty $l$ $/unit/day while stocked out.
Find. The non-linear average-daily-profit function $\pi(x,y)$ to be maximized over $x,y$. Do not solve (i.e. do not differentiate for the optimum).
Approach. Model one repeating cycle of length $y$: $x$ units raise on-hand stock to a level $x$ right after the run, which depletes at rate $d$; because total demand over a cycle ($dy$) must, by conservation, equal total units produced over the cycle, a backorder of $dy-x$ builds up for the remainder of the cycle before the next run clears it. Cost/revenue each cycle is then divided by $y$ to get an average daily rate.
One repeating cycle of the backorder inventory model: positive stock for $x/d$ days, then a growing backorder for the rest of the $y$-day cycle.
Cycle conservation. Over one $y$-day cycle, total units sold equal total demand $dy$; since replenishment supplies these units (some as fresh stock, some to pay off the prior cycle's backorder), the run size and the split within the cycle satisfy $0\le x\le dy$, with $x$ = the on-hand level immediately after the run and $dy-x$ = the backorder accumulated by the time the next run arrives.
Time split. Positive stock depletes from $x$ to 0 at rate $d$, taking $t_1=x/d$ days; the backorder then builds from 0 to $dy-x$ over the remaining $t_2=y-t_1=(dy-x)/d$ days.
Holding and shortage costs per cycle (triangular area × rate): holding $=h\cdot\dfrac{x\cdot t_1}{2}=\dfrac{h\,x^2}{2d}$; shortage $=l\cdot\dfrac{(dy-x)\,t_2}{2}=\dfrac{l\,(dy-x)^2}{2d}$.
Revenue and production cost per cycle. All $dy$ units demanded in the cycle are eventually sold (backordered sales are delayed, not lost): revenue $=r\,dy$; production cost (total units actually manufactured in the run always equals the cycle's demand, by Step 1) $=c\,dy$; set-up cost $=k$.
Average daily profit. Divide net cycle profit by the cycle length $y$:
$$\boxed{\pi(x,y)=\frac{r\,dy-c\,dy-k-\dfrac{h\,x^2}{2d}-\dfrac{l\,(dy-x)^2}{2d}}{y}=(r-c)d-\frac{k}{y}-\frac{h\,x^2+l\,(dy-x)^2}{2\,d\,y}}$$
to be maximized over $x,y$ subject to $0\le x\le dy$, $y>0$ (not solved).
Element
Formulation
Decision variables
$x$ (on-hand level after each run), $y$ (cycle length, days)
The source does not explicitly state how $x$ (units produced) relates to the depletion timing within a cycle. Conservation forces total production per cycle to equal total demand $dy$ regardless of the holding/shortage split, so $x$ is interpreted here as the on-hand level achieved right after each run (the standard backorder-EOQ decision variable), with the backorder level $dy-x$ following directly — this is the reading that makes both $h$ and $l$ appear in the objective, as the question's parameter list requires.
