Getting a Demand Distribution, and Solving the Final Buy

Agenda

• The belt’s 104 weeks of demand: look, summarize, then hypothesize, estimate, test
• What a goodness-of-fit test does not settle
• Intermittent demand, and why forecasting it is a different question
• The final buy solved with the fitted distribution
• How flat the optimum is, and what the choice of family costs

The Distribution Carries More Than the Formula

Every model from here on takes a demand distribution as an input, and the distribution carries more of the answer than the policy formula built on it. The procedure is Appendix B of Simulation Modeling using the KSL, applied with one difference made explicit: demand for a spare part is a count, so every tool is discrete.

The distributor has 104 weeks of demand, recorded after the deck went out of production, in data/ch7-weekly-demand.csv:

val df = DataFrame.readCsv(
    File("data/ch7-weekly-demand.csv"),
    colTypes = mapOf("week" to ColType.Int, "demand" to ColType.Int)
)
val history: IntArray = df.getColumn("demand").toList().map { it as Int }.toIntArray()

Notice that the column types are named, not inferred. A column of counts silently read as text or as real numbers fails later, and somewhere else.

Look Before You Model: What Shape Is It?

The Same Process, and Independent Weeks

Thus the 104 numbers are independent, identically distributed, and strongly right-skewed: 104 draws from one distribution.

Summarize It

Step 1, Hypothesize; Step 2, Estimate

Hypothesize. The Poisson fixes \(\mathit{VMR} = 1\); the binomial forces \(\mathit{VMR} < 1\). A ratio of 18.44 is near neither, and no parameter repairs that. What remains is the negative binomial: any \(\mathit{VMR} > 1\), a Poisson whose rate varies from week to week. A plausible story for a spare part.

Estimate. PMFModeler estimates every discrete family it knows:

RV Type = NegativeBinomial
Double Parameters {probOfSuccess=0.05422232292974839, numSuccesses=1.9057024844439698}
RV Type = Poisson
Double Parameters {mean=33.24038461538462}
FAILED: Cannot match moments when sample average <= estimated variance
RV Type = Binomial
Double Parameters {probOfSuccess=0.29133420825260514, numTrials=114.09708737864078}

The most useful line is the failure. The moment-matching binomial refuses, for the reason \(\mathit{VMR}\) gave: the software performing Step 1. The other binomial matched the largest observation and reports 114.1 trials for a quantity with no trials in it. An estimate returned is not a family appropriate.

The Estimate, by Hand

Matching the negative binomial’s two moments to the sample mean \(\bar{x}\) and variance \(s^{2}\):

\[ \begin{aligned} \hat{p} &= \frac{\bar{x}}{s^{2}} = \frac{33.2404}{613.0387} = 0.054222 &&\text{mean over variance}\\ \hat{r} &= \frac{\bar{x}^{2}}{s^{2} - \bar{x}} = \frac{33.2404^{2}}{613.0387 - 33.2404} = 1.9057 &&\text{method of moments} \end{aligned} \]

which is what the KSL reported. Notice that \(1/\hat{p} = 18.44\) is the variance to mean ratio we started with: the fit is a restatement of two summary numbers in the parameters of a family.

Step 3, Test

What the Test Does Not Settle

Failing to reject is not proof. With 104 observations over twenty bins, several bins expect fewer than five, where the chi-squared approximation is least trustworthy; the KSL flags each. The test and the ratio narrow the field, and the choice among survivors is yours to defend. Where survivors disagree enough to change the decision, compute the answer under both and report the spread.

What the section did not do:

• Deseasonalize. There is nothing to remove, and the season is over.
• Forecast. A different job from fitting a distribution, and outside this book.
• Correct for censoring. The distributor does not record lost sales: the limitation the first session warned about, which no care here repairs.

Intermittent and Slow-Moving Demand

Recall the two statistics that classify a history: the average interval between demands, \(\mathit{ADI}\), and the squared coefficient of variation of the nonzero sizes, \(\mathit{CV}^{2}\). \(\mathit{ADI} > 1.32\) is intermittent; with \(\mathit{CV}^{2} > 0.49\) too, lumpy. Syntetos and colleagues derive the cut-offs as the points where one forecasting method overtakes another.

For the belt, \(\mathit{ADI} = 104/102 = 1.02\), comfortably under: the check that licenses everything so far.

Forecasting Intermittent Demand Is a Different Question

This book does not forecast intermittent series. The literature begins with Croston (1972).

Croston’s insight. Exponentially smoothing an intermittent series directly jumps at every demand and decays through every empty period. The estimate is highest immediately after a demand, when stock has just been drawn down and a replenishment is most likely to be triggered: unbiased across the series, biased upward at the only moments anyone reads it. He smoothed the size and the interval separately.

Syntetos and Boylan showed (2001) that combining the two in the obvious way is biased high, because the ratio of two smoothed estimates does not estimate the ratio of their means, and gave the correction (2005) that carries their names.

The Final Buy, Solved

Let \(X\) represent total demand over the fourteen weeks and \(X_i\) demand in week \(i\), negative binomial with mean 33.2404 and variance 613.0387. Weeks are independent, so means and variances add:

\[ \begin{aligned} E[X] &= 14(33.2404) = 465.37, \quad \mathit{Var}[X] = 14(613.0387) = 8582.54 &&\text{independent weeks}\\ \sigma &= \sqrt{8582.54} = 92.64, \quad c_X = 92.64/465.37 = 0.1991 &&\text{spread of the total} \end{aligned} \]

Two approximations, both earned. Continuous: one belt is \(1/92.64 = 1.1\%\) of \(\sigma\), so rounding cannot decide the answer; at the weekly level it is 4%. Normal: the weekly coefficient was 0.7449, but the total’s is 0.1991, where a normal puts essentially nothing below zero, and a sum of many independent terms tends toward the normal.

The Quantity, the Consequences, the Check

\[ \begin{aligned} \frac{c_u}{c_u + c_o} &= \frac{45}{83} = 0.542169, \quad z = \Phi^{-1}(0.542169) = 0.1059 &&\text{critical ratio}\\ Q^{*} &= 465.37 + 0.1059(92.64) = 475.18 \to \textbf{475 belts} &&\mu + z\sigma\\ E[\text{shortage}] &= 92.64\,\Phi^{1}(0.1059) = 92.64(0.3482) = 32.26 &&\text{normal loss, scaled}\\ E[\text{leftover}] &= 32.26 + 475.18 - 465.37 = 42.07 &&\text{shortage} + Q - E[X]\\ E[C(Q^{*})] &= 38(42.07) + 45(32.26) = 1598.71 + 1451.72 = \$3{,}050.43 &&\text{the cost} \end{aligned} \]

The check. The other form: \((45 + 38)(32.26) + 38(475.18 - 465.37) = 2677.62 + 372.81 = \$3{,}050.43\). It catches a missing \(\sigma\) in the loss function immediately.

In business terms. Expected profit \((95 - 50)(465.37) - 3050.43 = \$17{,}891.01\): commit $23,750 for 475 belts and expect to clear about $17,900. Scrap about 42; turn away about 32 dealers.

The Optimum Is Flat

What does ordering the mean, rounded to 465 belts, cost against the optimum’s $3,050.43?

What the Choice of Family Costs

A sum of independent negative binomials sharing \(p\) is negative binomial with the same \(p\) and the successes added. So the weekly fit gives the fourteen-week distribution exactly: \(\mathit{NB}(0.054222,\ 14 \times 1.9057 = 26.6798)\). Its smallest \(Q\) with \(F(Q) \ge 0.5422\) is 469: \(F(468) = 0.5392\), \(F(469) = 0.5434\).

Read it as a bound, not a ranking. The worst family costs 0.19% more, on a decision worth about $3,050, because the total’s coefficient of variation is 0.1991. At the weekly 0.7449 it would not be small.

The normal’s defect is real and did not bite. It puts \(\Phi(-5.02)\), about one in four million, below zero here; on the weekly data, about 9%.

What We Did Today

Look, summarize, then fit. Three plots showed the weeks independent, identically distributed and right-skewed.

The variance to mean ratio chose the family, 18.44, negative binomial; the estimator’s failure and the test confirmed it.

Intermittent demand needs other tools, and forecasting it is a different question from fitting it.

The final buy is 475 belts by the normal approximation, $17,891.01 expected profit, checked two ways.

The optimum is flat, and the family costs at most 0.19% here, because fourteen weeks aggregate away the weekly skew.

For Next Time

Read before next session: Discrete Demand, Extensions and Estimating the Answer by Simulation.

Today’s material was Getting a Demand Distribution, Intermittent and Slow-Moving Demand and Continuous Demand.

Next session:

• Demand counted in small whole numbers, and the staircase rule
• A penalty beyond the lost margin, stock already on hand, and when the model refuses
• Estimating the answer by simulation, and when it is worth it