Analytic Results for the Basestock Policy

1. Analytic Results for the Basestock Policy

In this section, we establish a set of formulas to compute performance measures for the single-item basestock inventory system of \cref{sec:single-item-invent}.¹¹ Like this we do not have to simulate to see how performance depends on the order-up-to level $S$.

We introduce some notation: The period demands $\{D_{i}\}_{i=1}^{\infty}$ are assumed to be iid rvs²² Why did we not need to make this assumption earlier? ; for notational ease we take $D_{i} = 0$ for $i\leq 0$. Write

\begin{align*} D(i, j] &= \sum_{k=i+1}^{j} D_k, & D[i, j] &= D_{i} + D(i, j], & D[i, j) &= D[i, j] - D_{j}, \end{align*}

and likewise for $Q(i, j]$ for the total outstanding reorders.

We assume that the lead time $\leadtime$ is constant and a multiple of the period duration. Define $F_X(x) = \P{\sum_{i=1}^{\leadtime} D_{i} \leq x}$; write $X\sim F_X$ for the lead time demand, $G_X = 1 -F_X$ for the survivor function, and $f_{X}$ for the pmf. Clearly, $\E X = \leadtime \E D$. Since the period demands are iid rvs, all rvs $\{D[k-\leadtime, k)\}_{k=\leadtime+1}^{\infty}$ have the same distribution $F_X$.

We use the recursions of \cref{sec:single-item-invent} to derive expressions for the long-run performance measures. The first step is to relate the outstanding orders to the demand.

Under a basestock policy with order-up-to level $S$, $Q(1, k] = D[1, k)$ for all periods $k\geq 1$.

Proof

Using the recursions for the inventory position under the basestock policy repeatedly,

\begin{align*} \IP_k &= \IP_{k-1}'+ Q_k = \IP_{k-1} -D_{k-1}+ Q_k \\ &= \IP_{k-1} -D[k-1, k) + Q(k-1, k] \\ &= \IP_1 - D[1, k) + Q(1, k]. \end{align*}

The proof is finished by noting that $\IP_{k} = S$ for all $k\geq 1$.

We can use this lemma to express $I_{k}$ in terms of $S$ and the lead time demand.

Under a basestock policy, if $\IL_{1} = S$ and there are no outstanding orders of period $k \leq 1$, then for all periods $k\geq 1$,

\begin{align*} \IL_k&= S - D[k-\leadtime, k), & \IL_k'&= \IL_{k} - D_{k} = S - D[k-\leadtime, k]. \end{align*}

Proof

Using the recursion for the inventory level and the assumptions,

\begin{align*} \IL_{k} &= \IL_{k-1} -D_{k-1}+ Q_{k-\leadtime} \\ &= \IL_{k-1}- D[k-1, k) + Q(k-\leadtime-1, k-\leadtime] \\ &= \IL_{1}- D[1, k) + Q(1-\leadtime, k-\leadtime] \\ &\stackrel1= S - D[1, k) + Q(1, k-\leadtime] \\ &\stackrel2= S - D[k-\leadtime, k); \end{align*}

1 uses the assumptions, 2 follows from the above lemma and $D[k-\leadtime, k) = D[1, k) - D[1, k-l)$.

This next theorem, which allows us to define the inventory level as an rv $\IL$, follows directly from the above lemmas.

Theorem 1.

The rvs $\{\IL_{k}\}_{k=\leadtime+1}^{\infty}$ are identically³³ But not necessarily independent. distributed as the rv

\begin{equation*} \IL = S - X. \end{equation*}

Proof

As observed earlier, each rv $D[k-\leadtime, k) \sim X$. As $S$ is constant, the rvs $I_{k} = S - D[k-\leadtime, k) \sim S - X$ for all $k\geq \leadtime+1$.

Now that we have characterized the distribution of $\IL$ in terms of the control parameter $S$ and the lead-time demand $X$, we can find expressions for the performance measures. First, for the ready rate,

\begin{equation} \label{eq:5} \alpha \stackrel1= \P{I'\geq 0} \stackrel2= \P{I - D \geq 0} \stackrel3= \P{S-X - D\geq 0} = \P{X+D\leq S}, \end{equation}

where 1 follows from the definition of $\alpha$, 2 from $I' = I- D$, 3 from $I=S-X$. Second, for the fill rate, by taking expectations in \cref{eq:c24}, we obtain

\begin{equation}\label{eq:ic3} \beta = \frac{\E{\min\{D, I^{+}\}}}{\E D}. \end{equation}

For the third performance measure, the cycle service level, I have not yet tried to find a simple expression in terms of expectations.

The mean inventory level is

\begin{equation*} \E{\IL} = \E{S-X} = S - \E X = S - \leadtime\E D. \end{equation*}

Next, noting that $I^+ - I^- = I$, the average on-hand inventory can be expressed in terms of the average backlog:

\begin{equation}\label{eq:27} \E{\IL^+} = \E{I} + \E{I^-} = S - \leadtime \E D + \E{I^-}. \end{equation}

From the definition of the backlog,

\begin{equation}\label{eq:ia22} \E{\IL^-} = \E{[-I]^{+}} = \E{[X-S]^{+}} = \sum_{k=0}^\infty (k- S)^+f(k). \end{equation}

This summation runs to $\infty$ if the support of $D$ is unbounded. For numerical purposes, we rewrite it as follows.

The average backlog is equal to

\begin{equation*} \E{\IL^-} = \leadtime \E D - \sum_{j=0}^{S-1} G_X(j). \end{equation*}

Proof

Since $\sum_{j=0}^\infty \1{j< k} = k$, we find from \cref{eq:ia22}

\begin{align*} \sum_{k=S}^\infty (k-S) f(k) &= \sum_{k=S}^\infty\sum_{j=0}^\infty \1{j < k-S}\, f(k) = \sum_{j=0}^\infty \sum_{k=S}^\infty \1{k > j +S}\, f(k)\\ &= \sum_{j=0}^\infty G_X(j+S) = \sum_{j=0}^\infty G_X(j) - \sum_{j=0}^{S-1} G_X(j). \end{align*}

Since $\sum_{i=0}^\infty G_X(i) = \leadtime \E D$, the claim follows.

Recall from \cref{th:5} that $I$, hence $I^{+ }$ and $I^{-}$, are functions of the reorder level $S$. Therefore, the average inventory cost $c(S) = h \E{I^{+}} + b \E{I^-}$ is also a function of $S$.

Here we illustrate the Lighthouse Company example of \cref{sec:single-item-invent}. With the RV class from \cref{sec:prob_arithmetic}, our code stays nearly in one-to-one correspondence with the mathematical definitions above.

from functools import cache
import numpy as np

import random_variable as rv
from functions import Plus, Min

from lighthouse_case import D, K, N, l, h, b

The basestock policy follows directly once we note that it suffices to specify the rv $X$; the other rvs then follow from $X$. Note that the performance measures depend on the control parameter $S$.

class Basestock:
    """Implements KPIs for a basestock inventory system."""

    def __init__(self, D, l, h, b):
        self.D = D
        self.l = l
        self.h = h
        self.b = b
        self.X = sum(self.D for _ in range(l)) if l > 0 else rv.RV({0: 1})

    @cache
    def IP(self, S):
        return S

    @cache
    def IL(self, S):
        return self.IP(S) - self.X

    @cache
    def Imin(self, S):
        return rv.apply_function(lambda x: Min(x), self.IL(S))

    @cache
    def Iplus(self, S):
        return rv.apply_function(lambda x: Plus(x), self.IL(S))

    def cost(self, S):
        return self.b * self.Imin(S).mean() + self.h * self.Iplus(S).mean()

    def alpha(self, S):
        return (self.IL(S) - self.D).sf(-1)

    def beta(self, S):
        m = rv.compose_function(lambda x, y: min(x, y), self.D, self.Iplus(S))
        return m.mean() / self.D.mean()

Here are some tests to verify the implementation and the preceding algebra.

S = 5

base = Basestock(D, l, h, b)

theta = l * D.mean()
print(np.isclose(base.IL(S).mean(), S - theta))
print(
    np.isclose(
        base.Imin(S).mean(), theta - sum(base.X.sf(j) for j in range(0, S))
    )
)
print(np.isclose(base.alpha(S), (base.X + D).cdf(S)))

True
True
True

The following shows how to use the code. We see that, as a function of $S$, the service levels increase, but the cost first decreases and then increases.⁴⁴ High service levels do not come for free.

for S in range(5, 12):
    print(f"{S=}: {base.alpha(S)=:.2f}, {base.beta(S)=:.2f}, {base.cost(S)=:.2f}")

S=5: base.alpha(S)=0.34, base.beta(S)=0.33, base.cost(S)=15.84
S=6: base.alpha(S)=0.47, base.beta(S)=0.47, base.cost(S)=9.42
S=7: base.alpha(S)=0.60, base.beta(S)=0.61, base.cost(S)=5.65
S=8: base.alpha(S)=0.72, base.beta(S)=0.74, base.cost(S)=4.08
S=9: base.alpha(S)=0.82, base.beta(S)=0.84, base.cost(S)=3.54
S=10: base.alpha(S)=0.89, base.beta(S)=0.91, base.cost(S)=3.63
S=11: base.alpha(S)=0.94, base.beta(S)=0.95, base.cost(S)=4.30

TF 1:

In a basestock model, the ready rate is given by $\alpha = \P{I - D \geq 0}$. Claim: The next code implements this correctly.

class Basestock:
    def __init__(self, D, L, h, b):
        self.D = D
        self.L = L
        self.h = h
        self.b = b
        self.X = sum(self.D for _ in range(self.L))

    def IL(self, S):
        return S - self.X

    def alpha(self, S):
        return (self.IL(S) - self.D).sf(0)

Solution

Did you actually try? Maybe see the ‘hints’ above!:

Solution, for real

False. It should be (self.IL(S) - self.D).sf(-1), so $-1$ instead of $0$.

TF 2:

Claim: In a basestock model with lead time demand $X$, the backlog is given by $\E{[S-X]^+}$.

Solution

Did you actually try? Maybe see the ‘hints’ above!:

Solution, for real

True.

Exercise 3:

Suppose $l=0$. What values do $X$, $\E\IL$, $\E{\IL^{+}}$ and $\E{\IL^{-}}$ take?

Solution

Did you actually try? Maybe see the ‘hints’ above!:

Solution, for real

If $\leadtime=0$, then $X=0$, hence $f_0 = \P{X=0} = 1$. Consequently, $\E I = S$, and $\E{\IL^{-}} = 0$ from \cref{eq:ia22}. Then from \cref{eq:27} it follows that $\E{\IL^{+}} = S$.

Exercise 4:

Why does the definition of $\beta$ follow from \cref{eq:c24}?

Solution

Did you actually try? Maybe see the ‘hints’ above!:

Solution, for real

$D_{k}$ and $I_{k}$ are independent, but $D_{k}$ and $I_{k}'$ not.

Exercise 5:

The cost function $S \mapsto c(S)$ is convex and coercive.A function $f$ is coercive if $\lim_{x\to \pm \infty} f(x) = \infty$. Prove that there exists an order-up-to level $S^{*}$ that minimizes the average cost.

Hint

Use the following properties for convex functions.

$f$ convex, $\alpha\geq 0$ $\implies$ $\alpha f$ convex.
$f, g$ convex $\implies$ $f+g$ convex.
$f$ convex, $R(x) = -x$ $\implies$ $f\circ R$ convex.

Solution

Did you actually try? Maybe see the ‘hints’ above!:

Solution, for real

Fix $i$. The function $S\to [S-i]^{+ }$ is convex. Since a weighted sum of convex functions is still convex, $\sum_{i=0}^{\infty} f(i) [S-i]^{+}$ is also convex. But $\sum_{i=0}^{\infty} f(i) [S-i]^{+} = \E{[S-X]^{+}} = \E{I^{+}}$, and therefore $h\E{I^{+}}$ is a convex function of $S$. Similarly, $S\to [i-S]^{+}$ is convex, and therefore $\E{I^{-}}$ is convex. Consequently, $c(S)$, being the sum of two convex functions, is convex. As $\E{I^{+}} \to \infty$ when $S\to\infty$, and $\E{I^{-}} \to \infty$ when $S\to-\infty$, $c(\cdot)$ attains a minimum.

Exercise 6:

For a nonnegative discrete rv $X$, use indicator functions to prove:

$\E X = \sum_{k=0}^\infty G(k)$,
$\sum_{i=0}^\infty i G(i) = \E{X^2}/2 - \E{X}/2$.

Hint

For (1) write $\sum_{k=0}^\infty G(k) = \sum_{k=0}^\infty \sum_{m=k+1}^\infty \P{X=m}$, reverse the summations. Then note that \(\sum_{k=0}^\infty \1{k \begin{equation*} \sum_{i=0}^\infty i G(i) = \sum_{n=0}^\infty \P{X=n} \sum_{i=0}^\infty i \1{n\geq i+1}, \end{equation*}

and reverse the summations.

Solution

Did you actually try? Maybe see the ‘hints’ above!:

Solution, for real

For (1): observe first that $\sum_{k=0}^\infty \1{m>k} = m$, since $\1{m>k}=1$ if $kk} = 0$ if $k\geq m$. With this,

\begin{align*} \sum_{k=0}^\infty G(k) &= \sum_{k=0}^\infty \P{X>k} = \sum_{k=0}^\infty \sum_{m=k+1}^\infty \P{X=m} \\ & = \sum_{k=0}^\infty \sum_{m=0}^\infty \1{m>k} \P{X=m} = \sum_{m=0}^\infty \sum_{k=0}^\infty \1{m>k} \P{X=m} \\ &= \sum_{m=0}^\infty m\P{X=m} = \E X. \end{align*}

Since all summands are nonnegative, interchanging the summations is valid.

For (2):

\begin{align*} \sum_{i=0}^\infty i G(i) &= \sum_{i=0}^\infty i \sum_{n=i+1}^\infty \P{X=n} = \sum_{n=0}^\infty \P{X=n} \sum_{i=0}^\infty i \1{n\geq i+1} \\ &= \sum_{n=0}^\infty \P{X=n} \sum_{i=0}^{n-1}i = \sum_{n=0}^\infty \P{X=n} \frac{(n-1)n}{2} \\ &= \sum_{n=0}^\infty \frac{n^2}{2} \P{X=n} - \frac{\E X}{2} = \frac{\E{X^2}}{2} - \frac{\E X}{2}. \end{align*}

This work is licensed by the University of Groningen under a Creative Commons Attribution-ShareAlike 4.0 International License.