5.7 Analytic Results for the Base-Stock Policy

In this section, we establish a set of formulas to compute performance measures for the single-item base-stock inventory system of Section 2.5.¹¹This way, 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 rvs²²Why 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 Section 2.5 to derive expressions for the long-run performance measures. The first step is to relate the outstanding orders to the demand.

Proof

Using the recursions for the inventory position under the base-stock 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.

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 \(D[k-\leadtime, k) = D[1, k) - D[1, k-\leadtime)\).

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

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*} \alpha \stackrel1= \P{I'\geq 0} \stackrel2= \P{I - D \geq 0} \stackrel3= \P{S-X - D\geq 0} = \P{X+D\leq S}, \tag{5.7.1} \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 (2.5.4), we obtain

\begin{equation*} \beta = \frac{\E{\min\{D, I^{+}\}}}{\E D}. \tag{5.7.2} \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*} \E{\IL^+} = \E{I} + \E{I^-} = S - \leadtime \E D + \E{I^-}. \tag{5.7.3} \end{equation*}

From the definition of the backlog,

\begin{equation*} \E{\IL^-} = \E{[-I]^{+}} = \E{[X-S]^{+}} = \sum_{k=0}^\infty (k- S)^+f(k). \tag{5.7.4} \end{equation*}

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

Proof

Since \(\sum_{j=0}^\infty \1{j< k} = k\), we find from (5.7.4)

\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 Th. 5.7.3 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 \(S\).

For the Lighthouse Company case of Section 2.5, Alg. 5.7.4 computes the ready rate, the fill rate and the cost for various values of \(S\) of the base-stock policy. It relies heavily on the algorithms of Section 1.4.

\begin{algorithm}
\caption{Compute the ready rate of a base-stock policy.}
\begin{algorithmic}[1]
\State \textbf{Input:} inventory and demand pmf maps \(p_I,p_D\)
\State \textbf{Output:} ready rate
\Procedure{ReadyRate}{\(p_I,p_D\)}
    \State \(p_{I-D}\gets\pr{Subtract}\rb{p_I,p_D}\)
    \Return \(\pr{Expectation}\rb{p_{I-D},\lambdafun{z}{\1{z\geq0}}}\)
\EndProcedure
\end{algorithmic}

\end{algorithm}

\begin{algorithm}
\caption{Compute the fill rate of a base-stock policy.}
\begin{algorithmic}[1]
\State \textbf{Input:} inventory and demand pmf maps \(p_I,p_D\)
\State \textbf{Output:} fill rate
\Procedure{FillRate}{\(p_I,p_D\)}
    \State \(p_{I^+}\gets\pr{PositivePart}\rb{p_I}\)
    \State \(p_{\min\{D,I^+\}}\gets\pr{Minimum}\rb{p_D,p_{I^+}}\)
    \Return \(\pr{Mean}\rb{p_{\min\{D,I^+\}}}/\pr{Mean}\rb{p_D}\)
\EndProcedure
\end{algorithmic}

\end{algorithm}

\begin{algorithm}
\caption{Compute the average cost of a base-stock policy.}
\begin{algorithmic}[1]
\State \textbf{Input:} inventory pmf map \(p_I\), costs \(h,b\)
\State \textbf{Output:} average inventory cost
\Procedure{InventoryCost}{\(p_I,h,b\)}
    \State \(p_{I^+}\gets\pr{PositivePart}\rb{p_I}\)
    \State \(p_{I^-}\gets\pr{NegativePart}\rb{p_I}\)
    \Return \(h\pr{Mean}\rb{p_{I^+}}+b\pr{Mean}\rb{p_{I^-}}\)
\EndProcedure
\end{algorithmic}

\end{algorithm}

\begin{algorithm}
\caption{Evaluate a base-stock policy.}
\begin{algorithmic}[1]
\State \textbf{Input:} demand pmf map \(p_D\), lead time \(\leadtime\), costs \(h,b\), base-stock level \(S\)
\State \textbf{Output:} ready rate \(\alpha\), fill rate \(\beta\), and average cost \(c\)
\Procedure{EvaluateBaseStock}{\(p_D,\leadtime,h,b,S\)}
    \State \(p_X\gets\pr{Convolution}\rb{p_D,\leadtime}\)
    \State \(p_I\gets\pr{Subtract}\rb{\pr{Constant}\rb{S},p_X}\)
    \State \(\alpha\gets\pr{ReadyRate}\rb{p_I,p_D}\)
    \State \(\beta\gets\pr{FillRate}\rb{p_I,p_D}\)
    \State \(c\gets\pr{InventoryCost}\rb{p_I,h,b}\)
    \Return \(\alpha,\beta,c\)
\EndProcedure
\end{algorithmic}

\end{algorithm}

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

Thus, as a function of \(S\), the service levels increase, but the cost first decreases and then increases.⁴⁴High service levels do not come for free.

The cost function \(S \mapsto c(S)\) is convex and coercive.⁵⁵A function \(f\) is coercive if \(\lim_{x\to \pm \infty} f(x) = \infty\).