Analytic Results for the $(Q,r)$ Policy

1. Analytic Results for the $(Q,r)$ Policy

In this section, we extend the analysis of a single-item inventory system under a basestock policy to a $(Q,r)$ inventory system.

There is a convenient way to relate the $(Q,r)$ system to a number of basestock systems. To see this, imagine two animals, a squirrel that observes the $(Q,r)$ system at the end of each demand period, and a bear that only wakes up when the inventory position right at the start of the interval equals some fixed $i$, where $i$ must lie in the set $\{r+1, \ldots r+Q\}$; if not, the bear hibernates. Clearly, from the bear’s point of view, the system behaves as a basestock system with order-up-to level $i$, because whenever the bear wakes, the inventory position is $i$, and the position just before the bear falls asleep is $i-D$. More generally, we can associate a different bear with each position in the set $\{r+1, \ldots r+Q\}$, so exactly one bear is awake at any time. To combine the statistics observed by the bears and the statistics observed by the squirrel, we need, for each $i$, the long-run fraction of periods in which the inventory position $\IP$ equals $i$. For instance, if the squirrel knows¹¹ We assume that the limit over all sample paths $\{\IP_{k}\}$ exists almost surely, so that we can invoke the strong law of large numbers.

\begin{equation*} p(i) = \lim_{T\to \infty} \frac{1}{T} \sum_{k=1}^{T}\1{\IP_k = i}, \end{equation*}

for all $i\in \{r+1, \ldots, r+Q\}$, then it can retrieve the ready rate $\sum_{i=r+1}^{Q+r} p(i) \alpha(i)$ from the ready rate $\alpha(i)$ observed by the bear with order-up-to level $i$.²² We label the ready rate of the basestock model by the order-up-to level.

Theorem 2.

Under the $(Q,r)$ policy, $p(i) = \P{\IP = i} = 1/Q$ for $i=r+1, \ldots, r+Q$, in the long-run time limit. Consequently, $\IL$ is distributed as $\IP-X$.

Proof

For simplicity take $Q=3, r=0$, so that the inventory position $\IP$ cycles between the levels $1, 2$ and $3$.³³ The proof is the same for general $Q$ and $r$, but needs a bit more notation. We apply level-crossing arguments, as follows. Write $f_{i} = \P{D\in \{i, i+3, i+6, \ldots\}}$ for the probability that the period demand equals a multiple of $3$ plus $i$, $i=0, 1, 2$. Suppose that, at the start of period $k$, the position $\IP_{k} = 3$. Since we order in multiples of $Q=3$, the probability $\P{\IP_{k+1} =3|\IP_{k}=3} = f_{0}$, i.e., equal to the probability that $D_{k}$ is a multiple of $3$. Likewise, $\P{\IP_{k+1} = 2|\IP_k=3} = f_{1}$, and $\P{\IP_{k+1} = 1|\IP_k=3} = f_{2}$. Next, since we again order in multiples of $Q$, $\P{\IP_{k+1} = 1|\IP_k=2} = f_{1}$, and similarly for the other possibilities. Therefore, we can use level crossing to conclude that in the long run, $(f_{1}+f_2)p(3) = f_1p(2) + f_2 p(1)$. However, because of the $(Q,r)$ ordering rule, we can cyclically change $p(1), p(2)$ and $p(3)$ in this equality. By symmetry, the only possible solution is $p(i) = 1/3$ for all $i$.

With the above theorem the ready rate for the $(Q,r)$ model can be computed with the formula

\begin{align} \alpha = \frac1Q \sum_{i=r+1}^{r+Q} \alpha(i) = \frac{1}{Q} \sum_{i=r+1}^{r+Q} \P{X+D\leq i}, \end{align}

where $\alpha(i)$ is the ready rate of the basestock model with reorder level $i$.

For the average inventory level, we write $I(i)$ for the inventory level of the basestock model with order-up-to level $i$, and get

\begin{align*} \E{I} &= \frac1Q\sum_{i=r+1}^{r+Q} \E{I(i)} = \frac1Q\sum_{i=r+1}^{r+Q} (i - \E X) = \frac{Q+1}2 + r - \E X. \end{align*}

The expected number of back-orders $\E{I^-}$ and expected on-hand inventory $\E{I^+}$ can be found by equivalent summations:

\begin{align*} \E{I^{+}} &= \frac1Q\sum_{i=r+1}^{r+Q} \E{I^{+}(i)}, & \E{I^{-}} &= \frac1Q\sum_{i=r+1}^{r+Q} \E{I^{-}(i)}, \end{align*}

using \cref{eq:27} and \cref{eq:ia22} of the basestock model.

Finally, with $h$ and $b$ the holding and backorder cost per item per period, the total average cost becomes

\begin{equation*} c(Q,r) = K\frac{\E D}{Q} + h \E{I^{+}} + b \E{I^-}. \end{equation*}

With regard to the first term, note that $\E D$ is the average demand per period, so $\E D / Q$ is the order frequency.

It remains to find optimal values for $r$ and $Q$. Fast algorithms exist to compute optimal $r$ and $Q$, but we omit this. A simple workaround is to perform a complete grid search over a reasonable set of pairs of $r$ and $Q$. Another simple heuristic is to use the EOQ⁴⁴ See \cref{ex:qr2}, and set $d=\E D$. formula to set $Q = \sqrt{2\E{D}K/h}$ and then search for $r$ such that the total average cost is minimal.\sidenote{For that value of $Q$!}

With similar methods we can search for an $r$ such that the $\alpha$ or $\beta$ service level criteria are met.

To code the above formulas we use the RV class again so that our implementation can stay on a high level. Note in particular that the code relies on \cref{th:3}.

from functools import cache
import numpy as np

import random_variable as rv
from functions import Plus, Min, grid_search
from lighthouse_case import D, K, N, l, h, b


class Qr:
    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, Q, r):
        return rv.RV({i: 1 / Q for i in range(r + 1, r + Q + 1)})

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

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

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

    def order_freq(self, Q,r):
        return self.D.mean()/Q

    def cost(self, Q, r):
        C = self.K*self.order_freq(Q)
        C += self.b * self.Imin(Q, r).mean() + self.h * self.Iplus(Q, r).mean()
        return C

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

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

2.275

Here is an example of how to use this class.

qr = Qr(D, l, h, b)
Q, r = 3, 10
print(qr.IL(Q, r).mean())
print((Q + 1) / 2 + r - qr.X.mean())

7.450000000000001
7.45

TF 1:

Claim: The ($Q, r$) inventory policy can be analyzed by comparing it with multiple basestock systems.

Solution

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

Solution, for real

True.

TF 2:

Claim: In a ($Q, r$) policy system, the inventory position $P$ at the beginning of an interval must always be equal to the reorder level $r$.

Solution

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

Solution, for real

False.

TF 3:

Consider the following Python code.

class Qr:
    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))

qr_instance = Qr(D=2, l=3, h=5, b=1)

Claim: The value of \texttt{qr\_instance.X} will be 6.

Solution

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

Solution, for real

True. The sum operation in the constructor computes $2 + 2 + 2 = 6$ based on $D=2$ and $l=3$.

TF 4:

The fill rate for the $(Q,r)$ inventory system is defined as

\begin{equation*} \beta = \E{\min\{D, I^{+}\}}/\E D. \end{equation*}

Claim: The following Python code implements this correctly:

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

Solution

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

Solution, for real

True.

TF 5:

Claim: The order frequency of the ($Q, r$) is equal to $Q/\E D$.

Solution

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

Solution, for real

False.

TF 6:

Claim: this Python code computes the average cost under a $(Q,r)$ policy.

C = K * Q / D.mean()
C += b * Imin(Q, r).mean() + h * Iplus(Q, r).mean()

Claim: The value of \texttt{qr\_instance.X} will be 6.

Solution

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

Solution, for real

True. The sum operation in the constructor computes $2 + 2 + 2 = 6$ based on $D=2$ and $l=3$.

Exercise 7:

Perhaps the most important formula in practical inventory management is the Economic Order Quantity (EOQ) that determines the optimal order quantity in case demand is constant, i.e., $D\equiv d$, the lead time $\leadtime=0$ and backlogging is not allowed. Use the $(Q,r)$ model to show that it is optimal to take $Q^{2} = 2 d K/h$.

Hint

No backlogging and $l=0$ $\implies r=-1$. Use \cref{ex:bs1}.

Solution

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

Solution, for real

When $\leadtime = 0$, we have $\E X = 0$ so that for the basestock model, $\E{\IL^{+}(i)} = i$. Hence, for the $(Q,r)$ model, $\E{\IL^{+}} = Q(Q-1)/2Q = (Q-1)/2$. Since demand arrives at a constant rate $d$, the average cost becomes $Kd/Q + h Q/2 - h/2$. Differentiating with respect to $Q$ and setting the derivative to zero yields the result.

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1. Analytic Results for the \((Q,r)\) Policy