Two-Echelon System

Abstract

Chapter 6 is devoted to a multi-item, two-echelon spare parts inventory system. The system consists of a central warehouse and multiple local warehouses, and there is a target for the aggregate mean waiting time per local warehouse. We describe an exact recursion for the evaluation of a given basestock policy, leading to an exact evaluation method, and we give approximate evaluation methods based on two-moment and one-moment fits, respectively. Next, we describe a greedy heuristic and the Dantzig-Wolfe heuristic, together with the Dantzig-Wolfe lower bound on the optimal total costs. Both heuristics can be combined with a local search, which gives four heuristics in total. The quality of the heuristics, all based on exact evaluations, is tested in a computational experiment. The greedy heuristic (without local search) performs most satisfactorily. It is accurate, easy to implement, and the computational requirements are limited. Its computational efficiency can be increased by using approximate evaluation based on two-moment fits. That results in a good feasible algorithm for real-life cases with many SKU’s and local warehouses.

Appendix: Fitting Discrete Distributions on the First Two Moments

In this appendix, we discuss two-moment fits of discrete probability distributions on the first two moments E{X} and E{X ²} of a random variable X with support $\mathbb{N}_{0}$. We first discuss a number of discrete probability distributions. Then we discuss the fit of a negative binomial distribution on the given first two moments, where the so-called stopping parameter k of the negative binomial distribution is allowed to be non-integer. After that we discuss the two-moment fit of Adan et al. [1].

6.1.1 Discrete Probability Distributions

A Bernoulli trial is an experiment with two possible outcomes, “success” and “failure”. Let p, 0 ≤ p < 1, be the probability for the outcome “success”. Let B(p) be the random variable that is equal to 1 if the outcome is “success” and 0 if the outcome is “failure”. So,

$$\displaystyle{P\{B(p) = 0\} = 1 - p,\ \ P\{B(p) = 1\} = p.}$$

The distribution of B(p) is called a Bernoulli distribution with success probability p. The first two moments and variance are equal to:

$$\displaystyle{E\{B(p)\} = p,\ \ E\{(B(p))^{2}\} = p,\ \ V ar\{B(p)\} = p(1 - p).}$$

Next, suppose that one executes a sequence of k, $k \in \mathbb{N}$, Bernoulli trials, where each trial is independent of earlier trials and each trial has the same probability p for the outcome “success”. Let BIN(k, p) be the number of successes. Then

$$\displaystyle{P\{BIN(k,p) = x\} = \binom{k}{x}p^{x}(1 - p)^{k-x},\ \ x \in \mathbb{N}_{ 0}.}$$

The distribution of BIN(k, p) is called a binomial distribution with k trials and success probability p per trial. The first two moments and variance are equal to:

$$\displaystyle\begin{array}{rcl} & & E\{BIN(k,p)\} = kp,\ \ E\{(BIN(k,p))^{2}\} = kp[(k - 1)p + 1],{}\end{array}$$

(6.32)

$$\displaystyle\begin{array}{rcl} & & V ar\{BIN(k,p)\} = kp(1 - p).{}\end{array}$$

(6.33)

Suppose that one executes a sequence of Bernoulli trials, with success probability p per trial, and this sequence is stopped as soon as one gets a failure. Let GEO(p) be the random variable that denotes the number of successes. Then

$$\displaystyle{P\{GEO(p) = x\} = p^{x}(1 - p),\ \ x \in \mathbb{N}_{ 0}.}$$

The distribution of GEO(p) is called a geometric distribution with success probability p per trial. The first two moments and variance are equal to:

$$\displaystyle\begin{array}{rcl} & & E\{GEO(p)\} = \frac{p} {1 - p},\ \ E\{(GEO(p))^{2}\} = \frac{p(1 + p)} {(1 - p)^{2}}, {}\\ & & V ar\{GEO(p)\} = \frac{p} {(1 - p)^{2}}. {}\\ \end{array}$$

Let us now consider the execution of a sequence of Bernoulli trials, with success probability p per trial, and assume that this sequence is stopped as soon as one gets the k-th failure. Let NB(k, p) be the random variable that denotes the total number of successes. NB(k, p) is the sum of k independent, geometrically distributed random variables with success probability p, and its distribution is called a negative binomial distribution with stopping parameter k and success probability p. The variable NB(k, p) has the above interpretation for strictly positive, integer values of k, but NB(k, p) is also well-defined for strictly positive, non-integer values (i.e., for k > 0). The distribution of NB(k, p) is given by

$$\displaystyle{P\{NB(k,p) = x\} = \binom{x + k - 1}{x}p^{x}(1 - p)^{k},\ \ x \in \mathbb{N}_{ 0},}$$

where the binomial coefficient may be computed by the following formula (for all k > 0):

$$\displaystyle{\binom{x + k - 1}{x} = \frac{(x + k - 1) \cdot (x + k - 2) \cdot \ldots \cdot k} {x!}.}$$

The first two moments and variance are equal to:

$$\displaystyle\begin{array}{rcl} & & E\{NB(k,p)\} = \frac{kp} {1 - p},\ \ E\{(NB(k,p))^{2}\} = \frac{kp(1 + kp)} {(1 - p)^{2}}, {}\\ & & V ar\{NB(k,p)\} = \frac{kp} {(1 - p)^{2}}. {}\\ \end{array}$$

Finally, we consider the Poisson distribution. Let POIS(λ) be a random variable that is Poisson distributed with parameter λ > 0. Then:

$$\displaystyle\begin{array}{rcl} & & P\{POIS(\lambda ) = x\} = \frac{\lambda ^{x}} {x!}e^{-\lambda },\ \ x \in \mathbb{N}_{ 0},{}\end{array}$$

(6.34)

$$\displaystyle\begin{array}{rcl} & & E\{POIS(\lambda )\} =\lambda,\ \ E\{(POIS(\lambda ))^{2}\} =\lambda (\lambda +1),{}\end{array}$$

(6.35)

$$\displaystyle\begin{array}{rcl} & & V ar\{POIS(\lambda )\} =\lambda.{}\end{array}$$

(6.36)

6.1.2 Two-Moment Fit by a Negative Binomial Distribution

Let X be a random variable with support $\mathbb{N}_{0}$ and given first two moments E{X} and E{X ²}. The variance of X is given by $V ar\{X\} = E\{X^{2}\} - (E\{X\})^{2}$. Assume that

$$\displaystyle{ V ar\{X\} > E\{X\}. }$$

(6.37)

Notice that this implies that E{X} > 0 (E{X} = 0 is only obtained if X = 0 with probability 1, but then Var{X} would be equal to 0 as well).

Under the assumption of (6.37), we can fit random variable NB(k, p) on the first two moments of X by choosing k and p equal to

$$\displaystyle{p = \frac{V ar\{X\} - E\{X\}} {V ar\{X\}},\ \ k = \frac{1 - p} {p} E\{X\}}$$

(notice that 0 < p < 1 and k > 0). It is easily seen that then E{NB(k, p)} = E{X} (this follows immediately from the expression for k) and Var{NB(k, p)} = Var{X} (use $V ar\{NB(k,p)\} = E\{NB(k,p)\}/(1 - p) = E\{X\}/(1 - p)$ and the expression for p), and thus it also holds that $E\{(NB(k,p))^{2}\} = E\{X^{2}\}$.

6.1.3 Two-Moment Fit of Adan et al. [1]

Let X be a random variable with support $\mathbb{N}_{0}$ and given first two moments E{X} and E{X ²}. We now describe the method of Adan et al. [1] for the fit of a discrete random variable Y (also with support $\mathbb{N}_{0}$) on these first two moments.

First, determine the squared coefficient of variation $c_{X}^{2} = V ar\{X\}/(E\{X\})^{2}$, and the parameter

$$\displaystyle{a = c_{X}^{2} - \frac{1} {E\{X\}}.}$$

It holds that a ≥ −1 (cf. Lemma 2.1 of [1]). We define Y as follows:

• (a) If − 1 ≤ a < 0, then choose $k \in \mathbb{N}$ such that $-1/k \leq a < -1/(k + 1)$, and

  $$\displaystyle{Y:= \left \{\begin{array}{ll} BIN(k,p) &\mbox{ w.p. }q;\\ BIN(k + 1, p)\ \ &\mbox{ w.p. } 1 - q, \end{array} \right.}$$

  where

  $$\displaystyle{q = \frac{1 + a(1 + k) + \sqrt{-ak(1 + k) - k}} {1 + a},\ \ p = \frac{E\{X\}} {k + 1 - q}.}$$

• (b) If a = 0, then Y: = POIS(λ) with λ = E{X}.

• (c) If 0 < a < 1, then choose $k \in \mathbb{N}$ such that $1/(k + 1) \leq a < 1/k$, and

  $$\displaystyle{Y:= \left \{\begin{array}{ll} NB(k,p) &\mbox{ w.p. }q;\\ NB(k + 1, p)\ \ &\mbox{ w.p. } 1 - q, \end{array} \right.}$$

  where

  $$\displaystyle{q = \frac{a(k + 1) -\sqrt{(k + 1)(1 - ak)}} {1 + a},\ \ p = \frac{E\{X\}} {k + 1 - q + E\{X\}}.}$$

• (d) If a ≥ 1, then

  $$\displaystyle{Y:= \left \{\begin{array}{ll} GEO(p_{1})&\mbox{ w.p. }q; \\ GEO(p_{2})\ \ &\mbox{ w.p. }1 - q, \end{array} \right.}$$

  where

  $$\displaystyle{q = \frac{1} {1 + a + \sqrt{a^{2 } - 1}},\ \ p_{1} = \frac{E\{X\}} {2q + E\{X\}},\ \ p_{2} = \frac{E\{X\}} {2(1 - q) + E\{X\}}.}$$

Then the first two moments of Y are equal to the first two moments of X (cf. Lemma 2.2 of [1]).

The cases (c) and (d) are used when a > 0, which is equivalent to (6.37). So, the fits of these cases form an alternative of the two-moment fit by NB(k, p) with a real-valued stopping parameter k > 0. The advantage of the fits used in the cases (c) and (d) is that they allow a simple interpretation in terms of mixtures and sums of geometric distributions.

Problems

6.1. Consider Example 6.1. In this example, $W_{i,1}(\mathbf{S}_{i})$ has been determined via the exact evaluation procedure, the approximate evaluation procedure based on two-moment fits, and the METRIC approach.

• (a) Determine W _i, 2(S _i ) via the exact evaluation procedure.

• (b) Determine $W_{i,2}(\mathbf{S}_{i})$ via the approximate evaluation procedure based on two-moment fits. Use a negative binomial distribution for the two-moment fit for $X_{i,2}(S_{i,0})$. How large is the difference with the exact outcome?

• (c) Determine $W_{i,2}(\mathbf{S}_{i})$ via the METRIC approach. How large is the difference with the exact outcome?

6.2. For a random variable X with support $\mathbb{N}_{0}$, it is given that E{X} = 1 and Var{X} = 1. 5.

• (a) Fit a negative binomial distribution on the first two moments of X. This gives a negative binomial distribution with specific values for the stopping parameter k and the success probability p. Use the formulas of Appendix “Discrete Probability Distributions” to compute the mean and variance of this negative binomial distribution.

• (b) Use the method of Adan et al. to fit a distribution on the first two moments of X (cf. Appendix “Two-Moment Fit of Adan et al. [1]”). Use the formulas of Appendix “Discrete Probability Distributions” to compute the mean and variance of the obtained distribution.

6.3. For the model in this chapter, we have assumed an aggregate mean waiting time constraint per local warehouse. Suppose that the whole set of local warehouses J ^loc can be partitioned in subsets $J_{1}^{\text{loc}},\ldots,J_{K}^{\text{loc}}$ and that we have an aggregate mean waiting time constraint per subset of local warehouses. This could be relevant when subsets of local warehouses are located in the same region and one wants to control the performance per region rather than per local warehouse. This gives a generalized model.

• (a) Formulate the aggregate mean waiting time constraints for this generalized model.

• (b) It is easily seen that the evaluation procedures remain applicable without any changes. This does not hold for the heuristics. Formulate an adapted version of the greedy heuristic for the generalized problem.

6.4. In Sect. 6.4.1, it is described that $\varDelta _{i,j}(\mathbf{S}) \geq 0$ for all $S \in \mathcal{S}$, i ∈ I and j ∈ J; see the paragraph after Eqs. (6.29)–(6.30). This can be shown via monotonicity properties for the mean waiting times W _i, j (S _i ). In this problem, you are asked to prove these monotonicity properties. Let i ∈ I.

• (a) Let j ∈ J ^loc, $\mathbf{S} \in \mathcal{S}$, and $\mathbf{S}^{{\prime}} = \mathbf{S} + \mathbf{E}_{i,j}$. Show that

  • $W_{i,j}(\mathbf{S}^{{\prime}}) \leq W_{i,j}(\mathbf{S})$;

  • $W_{i,j'}(\mathbf{S}^{{\prime}}) = W_{i,j'}(\mathbf{S})$ for all j′ ∈ J ^loc, j′ ≠ j.

• (b) Let $\mathbf{S} \in \mathcal{S}$, and $\mathbf{S}^{{\prime}} = \mathbf{S} + \mathbf{E}_{i,0}$. Show that $W_{i,j}(\mathbf{S}^{{\prime}}) \leq W_{i,j}(\mathbf{S})$ for all j ∈ J ^loc.

Hint: Use the recursive expressions of Sect. 6.3.1.

6.5. Consider a given SKU i ∈ I and assume that at the central warehouse an (s, Q)-policy instead of a basestock policy is used for this SKU. Its reorder level is given by s _i and the fixed batch size by Q _i . We study how the exact evaluation method can be adapted for this generalized inventory policy.

• (a) Derive adapted formulas for the distribution of the on-hand stock OH _i, 0 and number of backorders BO _i, 0 at the central warehouse. Hint: Use the results of Sect. 2.10.3

• (b) Denote what needs to be adapted in the rest of the exact evaluation procedure.

6.6. Consider an OEM with a two-echelon spare parts system as described in this chapter. The OEM has service contracts with many customers regarding the spare parts supply. Each customer is coupled to a local warehouse and the OEM ensures that the aggregate mean waiting time per local warehouse stays within a certain limit. Customers without service contracts can also buy spare parts, but they are served from the central warehouse. Their orders constitute a so-called direct demand stream at the central warehouse. For each SKU i ∈ I, this extra demand stream is assumed to be a Poisson stream and its demand rate is denoted by $\hat{m}_{i}$. At the central warehouse, one treats these demands in the same way as replenishment orders from the local warehouses.

The above situation leads to a generalized model with a direct demand stream at the central warehouse. The formulation of Problem (P) remains the same under this generalization. Describe what needs to be changed in the exact evaluation procedure to deal with this generalization.

6.7. Six European countries plan to buy the same military system. We consider a single, expensive repairable for which all repairs are executed by the original manufacturer from the US. We consider the following two scenarios:

• Scenario 1: Each country organizes its own spare parts inventory. In that case, each country has a own stockpoint and failed parts are sent back to the manufacturer in the US. After repair, the same parts come back as ready-for-use parts. The leadtime for this repair loop is 6 months.

• Scenario 2: The countries create pooling via a joint central warehouse. Countries send their failed parts back to the central warehouse (within a negligibly small leadtime), and request a ready-for-use part from there. The central warehouse sends the failed parts back to the manufacturer in the US, and after repair they are returned to the central warehouse as ready-for-use parts. This repair loop has a leadtime of 6 months (as in scenario 1). The leadtime for sending parts from the central warehouse to a local warehouse in a country is 1 week ( = 0.02 years, say).

For Scenario 2, further assumptions are such that the model of Chap. 6 applies (with 1 SKU instead of a general number of SKU’s). For Scenario 1, similar assumptions are made. Per country, the total failure rate for the repairable is 2 failures per year. Each country has a target of 0.01 years for the mean waiting time per failure.

The minimum amount of spare parts that each country has to take on stock under Scenario 1 is denoted by $\hat{S}$. The total amount of stock under Scenario i = 1, 2 is denoted by S _tot ⁽ⁱ⁾.

Under Scenario 2, one has more transportation movements and more handling because of the central warehouse. However, the extra costs are limited because the central warehouse also leads to consolidation of the transport to and from the US. Further, the repairable is expensive and thus inventory holding costs dominate. Hence, it is justified to compare both scenarios on the basis of the total stock that is required to meet the mean waiting time constraints.

• (a) For Scenario 1, determine $\hat{S}$ and $S_{\text{tot}}^{(1)}$. What is the mean waiting time under these values?

• (b) Next, we analyze Scenario 2, and assume that a basestock level of 1 is used in each local warehouse. Initially, we consider this scenario with a total stock S _tot ⁽²⁾ that is equal to the total stock S _tot ⁽¹⁾ found under (a), i.e., a basestock level equal to $S_{\text{tot}}^{(1)} - 6$ is used at the central warehouse. Determine the mean waiting time that is obtained at each of the local warehouses under this basestock policy?

• (c) Suppose that one sticks to a basestock level of 1 at each of the local warehouses. What is the minimum basestock level at the central warehouse that one needs to meet the mean waiting time constraints? What is the total stock under this value, and how large is the reduction in total stock compared to the total stock required under scenario 1?