Service-oriented decisions on inventory levels in the case of incomplete demand information

Abstract

New products without historical demand information or slow-moving items with little such information cause difficulties in defining inventory management policies facing demand uncertainty. The classical approach using the Normal distribution for describing the random demand during lead time might lead to a degraded level of customer service. But the choice for other types of distributions is also no option, so it is realistic that the full functional form of the distribution is unknown, but the decision-maker has some but not incomplete information on the demand distribution during lead time. As the distribution is only partially specified, several distributions satisfy the known information. Customer service measures therefore also take values in an interval between a lower and an upper bound. In this paper, upper and lower bounds are determined for two performance measures: the number of stock-out units and the stock-out probability per replenishment cycle, given incomplete information about the demand distribution, that is only the first two moments and the range, are known. Based on these results, the optimal inventory level given the desired maximum number of stock-out units or the desired maximum stock-out probability is calculated for the case where only the first two moments are known. The results of our approach are compared to the more traditional approach where a Normal distribution of demand during lead time is assumed. Comparisons with the Gamma, Uniform and symmetric triangular distribution are made. Furthermore, the robustness of our bounds to uncertainty in the parameters is tested.

Appendix

1.1 Number of stock-out units

1.1.1 Upper bounds

As already stated before, the problem is to find:

$$ \sup_{F\in\Upphi}\int\limits_{0}^{b}{f(x)\,{\rm d}F(x)} $$

where $\Upphi$ is the class of all distribution functions with range [0,b] and moments μ₁ and μ₂ and where f(x) = (x − d)₊.

The two-point distribution that will be used depends on the position of the parameter d in the interval [0,b]. Three situations can be distinguished.

1.1.1.1 Parabola through (0,0) and (0′,f(0′))

There exists a two-point distribution with moments μ₁ and μ₂ in (0,0′). The formula for the parabola can be used with u = 0, v = 0′ and f(0) = 0.

$$ g(x) = \frac{1}{0^{\prime 2}}[f(0^{\prime})0^{\prime}x+(f^{\prime}(0^{\prime})0^{\prime}-f(0^{\prime}))x(x-0^{\prime})]. $$

To assure that g ≥ 0 on [0, d], we impose g′(0) ≥ 0, which means that

$$ f'(0') \leq \frac{2f(0^{\prime})}{0^{\prime}} $$

or

$$ d \leq \frac{0^{\prime}}{2}. $$

The best upper bound is q_0'f(0′) or

$$ \frac{\mu_1}{\mu_2}(\mu_2-\mu_1d). $$

1.1.1.2 Parabola through (r,0) and (r′,f(r′))

The formula for the parabola is used with v = r, u = r′, f(v) = 0 and f′(v) = 0. This gives us:

$$ g(x) = \frac{f(r^{\prime})(x-r)^2}{(r^{\prime}-r)^2}. $$

The condition g′(u) = f′(r′) leads to

$$ 2f(r^{\prime}) = (r^{\prime}-r)f^{\prime}(r^{\prime}) $$

or

$$ d = \frac{r+r^{\prime}}{2}. $$

A unique solution (r,r′) can be assured by imposing the condition

$$ \frac{0^{\prime}}{2} \leq d \leq \frac{b+b^{\prime}}{2}. $$

Under this condition, the best upper bound is q_r'f(r′) or

$$ \frac{\mu_1-d+\sqrt{(\mu_2-\mu_1^2)+(d-\mu_1)^2}}{2}. $$

1.1.1.3 Parabola through (b′,0) and (b,f(b))

In this case, we take u = b, v = b′, f(v) = 0 and f′(v) = 0 and obtain

$$ g(x) = \frac{f(b)(x-b^{\prime})^2}{(b-b^{\prime})^2}. $$

To assure g ≥ f, we impose g′(b) ≤ f′(b) or

$$ 2f(b) \leq (b-b^{\prime})f^{\prime}(b) $$

or

$$ d \geq \frac{b+b^{\prime}}{2}. $$

In that case, the upper bound is q _b f(b) or

$$ \frac{(\mu_2-\mu_1^2)(b-d)}{(\mu_2-\mu_1^2)+(b-\mu_1)^2}. $$

1.1.2 Lower bounds

The problem is to find:

$$ \inf_{F\in\Upphi}\int\limits_{0}^{b}{f(x)\,{\rm d}F(x)} $$

where $\Upphi$ is the class of all distribution functions with range [0,b] and moments μ₁ and μ₂ and where f(x) = (x − d)₊.

Here, also three situations can be distinguished, depending on the position of d in the interval [0,b].

1.1.2.1 0 ≤ d ≤ b′

A solution is found when P is the straight line through (d, 0), (μ₁, f(μ₁)) and (b, f(b)). The three-point distribution will have masses:

$$ q_d=\frac{\mu_2-\mu_1^2}{(d-\mu_1)(d-b)}; q_{\mu_1}=\frac{\mu_2-\mu_1^2+(\mu_1-d)(\mu_1-b)}{(\mu_1-d)(\mu_1-b)}; q_b=\frac{\mu_2-\mu_1^2}{(b-d)(b-\mu_1)} $$

The lower bound equals q_{μ_1}f(μ₁) + q _b f(b) or

$$ \mu_1-d. $$

1.1.2.2 b′ < d < 0′

In this case, P is the parabola through (0, 0), (d,0) and (b, f(b)). The best lower bound is q _b f(b) or

$$ \frac{\mu_2-\mu_1d}{b}. $$

1.1.2.3 0′ ≤ d ≤ b

Here, a solution is found when P is the straight line through (0, 0), (μ, 0) and (d,0). The best lower bound is equal to 0.

1.2 Stock-out probability

When calculating bounds on tail probabilities [2], the problem is to find:

$$ \sup_{F\in\Upphi}\int\limits_{0}^{b}{f(x)\,{\rm d}F(x)} $$

and

$$ \inf_{F\in\Upphi}\int\limits_{0}^{b}{f(x)\,{\rm d}F(x)} $$

where $\Upphi$ is the class of all distribution functions with range [0,b] and with moments μ₁ and μ₂ known and where

$$ f(x) = \left\{ \begin{array}{ll} 0 & \hbox{ if } x \leq d;\\ 1 & \hbox{ if } x > d. \end{array} \right. $$

1.2.1 Upper bounds

1.2.1.1 0 ≤ d ≤ b′

A solution is found when P is the straight line through (b′,1) and (b,1). The upper bound is equal to q_b'f(b′) + q _b f(b) = 1.

1.2.1.2 b′ < d ≤ 0′

In this case, P is the parabola through (0, 0), (d, 1) and (b,1). According to Lemma 2, the three-point distribution in (0, d, b) will have masses:

$$ q_d=\frac{b\mu_1-\mu_2}{d(b-d)}, q_b=\frac{\mu_2-\mu_1d}{b(b-d)}, q_0=1-q_d-q_b. $$

The upper bound is q _d f(d) + q _b f(b) or

$$ \frac{(b+d)\mu_1-\mu_2}{bd}. $$

1.2.1.3 0′ < d ≤ b

Here, the solution is the parabola through (d′,0) and (d,1) and tangent tof(x) in d′. The best upper bound is q _d f′(d) + q_d'f(d′) or

$$ \frac{\mu_2-\mu_1^2}{\mu_2-\mu_1^2+(\mu_1-d)^2}. $$

1.2.2 Lower bounds

1.2.2.1 0 ≤ d ≤ b′

In this case, P is the parabola through (d,0) and d′,1) and tangent to f(x) at d′. The lower bound equals q _d f(d) + q_d'f(d′) or

$$ \frac{(\mu_1-d)^2}{\mu_2-\mu_1^2+(\mu_1-d)^2}. $$

1.2.2.2 b′ < d ≤ 0′

A solution is found when P is the parabola through (0, 0), (d,0) and (b,1). The masses of the three-point distribution give a lower bound of

$$ \frac{\mu_2-\mu_{1}d}{b(b-d)}. $$

1.2.2.3 0′ < d ≤ b

Here, P is the line through (0,0) and (0′,0). The lower bound is 0.