A risk-averse multi-item inventory problem with uncertain demand

Abstract

The observed values of demands in real-life inventory problems are sometimes imprecise due to the lack of information and historical data, thus a growing research is committed to study the properties of risk measures in fuzzy inventory optimization problems. In this paper, a risk-averse fuzzy optimization method is adopted for the multi-item inventory problem, in which the demands are described by common possibility distributions. Firstly, three classes of fuzzy inventory optimization models are built by combining the absolute semi-deviation with expected value operator and then model analysis is given for the min-max inventory models. To make the inventory problem tractable and computable, the equivalent forms of the proposed optimization models are discussed. Subsequently, several useful absolute semi-deviation formulas are presented under triangular, trapezoidal and Erlang possibility distributions. Finally, some numerical experiments are performed to highlight the modeling idea, and the computational results demonstrate the effectiveness of the solution method.

Appendix

1.1 Proof of Theorem 1

Proof

If we denote η = 1/ξ, then the credibility distribution of η reads that

$$ \begin{array}{@{}rcl@{}} \text{Cr}\{\eta \leq r\}=\left\{ \begin{array}{llll} &0,&& 0<r<\frac{1}{r_{3}}\\ &\frac{r_{3}-\frac{1}{r}}{2(r_{3}-r_{2})},&& \frac{1}{r_{3}}\leq r<\frac{1}{r_{2}}\\ &\frac{2r_{2}-r_{1}-\frac{1}{r}}{2(r_{2}-r_{1})},&& \frac{1}{r_{2}}\leq r<\frac{1}{r_{1}}\\ &1,&& r\geq\frac{1}{r_{1}}. \end{array} \right. \end{array} $$

Since the expected value $m=\frac {1}{2(r_{3}-r_{2})}\ln {\frac {r_{3}}{r_{2}}}+\frac {1}{2(r_{2}-r_{1})}\ln {\frac {r_{2}}{r_{1}}}$ has computed in Li and Liu (2016), then if $m\in [\frac {1}{r_{3}},\frac {1}{r_{2}})$, the absolute semi-deviation value E[(η − m)⁺] is as follows:

$$ \begin{array}{@{}rcl@{}} \mathrm{E}[(\eta-m)^{+}]&=&{\int}_{(m,+\infty)}(r-m)\mathrm{d}\text{Cr}\{\eta\leq r\}\\ &=&{\int}_{\left( m,\frac{1}{r_{2}}\right)}(r-m)\mathrm{d}\frac{r_{3}-\frac{1}{r}}{2(r_{3}-r_{2})}\\ &&+{\int}_{\left( \frac{1}{r_{2}},\frac{1}{r_{1}}\right)}(r-m)\mathrm{d}\frac{2r_{2}-r_{1}-\frac{1}{r}}{2(r_{2}-r_{1})}\\ &&+{\int}_{\left( \frac{1}{r_{1}},+\infty\right)}(r-m)\d1\\ &=&\frac{1}{2(r_{3}-r_{2})}\ln \frac{1}{r_{2}m}+\frac{1}{2(r_{2}-r_{1})}\ln \frac{r_{2}}{r_{1}}\\ &&+\frac{m(2r_{2}-r_{3})-1}{2(r_{3}-r_{2})}. \end{array} $$

Similarly, if $m\in \left [\frac {1}{r_{2}},\frac {1}{r_{1}}\right ]$, the absolute semi-deviation value E[(η − m)⁺] is computed by

$$ \mathrm{E}[(\eta-m)^{+}]=\frac{1}{2(r_{2}-r_{1})}\ln \frac{1}{r_{1}m}+\frac{mr_{1}-1}{2(r_{2}-r_{1})}. $$

We next to show that when r₁ = 0, the absolute semi-deviation value E[(η − m)⁺] does not exist. In fact, the credibility distribution of η in this case reads that

$$ \begin{array}{@{}rcl@{}} \text{Cr}\{\eta \geq r\}=\left\{ \begin{array}{llll} &1, &&0<r<\frac{1}{r_{3}}\\ &\frac{(r_{3}-2r_{2})r+1}{2r(r_{3}-r_{2})},&& \frac{1}{r_{3}}\leq r<\frac{1}{r_{2}}\\ &\frac{1}{2r_{2}r},&& r\geq \frac{1}{r_{2}}. \end{array} \right. \end{array} $$

Thus, the expected value m is computed by

$$ \begin{array}{@{}rcl@{}} m&=&{\int}_{0}^{+\infty}\text{Cr}\{\eta \geq r\}\text{dr}\\ &=&\frac{1}{2(r_{3}-r_{2})}\ln{\frac{r_{3}}{r_{2}}}+\frac{1}{2r_{2}}(1+\lim_{d\rightarrow+\infty}\ln{dr_{2}}), \end{array} $$

which implies the expected value m does not exist, so the absolute semi-deviation value E[(η − m)⁺] does not exist either. The proof of theorem is complete. □

1.2 Proof of Theorem 2

Proof

If we denote η = 1/ξ, then the credibility distribution of η reads that

$$ \begin{array}{@{}rcl@{}} \qquad &\text{Cr}\{\eta \leq r\}=\left\{ \begin{array}{llll} &0, && 0<r<\frac{1}{r_{4}}\\ &\frac{r_{4}-\frac{1}{r}}{2(r_{4}-r_{3})},&& \frac{1}{r_{4}}\leq r<\frac{1}{r_{3}}\\ &\frac{1}{2}, &&\frac{1}{r_{3}}\leq r<\frac{1}{r_{2}}\\ &\frac{2r_{2}-r_{1}-\frac{1}{r}}{2(r_{2}-r_{1})}, &&\frac{1}{r_{2}}\leq r<\frac{1}{r_{1}}\\ &1, && r\geq\frac{1}{r_{1}}. \end{array} \right. \end{array} $$

Since the expected value $m=\frac {1}{2(r_{4}-r_{3})}\ln {\frac {r_{4}}{r_{3}}}+\frac {1}{2(r_{2}-r_{1})}\ln {\frac {r_{2}}{r_{1}}}$ has computed in Li and Liu (2016), if $m\in \left [\frac {1}{r_{4}},\frac {1}{r_{3}}\right )$, the absolute semi-deviation value E[(η − m)⁺] is as follows:

$$ \begin{array}{@{}rcl@{}} \mathrm{E}[(\eta-m)^{+}]&=&{\int}_{(m,+\infty)}(r-m)\mathrm{d}\text{Cr}\{\eta \leq r\}\\ &=&\frac{1}{2(r_{4}-r_{3})}\ln \frac{1}{r_{3}m}+\frac{1}{2(r_{2}-r_{1})}\ln \frac{r_{2}}{r_{1}}\\ &&+\frac{m(2r_{3}-r_{4})-1}{2(r_{4}-r_{3})}. \end{array} $$

Similarly, if $m\in \left [\frac {1}{r_{3}},\frac {1}{r_{2}}\right )$, the absolute semi-deviation value E[(η − m)⁺] is computed by

$$ \mathrm{E}[(\eta-m)^{+}]=\frac{1}{2(r_{2}-r_{1})}\ln \frac{r_{2}}{r_{1}}-\frac{m}{2}, $$

if $m\in \left [\frac {1}{r_{2}},\frac {1}{r_{1}}\right ]$, the absolute semi-deviation value E[(η − m)⁺] is computed by

$$ \mathrm{E}[(\eta-m)^{+}]=\frac{1}{2(r_{2}-r_{1})}\ln \frac{1}{r_{1}m}+\frac{mr_{1}-1}{2(r_{2}-r_{1})}. $$

The rest of theorem is similar to prove. The proof of theorem is complete. □

1.3 Proof of Theorem 3

Proof

If demand ξ is an Erlang fuzzy variable Er(λ, r), then the credibility distribution of η reads that

$$ \begin{array}{@{}rcl@{}} \text{Cr}\{\eta \leq x\}=\left\{ \begin{array}{llll} &\frac{1}{2}\left( \frac{1}{\lambda rx}\right)^{r}e^{r-\frac{1}{\lambda x}},&& \frac{1}{r_{2}}\leq x< \frac{1}{\lambda r}\\ &1-\frac{1}{2}\left( \frac{1}{\lambda rx}\right)^{r}e^{r-\frac{1}{\lambda x}},&& \frac{1}{\lambda r}\leq x\leq \frac{1}{r_{1}}. \end{array} \right. \end{array} $$

The expected value m of η is computed by

$$ \begin{array}{@{}rcl@{}} m&=&{\int}_{0}^{+\infty}\text{Cr}\{\eta \geq x\}\text{dx}\\ &=&{\int}_{\frac{1}{r_{2}}}^{\frac{1}{\lambda r}}\left[1-\frac{1}{2}\left( \frac{1}{\lambda rx}\right)^{r}e^{r-\frac{1}{\lambda x}}\right]\text{dx}+{\int}_{\frac{1}{\lambda r}}^{\frac{1}{r_{1}}}\frac{1}{2}\left( \frac{1}{\lambda rx}\right)^{r}e^{r-\frac{1}{\lambda x}}\text{dx},\\ &=&\frac{e^{r}}{2\lambda}\left( \frac{1}{r}\right)^{r}\left( {\int}_{\frac{r_{2}}{\lambda}}^{r}t^{r-2}e^{-t}\text{dt}-{\int}_{r}^{\frac{r_{1}}{\lambda}}t^{r-2}e^{-t}\text{dt}\right)+\frac{1}{\lambda r}-\frac{1}{r_{2}}, \end{array} $$

we find that the above formulation has different computing forms according to the values of r, if r = 1, then the expected value of η is computed by the formulation (14); if r = n, (n > 1), then the expected value of η is computed by the formulation (16).

Therefore, if $m\in \left [\frac {1}{r_{2}},\frac {1}{\lambda r}\right )$, then the absolute semi-deviation value E[(η − m)⁺] is as follows:

$$ \begin{array}{@{}rcl@{}} \mathrm{E}[(\eta-m)^{+}]&=&{\int}_{(m,+\infty)}(x-m)\mathrm{d}\text{Cr}\{\eta \leq x\}\\ &=&{\int}_{(m,\frac{1}{\lambda r})}(x-m)\mathrm{d}\frac{1}{2}\left( \frac{1}{\lambda rx}\right)^{r}e^{r-\frac{1}{\lambda x}}\\ &&+{\int}_{(\frac{1}{\lambda r},\frac{1}{r_{1}})}(x-m)\mathrm{d}\left[1-\frac{1}{2}\left( \frac{1}{\lambda rx}\right)^{r}e^{r-\frac{1}{\lambda x}}\right]. \end{array} $$

(19)

By equation (19), we have the following result

$$ \begin{array}{@{}rcl@{}} \mathrm{E}[(\eta-m)^{+}]&=&\left( \frac{1}{r_{1}}-m\right)\left[1-\frac{1}{2}\left( \frac{r_{1}}{\lambda r}\right)^{r}e^{r-\frac{r_{1}}{\lambda}}\right]-\left( \frac{1}{r_{1}}-\frac{1}{\lambda r}\right)\\ &&+\frac{1}{2\lambda}\left( \frac{1}{r}\right)^{r}e^{r}\left( {\int}_{(\frac{1}{\lambda m},r)}t^{r-2}e^{-t}\text{dt}-{\int}_{(r,\frac{r_{1}}{\lambda})}t^{r-2}e^{-t}\text{dt}\right). \end{array} $$

Next, according to the different values of r, we present two cases to calculate the value of E[(η − m)⁺] when $m\in \left [\frac {1}{r_{2}},\frac {1}{\lambda r}\right ]$.

• (i) If r = 1, then the absolute semi-deviation value E[(η − m)⁺] is computed by the formulation (13).

• (ii) If r = n, (n > 1), then the absolute semi-deviation value E[(η − m)⁺] is computed by the formulation (15).

Similarly, if $m\in [\frac {1}{\lambda r},\frac {1}{r_{1}}]$, the absolute semi-deviation value E[(η − m)⁺] is as follows.

• (i) If r = 1, then the absolute semi-deviation value E[(η − m)⁺] is computed by the formulation (17).

• (ii) If r = n, (n > 1), then the absolute semi-deviation value E[(η − m)⁺] is computed by the formulation (18).

The rest of theorem is similar to prove. The proof of theorem is complete. □