Single versus dynamic lead-time quotations in make-to-order systems with delay-averse customers

Abstract

We develop a model for lead-time quotation in a Markovian Make-To-Order production or service system with strategic customers who exhibit risk aversion. Based on a CARA utility function, customers make individual decisions to join the system or balk after observing the state of the queue. The decisions of arriving customers result in a symmetric join/balk game. The provider announces a lead-time quotation for each state and a respective balking threshold. There is a fixed entrance fee and compensation rate for the part of a customer’ delay exceeding the quoted lead-time. We also consider the problem from the point of view of a social optimizer who maximizes the total system utility. We analyze the provider’s and social optimizer’s optimization problems and consider two classes of lead-time quotation policies, dynamic and single. We identify the optimal entrance thresholds in each case. Through computational experiments we quantify the effect of risk aversion on the profits and the flexibility the compensation policy offers. We show that the detrimental effects of risk aversion can be addressed more efficiently for the provider’s problem compared to the social optimizer’s one. Furthermore, the profit loss when setting a single lead-time quote is generally small compared to the optimal dynamic quotation policy.

Appendices

Appendix A : Proofs for the profit maximization problem

Proof of Lemma 1

(i) The expected customer utility is equal to :

$$\begin{aligned} B_n(d)= & {} \displaystyle \int _0^d \frac{1-e^{-r(R-p-cx)}}{r} f_{n,\mu }(x)\,dx +\int _d^\infty \frac{1-e^{-r(R-p-ld-(c-l)x)}}{r} f_{n,\mu }(x)\,dx. \\= & {} \frac{1}{r} \left[ 1 - e^{-r(R-p)} \left( K_1(n,d)+K_2(n,d) \right) \right] , \end{aligned}$$

where

$$\begin{aligned} K_1(n,d) = \int _0^d \frac{\mu ^{n+1} x^n e^{-(\mu -rc)x}}{n!}dx \end{aligned}$$

and

$$\begin{aligned} K_2(n,d) = e^{rld}\int _d^{\infty } \frac{\mu ^{n+1} x^n e^{-(\mu -r(c-l))x}}{n!}dx. \end{aligned}$$

The first integral is finite for any $d\ge 0$. When $\mu =rc$ it immediate that $K_1(n,d)= \frac{(\mu d)^{n+1}}{(n+1)!}$, whereas for $\mu \ne rc$ it can be shown by induction on n that

$$\begin{aligned} K_1(n,d)= \left( \frac{\mu }{\mu -rc} \right) ^{n+1} \left[ 1 - e^{-(\mu -rc)d} \sum _{k=0}^n \frac{((\mu -rc)d)^k}{k!} \right] . \end{aligned}$$

The second integral is infinite if $\mu \le r(c-l)$. When $\mu > r(c-l)$, it also follows by induction or standard properties that relate the Gamma and Poisson distributions that

$$\begin{aligned} K_2(n,d)= \left( \frac{\mu }{\mu -r(c-l)} \right) ^{n+1} e^{-(\mu -rc)d} \sum _{k=0}^n \frac{((\mu -r(c-l))d)^k}{k!}. \end{aligned}$$

(ii) From known properties of the Gamma distribution it follows $X_n \le _{st} X_{n+1}$, where the inequality denotes stochastic ordering. In addition U(X) is decreasing in X since $l \le c$. Thus, $E\left( U(X_{n+1})\right) \le E\left( U(X_{n})\right)$ and $B_{n+1}(d) \le B_n(d)$, for all d. The monotonicity of $B_n(d)$ in terms of d for a fixed n is immediate. $\square$

Proof of Proposition 1

Let a fixed $D=(d_0,d_1, \ldots )$. From Lemma 1 we know that $B_n(d)$ is decreasing in d and n which implies that: $\lim _{d \rightarrow \infty } B_n(d) \le B_n(d_n) \le B_n(0).$ Therefore, $\underline{n} \le n_0(d) \le \overline{n}$, where

$$\begin{aligned} \displaystyle \underline{n}=\min \{ n:\lim _{d \rightarrow \infty } B_n(d) <0 \} \end{aligned}$$

and

$$\begin{aligned} \overline{n}=\min \{ n:B_n(0) <0 \}. \end{aligned}$$

To derive the analytic expressions for the two bounds, we first consider $\lim _{d \rightarrow \infty } B_n(d)$. From Lemma 1 we have the following cases: If $\mu \le r(c-l)$, then $\lim _{d \rightarrow \infty } B_n(d) = - \infty$, for all $n\ge 0$. Similarly, when $r(c-l) <\mu \le rc$, it is true that $\lim _{d \rightarrow \infty } K_1(n,d) =\infty$ and since $K_2(n,d)\ge 0$, it follows that $\lim _{d \rightarrow \infty } B_n(d) = - \infty$, for all $n\ge 0$ in this range as well. Therefore, if $\mu \le rc$, then $\underline{n}=0$. Finally, when $\mu >rc$, we obtain from Lemma 1 that $\lim _{d \rightarrow \infty } K_1(n,d) = \left( \frac{\mu }{\mu -rc} \right) ^{n+1}$ and $\lim _{d \rightarrow \infty } K_2(n,d) = 0,$ thus

$$\begin{aligned} \lim _{d \rightarrow \infty } B_n(d)=\frac{1-e^{-r(R-p)}\left( \frac{\mu }{\mu -rc}\right) ^{n+1}}{r} \end{aligned}$$

The above expression is decreasing in n and becomes negative for

$$\begin{aligned} n+1 > {\frac{r(R-p)}{\ln \left( \frac{\mu }{\mu -rc}\right) }}. \end{aligned}$$

Therefore, in this case

$$\begin{aligned} \underline{n} = \big \lfloor {\frac{r(R-p)}{\ln \left( \frac{\mu }{\mu -rc}\right) }}\big \rfloor . \end{aligned}$$

We next consider $B_n(0)$. If $\mu \le r(c-l)$, then $B_n(0) = \infty$ for all $n\ge 0$, thus $\overline{n}=0$. If $\mu \le r(c-l)$, then from Lemma 1, $K_1(n,0) = 0$ and $K_2(n,0) = \left( \frac{\mu }{\mu -r(c-l)} \right) ^{n+1}$, thus

$$\begin{aligned} B_n(0)=\frac{1-e^{-r(R-p)}\left( \frac{\mu }{\mu -r(c-l)}\right) ^{n+1}}{r}, \end{aligned}$$

from which it follows that

$$\begin{aligned} \overline{n} = \big \lfloor {\frac{r(R-p)}{\ln \left( \frac{\mu }{\mu -r(c-l)}\right) }}\big \rfloor . \end{aligned}$$

$\square$

Proof of Lemma 2

(i) For a fixed d, since the function $(X-d)^{+} \text { is increasing in}$X$\text {and}$ $X_n \le _{st} X_{n+1},$ it follows that $E(X_n-d)^{+} \le E(X_{n+1}-d)^{+}.$ Therefore $L_n(d) \le L_{n+1}(d),$ for all d. The monotonicity of $L_n(d)$ in terms of d for a fixed n is immediate.

(ii) We know that $G_n(d)=p-lL_n(d)$. Therefore the proof follows from (i.).

(iii) Since, from Lemma 1, $B_n(d)$ is decreasing in d, we have that $B_n(d) \ge 0 \text { if and only if } d \le \tilde{d}^{P}_{n}.$ For any $d \le \tilde{d}^{P}_{n+1}$ it follows that $B_{n+1}(d) \ge 0$ and since $B_n(d)$ is decreasing in n from Lemma 1, it follows $B_n(d) \ge 0,$ thus $d \le \tilde{d}^{P}_{n}.$ We thus see that $\tilde{d}^{P}_{n+1} \le \tilde{d}^{P}_{n}.$

(iv) For $n < \underline{n}, \, \lim _{d \rightarrow \infty } B_n(d) > 0,$ which implies that $\sup \lbrace d \ge 0: B_n(d) \ge 0\rbrace =\infty .$ Therefore, $\tilde{d}^{P}_{n}=\infty .$ For $n=\underline{n}, \, \lim _{d \rightarrow \infty } B_n(d) \ge 0.$

(v) From (i), $L_n(d)$ increasing in n and decreasing in d. We have also proved that $\tilde{d}^{P}_{n+1} \le \tilde{d}^{P}_{n}.$ Therefore, $L_n(\tilde{d}^{P}_{n}) \le L_{n+1}(\tilde{d}^{P}_{n}) \le L_{n+1}(\tilde{d}^{P}_{n+1}).$ Thus, $L_n(\tilde{d}^{P}_{n})$ is increasing in n and $G_n(\tilde{d}^{P}_{n})$ is decreasing in n. $\square$

Proof of Proposition 2

Let $n_0$ be fixed and $D_{n_0}=(d_0, d_1, d_2,\ldots ,d_{n_0-1}, d_{n_0})$ be a feasible solution of (5), with acceptance threshold $n_0,$ i.e., $B_n(d_n)\ge 0$ for $n=0,1,\ldots ,n_0-1$ and $B_{n_0}(d_{n_0})<0$ with $d_n\ge 0$ for $n=0, 1,\ldots , n_0$.

Since $B_n(d_n)$ is decreasing from Lemma 1, it follows $d_n \le \tilde{d}^{P}_{n}.$ Assume that $d_k < \tilde{d}^{P}_{k}$ for some $k \le n_0-1.$ We will show that $D_{n_0}$ cannot be an optimal solution to (5). To see this, define another solution $D^{'}_{n_0}=(d^{'}_0, d^{'}_1, d^{'}_2,\ldots , d^{'}_k,\ldots ,d^{'}_{n_0-1}, d^{'}_{n_0})$ with $d^{'}_k = \tilde{d}^{P}_{k}$ and $d^{'}_n=d_n$ for all $n \ne k$.

Then $B_n(d^{'}_n)\ge 0$ for $n=0,1,...,n_0-1$ and $B_{n_0}(d^{'}_{n_0})<0$ with $d^{'}_n \ge 0$ for $n=0, 1,.., n_0$. Thus, $D^{'}_{n_0}$ is feasible. Also $F_k(d^{'}_k)>F_k(d_k)$, therefore $D_{n_0}$ is not optimal. $\square$

Proof of Theorem 1

Letting $G_n=G_n(\tilde{d}^{P}_{n})$, $H(n_0)$ can be written as:

$$\begin{aligned} H(n_0)=\lambda \frac{\displaystyle \sum \nolimits _{n=0}^{n_0-1} \rho ^{n} G_n}{\displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n}}. \end{aligned}$$

For any $\underline{n}\le n_0\le \overline{n}-1$ we have,

$$\begin{aligned} H(n_0+1)-H(n_0)&=\lambda \frac{\displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n} \displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n} G_n -\displaystyle \sum \nolimits _{n=0}^{n_0+1} \rho ^{n} \displaystyle \sum \nolimits _{n=0}^{n_0-1} \rho ^{n} G_n }{\displaystyle \sum \nolimits _{n=0}^{n_0+1} \rho ^{n}\displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n}} \nonumber \\&=\lambda \frac{\rho ^{n_0} G_{n_0} \displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n} -\rho ^{n_0+1}\displaystyle \sum \nolimits _{n=0}^{n_0-1} \rho ^{n} G_n}{\displaystyle \sum \nolimits _{n=0}^{n_0+1} \rho ^{n}\displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n}} \nonumber \\&=\lambda \frac{\rho ^{n_0}}{\displaystyle \sum \nolimits _{n=0}^{n_0+1} \rho ^{n}\displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n}} \left( G_{n_0} \displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n}-\rho \displaystyle \sum \nolimits _{n=0}^{n_0-1} \rho ^{n} G_n \right) \nonumber \\&=\lambda \frac{\rho ^{n_0}}{\displaystyle \sum \nolimits _{n=0}^{n_0+1} \rho ^{n}\displaystyle \sum \nolimits _{n=0}^{n_0} \rho ^{n}}A(n_0). \end{aligned}$$

(A1)

We next show that $A(n_0)$ is decreasing in $n_0$. For any $\underline{n}\le n_0\le \overline{n}-2$,

$$\begin{aligned} A(n_0+1)-A(n_0)&=G_{n_0+1}\displaystyle \sum _{n=0}^{n_0+1} \rho ^{n}-\rho \displaystyle \sum _{n=0}^{n_0} \rho ^{n} G_n-G_{n_0}\displaystyle \sum _{n=0}^{n_0} \rho ^{n}+\rho \displaystyle \sum _{n=0}^{n_0-1} \rho ^{n} G_n \nonumber \\&= \displaystyle \sum _{n=0}^{n_0} \rho ^{n} \left( G_{n_0+1}-G_{n_0} \right) +\rho ^{n_0+1} \left( G_{n_0+1}-G_{n_0} \right) \nonumber \\&= \left( G_{n_0+1}-G_{n_0} \right) \displaystyle \sum _{n=0}^{n_0+1} \rho ^{n} \nonumber \\&= \left( G_{n_0+1}-G_{n_0} \right) \frac{1-\rho ^{n_0+2}}{1-\rho } \le 0, \end{aligned}$$

(A2)

since $G_n$ decreasing in n from Lemma 2.

From (A1) and (A2) it follows that $H(n_0+1)-H(n_0) \ge 0$ for $n<\tilde{n}_P$ and $H(n_0+1)-H(n_0) < 0$ for $n \ge \tilde{n}_P$. Thus, $\tilde{n}_P$ is an optimal value that maximizes $G(n_0).$ $\square$

Appendix B : Proofs for the social benefit maximization problem

We first prove some intermediate properties on the distribution of the sojourn time in a finite capacity queue, that are necessary for the analysis in Sect. 6.

Let N, X denote the number of customers that an arriving customer finds upon arrival and the sojourn time of this customer respectively, in a $M/M/1/n_0$ queue with service rate $\mu$ in steady-state.

From the PASTA property the distribution of N is identical to the steady-state distribution of the queue length, i.e.,

$$\begin{aligned} q(n;n_0)=\frac{\rho ^{n}(1-\rho )}{1-\rho ^{n_0+1}}. \end{aligned}$$

The conditional distribution of X given $N=n$ is $Gamma(n+1,\mu )$, for $n=0,\ldots , n_0-1$. Let $f_{n,\mu }(x)$ and $\overline{F}_{n,\mu }(x)$ be the corresponding pdf and tail probability respectively.

The marginal distribution of X corresponds to the sojourn time of a random customer. We obtain:

$$\begin{aligned} \displaystyle f_{\mu }(x)&=\sum _{n=0}^{n_0-1} q(n;n_0) f_{n,\mu }(x)\\&=\lambda \frac{(1-\rho )e^{-(\mu -\lambda )x}}{\rho (1-\rho ^{n_0})} \sum _{n=0}^{n_0-1} \frac{(\lambda x)^{n} e^{-\lambda x}}{n!}\\&=(\mu -\lambda ) e^{-(\mu -\lambda )x} \frac{Pois(n_0-1;\lambda x)}{1-\rho ^{n_0}} \end{aligned}$$

and

$$\begin{aligned} \overline{F}_{\mu }(x)=\int \limits _{x}^\infty (\mu -\lambda ) e^{-(\mu -\lambda )x} \frac{Pois(n_0-1;\lambda x)}{1-\rho ^{n_0}}\mathrm {d}x, \end{aligned}$$

where $Pois(n_0-1;\lambda x)$ the cdf of a Poisson distribution with rate $\lambda x$.

The corresponding hazard rates of the conditional and the marginal distribution of X are:

$$\begin{aligned} h_{\mu }(x)= & {} \frac{f_{\mu }(x)}{\overline{F}_{\mu }(x)}, \\ h_{n,\mu }(x)= & {} \frac{f_{n,\mu }(x)}{\overline{F}_{n,\mu }(x)}. \end{aligned}$$

1.1 B.1 Dynamic lead-time quotation

Lemma 5

For $n \le n_0-1:$

• (i) $(G_n+B_n)(\tilde{d}^{S}_{n})$ is decreasing in n.

• (ii) $B_n(\tilde{d}^{P}_{n})\le B_n(\tilde{d}^{S}_{n})$.

Proof

(i) From Lemma 1, $(G_{n+1}+B_{n+1})(d) \le (G_n+B_n)(d)$ for all $d \ge 0$. From the definition of $\tilde{d}^{S}_{n}$, it implies that $(G_n+B_n)(d) \le (G_n+B_n)(\tilde{d}^{S}_{n})$ for all $d \le \tilde{d}^{P}_{n}$.

Let $d=\tilde{d}^{S}_{n+1}$. From the above inequalities we have: $(G_{n+1}+B_{n+1})(\tilde{d}^{S}_{n+1}) \le (G_n+B_n)(\tilde{d}^{S}_{n})$.

(ii) It follows from lemma 1 and the fact that $\tilde{d}^{P}_{n} \ge \tilde{d}^{S}_{n}$. $\square$

Proof of Theorem 3

The proof is along similar lines as in Theorem 1, using that $(G_n+B_n)(\tilde{d}^{S}_{n})$ decreasing in n from Lemma 5. $\square$

Lemma 6

For $X\sim Gamma(n,\mu )$ and $Y\sim Gamma(n,v)$ with $n\in \mathbb {N}$ and $v<\mu$, it is true that:

• (i) The hazard rate $h_{n,\mu }(d)$ is increasing in $\mu$.

• (ii) $\frac{\overline{F}_{n,\mu }(d)}{\overline{F}_{n,v}(d)}$ is decreasing in d.

Proof

(i.) Using the relationship between Gamma and Poisson distribution it follows that:

$$\begin{aligned} \overline{F}_{n,\mu }(d)=P(X_1+\ldots +X_n>d)=Pois(n_0-1;\mu d)=\sum _{k=0}^{n-1} e^{-\mu d} \frac{(\mu d)^k}{k!}. \end{aligned}$$

As a result,

$$\begin{aligned} \frac{\overline{F}_{n,\mu }(d)}{f_{n,\mu }(d)}=\frac{e^{-\mu d} \sum _{k=0}^{n-1} \frac{(\mu d)^k}{k!} }{e^{-\mu d}\frac{\mu ^{n}d^{n-1}}{(n-1)!}}=\frac{(n-1)!}{d^{n-1}} \sum _{k=0}^{n-1} \mu ^{k-n}\frac{d^k}{k!}. \end{aligned}$$

Since $k-n<0$, the last term is decreasing in $\mu$, thus $h_{n,\mu }(d)$ is increasing in $\mu$.

(ii) The first derivative of $\frac{\overline{F}_{n,\mu }(d)}{\overline{F}_{n,v}(d)}$ with respect to d is:

$$\begin{aligned} \left( \frac{\overline{F}_{n,\mu }(d)}{\overline{F}_{n,v}(d)}\right) ^{'}=\frac{-f_{n,\mu }(d)\overline{F}_{n,v}(d)+f_{n,v}(d)\overline{F}_{n,\mu }(d)}{(\overline{F}_{n,v}(d))^2}. \end{aligned}$$

From (i.) it follows that $h_{n,v}(d)<h_{n,\mu }(d)$, therefore the numerator of the above fraction is negative and the proof is complete. $\square$

Proof of Proposition 3

We fix $n \in \left\{ 0,1, \ldots ,\overline{n}-1 \right\}$, and consider the problem of maximizing $G_n(d_n)+B_n(d_n)$ in $d_n\in [0,\tilde{d}^{P}_{n}]$. The first derivative of $G_n(d_n)+B_n(d_n)$ is:

$$\begin{aligned} (G_n(d_n)+B_n(d_n))^{'}&=l\int \limits _{d_n}^\infty \left( 1-e^{-r(R-p-(c-l)x-ld_n)} \right) f_{n,\mu }(x)\mathrm {d}x\\&=l\left( \int \limits _{d_n}^\infty f_{n,\mu }(x)\mathrm {d}x -e^{-r(R-p)} e^{rld_n} \int \limits _{d_n}^\infty e^{r(c-l)x} f_{n,\mu }(x)\mathrm {d}x \right) . \end{aligned}$$

Since,

$$\begin{aligned} \int \limits _{d_n}^\infty e^{r(c-l)x} f_{n,\mu }(x)\mathrm {d}x&=\left( \frac{\mu }{v}\right) ^{n+1} \int \limits _{d_n}^\infty \frac{(v)^{n+1}x^n e^{-(v)x}}{n!} \mathrm {d}x\\&=\left( \frac{\mu }{v}\right) ^{n+1} \int \limits _{d_n}^\infty f_{n,v}(x)\mathrm {d}x\\&=\left( \frac{\mu }{v}\right) ^{n+1} \overline{F}_{n,v}(d_n), \end{aligned}$$

it follows that:

$$\begin{aligned} (G_n(d_n)+B_n(d_n))^{'}&=l\left( \overline{F}_{n,\mu }(d_n)-e^{-r(R-p)} e^{rld_n} \left( \frac{\mu }{v}\right) ^{n+1} \overline{F}_{n,v}(d_n) \right) \nonumber \\&=le^{rld_n}\overline{F}_{n,v}(d_n) \left( \frac{\overline{F}_{n,\mu }(d_n)}{\overline{F}_{n,v}(d_n)}e^{-rld}-\left( \frac{\mu }{v}\right) ^{n+1} e^{-r(R-p)} \right) \nonumber \\&=le^{rld_n}\overline{F}_{n,v}(d_n) a(d_n), \end{aligned}$$

(B3)

where $a(d_n)$ is defined in (15). We will show that the derivative in (15) is either minimized at a unique point in $[0,\tilde{d}^{P}_{n}]$ or it is always positive. From Lemma 6 it follows that $a(d_n)$ is decreasing in $d_n$.

For $d_n=0$, we have $\overline{F}_{n,\mu }(0)=\overline{F}_{n,v}(0)=1$. Thus:

$$\begin{aligned} a(0)=1-e^{-r(R-p)} \left( \frac{\mu }{v}\right) ^{n+1} \ge 0. \end{aligned}$$

For $d_n \rightarrow \infty$, by applying de L’ Hospital’s rule,

$$\begin{aligned} \lim _{d_n \rightarrow \infty } \frac{\overline{F}_{n,\mu }(d_n)e^{-rld_n}}{\overline{F}_{n,v}(d_n)}&=\lim _{d_n \rightarrow \infty } \frac{f_{n,\mu }(d_n)e^{-rld_n}+rle^{-rld_n}\overline{F}_{n,\mu }(d_n)}{f_{n,v}(d_n)}\\&=\lim _{d_n \rightarrow \infty } \frac{f_{n,\mu }(d_n)e^{-rld_n}}{f_{n,v}(d_n)} \left( 1+\frac{1}{h_{n,\mu }(d_n)} \right) \end{aligned}$$

and also,

$$\begin{aligned} \lim _{d_n \rightarrow \infty } \frac{f_{n,\mu }(d_n)e^{-rld_n}}{f_{n,v}(d_n)}= \left( \frac{\mu }{v}\right) ^{n+1} e^{-rcd_n}=0 \end{aligned}$$

and

$$\begin{aligned} \lim _{d_n \rightarrow \infty } \frac{1}{h_{n,\mu }(d_n)}>0, \end{aligned}$$

since the hazard rate $h_{n,\mu }(d_n)$ of Gamma distribution is increasing in $d_n$.

It follows that:

$$\begin{aligned} \lim _{d_n \rightarrow \infty } a(d_n)=-\left( \frac{\mu }{v}\right) ^{n+1} e^{-r(R-p)}<0. \end{aligned}$$

Since, $a(0) \ge 0$ and $\lim _{d_n \rightarrow \infty } a(d_n)<0$ and $a(d_n)$ is decreasing in $d_n$, there is a unique $d^0<\infty$ such that $a(d^0)=0$. Furthermore, $a(d_n) \ge 0$ for all $d_n \le d^0$ and $a(d_n) < 0$ otherwise.

Returning to (B3), we consider the following cases regarding $\tilde{d}^{S}_{n}$:

• (i) If $\tilde{d}^{P}_{n}=\infty$, then from the previous discussion on $a(d_n)$, it follows that $(G_n(d_n)+B_n(d_n))^{'} \ge 0$ for all $d_n \le d^0$ and negative for $d_n > d^0$. Therefore, the maximum of $G_n(d_n)+B_n(d_n)$ in $d_n\in [0,\infty ]$ is $\tilde{d}^{S}_{n}=d^0.$

• (ii) If $\tilde{d}^{P}_{n}<\infty$, then the maximum point of $G_n(d_n)+B_n(d_n)$ depends on the sign of $a(\tilde{d}^{P}_{n})$. If $a(\tilde{d}^{P}_{n})<0$ then $\tilde{d}^{S}_{n}=d^0\in [0,\tilde{d}^{P}_{n}]$, otherwise $\tilde{d}^{S}_{n}=\tilde{d}^{P}_{n}$.

$\square$

1.2 B.2 Single lead-time quotation

Lemma 7

Let X, Y be the sojourn times of a customer in two finite capacity $M/M/1/n_0$ queues in steady state, with service rates $\mu$ and v, respectively, and $v<\mu$. Then:

• (i) The hazard rate $h_{\mu }(d)$ is increasing in $\mu$.

• (ii) $\frac{\overline{F}_{\mu }(d)}{\overline{F}_{v}(d)}$ is decreasing in d.

Proof

(i) It is immediate that $h_{\mu }(d)$ is increasing in $\mu$ since:

$$\begin{aligned} h_{\mu }(d)=\frac{Pois(n_0-1;\mu d)}{\int \limits _{d}^\infty e^{(\mu -\lambda )(d-x)} \frac{Pois(n_0-1;\mu d)}{1-\rho ^{n_0}}\mathrm {d}x}. \end{aligned}$$

(ii) The proof follows by differentiation of $\frac{\overline{F}_{\mu }(d)}{\overline{F}_{v}(d)}$ as in the proof of Lemma 6 (ii). $\square$

Proof of Proposition 5

Using Lemma 3, we consider three cases regarding $n_0$. (i) We first fix $n_0 \in \big \lbrace \underline{n}+1, \ldots ,\overline{n}-1 \big \rbrace$, and consider the problem of maximizing $S_c(d)$, for $d\in (\tilde{d}^{P}_{n_0},\tilde{d}^{P}_{n_0-1}]$ where:

$$\begin{aligned} S_c(d)=\lambda \sum _{n=0}^{n_0-1} q(n;n_0)(G_n(d)+B_n(d)). \end{aligned}$$

In the proof of Proposition 3, it was shown that:

$$\begin{aligned} \left( G_n(d)+B_n(d)\right) ^{'}=l\left( \overline{F}_{n,\mu }(d)-e^{-r(R-p)} e^{rld} \left( \frac{\mu }{v}\right) ^{n+1} \overline{F}_{n,v}(d) \right) . \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\partial {S}_{c(d)}}{\partial {d}}&=\lambda l \left( \sum _{n=0}^{n_0-1} q(n;n_0)\overline{F}_{n,\mu }(d)- e^{-r(R-p)} e^{rld} \sum _{n=0}^{n_0-1} q(n;n_0) \left( \frac{\mu }{v}\right) ^{n+1} \overline{F}_{n,v}(d)\right) . \end{aligned}$$

We have that

$$\begin{aligned} \sum _{n=0}^{n_0-1} q(n;n_0)\overline{F}_{n,\mu }(d)=\overline{F}_{\mu }(d), \end{aligned}$$

and

$$\begin{aligned}&\sum _{n=0}^{n_0-1} q(n;n_0) \left( \frac{\mu }{v}\right) ^{n+1} \overline{F}_{n,v}(d) \\&\qquad =\sum _{n=0}^{n_0-1} \frac{(1-\rho ) \rho ^n}{1-\rho ^{n_0+1}} \left( \frac{\mu }{v}\right) ^{n+1} \overline{F}_{n,v}(d) \\&\qquad =\left( \frac{\mu -\lambda }{1-\rho ^{n_0+1}}\right) \left( \frac{1-\left( \frac{\lambda }{v}\right) ^{n_0+1}}{v-\lambda }\right) \sum _{n=0}^{n_0-1} \frac{\left( \frac{\lambda }{v}\right) ^{n} \left( 1-\frac{\lambda }{v}\right) }{1-\left( \frac{\lambda }{v}\right) ^{n_0+1}} \overline{F}_{n,v}(d)\\&\qquad =\beta \overline{F}_{v}(d), \end{aligned}$$

with $\beta >0$.

Thus, the first derivative of $S_c(d)$ with respect to d is equal to:

$$\begin{aligned} \frac{\partial S_c(d)}{\partial d}&=\lambda l \overline{F}_{v}(d) e^{rld} \left( \frac{\overline{F}_{\mu }(d)}{\overline{F}_{v}(d)} e^{-rld}-\beta e^{-r(R-p)} \right) \nonumber \\&=\lambda l \overline{F}_{v}(d) e^{rld} a_c(d), \end{aligned}$$

(B4)

where $a_c(d)$ is defined in (16). From Lemma 7 it follows that $a_c(d)$ is decreasing in d.

Therefore, either $a_c(d) \ge 0$ for all $d\in (\tilde{d}^{P}_{n_0},\tilde{d}^{P}_{n_0-1}]$, in which case ${d}^{S_c}_{n}=\tilde{d}^{P}_{n_0-1}$, or $a_c(d) \le 0$ for all $d\in (\tilde{d}^{P}_{n_0},\tilde{d}^{P}_{n_0-1}]$, in which case there exist $\epsilon$-optimal lead-time quotes which approach the sup in (13) arbitrary close, since $a_c(d)$ is continuous and decreasing and the lower bound cannot be attained by any d, or the maximizing value ${d}^{S_c}_{n_0}\in (\tilde{d}^{P}_{n_0},\tilde{d}^{P}_{n_0-1}]$ is the unique solution of $a_c(d)=0$.

(ii) When $n_0=\underline{n}$, we maximize $S_c(d)$ for $d\in (\tilde{d}^{P}_{n_0},\infty ]$.

For $d \rightarrow \infty$, by applying de L’Hospital’s rule,

$$\begin{aligned} \lim _{d \rightarrow \infty } \frac{\overline{F}_{\mu }(d)}{\overline{F}_{v}(d)}e^{-rld}&=\lim _{d \rightarrow \infty } \frac{f_{\mu }(d)e^{-rld}}{f_{v}(d)}+rl\lim _{d \rightarrow \infty } \frac{e^{-rld}\overline{F}_{\mu }(d)}{f_{v}(d)}\\&=\lim _{d \rightarrow \infty } \frac{f_{\mu }(d)e^{-rld}}{f_{v}(d)} \left( 1+\frac{1}{h_{\mu }(d)} \right) \end{aligned}$$

and also,

$$\begin{aligned} \lim _{d \rightarrow \infty } \frac{f_{\mu }(d)e^{-rld}}{f_{v}(d)}= \beta e^{-rcd}=0 \end{aligned}$$

and

$$\begin{aligned} \lim _{d \rightarrow \infty } \frac{1}{h_{\mu }(d)}>0, \end{aligned}$$

since the hazard rate $h_{\mu }(d)$ is increasing in d.

It follows that:

$$\begin{aligned} \lim _{d \rightarrow \infty } a_c(d)=-\beta e^{-r(R-p)}<0. \end{aligned}$$

Since $\lim _{d \rightarrow \infty } a_c(d)<0$ and $a_c(d)$ is decreasing in d, returning to (B4), the maximum point of $G_n(d)+B_n(d)$ depends on the sign of $a_c(\tilde{d}^{P}_{\underline{n}})$. If $a_c(\tilde{d}^{P}_{\underline{n}}) >0$, then ${d}^{S_c}_{\underline{n}}$ is the unique solution of $a_c(d)=0$. Otherwise, there exist $\epsilon$-optimal lead-time quotes which approach the sup in (13) arbitrary close.

(iii) When $n_0=\overline{n}$, we maximize $S_c(d)$ for $d\in [0,\tilde{d}^{P}_{n_0-1}]$.

For $d=0$, we have $\overline{F}_{\mu }(0)=\overline{F}_{v}(0)=1$. Thus:

$$\begin{aligned} a_c(0)=1- \beta e^{-r(R-p)} , \end{aligned}$$

where $a_c(0)$ can be either nonpositive or nonnegative.

In this case there is the closed interval $[0, \tilde{d}^{P}_{\overline{n}}]$. The only difference with the approach of case (i.) is that when $a_c(d) \le 0$ for all $d\in [0, \tilde{d}^{P}_{\overline{n}}]$, the optimal quote can be attained and it is ${d}^{S_c}_{\overline{n}}=0$. $\square$

Proof of Lemma 4

Let $n_0$ be a balking threshold with $a_c(\tilde{d}^{P}_{n_0})\le 0$. We know that $Z_{n_0}(d)$ is decreasing in $n_0$ and that $a_c(d)$ is decreasing in d. Moreover the sign of $Z^{'}_{n_0}(d)$ is determined by the sign of $a_c(d)$. Therefore $Z^{'}_{n_0}(d)\le 0$ and $Z_{n_0}(\tilde{d}^{P}_{n_0})\ge Z_{n_0}(d)$ for any $d\in (\tilde{d}^{P}_{n_0},\tilde{d}^{P}_{n_0-1}]$.

However, $\tilde{d}^{P}_{n_0}$ is not a feasible value when the balking threshold is $n_0$ but it is feasible when the balking threshold is $n_0-1$. Additionally, $Z_{n_0-1}(\tilde{d}^{P}_{n_0})\ge Z_{n_0}(\tilde{d}^{P}_{n_0})$.

From the above, it follows that $Z_{n_0-1}(\tilde{d}^{P}_{n_0})\ge Z_{n_0}(d)$ for any $d\in [\tilde{d}^{P}_{n_0},\tilde{d}^{P}_{n_0-1}]$ thus $n_0$ cannot be an optimal balking threshold. $\square$