Compensation policy for delivery delay in online retailing

Abstract

Financial compensation is an effective marketing tool to deal with delivery delay issues in online retailing. In this paper, we develop an analytical model to study two different compensation policies prevalent in practice: uniform compensation and discriminated compensation. By endogenizing the customers’ choice behavior, we first characterize the optimal retail price and compensation level for these different compensation cases. Then, we conduct a sensitivity analysis of the optimal solutions with respect to driving factors. The results show how the retailer’s delivery quality, as well as the product characteristics, affect the optimal compensation and pricing policies, leading to interesting insights. For example, we show that the retailer optimally adopts a more generous compensation policy when the relative delay time becomes longer under the uniform compensation policy, but a lower retail price with a constrictive compensation policy can be a good choice for the retailer under discriminated compensation case. Through comparison to the benchmark case, we identify the impacts of the delay compensation policies from three aspects: the profit margin effect, total demand effect, and return rate effect. Then, the aggregate effect analysis verifies the profit implications of the compensation policies. The results show that the retailer is worse off due to deploying a compensation policy when both the cost-value rate and delay probability are sufficiently large. Furthermore, we also compare the optimal solutions, customer choice behavior, and retailer performance for these two compensation cases. The analytical and numerical insights can serve as a reference for internet retailers facing delivery delay problems.

Appendix

Proof of Lemma 1

By equating $U_a=U_r$, we can derive $\theta _0$ directly. For $\theta \le \theta _0$, $U_a\ge U_r$, the customers choose to wait the delayed product. Otherwise, for $\theta >\theta _0$, the expected utility to buy the product is

$$\begin{aligned} EU(Buy)=(1-f)(v-p)-\theta d. \end{aligned}$$

(26)

Comparing this equation to the outside option 0, we can also obtain $\theta _2$ directly. Since $1-f-\beta >0$ is assumed, i.e., $\theta _0<\theta _2$, we know that the customers choose to return the product for $\theta _0<\theta <\theta _2$, and the others choose to buy from the outside option. $\square$

Proof of Lemma 2

Taking the second-order condition with respect to p, we have

$$\begin{aligned} \frac{\partial ^2 \Pi _B}{\partial p^2}=-\frac{2(1-f)^2}{d}-\frac{2f}{d_0}<0. \end{aligned}$$

(27)

Thus, the profit in Eq. (5) is a concave function. From the first-order condition, we can derive the optimal price as

$$\begin{aligned} p_B^*= \;& {} \frac{d_0(1-f)[v+c-f(v+s)]+df(v+s)}{2d_0(1-f)^2+2df}. \end{aligned}$$

(28)

$\square$

Proof of Lemma 3

By equating $U_r =U_c$ in (7) and $EU(Buy)=0$ in (9), we can derive $\theta _1$, $\theta _2$, $\theta _3$ directly. Then, we have two cases:

• (1) If $\theta _1\le \theta _3$, we have $\theta _2\ge \theta _3$. Thus, when $\theta \le \theta _1$, $U_c\ge U_r$, the customers accept the compensation and wait for the delayed product; when $\theta _1<\theta \le \theta _2$, $U_c< U_r$, the customers return the delayed product; when $\theta >\theta _2$, the customers buy from the outside option.

• (2) If $\theta _1>\theta _3$, we have $\theta _2<\theta _3$. Thus, when $\theta \le \theta _3$, $U_c\ge U_r\ge 0$, the customers accept the compensation and wait for the delayed product; when $\theta >\theta _3$, the customers buy from the outside option.

Combining the two cases, we can derive the results in Lemma 3. $\square$

Proof of Proposition 1

To solve the retailer’s optimization problem in Eqs. (11) and (12), we first need to determine the feasible regions for different constraints. With simplification, we find that the constraints $\theta _1\le (>) \theta _3$ are equivalent to the conditions,

$$\begin{aligned} p+\frac{dr}{d_0(1-f)-d}-v\le (>) 0, \end{aligned}$$

(29)

which are shown in the following figure. As illustrated, the optimization problem should be solved within their feasible regions.

Fig. 7

Feasible regions for different constraints

Next, we take the first partial derivatives of Eqs. (11) and (12) with respect to p and r respectively, and have two cases:

• (1) For $\theta _1\le \theta _3$,

  $$\begin{aligned} \frac{\partial \Pi _U}{\partial p}= \;& {} \frac{(1-f)^2(v-2p+c)-f(1-f)(s-c)}{d}+\frac{f(v-2p+2r+s)}{d_0}, \end{aligned}$$

  (30)
  $$\begin{aligned} \frac{\partial \Pi _U}{\partial r}= \;& {} \frac{-f(v-2p+2r+s)}{d_0}. \end{aligned}$$

  (31)

  Taking the second-order partial derivatives of $\Pi _U$ with respect to p and r and using simplification, we know that the Hessian matrix is negative definite. Thus, $\Pi _U$ is strictly jointly concave in p and r, and the optimal solutions without considering the constraint is

  $$\begin{aligned} p_U^* = \frac{(1-f)(v+c)-f(s-c)}{2(1-f)},\ \ \ \ r_U^* = \frac{c-s}{2(1-f)}. \end{aligned}$$

  (32)

  Substituting the above optimal solutions into the constraint, we can derive that $(p_U^*,r_U^*)$ is the feasible optimal solution to the optimization problem (11) only when $\alpha \le 1-f-\beta$. Otherwise, the optimal solutions must occur on the boundary. In this case, we substitute $\theta _1= \theta _3$ into the profit function, and with optimization, we can derive the optimal solutions for $\alpha > 1-f-\beta$ as follows

  $$\begin{aligned} p_U^* = \frac{\bigl [2(1-f)(d+fd_0)-d\bigr ]v+dc}{2(1-f)(d+fd_0)},\ \ \ \ r_U^* = \frac{\bigl [d_0(1-f)-d\bigr ](v-c)}{2(1-f)(d+fd_0)}. \end{aligned}$$

  (33)

• (2) For $\theta _1>\theta _3$,

  $$\begin{aligned} \frac{\partial \Pi _U}{\partial p}=\frac{v-2p+c+2fr}{d+fd_0},\ \ \ \ \frac{\partial \Pi _U}{\partial r}=\frac{-f(v-2p+c+2fr)}{d+fd_0}, \end{aligned}$$

  (34)

  Also, we can show that the Hessian matrix is negative semi-definite. Combining the first-order conditions, the possible extreme point must occur on the red line in Fig. 7. Since each solution on the red line brings the same profit level to the retailer, we select a specific point on the constraint boundary and compare it to the region that $\theta _1\le \theta _3$. Recall that when $\theta _1\le \theta _3$, if the optimal solutions occur in the interior region, it must dominate any solutions on the boundary, including optimal solutions on boundary for $\theta _1> \theta _3$. If optimal solutions occur on the boundary, they would also perform equal to or better than any specific optimal point for $\theta _1> \theta _3$. Hence, we cannot derive the global optimal solutions in this case.

$\square$

Proof of Proposition 2

Considering the different optimal solutions in Eqs. (13) and (14), we conduct the sensitivity analysis in two cases:

• (1) When $\alpha \le t_u$, we have

  $$\begin{aligned}&\frac{\partial p_U^*}{\partial \beta }=0, \ \ \ \ \frac{\partial r_U^*}{\partial \beta }=0;\\&\frac{\partial p_U^*}{\partial f}=\frac{c-s}{2(1-f)^2}>0,\ \ \ \ \frac{\partial r_U^*}{\partial f}=\frac{c-s}{2(1-f)^2}>0;\\&\frac{\partial p_U^*}{\partial c}=\frac{1}{2(1-f)}>0,\ \ \ \ \frac{\partial r_U^*}{\partial c}=\frac{1}{2(1-f)}>0;\\&\frac{\partial p_U^*}{\partial s}=\frac{-f}{2(1-f)}<0,\ \ \ \ \frac{\partial r_U^*}{\partial s}=-\frac{1}{2(1-f)}<0;\\&\frac{\partial p_U^*}{\partial v}=\frac{1}{2}>0,\ \ \ \ \frac{\partial r_U^*}{\partial v}=0. \end{aligned}$$

• (2) When $\alpha > t_u$, we have

  $$\begin{aligned}&\frac{\partial p_U^*}{\partial \beta }=-\frac{f(v-c)}{2(1-f)(f+\beta )^2}<0, \ \ \ \ \frac{\partial r_U^*}{\partial \beta }=-\frac{v-c}{2(1-f)(f+\beta )^2}<0;\\&\frac{\partial p_U^*}{\partial f}=\frac{d(d_0-2fd_0-d)(v-c)}{2(1-f)^2(d+fd_0)^2}>0\ \ \text {for} \\&f\in \Bigl (0,\frac{1}{2}-\beta \Bigr ), \le 0 \ \ \text {for}\ \ f\in \Bigl [\frac{1}{2}-\beta ,1-\beta \Bigr ),\\&\frac{\partial r_U^*}{\partial f}=-\frac{(v-c)(\beta ^2+f^2+2\beta f-\beta -2f+1)}{2(1-f)^2(f+\beta )^2}<0;\\&\frac{\partial p_U^*}{\partial c}=\frac{d}{2(1-f)(d+fd_0)}>0,\ \ \ \ \frac{\partial r_U^*}{\partial c}=-\frac{d_0(1-f)-d}{2(1-f)(d+fd_0)}<0;\\&\frac{\partial p_U^*}{\partial s}=0,\ \ \ \ \frac{\partial r_U^*}{\partial s}=0;\\&\frac{\partial p_U^*}{\partial v}=\frac{2(1-f)(d+fd_0)-d}{2(1-f)(d+fd_0)}>0,\ \ \ \ \frac{\partial r_U^*}{\partial v}=\frac{d_0(1-f)-d}{2(1-f)(d+fd_0)}>0. \end{aligned}$$

Combining the two cases, we can derive the results in Proposition 2. $\square$

Proof of Proposition 3

Note that

$$\begin{aligned} p_U^*-p_B^*=\left\{ \begin{array}{ll} \frac{\beta f (c-s)}{2(1-f)(f^2+\beta f-2f+1)^2}>0 \quad &{} \text {for}\quad \alpha \le t_u,\,\\ \frac{f(1-\beta -f)[(f^2+\beta f-2f+1)(v-s)-(1+\beta -f)(c-s)]}{2(1-f)(f+\beta )(f^2+\beta f-2f+1)}>0 \quad &{} \text {for}\quad t_u<\alpha < t_1^p, \ \le 0 \ \text {for}\quad \alpha \ge t_1^p. \end{array}\right. \end{aligned}$$

(35)

Combining these results, we derive that: $p_U^*>p_B^*$ for $\alpha <t_1^p$, otherwise $p_U^*\le p_B^*$ for $\alpha \ge t_1^p$.

Since we have

$$\begin{aligned} \theta _{2U}^*-\theta _{2B}^* =\frac{(1-f)(v-p_U^*)}{d}-\frac{(1-f)(v-p_B^*)}{d}, \end{aligned}$$

(36)

the change of total purchase rates must be opposite to that of the prices. Therefore, we can directly derive the comparison results of the purchase rates.

The return rates can be written as

$$\begin{aligned} \bigl [\theta _{2U}^*-\theta _{0U}^*\bigr ]- \bigl [\theta _{2B}^*-\theta _{0B}^*\bigr ]=\left\{ \begin{array}{cl} \frac{-(c-s)}{2(f^2+\beta f-2f+1)}<0 \quad &{} \text {for}\quad \alpha \le t_u,\\ \frac{-(1-\beta -f)[(f^2+\beta f-2f+1)(v-s)-(1-f)(c-s)]}{2\beta (f^2+\beta f-2f+1)}<0 \quad &{} \text {for}\quad \alpha \ge t_u. \end{array}\right. \end{aligned}$$

(37)

Thus, the return rate under uniform compensation policy is lower.

For $\alpha \le t_u$, we have

$$\begin{aligned} \Pi _{U}^*-\Pi _{B}^*=\frac{f(c-s)^2(\beta f^2+\beta ^2 f-2\beta f+\beta )}{4d(f^2+\beta f-2f+1)^2}>0, \end{aligned}$$

(38)

which implies that $\Pi _{U}^*>\Pi _{B}^*$ in this case. $\square$

Proof of Lemma 4

As stated in the main text, we first derive the values of $\theta _0$, $\theta _1$, $\theta _2$, and $\theta _4$ in this case. Since $1-f-\beta <0$, we know that $\theta _0<\theta _1<\theta _4<\theta _2$ must hold. Thus, when $\theta \le \theta _0$, $U_a\ge U_r$, the customers do not claim for return and wait to accept the delayed product. When $\theta _0<\theta \le \theta _1$, we have $U_a< U_r$ and $U_c\ge U_r$, so the customers accept the compensation and wait for the delayed product. When $\theta _1<\theta \le \theta _2$, we have $U_c< U_r$, so the customers return the delayed product When $\theta _2<\theta \le 1$, we have $EU(Buy)< 0$, so the customers buy from the outside option. $\square$

Proof of Proposition 4

Taking the first partial derivatives of $\Pi _D$ in Eq. (18) with respect to p and r, we have

$$\begin{aligned} \frac{\partial \Pi _D}{\partial p}=\frac{(1-f)^2(v-2p+c)-f(1-f)(s-c)}{d}+\frac{f(v-2p+r+s)}{d_0},\ \ \ \ \frac{\partial \Pi _D}{\partial r}=\frac{f(p-2r-s)}{d_0}, \end{aligned}$$

(39)

By taking the second partial derivatives, we can show that $\frac{\partial ^2\Pi }{\partial p^2}<0$, and the Hessian matrix is negative definite. Thus, the profit function is jointly concave in p and r. Without considering the constraint $\theta _1\le \theta _2$, we first derive the optimal solutions as

$$\begin{aligned}&p_D^* = \frac{2 d_0(1-f)[v+c-f(v+s)]+df(2v+s)}{4 d_0 (1-f)^2+3df},\nonumber \\&r_D^* = \frac{d_0(1-f)[v+c-2s-f(v-s)]+df(v-s)}{4 d_0 (1-f)^2+3df}. \end{aligned}$$

(40)

Substituting these expressions into the constraint, we find that these results are the global optimal solutions for $\alpha \le t_d$. Otherwise, for $\alpha > t_d$, the feasible optimal solutions must occur on the boundary. Thus, by equating $\theta _1= \theta _2$ and solving the optimization problem, we can derive the optimal solutions for $\alpha > t_d$ as

$$\begin{aligned}&p_D^* = \frac{d d_0(1-f)(v+c)+2f[d-d_0(1-f)]^2 v}{2d d_0(1-f)+2f[d-d_0(1-f)]^2}, \nonumber \\&r_D^* = \frac{d_0 (1-f)(v-c)[d_0(1-f)-d]}{2d d_0(1-f)+2f[d-d_0(1-f)]^2}. \end{aligned}$$

(41)

$\square$

Proof of Proposition 5

Considering the different optimal solutions in Eqs. (20) and (21), we conduct the sensitivity analysis in two cases:

• 1 When $\alpha \le t_d$, we have

$$\begin{aligned}&\frac{\partial p_D^*}{\partial \beta }=\frac{2f(1-f)[(1-f)(v-s)-3(c-s)]}{(4f^2+3\beta f-8f+4)^2}>0\ \ \text {for}\ \ f<1-3\alpha ,\ \le 0 \ \ \text {for}\ \ 1-3\alpha \le f<1-\beta ; \\&\frac{\partial r_D^*}{\partial \beta }=\frac{f(1-f)[(1-f)(v-s)-3(c-s)]}{(4f^2+3\beta f-8f+4)^2}>0\ \ \text {for}\ \ f<1-3\alpha ,\ \le 0 \ \ \text {for}\ \ 1-3\alpha \le f<1-\beta ;\\&\frac{\partial p_D^*}{\partial c}=\frac{2(1-f)}{4(1-f)^2+3\beta f}>0,\ \ \ \ \frac{\partial r_D^*}{\partial c}=\frac{1-f}{4(1-f)^2+3\beta f}>0;\\&\frac{\partial p_D^*}{\partial s}=\frac{f(2f+\beta -2)}{4(1-f)^2+3\beta f}<0,\ \ \ \ \frac{\partial r_D^*}{\partial s}=-\frac{(1-f)(2-f)+\beta f}{4(1-f)^2+3\beta f}<0;\\&\frac{\partial p_D^*}{\partial v}=\frac{2(1-f)^2+2\beta f}{4(1-f)^2+3\beta f}>0,\ \ \ \ \frac{\partial r_D^*}{\partial v}=\frac{(1-f)^2+\beta f}{4(1-f)^2+3\beta f}>0;\\&\frac{\partial p_D^*}{\partial f}=\frac{2[\beta (1-f^2)(v-s)+(4f^2-3\beta -8f+4)(c-s)]}{(4f^2+3\beta f-8f+4)^2},\\&\frac{\partial r_D^*}{\partial f}=\frac{\beta (1-f^2)(v-s)+(4f^2-3\beta -8f+4)(c-s)}{(4f^2+3\beta f-8f+4)^2}. \end{aligned}$$

Define $G(f)=\beta (1-f^2)(v-s)+(4f^2-3\beta -8f+4)(c-s)$, which is a decreasing function of f. Note that

$$\begin{aligned} \lim _{f\rightarrow 0} G(f)= \;& {} \beta (v-s)+(4-3\beta )(c-s)>0,\end{aligned}$$

(42)

$$\begin{aligned} \lim _{f\rightarrow 1-\beta } G(f)= \;& {} \beta (4\beta -3). \end{aligned}$$

(43)

Thus, we have two subcases: First, for $\beta \ge \frac{3}{4}$, we have $\frac{\partial p_D^*}{\partial f}\ge 0$, $\frac{\partial r_D^*}{\partial f}\ge 0$. Second, for $\beta < \frac{3}{4}$, there must exist a unique threshold $f_0\in (0,1-\beta )$ such that $\frac{\partial p_D^*}{\partial f}\ge 0$, $\frac{\partial r_D^*}{\partial f}\ge 0$ for $f\in (0,f_0)$, and $\frac{\partial p_D^*}{\partial f}< 0$, $\frac{\partial r_D^*}{\partial f}< 0$ for $f\in [f_0,1-\beta )$.

• (2) When $\alpha > t_d$, we have

$$\begin{aligned}&\frac{\partial p_D^*}{\partial \beta }=-\frac{f(1-f)(1-f-\beta )(1-f+\beta )(v-c)}{2[\beta (1-f)+f(1-f-\beta )^2]^2}<0, \\&\frac{\partial r_D^*}{\partial \beta }=-\frac{(1-f)[\beta f(1-f-\beta )+(1-f)(f^2+\beta f-2f+1)](v-c)}{2[\beta (1-f)+f(1-f-\beta )^2]^2}<0;\\&\frac{\partial p_D^*}{\partial f}=\frac{\beta (1-f-\beta )(2f^2-\beta -3f+1)(v-c)}{2[\beta (1-f)+f(1-f-\beta )^2]^2}>0\ \text {for} \ f<\frac{3-\sqrt{8\beta +1}}{4}, \le 0 \ \text {for}\ f\ge \frac{3-\sqrt{8\beta +1}}{4},\\&\frac{\partial r_D^*}{\partial f}=\frac{(v-c)(-\beta ^2 f^2-2\beta f^3-f^4+\beta ^3+4\beta ^2 f+6\beta f^2+4f^3-3\beta ^2-6\beta f-6f^2+2\beta +4f-1)}{2[\beta (1-f)+f(1-f-\beta )^2]^2}<0;\\&\frac{\partial p_D^*}{\partial c}=\frac{dd_0(1-f)}{2d d_0(1-f)+2f[d-d_0(1-f)]^2}>0,\ \ \ \\&\frac{\partial r_D^*}{\partial c}=-\frac{d_0(1-f)[d_0(1-f)-d]}{2d d_0(1-f)+2f[d-d_0(1-f)]^2}<0;\\&\frac{\partial p_D^*}{\partial s}=0,\ \ \ \ \frac{\partial r_D^*}{\partial s}=0;\\&\frac{\partial p_D^*}{\partial v}=\frac{dd_0(1-f)+2f[d-d_0(1-f)]^2}{2d d_0(1-f)+2f[d-d_0(1-f)]^2}>0,\ \ \ \\&\frac{\partial r_D^*}{\partial v}=\frac{d_0(1-f)[d_0(1-f)-d]}{2d d_0(1-f)+2f[d-d_0(1-f)]^2}>0. \end{aligned}$$

Combining the two cases, we can derive the results in Proposition 5. $\square$

Proof of Proposition 6

We have

$$\begin{aligned} p_D^*-p_B^*=\left\{ \begin{array}{cl} \frac{\beta f [(f^2+\beta f-2f+1)(v-s)+(1-f)(c-s)]}{2(4f^2+3\beta f-8f+4)(f^2+\beta f-2f+1)}>0 \quad &{} \text {for}\quad \alpha \le t_d,\\ \frac{f(1-\beta -f)[(1-\beta -f)(f^2+\beta f-2f+1)(v-s)-(1-f)^2(c-s)]}{2[\beta (1-f)+f(1-f-\beta )^2](f^2+\beta f-2f+1)}>0 \quad &{} \text {for}\quad t_d<\alpha < t_2^p, \ \le 0 \ \text {for}\quad \alpha \ge t_2^p. \end{array}\right. \end{aligned}$$

(44)

Therefore, we derive that: $p_D^*>p_B^*$ if $\alpha <t_2^p$, and $p_U^*\le p_B^*$ for $\alpha \ge t_2^p$.

Note that

$$\begin{aligned} \theta _{2D}^*-\theta _{2B}^* =\frac{(1-f)(v-p_D^*)}{d}-\frac{(1-f)(v-p_B^*)}{d}, \end{aligned}$$

(45)

so the change of total purchase rates is opposite to that of the prices. Therefore, we can directly derive the comparison results of the purchase rates.

The return rates can be written as

$$\begin{aligned} \bigl [\theta _{2D}^*-\theta _{0D}^*\bigr ]- \bigl [\theta _{2B}^*-\theta _{0B}^*\bigr ]=\left\{ \begin{array}{cl} \frac{-(f^2+\beta f-3f+2)[(f^2+\beta f-2f+1)(v-s)+(1-f)(c-s)]}{2(4f^2+3\beta f-8f+4)(f^2+\beta f-2f+1)}<0 \quad &{} \text {for}\quad \alpha \le t_d,\,\\ \frac{-(1-\beta -f)[(f^2+\beta f-2f+1)(v-s)-(1-f)(c-s)]}{2\beta (f^2+\beta f-2f+1)}<0 \quad &{} \text {for}\quad \alpha \ge t_d. \end{array}\right. \end{aligned}$$

(46)

Thus, the return rate under discriminated compensation policy is lower.

For $\alpha \le t_d$, we have

$$\begin{aligned} \Pi _{D}^*-\Pi _{B}^*=\frac{\beta f[(f^2+\beta f-2f+1)(v-s)+(1-f)(c-s)]^2}{4d(f^2+\beta f-2f+1)(4f^2+3\beta f-8f+4)}>0, \end{aligned}$$

(47)

which implies that $\Pi _{D}^*>\Pi _{B}^*$ in this case. $\square$

Proof of Proposition 7

Before analyzing, we note that the conditions $t_u\le t_d$ ($t_u> t_d$) is equivalent to $f^2+\beta f-f+\beta \le 0$ ($f^2+\beta f-f+\beta > 0$). Based on this result, we discuss two cases:

• (1) When $f^2+\beta f-f+\beta \le 0$, i.e, $t_u\le t_d$, we have three subcases:

  • For $0<\alpha \le t_u$, we have

    $$\begin{aligned} p_D^*- p_U^*=\frac{\beta f[(1-f)(v-s)-3(c-s)]}{2(1-f)(4f^2+3\beta f-8f+4)}. \end{aligned}$$

    (48)

    Note that $t_u>\frac{1-f}{3}$ can be proved in this case, so we derive that $p_D^*\ge p_U^*$ for $0<\alpha \le \frac{1-f}{3}$, and $p_D^*< p_U^*$ for $\frac{1-f}{3}<\alpha \le t_u$;

    $$\begin{aligned} r_D^*- r_U^*=\frac{2(1-f)(f^2+\beta f-2f+1)(v-s)-(2f^2+3\beta f-4f+2)(c-s)}{2(1-f)(4f^2+3\beta f-8f+4)}. \end{aligned}$$

    (49)

    Denoting

    $$\begin{aligned} t_1^r=\frac{2(1-f)(f^2+\beta f -2f+1)}{2f^2+3\beta f-4f+2}, \end{aligned}$$

    (50)

    we get that $r_D^*\ge r_U^*$ for $0<\alpha \le \min \{t_1^r,t_u\}$, and $r_D^*< r_U^*$ for $\min \{t_1^r,t_u\}<\alpha \le t_u$.

  • For $t_u<\alpha \le t_d$, we have

    $$\begin{aligned} p_D^*- p_U^*=\frac{f[(-2\beta ^2f-6\beta f^2-4f^3-\beta ^2+10\beta f+12f^2-4\beta -12f+4)(v-s)+(3\beta ^2-4f^2+8f-4)(c-s)] }{-2(4f^2+3\beta f-8f+4)(1-f)(\beta +f)}. \end{aligned}$$

    (51)

    Since

    $$\begin{aligned}&\frac{-2\beta ^2f-6\beta f^2-4f^3-\beta ^2+10\beta f+12f^2-4\beta -12f+4}{-(3\beta ^2-4f^2+8f-4)}-t_d\end{aligned}$$

    (52)
    $$\begin{aligned}&\quad =\frac{2\beta ^2(4f^2+3\beta f-8f+4)(1-f-\beta )}{(3\beta ^2-4f^2+8f-4)(1-f)(2f+\beta -2)}>0, \end{aligned}$$

    (53)

    we have $p_D^*< p_U^*$ in this case.

    $$\begin{aligned} r_D^*- r_U^*= & {} \frac{(2\beta ^2f^2+4\beta f^3+2f^4-5\beta ^2 f-15\beta f^2-10f^3+17\beta f+18 f^2-6\beta -14f+4)(v-s)}{-2(4f^2+3\beta f-8f+4)(1-f)(\beta +f)}\nonumber \\&+\frac{(3\beta ^2f+5\beta f^2+2f^3-7\beta f-8f^2+2\beta +10f-4)(c-s)}{-2(4f^2+3\beta f-8f+4)(1-f)(\beta +f)}. \end{aligned}$$

    (54)

    Denoting

    $$\begin{aligned} t_2^r=\frac{2\beta ^2 f^2 +4\beta f^3 +2f^4-5\beta ^2 f-15\beta f^2-10f^3+17 \beta f+18f^2-6\beta -14f+4}{-(3\beta ^2 f+5\beta f^2+2f^3-7\beta f-8f^2+2\beta +10f-4)}, \end{aligned}$$

    (55)

    we know that

    $$\begin{aligned} t_2^r-t_d=\frac{2\beta f(4f^2+3\beta f-8f+4)(1-f-\beta )^2}{(3\beta ^2 f+5\beta f^2+2f^3-7\beta f-8f^2+2\beta +10f-4)(1-f)(2-2f-\beta )}<0, \end{aligned}$$

    (56)

    i.e., $t_2^r<t_d$. Therefore, we can derive that $r_D^*> r_U^*$ for $\max \{t_2^r,t_u\}<\alpha \le t_d$, and $r_D^*\le r_U^*$ for $t_u<\alpha \le \max \{t_2^r,t_u\}$.

  • For $t_d\le \alpha <1$, we have

    $$\begin{aligned} p_D^*-p_U^*=\frac{-\beta ^2 f(1-f_\beta )(v-c)}{2(\beta ^2f+2\beta f^2+f^3-3\beta f-2f^2+\beta +f)(1-f)(\beta +f)}<0, \end{aligned}$$

    (57)

    i.e., $p_D^*<p_U^*$.

    $$\begin{aligned} r_D^*-r_U^*=\frac{\beta f(1-f-\beta )^2(v-c)}{2(\beta ^2f+2\beta f^2+f^3-3\beta f-2f^2+\beta +f)(1-f)(\beta +f)}>0, \end{aligned}$$

    (58)

    i.e., $r_D^*>r_U^*$.

• (2) When $f^2+\beta f-f+\beta > 0$, i.e, $t_u> t_d$, we have three subcases:

  • For $0<\alpha \le t_d$, we have

    $$\begin{aligned} p_D^*- p_U^*=\frac{\beta f[(1-f)(v-s)-3(c-s)]}{2(1-f)(4f^2+3\beta f-8f+4)}. \end{aligned}$$

    (59)

    Then, we can derive that $p_D^*\ge p_U^*$ for $0<\alpha \le \min \{\frac{1-f}{3},t_d\}$, and $p_D^*< p_U^*$ for $\min \{\frac{1-f}{3},t_d\}<\alpha \le t_d$.

    $$\begin{aligned} r_D^*- r_U^*=\frac{2(1-f)(f^2+\beta f-2f+1)(v-s)-(2f^2+3\beta f-4f+2)(c-s)}{2(1-f)(4f^2+3\beta f-8f+4)}. \end{aligned}$$

    (60)

    Since

    $$\begin{aligned} \frac{2(1-f)(f^2+\beta f-2f+1)}{2f^2+3\beta f-4f+2}-t_d>0, \end{aligned}$$

    (61)

    we know that $r_D^*> r_U^*$.

  • For $t_d<\alpha \le t_u$, we have

    $$\begin{aligned} p_D^*- p_U^*=\frac{f[(1-f)(1-f-\beta )^2(v-s)-(\beta ^2+\beta f+f^2-\beta -2f+1)(c-s)}{2(1-f)(\beta ^2f+2\beta f^2+f^3-3\beta f-2f^2+\beta +f)}. \end{aligned}$$

    (62)

    Define

    $$\begin{aligned} t_3^p=\frac{(1-f)(1-f-\beta )^2}{\beta ^2+\beta f+f^2-\beta -2f+1}, \end{aligned}$$

    (63)

    we know that

    $$\begin{aligned} t_3^p-t_u=\frac{-\beta ^2(1-f-\beta )}{\beta ^2+\beta f+f^2-\beta -2f+1}<0, \end{aligned}$$

    (64)

    i.e., $t_3^p<t_u$. Then, we can derive that $p_D^*\ge p_U^*$ for $t_d<\alpha \le \max \{t_3^p,t_d\}$, and $p_D^*< p_U^*$ for $\max \{t_3^p,t_d\}<\alpha \le t_u$;

    $$\begin{aligned} r_D^*- r_U^*=\frac{(1-f)^2(1-f-\beta )(v-s)-(\beta ^2 f+\beta f^2-\beta f+f^2-2f+1)(c-s)}{2(1-f)(\beta ^2f+2\beta f^2+f^3-3\beta f-2f^2+\beta +f)}. \end{aligned}$$

    (65)

    Since

    $$\begin{aligned}&\frac{(1-f)^2(1-f-\beta )}{\beta ^2 f+\beta f^2-\beta f+f^2-2f+1}-t_u\nonumber \\&\quad =\frac{\beta f (1-\beta -f)^2}{\beta ^2 f+\beta f^2-\beta f+f^2-2f+1}>0, \end{aligned}$$

    (66)

    we know that $r_D^*> r_U^*$;

  • For $t_u\le \alpha <1$, the proof is the same as the third subcase in case (1), i.e., $p_D^*<p_U^*$, $r_D^*>r_U^*$. We avoid the repetition here.

$\square$

Moreover, by comparing the optimal profits in the two compensation cases, we have

$$\begin{aligned} \Pi _D^*(p_D^*,r_D^*)-\Pi _U^*(p_U^*,r_U^*)\ge & {} \Pi _D(p_U^*,r_U^*)-\Pi _U^*(p_U^*,r_U^*)\nonumber \\= & {} \frac{f(v-p_U^*)r_U^*}{d_0}\ge 0, \end{aligned}$$

(67)

which implies that $\Pi _D^*\ge \Pi _U^*$.