Effects of production capacity and substitutability on optimal pricing and inventory policies

Abstract

Manufacturers produce substitutable products to meet the different needs of consumers. Meanwhile, to formulate the optimal production strategy, manufacturers have to deal with demand uncertainty and capacity constraints. This study employs a stochastic model of a monopolistic manufacturer with limited capacity to sell two types of substitutable products: high- and low-end products. We then develop an expected profit function and solve for the optimal prices, safety stocks, and expected profit. Additionally, this study uses cost functions related to the substitutability (quality) to extend our basic model and investigate how the manufacturer makes decisions. The optimal results are derived under capacity constraints. The findings show that when the cost of low-end products is high, with an increasing substitutability, the manufacturer should reduce the price of high-end products and raise the price of low-end products. Otherwise, the manufacturer should conduct an opposite modification. In addition, different capacity constraints moderate the effects of demand uncertainty on the safety stocks, and the safety stock of high-end products may decrease in demand uncertainty. The results also reveal that, the manufacturer should raise the allocation ratio of high-end products if the capacity increases. With a higher substitutability, the manufacturer may allocate more capacity to high-end products. In the case of low-end products’ quality decision, a tighter capacity results in a higher optimal substitutability. Considering two products’ quality decisions, however, a moderately tight capacity leads to the highest quality level of low-end products, and the quality of high-end products always remains at a very high level.

Appendix

Proof of Lemma 1

For given ${{\mathrm{s}}_{1}}^{*},{{\mathrm{s}}_{2}}^{*}$, we can get the Hessian matrix $H$ of ${\mathbb{E}}(\Pi \left({p}_{1},{p}_{2}\right))$ is

$$\left[\begin{array}{cc}-2\left(m+{b}_{1}\right)& 2m\\ 2m& -2\left(m+{b}_{2}\right)\end{array}\right].$$

Then, we have the first-order determinant of $H$ is $-2\left(m+{b}_{1}\right)<0$, the second-order determinant is $4\left(m+{b}_{1}\right)\left(m+{b}_{2}\right)-4{m}^{2}>0$.Thus, matrix $H$ is negative definite, and ${\mathbb{E}}(\Pi \left({p}_{1},{p}_{2}\right))$ is an concave function in ${p}_{1}$ and ${p}_{2}$ for given ${{\mathrm{s}}_{1}}^{*}$ and ${{\mathrm{s}}_{2}}^{*}$. Based on the features of multivariate function, ${\mathbb{E}}(\Pi \left({p}_{1},{p}_{2}\right))$ has a unique optimal solution. Then, according to $\frac{{\mathbb{E}}(\Pi \left({p}_{1},{p}_{2}\right))}{\partial {p}_{i}}=0$, we have

$${{p}_{i}}^{*}=\frac{{a}_{i}+{b}_{i}{c}_{i}+m\left({c}_{i}-{c}_{j}+2{{p}_{j}}^{*}\right)+{\mu }_{i}}{2({b}_{i}+m)}-\frac{{S}_{i}\left({{s}_{i}}^{*}\right)}{2\left({b}_{i}+m\right)}.$$

For given ${{p}_{1}}^{*},{{p}_{2}}^{*}$, we can get the Hessian matrix $\overline{H }$ of ${\mathbb{E}}(\Pi \left({s}_{1},{s}_{2}\right))$ is

$$\left[\begin{array}{cc}-\left({p}_{1}+h+{g}_{1}\right){f}_{1}\left({s}_{1}\right)& 0\\ 0& -\left({p}_{2}+h+{g}_{2}\right){f}_{2}\left({s}_{2}\right)\end{array}\right].$$

Then, we have the first-order determinant of $\overline{H }$ is $-\left({p}_{1}+h+{g}_{1}\right){f}_{1}\left({s}_{1}\right)<0$, the second-order determinant is $\left({p}_{1}+h+{g}_{1}\right){f}_{1}\left({s}_{1}\right)\left({p}_{2}+h+{g}_{2}\right){f}_{2}\left({s}_{2}\right)>0$.Thus, matrix $\overline{H }$ is negative definite, and the profit function ${\mathbb{E}}(\Pi \left({s}_{1},{s}_{2}\right))$ has a unique optimal solution. Based on $\frac{{\mathbb{E}}(\Pi \left({s}_{1},{s}_{2}\right))}{\partial {p}_{i}}=0$, we have

$$\frac{1-{F}_{i}({{s}_{i}}^{*})}{{F}_{i}({{s}_{i}}^{*})}=\frac{{c}_{i}+h}{{{p}_{i}}^{*}-{c}_{i}+{g}_{i}}.$$

In the case of low capacity, the optimal solution can be obtained on the bounded boundary. As ${{s}_{1}}^{*}+{{s}_{2}}^{*}+1-{b}_{1}{{p}_{1}}^{*}-{b}_{2}{{p}_{2}}^{*}>Q$, the optimal ${p}_{1},{p}_{2},{s}_{1},{s}_{2}$ must satisfy ${s}_{1}+{s}_{2}+1-{b}_{1}{p}_{1}-{b}_{2}{p}_{2}=Q$. Hence, we use the Lagrange function to change the original problem, and derive

$$\mathcal{L}({p}_{1},{p}_{2},{s}_{2},{s}_{2},\uplambda )={(p}_{1}-{c}_{1})({Q}_{1}+{\mu }_{1})-({c}_{1}+h){I}_{1}({s}_{1})$$

$$-{(p}_{1}-{c}_{1}+{g}_{1}){S}_{1}({s}_{1})+{(p}_{2}-{c}_{2})({Q}_{2}+{\mu }_{2})-\left({c}_{2}+h\right){I}_{2}\left({s}_{2}\right)$$

$$-{(p}_{2}-{c}_{2}+{g}_{2}){S}_{2}({s}_{2})+\uplambda ({ s}_{1}+{s}_{2}+1-{b}_{1}{p}_{1}-{b}_{2}{p}_{2}-q).$$

Since the conditions of the optimal solution are

$$\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2},\uplambda ))}{\partial {p}_{i}}=0$$

$$\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2},\uplambda ))}{\partial {s}_{i}}=0$$

$$\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2},\uplambda ))}{\partial\uplambda }=0,$$

we have

$${a}_{i}+m\left({c}_{i}-{c}_{j}-2{{p}_{i}}^{*}+2{{p}_{j}}^{*}\right)+{b}_{i}\left({c}_{i}-2{{p}_{i}}^{*}-\uplambda \right)+{\mu }_{1}-{S}_{i}\left({{s}_{i}}^{*}\right)=0$$

$$\uplambda -\left({{p}_{i}}^{*}+{g}_{i}-{c}_{i}\right)\left(1-F\left({{s}_{i}}^{*}\right)\right)+\left({c}_{i}+h\right)F\left({{s}_{i}}^{*}\right)=0$$

$${{s}_{1}}^{*}+{{s}_{2}}^{*}+1-{b}_{1}{{p}_{1}}^{*}-{b}_{2}{{p}_{2}}^{*}=Q.$$

Then, according to function characters that any positively liner combination of concave functions is concave. We can get $\mathcal{L}({p}_{1},{p}_{2},{s}_{2},{s}_{2},\uplambda )$ is a concave function, and it has a unique solution. Thus, there exist unique optimal ${{p}_{1}}^{*},{{p}_{2}}^{*}, {{s}_{1}}^{*},{{s}_{2}}^{*}$ to maximize the expected profit.

From these calculations, we can find out that, the capacity constraint ($\sum_{i=\mathrm{1,2}}{s}_{i}-{b}_{i}{p}_{i}+1\le Q)$ is ignored, if $\uplambda =0$. However, if $\uplambda >0$, the optimal solution is solved on the boundary. In other words, $\uplambda =0$ represents that, the capacity is ample, and $\uplambda >0$ represents that, the capacity is constrained. Based on the different value of $\uplambda$, the optimal solution can be got in the boundary or on the boundary. Then, there are always unique solutions to maximize the profit, and the value of $\uplambda$ is affected by $Q$. $\hfill\square$

Proof of Lemma 2

When the manufacturer only produces product 2, the expected profit function is.

${\Pi }_{2}({p}_{2},{s}_{2})={(p}_{2}-{c}_{2})({a}_{2}-{b}_{2}{p}_{2}+{\mu }_{2})-\left({c}_{2}+h\right){I}_{2}\left({s}_{2}\right) -{(p}_{2}-{c}_{2}+{g}_{2}){S}_{2}({s}_{2}$).

According to the Envelope Theorem, we derive

$$\frac{\partial ({\mathbb{E}}{\Pi }^{*}-{\mathbb{E}}{{\Pi }_{2}}^{*})}{\partial {a}_{1}}={\left.\frac{\partial ({\mathbb{E}}\Pi -{\Pi }_{2})}{\partial {a}_{1}}\right|}_{{p}_{1}={{p}_{1}}^{*},{p}_{2}={{p}_{2}}^{*},{s}_{1}={{s}_{1}}^{*},{s}_{2}={{s}_{2}}^{*}, {p}_{2}=\overline{{p }_{2}},{s}_{2}=\overline{{s }_{2}}}$$

$$=\overline{{p }_{2}}-{{p}_{2}}^{*}+{{p}_{1}}^{*}-{c}_{1,}$$

where $\overline{{p }_{1}}$ and $\overline{{s }_{1}}$ indicate the optimal solutions of ${\Pi }_{2}$. Since $\overline{{p }_{2}}$ is the price, which is not under competition, $\overline{{p }_{2}}>{{p}_{2}}^{*}$, and we have $\overline{{p }_{2}}-{{p}_{2}}^{*}+{{p}_{1}}^{*}-{c}_{1}>0$, ${\mathbb{E}}{\Pi }^{*}-{\mathbb{E}}{{\Pi }_{2}}^{*}$ monotonically decreases in ${a}_{1}$. If ${a}_{1}=0$, ${\Pi }^{*}<{{\Pi }_{2}}^{*}$, there is no demand for product 1, and producing product 1 hurts the manufacturer’s profit. If ${a}_{1}=1$, ${\Pi }^{*}>{{\Pi }_{2}}^{*}$, since the demand for product 2 is zero, and ${{\Pi }_{2}}^{*}$ is equal to zero. Hence, there exists a value of ${a}_{1}$(${a}_{11}$) such that ${\Pi }^{*}-{\mathbb{E}}{{\Pi }_{2}}^{*}=0$. Meanwhile, we have

$$ \frac{\partial ({\mathbb{E}}{\Pi }^{*}-{\mathbb{E}}{{\Pi }_{1}}^{*})}{\partial {a}_{1}}= {\left.\frac{\partial ({\mathbb{E}}\Pi -{\Pi }_{1})}{\partial {a}_{1}}\right|}_{{p}_{1} ={{p}_{1}}^{*},{p}_{2}={{p}_{2}}^{*},{s}_{1}={{s}_{1}}^{*},{s}_{2}={{s}_{2}}^{*}, {p}_{1} =\overline{{p }_{1}},{s}_{1}=\overline{{s }_{1}}}={c}_{2}-{{p}_{1}}^{*}+{{p}_{2}}^{*} +\overline{{p }_{1}}>0.$$

The same process can be employed to demonstrate that there exists a value of ${a}_{1}$(${a}_{12}$) such that ${\Pi }^{*}-{\mathbb{E}}{{\Pi }_{1}}^{*}=0$.

Moreover, when ${a}_{11}<{a}_{12}$, the intersection of ${\Pi }^{*}$ and function ${{\Pi }_{2}}^{*}$(i.e. ${a}_{11}$) is on the left of the intersection of ${\Pi }^{*}$ and function ${{\Pi }_{1}}^{*}$(i.e., ${a}_{12}$). Then, based on the properties of the three profit functions, the function values of ${\Pi }^{*}$ are greater than the those of ${{\Pi }_{1}}^{*}$ and ${{\Pi }_{2}}^{*}$, when ${a}_{1}$ is in the range of $[{a}_{11},{a}_{12}]$. $\hfill\square$

Proof of Proposition 1 (i)

Based on optimal values, we have.

$$\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {{p}_{1}}^{*}}=0$$

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {{{p}_{1}}^{*}}^{2}}=-2\left({b}_{1}+m\right)<0.$$

Thus, $\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {{p}_{1}}^{*}}$ decreases in ${{p}_{1}}^{*}$. Additionally, we have

$$\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))/\partial {p}_{1}}{\partial m}={c}_{1}-{c}_{2}-2{p}_{1}+2{p}_{2},$$

which means that, if ${c}_{1}-{c}_{2}-2{p}_{1}+2{p}_{2}>0$, $\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}}$ increases in $m$. Thus, an increasing $m$ leads to a higher value of $\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}}$. However, to ensure optimal values, namely, $\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {{p}_{1}}^{*}}=0$, a higher ${{p}_{1}}^{*}$ should be obtained to decrease the value of $\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}}$, since $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {{{p}_{1}}^{*}}^{2}}<0$. Conversely, if ${c}_{1}-{c}_{2}-2{p}_{1}+2{p}_{2}<0$, $\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}}$ decreases in $m$, a higher $m$ leads to a lower ${{p}_{1}}^{*}$. Hence, to analyse how substitutability affect the optimal prices, we use the following equation:

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}\partial m}={c}_{1}-{c}_{2}-2{p}_{1}+2{p}_{2},$$

where ${p}_{1}>{c}_{1}$, ${p}_{2}>{c}_{2}$. ${c}_{1}-{c}_{2}-2{p}_{1}+2{p}_{2}<0$ can be rewritten as ${p}_{1}-{p}_{2}<\frac{{c}_{1}-{c}_{2}}{2}$, and $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}\partial m}<0$. Thus, ${{p}_{1}}^{*}$ strictly decreases in $m$. When ${c}_{1}-{c}_{2}-2{p}_{1}+2{p}_{2}>0$, which can be rewritten as ${p}_{1}-{p}_{2}>\frac{{c}_{1}-{c}_{2}}{2}$, $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}\partial m}>0$, ${{p}_{1}}^{*}$ strictly increases in $m$. Using the same calculation, we obtain the change in the low-end product with respect to $m$ and the same result in the case of low capacity. $\hfill\square$

Proof of Proposition 1 (ii)

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}\partial {c}_{2}}=\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{2}\partial {c}_{1}}=-m,$$

where $-m<0$. Thus, $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}\partial {c}_{2}}=\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{2}\partial {c}_{1}}<0$. ${{p}_{1}}^{*}$ strictly decreases in ${c}_{2}$, and ${{p}_{2}}^{*}$ strictly decreases in ${c}_{1}$.$\hfill\square$

Proof of Proposition 2

By applying the Envelope Theorem, we have.

$$\frac{\partial {\mathbb{E}}(\Pi \left({{p}_{1}}^{*},{{p}_{2}}^{*},{{s}_{1}}^{*},{{s}_{2}}^{*}\right))}{\partial m}={\left.\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{2},{s}_{2})}{\partial m}\right|}_{{p}_{1}={{p}_{1}}^{*},{p}_{2}={{p}_{2}}^{*},{s}_{1}={{s}_{1}}^{*},{s}_{2}={{s}_{2}}^{*}}.$$

Then,

$$\frac{\partial {\mathbb{E}}(\Pi ({{p}_{1}}^{*},{{p}_{2}}^{*},{{s}_{1}}^{*},{{s}_{2}}^{*})}{\partial m}={\left.-\left({p}_{1}-{p}_{2}\right)(-{c}_{1}+{c}_{2}+{p}_{1}-{p}_{2})\right|}_{{p}_{1}={{p}_{1}}^{*},{p}_{2}={{p}_{2}}^{*},{s}_{1}={{s}_{1}}^{*},{s}_{2}={{s}_{2}}^{*}},$$

where ${{p}_{1}}^{*}-{{p}_{2}}^{*}>0$. When $-{c}_{1}+{c}_{2}+{{p}_{1}}^{*}-{{p}_{2}}^{*}<0$, we get ${{p}_{1}}^{*}-{c}_{1}<{{p}_{2}}^{*}-{c}_{2}$, and $-\left({{p}_{1}}^{*}-{{p}_{2}}^{*}\right)\left(-{c}_{1}+{c}_{2}+{{p}_{2}}^{*}-{{p}_{2}}^{*}\right)=\frac{\partial {\mathbb{E}}(\Pi ({{p}_{1}}^{*},{{p}_{2}}^{*},{{s}_{1}}^{*},{{s}_{2}}^{*})}{\partial m}>0$. Therefore, $\partial {\mathbb{E}}(\Pi ({{p}_{1}}^{*},{{p}_{2}}^{*},{{s}_{1}}^{*},{{s}_{2}}^{*})$ increases in $m$. When $-{c}_{1}+{c}_{2}+{{p}_{1}}^{*}-{{p}_{2}}^{*}>0$, which can be rewritten as ${{p}_{1}}^{*}-{c}_{1}>{{p}_{2}}^{*}-{c}_{2}$, $-\left({{p}_{1}}^{*}-{{p}_{2}}^{*}\right)\left(-{c}_{1}+{c}_{2}+{{p}_{2}}^{*}-{{p}_{2}}^{*}\right)=\frac{\partial {\mathbb{E}}(\Pi ({{p}_{1}}^{*},{{p}_{2}}^{*},{{s}_{1}}^{*},{{s}_{2}}^{*})}{\partial m}>0$, $\partial {\mathbb{E}}(\Pi ({{p}_{1}}^{*},{{p}_{2}}^{*},{{s}_{1}}^{*},{{s}_{2}}^{*})$ decreases in $m$.

Using the same calculation, we obtain the same process and result in low-capacity situations. $\hfill\square$

Proof of Proposition 3

When capacity is low, the optimal solution will be obtained on the boundary. Since ${s}_{1}+{s}_{2}+1-{b}_{1}{p}_{1}-{b}_{2}{p}_{2}=Q$, we can transfer the original function to ${\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1}))$. Then, we have.

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1})).}{\partial {p}_{1}\partial {b}_{2}}=-{b}_{1}{p}_{2}({p}_{2}+{g}_{2}+h){f}_{2}\left({s}_{2}\right)$$

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1})).}{\partial {p}_{1}\partial {g}_{2}}={b}_{1}(1-{F}_{2}\left({s}_{2}\right))$$

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1})).}{\partial {p}_{1}\partial h}=-{b}_{1}{F}_{2}\left({s}_{2}\right),$$

$1-{F}_{2}\left({s}_{2}\right)>0$. We get $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1})).}{\partial {p}_{1}\partial m}<0$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1})).}{\partial {p}_{1}\partial {g}_{2}}>0$, $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1})).}{\partial {p}_{1}\partial h}<0$; thus, ${{p}_{1}}^{*}$ strictly decreases in ${b}_{2}$, increases in ${g}_{2}$, and increases in $h$where , .

Similarly, since $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1}))}{\partial {s}_{1}\partial {g}_{2}}=-\left(1-{F}_{2}\left({s}_{2}\right)\right)<0$, ${{s}_{1}}^{*}$ strictly decreases in ${g}_{2}$.

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1}))}{\partial {s}_{1}\partial h}={F}_{2}\left({s}_{2}\right)-{F}_{1}\left({s}_{1}\right),$$

where ${s}_{2}=Q-{ s}_{1}-1+{b}_{1}{p}_{1}+{b}_{2}{p}_{2}$. When ${F}_{2}\left({s}_{2}\right)>{F}_{1}\left({s}_{1}\right)$, $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1}))}{\partial {s}_{1}\partial h}>0$, and ${{s}_{1}}^{*}$ increases in $h$. When ${F}_{2}\left({s}_{2}\right)<{F}_{1}\left({s}_{1}\right)$, $\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1}))}{\partial {s}_{1}\partial h}<0$, and ${{s}_{1}}^{*}$ decreases in $h$. $\hfill\square$

Proof of Proposition 4

If capacity is high, the profit function is ${\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))$. Then, we have.

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{1}\partial m}={c}_{1}-{c}_{0}-k-\frac{3}{2} {m}^{2}c-2\left({p}_{1}-{p}_{2}\right).$$

By defining.

$${\mathrm{G}}_{1}\left(m\right)={c}_{1}-{c}_{0}-k-\frac{3}{2} {m}^{2}c-2\left({p}_{1}-{p}_{2}\right),$$

we obtain

$${{\mathrm{G}}_{1}}^{\mathrm{^{\prime}}}\left(m\right)=-3mc<0,$$

$${\mathrm{G}}_{1}\left(0\right)={c}_{1}-{c}_{0}-k-2\left({p}_{1}-{p}_{2}\right),$$

$$\underset{m\to 1}{\mathrm{lim}}{\mathrm{G}}_{1}\left(m\right)={c}_{1}-{c}_{0}-\frac{3}{2}c-2\left({p}_{1}-{p}_{2}\right)-k<0.$$

Since $m\in [\mathrm{0,1})$, ${{G}_{1}}^{^{\prime}}\left(m\right)<0$ and ${G}_{1}\left(m\right)$ decreases in $m$, if ${G}_{1}\left(0\right)>0$, which yields ${c}_{1}-{c}_{0}-k-2\left({p}_{1}-{p}_{2}\right)>0$, we obtain ${m}_{1,}$ as ${G}_{1}\left({m}_{1}\right)=0$.

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1},{s}_{2}))}{\partial {p}_{2}\partial m}={c}_{1}-{c}_{0}+c{b}_{2}m+\frac{3}{2} {m}^{2}c+2\left({p}_{1}-{p}_{2}\right)+\theta .$$

By defining.

$${\mathrm{G}}_{2}\left(m\right)={c}_{1}-{c}_{0}+c{b}_{2}m+\frac{3}{2} {m}^{2}c+2\left({p}_{1}-{p}_{2}\right)+\theta ,$$

we have

$${{\mathrm{G}}_{2}}^{\mathrm{^{\prime}}}\left(m\right)=c\left({b}_{1}+3m\right)>0,$$

$${\mathrm{G}}_{2}\left(0\right)={c}_{0}-{c}_{1}+2\left({p}_{1}-{p}_{2}\right)+\theta ,$$

$$\underset{m\to 1}{\mathrm{lim}}{\mathrm{G}}_{1}\left(m\right)={c}_{0}-{c}_{1}-c{b}_{2}+\frac{3}{2}c+2\left({p}_{1}-{p}_{2}\right)>0.$$

In $m\in [\mathrm{0,1})$, ${{\mathrm{G}}_{2}}^{^{\prime}}\left(m\right)>0$; thus, ${\mathrm{G}}_{2}\left(m\right)$ increases in $m$. Hence, when ${G}_{2}\left(0\right)<0$(i.e. ${c}_{0}-{c}_{1}+2\left({p}_{1}-{p}_{2}\right)+\theta >0$), we obtain ${m}_{2}$, as ${G}_{2}\left({m}_{2}\right)=0$.

When capacity is low, the optimal solution will be obtained on the boundary. Since ${s}_{1}+{s}_{2}+1-{b}_{1}{p}_{1}-{b}_{2}{p}_{2}+(\theta -k)({\tau }_{1}+{\tau }_{2})=Q$, we can transfer the original function to ${\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1}))$.

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1}))}{\partial {p}_{1}\partial m}={c}_{1}-{c}_{0}-\frac{3}{2} {m}^{2}c-2\left({p}_{1}-{p}_{2}\right)-c{b}_{1}m-k$$

By defining.

$${{\mathrm{G}}_{1}\left(m\right)}^{\mathrm{^{\prime}}}={c}_{1}-{c}_{0}-\frac{3}{2} {m}^{2}c-2\left({p}_{1}-{p}_{2}\right)-c{b}_{1}m-k,$$

we obtain

$${{{\mathrm{G}}_{1}}^{\mathrm{^{\prime}}}\left(m\right)}^{\mathrm{^{\prime}}}=-c\left({b}_{1}+3m\right)<0,$$

$${{\mathrm{G}}_{1}\left(0\right)}^{\mathrm{^{\prime}}}={c}_{1}-{c}_{0}-2\left({p}_{1}-{p}_{2}\right)-k,$$

$$\underset{m\to 1}{\mathrm{lim}}{{\mathrm{G}}_{1}\left(m\right)}^{^{\prime}}=\left({c}_{1}-{c}_{0}-\frac{3}{2}c\right)-2\left({p}_{1}-{p}_{2}\right)-c{b}_{1}<0.$$

Since $m\in [\mathrm{0,1})$, ${{{\mathrm{G}}_{1}}^{^{\prime}}\left(m\right)}^{^{\prime}}<0$, and ${G}_{1}\left(m\right)$ decreases in $m$, if ${{\mathrm{G}}_{1}\left(0\right)}^{^{\prime}}>0$, which yields ${c}_{1}-{c}_{0}-2\left({p}_{1}-{p}_{2}\right)-k>0$, we get ${m}_{1},$ as ${G}_{1}\left({m}_{1}\right)=0$.

$$\frac{{\partial }^{2}{\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{1}))}{\partial {p}_{2}\partial m}={c}_{0}-{c}_{1}+\frac{3}{2} {m}^{2}c+2\left({p}_{1}-{p}_{2}\right)+\theta .$$

By defining

$${{\mathrm{G}}_{2}\left(m\right)}^{\mathrm{^{\prime}}}={c}_{0}-{c}_{1}+\frac{3}{2} {m}^{2}c+2\left({p}_{1}-{p}_{2}\right)+\theta ,$$

we get

$${{{G}_{2}}^{\mathrm{^{\prime}}}\left(m\right)}^{\mathrm{^{\prime}}}=3cm>0,$$

$${{\mathrm{G}}_{2}\left(0\right)}^{\mathrm{^{\prime}}}={c}_{0}-{c}_{1}+2\left({p}_{1}-{p}_{2}\right)+\theta ,$$

$$\underset{m\to 1}{\mathrm{lim}}{{\mathrm{G}}_{2}\left(m\right)}^{^{\prime}}={c}_{0}-{c}_{1}+\frac{3}{2}c+2\left({p}_{1}-{p}_{2}\right)+\theta >0.$$

In $m\in [\mathrm{0,1})$, ${{{G}_{2}}^{^{\prime}}\left(m\right)}^{^{\prime}}>0$; thus, ${{\mathrm{G}}_{2}\left(m\right)}^{^{\prime}}$ increases in $m$. Hence, when ${{\mathrm{G}}_{2}\left(0\right)}^{^{\prime}}<0$ (i.e. ${c}_{0}-{c}_{1}+2\left({p}_{1}-{p}_{2}\right)+\theta >0$), we obtain ${m}_{2},$ as ${{G}_{2}\left({m}_{2}\right)}^{^{\prime}}=0$.

Finally, we transfer ${c}_{1}-{c}_{0}-2\left({p}_{1}-{p}_{2}\right)-k>0$ and ${c}_{0}-{c}_{1}+2\left({p}_{1}-{p}_{2}\right)+\theta >0$ to $\left({p}_{1}-{p}_{2}\right)<\frac{{c}_{1}-{c}_{0}-k}{2}$ and $\left({p}_{1}-{p}_{2}\right)<\frac{{c}_{1}-{c}_{0}-\theta }{2}$. $\hfill\square$

Proof of Proposition 5

Based on the Envelope Theorem, we get

$$\frac{\partial {\mathbb{E}}(\Pi ({{p}_{1}}^{*},{{p}_{2}}^{*},{{s}_{1}}^{*},{{s}_{2}}^{*})}{\partial m}={\left.\frac{\partial {\mathbb{E}}(\Pi ({p}_{1},{p}_{2},{s}_{2},{s}_{2})}{\partial m}\right|}_{{p}_{1}={{p}_{1}}^{*},{p}_{2}={{p}_{2}}^{*},{s}_{1}={{s}_{1}}^{*},{s}_{2}={{s}_{2}}^{*}}.$$

Then,

$$\frac{\partial {\mathbb{E}}(\Pi ({{p}_{1}}^{*},{{p}_{2}}^{*},{{s}_{1}}^{*},{{s}_{2}}^{*})}{\partial m}=\left({c}_{1}-{p}_{1}\right)\left(k+{p}_{1}-{p}_{2}\right)+\left({p}_{2}-{c}_{0}-\frac{1}{2} {m}^{2}c\right)\left({p}_{1}-{p}_{2}+\theta \right)$$

$$-cm({a}_{2}-k+m{p}_{1}+{b}_{2}{p}_{2}-m{p}_{2}{u}_{2}+m\theta {|}_{{p}_{1}={{p}_{1}}^{*},{p}_{2}={{p}_{2}}^{*},{s}_{1}={{s}_{1}}^{*},{s}_{2}={{s}_{2}}^{*}}.$$

By defining

$${\mathrm{G}}_{3}\left(m\right)=\left({c}_{1}-{p}_{1}\right)\left(k+{p}_{1}-{p}_{2}\right)+\left({p}_{2}-{c}_{0}-\frac{1}{2} {m}^{2}c\right)\left({p}_{1}-{p}_{2}+\theta \right)-cm({a}_{2}-k+m{p}_{1}+{b}_{2}{p}_{2}-m{p}_{2}{u}_{2}+m\theta ,$$

we have

$${\mathrm{G}}_{3}^{\mathrm{^{\prime}}}\left(m\right)=-c[{a}_{2}+3m\left({p}_{1}-{p}_{2}+\theta \right)-{b}_{2}{p}_{2}+{s}_{2}+{\mu }_{2}-k]<0,$$

$${\mathrm{G}}_{3}\left(0\right)=\left({c}_{1}-{p}_{1}\right)\left({p}_{1}-{p}_{2}+k\right)-({c}_{0}-{p}_{2})\left({p}_{1}-{p}_{2}+\theta \right),$$

$$\underset{m\to 1}{\mathrm{lim}}{\mathrm{G}}_{3}\left(m\right)=\left({p}_{1}-{p}_{2}+k\right)\left({c}_{1}-{p}_{1}\right)+\left({p}_{2}-{c}_{0}-\frac{c}{2}\right)\left({p}_{1}-{p}_{2}+\theta \right)c-\left[{a}_{2}-k+{p}_{1}-{p}_{2}-{b}_{2}{p}_{2}-{s}_{2}+{\mu }_{2}+\theta \right]<0.$$

Since $m\in [\mathrm{0,1})$, ${\mathrm{G}}_{3}^{^{\prime}}\left(m\right)<0$, and ${\mathrm{G}}_{3}\left(m\right)$ decreases in $m$, if ${\mathrm{G}}_{3}\left(0\right)>0$, we get ${m}^{*},$ as $\mathrm{G}\left(m\right)=0$. Observing $\left({p}_{1}-{p}_{2}+k\right)\left({c}_{1}-{p}_{1}\right)-({c}_{0}-{p}_{2})\left({p}_{1}-{p}_{2}+\theta \right)$, we can obtain that ${\mathrm{G}}_{3}\left(0\right)$ strictly greater than 0 when $\left({c}_{1}-{p}_{1}\right)>({c}_{0}-{p}_{2})$, i.e., ${p}_{1}-{p}_{2}>{c}_{1}-{c}_{0}$.

If capacity is low, the optimal solution will be obtained in ${s}_{1}+{s}_{2}+1-{b}_{1}{p}_{1}-{b}_{2}{p}_{2}+(\theta -k)({\tau }_{1}+{\tau }_{2}))=Q$. By same calculations, we also can get that when there exists a ${m}^{*}$ to maximize the profit function when $\left({c}_{1}-{p}_{1}\right)>({c}_{0}-{p}_{2})$, i.e., ${p}_{1}-{p}_{2}>{c}_{1}-{c}_{0}$. $\hfill\square$

Proof of Lemma 3

For given ${{p}_{i}}^{*}({{s}_{i}}^{*} or {{\tau }_{i}}^{*})$, we can get the Hessian matrix ${H}_{1}$, ${H}_{2}$ ${H}_{3}$ respectively are.

$$\left[\begin{array}{cc}-2\left(m+{b}_{1}\right)& 2m\\ 2m& -2\left(m+{b}_{2}\right)\end{array}\right],$$

$$\left[\begin{array}{cc}-\left({p}_{1}+h+{g}_{1}\right){f}_{1}\left({s}_{1}\right)& 0\\ 0& -\left({p}_{2}+h+{g}_{2}\right){f}_{2}\left({s}_{2}\right)\end{array}\right],$$

$$\left[\begin{array}{cc}{N}_{1}& ck\left({\tau }_{1}+{\tau }_{2}\right)\\ ck\left({\tau }_{1}+{\tau }_{2}\right)& {N}_{2}\end{array}\right],$$

where ${N}_{i}=-c({a}_{i}\left({b}_{i}+m\right){p}_{i}-m{p}_{j}-k{\tau }_{j}+{u}_{i}+3\theta {\tau }_{j}+{I}_{i}\left({s}_{i}\right)-{S}_{i}({s}_{i})$, $i\ne j\in \{1, 2\}$. Obviously, ${H}_{1}$, ${H}_{2}$ are negative definite. The first-order determinant of ${H}_{3}$ is ${N}_{1}<0$, and the second-order determinant is ${N}_{1}{N}_{2}-ck{\left({\tau }_{1}+{\tau }_{2}\right)}^{2}>0$. Thus, matrix ${H}_{3}$ is negative definite as well.

Based on the constraint, we use the Lagrange function to change the ${\mathbb{E}}\left(\Pi \left({p}_{1},{p}_{2},{s}_{1},{s}_{2},{\tau }_{1},{\tau }_{2}\right)\right)$, and derive

$${\mathcal{L}}^{\mathrm{^{\prime}}}({p}_{1},{p}_{2},{s}_{1},{s}_{2},{\tau }_{1},{\tau }_{2},\uplambda )={(p}_{1}-{c}_{1})({Q}_{1}+{\mu }_{1})-({c}_{1}+h){I}_{1}({s}_{1})-{(p}_{1}-{c}_{1}+{g}_{1}){S}_{1}({s}_{1})$$

$$+{(p}_{2}-{c}_{2})({Q}_{2}+{\mu }_{2})-\left({c}_{2}+h\right){I}_{2}\left({s}_{2}\right)-{(p}_{2}-{c}_{2}+{g}_{2}){S}_{2}({s}_{2})$$

$$+\uplambda ({ s}_{1}+{s}_{2}+1-{b}_{1}{p}_{1}-{b}_{2}{p}_{2}-Q+(\theta -k)({\tau }_{1}+{\tau }_{2})).$$

Since the conditions of the optimal solution are

$$\frac{\partial {\mathbb{E}}\left(\Pi \left({p}_{1},{p}_{2},{s}_{1},{s}_{2},{\tau }_{1},{\tau }_{2}\right)\right)}{\partial {p}_{i}}=0$$

$$\frac{\partial {\mathbb{E}}\left(\Pi \left({p}_{1},{p}_{2},{s}_{1},{s}_{2},{\tau }_{1},{\tau }_{2}\right)\right)}{\partial {s}_{i}}=0$$

$$\frac{\partial {\mathbb{E}}\left(\Pi \left({p}_{1},{p}_{2},{s}_{1},{s}_{2},{\tau }_{1},{\tau }_{2}\right)\right)}{\partial {\tau }_{i}}=0$$

$$\frac{\partial \left(\Pi \left({p}_{1},{p}_{2},{s}_{1},{s}_{2},{\tau }_{1},{\tau }_{2}\right)\right)}{\partial\uplambda }=0,$$

we have

$$\begin{array}{c}{{p}_{i}}^{*}=\frac{2{a}_{i}+4m{{p}_{j}}^{*}+{b}_{i}\left(c{{{\tau }_{i}}^{*}}^{2}-2\lambda \right)+cm\left({{{\tau }_{i}}^{*}}^{2}-{{{\tau }_{j}}^{*}}^{2}\right)+2\left({u}_{i}-2k{{\tau }_{j}}^{*}+\theta {{\tau }_{j}}^{*}\right)-2{S}_{i}\left({{s}_{i}}^{*}\right)}{4({b}_{i}+m)},\\ \left(2\left({{p}_{i}}^{*}+{g}_{i}\right)-c{{{\tau }_{i}}^{*}}^{2}\right)\left(1-{F}_{i}\left({{s}_{i}}^{*}\right)\right)-\left({{{\tau }_{i}}^{*}}^{2}+2h\right){F}_{i}\left({{s}_{i}}^{*}\right)=\lambda ,\\ \frac{\begin{array}{c}ck{{\tau }_{j}}^{*}\left({{\tau }_{i}}^{*}+2{{\tau }_{j}}^{*}\right)-2{{\tau }_{j}}^{*}-2c{{\tau }_{i}}^{*}\left({a}_{i}-{b}_{i}{{p}_{i}}^{*}+m\left({{p}_{i}}^{*}-{{p}_{j}}^{*}\right)+{u}_{i}\right)+\\ \left(2{p}_{i}-3c{{{\tau }_{i}}^{*}}^{2}\right)\theta +2c{{\tau }_{i}}^{*}({S}_{i}\left({{s}_{i}}^{*}\right)+{I}_{i}\left({{s}_{i}}^{*}\right))\end{array}}{2(\theta -k)}=\lambda ,\\ {{s}_{1}}^{*}+{{s}_{2}}^{*}+1-{b}_{1}{{p}_{1}}^{*}-{b}_{2}{{p}_{2}}^{*}+\left(\theta -k\right)\left({{\tau }_{i}}^{*}+{{\tau }_{j}}^{*}\right)=Q.\end{array}$$

Finally, according to ${H}_{1}$, ${H}_{2}$ ${H}_{3}$ and the characters of ${\mathcal{L}}^{\mathrm{^{\prime}}}({p}_{1},{p}_{2},{s}_{1},{s}_{2},{\tau }_{1},{\tau }_{2},\uplambda )$, we can easily get there exist unique optimal ${{p}_{i}}^{*},{{s}_{i}}^{*}, {{\tau }_{i}}^{*}$ to maximize ${\mathbb{E}}\left(\Pi \left({p}_{1},{p}_{2},{s}_{1},{s}_{2},{\tau }_{1},{\tau }_{2}\right)\right)$ and the logic is similar to Lemma 1. $\hfill\square$