The optimal payment policy for a firm: cash sale versus credit sale

Abstract

Both cash and credit sale are common. Considering the upsides and downsides of these two payment policies, in this paper, we study how a firm should choose between them. In particular, based on the relationship between demand and the inventory level, four strategies are considered for the case of credit sale: only selling to high-quality buyers (S1); selling to high-quality buyers when demand is high and selling to both low- and high-quality buyers otherwise (S2); selling to both when the demand is moderate and selling only to high-quality buyers otherwise (S3); and selling to both when demand is low and selling only to high-quality buyers otherwise (S4). What we find is that credit sale strictly dominates cash sale except when the expected profit from selling to low-quality buyers is less than the salvage value and the demand coefficient of variation is extremely high. Further, under credit sale, two counterintuitive observations are obtained if the demand coefficient of variation is small. The first is that the analysis reveals that the firm may not serve the low-quality buyers despite that doing so offers profits, i.e., S3 is the most preferable strategy for the firm. The second is that the firm may also serve the low-quality buyers despite incurring a loss by doing so, i.e., S4 is the most attractive strategy. In addition, we demonstrate that when the credit sale is employed, the firm’s optimal expected profit decreases both in the proportion of high-quality buyers in the market and in the low-quality buyers’ payment probability when the demand coefficient of variation is significantly small, which is quite different from our conjecture. All of these results provide some distinct insights into the question of what the firm’s best payment policy is and how credit should be extended when credit sale is used.

Appendix: proofs

Proof to Theorem 1

In this case, it can be verified that the firm’s objective function is a quadratic function, with the second-order derivative being $\frac{\partial ^2 \Pi _0}{\partial Q^2}=-\frac{p-s}{\mu \Delta }\le 0$. Hence, the optimal solution exists and it is $Q_0^{*}=\frac{\mu \left[ \Delta (p-c)+k(p-s) \right] }{p-s} \ge 0$ and the firm’s expected profit is $\Pi _0^{*}=\frac{\mu \left[ \Delta (p-c)^2+2k(p-s)(p-c)\right] }{2(p-s)}\ge 0$. Further, we have

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial Q_0^{*}}{\partial k}=\mu>0\\ \displaystyle \frac{\partial Q_0^{*}}{\partial \Delta }=\frac{\mu (p-c)}{p-s}>0\\ \displaystyle \frac{\partial Q_0^{*}}{\partial s}=\frac{\mu \Delta (p-c)}{(p-s)^2}>0\\ \displaystyle \frac{\partial Q_0^{*}}{\partial \mu }=\frac{\Delta (p-c)+k(p-s)}{p-s}>0,\\ \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial \Pi _0^{*}}{\partial k}=\frac{2\mu (p-s)(p-c)}{2(p-s)}>0\\ \displaystyle \frac{\partial \Pi _0^{*}}{\partial \Delta }=\frac{\mu (p-c)^2}{2(p-s)}>0\\ \displaystyle \frac{\partial \Pi _0^{*}}{\partial s}=\frac{\Delta \mu (p-c)^2}{2(p-s)}>0\\ \displaystyle \frac{\partial \Pi _0^{*}}{\partial \mu }=\frac{\Delta (p-c)^2+2k(p-s)(p-c)}{2(p-s)}>0.\\ \end{array} \right. \end{aligned}$$

$\square$

Proof to Theorem 2

The firm’s objective function is a quadratic function, with the second-order derivative being $\frac{\partial ^2 \Pi _3}{\partial Q^2}=-\frac{\mu (p\gamma -s)+p(1-\gamma )}{\mu \Delta }\le 0$. Hence, optimal solution exists only when $\mu (p\gamma -s)+p(1-\gamma )\ge 0$. This is always true since if $\mu (p\gamma -s)+p(1-\gamma )<0$, then $s>\frac{(1+\mu \gamma -\gamma )p}{\mu }=p+\frac{(1-\mu )(1-\gamma )p}{\mu }{}_{\phantom {\frac{1}{1}}}$ can be derived, which is impossible. Hence, the optimal solution exists and it is $Q_2^{*}=\frac{\mu \left[ \Delta (p-c)+k(p-s) \right] }{\mu (p\gamma -s)+p(1-\gamma )} \ge 0{}_{\phantom {\frac{1}{1}}}$ and the firm’s expected profit is $\Pi _2^{*}=\frac{\mu \Delta ^2(p-c)^2+2\mu \Delta k(p-s)(p-c)-k^2(1-\mu )^2(p\gamma -s)(1-\gamma )p}{2\Delta \left[ \mu (p\gamma -s)+p(1-\gamma )\right] }$. Further, we have

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial Q_2^{*}}{\partial k}=\frac{\mu (p-s)}{\mu (p\gamma -s)+p(1-\gamma )}>0\\ \displaystyle \frac{\partial Q_2^{*}}{\partial \Delta }=\frac{\mu (p-c)}{\mu (p\gamma -s)+p(1-\gamma )}>0\\ \displaystyle \frac{\partial Q_2^{*}}{\partial s}=\frac{\mu \left[ \mu \Delta (p-c)-kp(1-\mu )(1-\gamma )\right] }{\left[ \mu (p\gamma -s)+p(1-\gamma )\right] ^2}\\ \displaystyle \frac{\partial Q_2^{*}}{\partial \mu }=\frac{p(\gamma -1)\left[ \Delta (c-p)-k(p-s)\right] }{\left[ \mu (p\gamma -s)+p(1-\gamma )\right] ^2}>0\\ \displaystyle \frac{\partial Q_2^{*}}{\partial \gamma }=\frac{\mu (1-\mu )p\left[ \Delta (p-c)+k(p-s)\right] }{\left[ \mu (p\gamma -s)+p(1-\gamma )\right] ^2}>0,\\ \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial \Pi _2^{*}}{\partial k}=\frac{\mu \Delta (p-s)(p-c)-k(1-\mu )^2(p\gamma -s)p(1-\gamma )}{\Delta \left[ \mu (p\gamma -s)+p(1-\gamma )\right] }\\ \displaystyle \frac{\partial \Pi _2^{*}}{\partial \Delta }=\frac{\mu \Delta ^2(p-c)^2+k^2 p(1-\mu )^2(1-\gamma )(p\gamma -s)}{2\Delta ^2\left[ \mu (p\gamma -s)+p(1-\gamma )\right] }\\ \displaystyle \frac{\partial \Pi _2^{*}}{\partial s}=\frac{\left[ k(1-\mu )p(1-\gamma )-\Delta \mu (p-c)\right] ^2}{2\Delta \left[ \mu (p\gamma -s)+p(1-\gamma )\right] ^2}\ge 0\\ \displaystyle \frac{\partial \Pi _2^{*}}{\partial \mu }=\frac{p(1-\gamma )\left[ \Delta (p-c)+k(p-s)\right] ^2+k^2 p(1-\gamma )\left[ p(1-\gamma )+\mu (p\gamma -s)\right] \left[ (1-\mu )(p\gamma -s)-(p-s)\right] }{2\Delta \left[ \mu (p\gamma -s)+p(1-\gamma )\right] ^2}\\ \displaystyle \frac{\partial \Pi _2^{*}}{\partial \gamma }=\frac{\mu \Delta (1-\mu )p(p-c)\left[ \Delta (p-c)+2k(p-s)\right] -k^2(1-\mu )^2p\left[ p^2(1-\gamma )^2-\mu (p\gamma -s)^2\right] }{2\Delta \left[ \mu (p\gamma -s)+p(1-\gamma )\right] ^2}.\\ \end{array} \right. \end{aligned}$$

$\square$

Proof to Theorem 3

The firm’s objective function is a quadratic function, with the second-order derivative being $\frac{\partial ^2 \Pi _3}{\partial Q^2}=\frac{(1-\mu )^2(p\gamma -s)-(p-s)}{\mu \Delta }\le 0$. Hence, the optimal solution exists and it is $Q_3^{*}=\frac{\mu \left[ \Delta (c-p)+k(s-p)\right] }{(p\gamma -s)(1-\mu )^2-(p-s)} \ge 0$ and the firm’s expected profit is

$$\begin{aligned} \Pi _3^{*}=\frac{\mu \left[ \Delta ^2(p-c)^2+2\Delta k(p-c)(p-s)+k^2(p-s)(1-\mu )^2(p\gamma -s)\right] }{2\Delta \left[ (p-s)-(1-\mu )^2(p\gamma -s)\right] }. \end{aligned}$$

Further, we have

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial Q_3^{*}}{\partial k}=-\frac{\mu (p-s)}{(p\gamma -s)(1-\mu )^2-(p-s)}\ge 0\\ \displaystyle \frac{\partial Q_3^{*}}{\partial \Delta }=-\frac{\mu (p-c)}{(p\gamma -s)(1-\mu )^2-(p-s)}\ge 0\\ \displaystyle \frac{\partial Q_3^{*}}{\partial s}=\frac{\mu \left[ kp(\gamma -1)(1-\mu )^2+\Delta \mu ^2(c-p)+2\Delta \mu (p-c)\right] }{\left[ (p\gamma -s)(1-\mu )^2-(p-s)\right] ^2}\\ \displaystyle \frac{\partial Q_3^{*}}{\partial \mu }=-\frac{\left[ \Delta (c-p)+k(s-p)\right] \left[ \mu ^2(p\gamma -s)+p(1-\gamma )\right] }{\left[ (p\gamma -s)(1-\mu )^2-(p-s)\right] ^2}\ge 0\\ \displaystyle \frac{\partial Q_3^{*}}{\partial \gamma }=-\frac{\mu \left[ \Delta (c-p)+k(s-p)\right] p(1-\mu )^2}{\left[ (p\gamma -s)(1-\mu )^2-(p-s)\right] ^2}\ge 0,\\ \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial \Pi _3^{*}}{\partial k}=\frac{\mu (p-s)\left[ \Delta (p-c)+k(1-\mu )^2(p\gamma -s)\right] }{\Delta \left[ p-s-(1-\mu )^2(p\gamma -s)\right] }\\ \displaystyle \frac{\partial \Pi _3^{*}}{\partial \Delta }=\frac{\mu \left[ \Delta ^2(p-c)^2-k^2(p-s)(1-\mu )^2(p\gamma -s)\right] }{2\Delta ^2\left[ p-s-(1-\mu )^2(p\gamma -s)\right] }\\ \displaystyle \frac{\partial \Pi _3^{*}}{\partial s}=\frac{2\mu \Delta k(p-c)(1-\mu )^2p(\gamma -1)+\mu ^2\Delta ^2(2-\mu )(p-c)^2+\mu k^2(1-\mu )^2\left[ (1-\mu )^2(p\gamma -s)^2-(p-s)^2\right] }{2\Delta \left[ p-s-(1-\mu )^2(p\gamma -s)\right] ^2}\\ \displaystyle \frac{\partial \Pi _3^{*}}{\partial \mu }=\frac{\left[ \Delta (p-c)+k(p-s)\right] ^2\left[ p-s-(1-\mu )(p\gamma -s)(1+\mu )\right] -k^2(p-s)\left[ p-s-(1-\mu )^2(p\gamma -s)\right] ^2}{2\Delta \left[ p-s-(1-\mu )^2(p\gamma -s)\right] ^2}\\ \displaystyle \frac{\partial \Pi _3^{*}}{\partial \gamma }=\frac{\mu p(1-\mu )^2\left[ \Delta (p-c)+k(p-s)\right] ^2}{2\Delta \left[ p-s-(1-\mu )^2(p\gamma -s)\right] ^2}\ge 0.\\ \end{array} \right. \end{aligned}$$

$\square$

Proof to Theorem 4

In this case, the firm’s objective function is also a quadratic function, with the second-order derivative being $\frac{\partial ^2 \Pi _4}{\partial Q^2}=\frac{\mu (1-\mu )(p\gamma -s)-(p-s)}{\mu \Delta }\le 0$. Hence, the optimal solution also exists and it is $Q_4^{*}=\frac{\mu \left[ \Delta (p-c)+k(p-s)\right] }{\mu (\mu -1) (p\gamma -s)+p-s}{}_{\phantom {\frac{1}{1}}} \ge 0$ and the firm’s expected profit is $\Pi _4^{*}=\frac{\mu \Delta ^2(p-c)^2+2\mu \Delta k(p-s)(p-c)-k^2(p\gamma -s)(1-\mu )^2\left[ (p-s)+\mu p(1-\gamma )\right] }{2\Delta \left[ p-s-\mu (1-\mu )(p\gamma -s)\right] }$. Further, we have

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial Q_4^{*}}{\partial k}=\frac{\mu (p-s)}{\mu (\mu -1)(p\gamma -s)+p-s}\ge 0\\ \displaystyle \frac{\partial Q_4^{*}}{\partial \Delta }=-\frac{\mu (c-p)}{\mu (\mu -1)(p\gamma -s)+p-s}\ge 0\\ \displaystyle \frac{\partial Q_4^{*}}{\partial s}=-\frac{\mu \left[ k\mu p(1-\mu )(1-\gamma )+\Delta (p-c)\left( \mu -1-\mu ^2\right) \right] }{\left[ \mu (\mu -1)(p\gamma -s)+p-s\right] ^2}\\ \displaystyle \frac{\partial Q_4^{*}}{\partial \mu }=\frac{\left[ \Delta (c-p)+k(s-p)\right] \left[ \mu ^2(p\gamma -s)-(p-s)\right] }{\left[ \mu (\mu -1)(p\gamma -s)+p-s\right] ^2}\ge 0\\ \displaystyle \frac{\partial Q_4^{*}}{\partial \gamma }=\frac{\mu ^2\left[ \Delta (c-p)-k(p-s)\right] (\mu -1)p}{\left[ \mu (\mu -1)(p\gamma -s)+p-s\right] ^2}\ge 0,\\ \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial \Pi _4^{*}}{\partial k}=\frac{\mu \Delta (p-s)(p-c)-k(p\gamma -s)(1-\mu )^2\left[ p-s+\mu p(1-\gamma )\right] }{\Delta \left[ p-s-\mu (1-\mu )(p\gamma -s)\right] }\\ \displaystyle \frac{\partial \Pi _4^{*}}{\partial \Delta }=\frac{\mu \Delta ^2(p-c)^2+k^2(p\gamma -s)(1-\mu )^2\left[ p-s+\mu p(1-\gamma )\right] }{2\Delta ^2\left[ p-s-\mu (1-\mu )(p\gamma -s)\right] }\ge 0\\ \displaystyle \frac{\partial \Pi _4^{*}}{\partial s}=\frac{\mu \left[ \mu \Delta (p-c)-k(1-\mu )p(1-\gamma )\right] ^2+\mu (1-\mu )\Delta ^2(p-c)^2+k^2(1-\mu )^2\left[ (p-s)^2-\mu (1-\mu )(p\gamma -s)^2\right] }{2\Delta \left[ p-s-\mu (1-\mu )(p\gamma -s)\right] ^2}\\ \displaystyle \frac{\partial \Pi _4^{*}}{\partial \mu }=\frac{\left[ \Delta (p-c)+k(p-s)\right] ^2\left[ p-s-\mu ^2(p\gamma -s)\right] -k^2p(1-\gamma )\left[ p-s-\mu (1-\mu )(p\gamma -s)\right] ^2}{2\Delta \left[ p-s-\mu (1-\mu )(p\gamma -s)\right] ^2}\\ \displaystyle \frac{\partial \Pi _4^{*}}{\partial \gamma }=\frac{\mu ^2(1-\mu )p\left[ \Delta (p-c)+k(p-s)\right] ^2-k^2(1-\mu )p\left[ p-s-\mu (1-\mu )(p\gamma -s)\right] ^2}{2\Delta \left[ p-s-\mu (1-\mu )(p\gamma -s)\right] ^2}.\\ \end{array} \right. \end{aligned}$$

$\square$

Proof to Proposition 1

From Theorems 1–4, we can see that the relationship between the expected profit from selling the products to the low-quality buyers and the unit salvage value are quite important.

• If $p\gamma -s>0$, then

  $$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \Pi _{4}^{*}-\Pi _2^{*}=-\frac{\mu (1-\mu )^2(p\gamma -s) \left[ \Delta (p-c)+k(p-s)\right] ^2}{2\Delta \left[ p-s-\mu (1-\mu )(p\gamma -s)\right] \left[ \mu (p\gamma -s)+p(1-\gamma )\right] } \le 0\\ \displaystyle \Pi _{3}^{*}-\Pi _1^{*}=\frac{\mu (1-\mu )^2(p\gamma -s) \left[ \Delta (p-c)+k(p-s)\right] ^2}{2\Delta (p-s)\left[ p-s-(1-\mu )^2(p\gamma -s) \right] } \ge 0.\\ \end{array} \right. \end{aligned}$$

  (EC.4)

  Therefore, only S3 and S2 are possible and $\Pi _{3}-\Pi _2$ matters, which is

  $$\begin{aligned} \Pi _{3}^{*}-\Pi _2^{*}=\frac{(1-\mu )(p\gamma -s)\left\{ -\mu ^2\left[ \Delta (p-c)+k(p-s)\right] ^2+k^2\left[ \mu (p\gamma -s)+p(1-\gamma )\right] \left[ p-s-(1-\mu )^2(p\gamma -s)\right] \right\} }{2\Delta \left[ (p-s)-(1-\mu )^2 (p\gamma -s)\right] \left[ \mu (p\gamma -s)+p(1-\gamma )\right] }. \end{aligned}$$

  (EC.5)

  That is, it depends on the sign of

  $$\begin{aligned} \displaystyle B_1= & {} -\mu ^2\left[ \Delta (p-c)+k(p-s)\right] ^2\nonumber \\&+\,k^2\left[ \mu (p\gamma -s)+p(1-\gamma )\right] \nonumber \\&\times \,\left[ p-s-(1-\mu )^2(p\gamma -s)\right] . \end{aligned}$$

  (EC.6)

  When $B_1\ge 0$, then S3 is the best strategy; otherwise, strategy S2 is the best one. But no matter which one is the best, we always have

  $$\begin{aligned} \displaystyle Q_{3}^{*}-Q_2^{*}= & {} \frac{\mu (1-\mu )(p\gamma -s)\left[ \mu \Delta (p-c)+\mu k(p-s)\right] }{\left[ (p\gamma -s)(1-\mu )^2-(p-s)\right] \left[ \mu (p\gamma -s)+p(1-\gamma )\right] }\nonumber \\\le & {} 0. \end{aligned}$$

  Further, the impacts of the parameters such as k, $\Delta$, s, $\mu$, and $\gamma$ are given as follows:

  $$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial B_1}{\partial k}=2\mu ^2\left[ \Delta (p-c)+k(p-s)\right] (p-s)\\ \quad -2k\left[ \mu (p\gamma -s)+p(1-\gamma )\right] \left[ p-s-(1-\mu )^2(p\gamma -s)\right] \\ \displaystyle \frac{\partial B_1}{\partial \Delta }=2\mu ^2\left[ \Delta (p-c)+k(p-s)\right] (p-c)\ge 0\\ \displaystyle \frac{\partial B_1}{\partial s}=-2\mu ^2\Delta k(p-c)+k^2\left\{ \mu (2-\mu )\right. \\ \quad \left[ p(1-\gamma )+\mu (p\gamma -s)\right] +\mu \left[ (1-2\mu )(p-s)\right. \\ \left. \left. \quad -(1-\mu )^2(p\gamma -s)\right] \right\} \\ \displaystyle \frac{\partial B_1}{\partial \mu }=2\mu \left[ \Delta (p-c)+k(p-s)\right] ^2-k^2(p\gamma -s)\\ \quad \left\{ p-s-(1-\mu )^2(p\gamma -s)+2(1-\mu )\right. \\ \left. \quad \left[ \mu (p\gamma -s)+p(1-\gamma )\right] \right\} .\\ \displaystyle \frac{\partial B_1}{\partial \gamma }=k^2(1-\mu )p\left[ (3\mu -1-2\mu ^2)(p\gamma -s)\right. \\ \left. \quad +p-s+p(1-\mu )(1-\gamma )\right] .\\ \end{array} \right. \end{aligned}$$

• If $p\gamma -s<0$, then

  $$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \Pi _{4}^{*}-\Pi _2^{*}=-\frac{\mu (1-\mu )^2(p\gamma -s)\left[ \Delta (p-c)+k(p-s)\right] ^2}{2\Delta \left[ p-s-\mu (1-\mu )(p\gamma -s)\right] \left[ \mu (p\gamma -s)+p(1-\gamma )\right] } \ge 0\\ \displaystyle \Pi _{3}^{*}-\Pi _1^{*}=\frac{\mu (1-\mu )^2(p\gamma -s)\left[ \Delta (p-c)+k(p-s)\right] ^2}{2\Delta (p-s)\left[ p-s-(1-\mu )^2(p\gamma -s)\right] } \le 0.\\ \end{array} \right. \end{aligned}$$

  (EC.7)

  Therefore, only S1 and S4 are possible and $\Pi _{4}-\Pi _1$ matters, which is

  $$\begin{aligned} \displaystyle \Pi _{4}^{*}-\Pi _1^{*}=\frac{(1-\mu )(p\gamma -s)\left\{ \mu ^2\left[ \Delta (p-c)+k(p-s)\right] ^2-k^2(p-s)\left[ p-s-\mu (1-\mu )(p\gamma -s)\right] \right\} }{2\Delta (p-s)\left[ p-s-\mu (1-\mu )(p\gamma -s)\right] }. \end{aligned}$$

  (EC.8)

  That is, it depends on the sign of

  $$\begin{aligned} \displaystyle B_2= & {} \mu ^2\left[ \Delta (p-c)+k(p-s)\right] ^2\nonumber \\&-k^2(p-s) \left[ p-s-\mu (1-\mu )(p\gamma -s)\right] .\nonumber \\ \end{aligned}$$

  (EC.9)

  When $B_2\ge 0$, then S1 is the best one; otherwise, S1 is the best one. However, no matter which strategy is the best, we always have

  $$\begin{aligned} \displaystyle Q_{4}^{*}-Q_1^{*}= & {} \frac{\mu (1-\mu )(p\gamma -s)\left[ \mu \Delta (p-c)+\mu k(p-s)\right] }{(p-s)\left[ p-s-\mu (1-\mu )(p\gamma -s)\right] }\nonumber \\\le & {} 0. \end{aligned}$$

  (EC.10)

  Further, we have:

  $$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{\partial B_2}{\partial k}=2\mu ^2\Delta (p-c)(p-s)-2k(p-s)(1-\mu )\\ \qquad \,\,\,\, \left[ p-s+\mu p(1-\gamma )\right] \\ \displaystyle \frac{\partial B_2}{\partial \Delta }=2\mu ^2\left[ \Delta (p-c)+k(p-s)\right] (p-c)\ge 0\\ \displaystyle \frac{\partial B_2}{\partial s}=-2\mu ^2\Delta k(p-c)+k^2\left[ (2-\mu -\mu ^2)(p-s)\right. \\ \qquad \qquad \left. -\,\mu (1-\mu )(p\gamma -s)\right] \\ \displaystyle \frac{\partial B_2}{\partial \mu }=2\mu \left[ \Delta (p-c)+k(p-s)\right] ^2\\ \qquad \,\,\,\, -\,k^2(2\mu -1)(p-s)(p\gamma -s)\\ \displaystyle \frac{\partial B_2}{\partial \gamma }=\mu (1-\mu )pk^2(p-s)\ge 0.\\ \end{array} \right. \end{aligned}$$

  (EC.11)

$\square$