Product quantity and supplier format choice under platform information sharing

Article Content

Abstract

Under uncertain demand, this paper examines a situation where a supplier distributes products on a retail platform through reselling or agency selling in the presence of pricing and quantity decisions. Under reselling, the platform can observe the actual demand and set retail prices contingent on demand realization; under agency selling, the supplier can only price responsively if the platform shares demand information, otherwise the supplier needs to make retail prices based on random demand. However, product quantity should be pre-determined regardless of the selling format and information policy. Interestingly, we find that the product quantity may be higher under reselling than under agency selling, and that information sharing does not necessarily increase the quantity in the case of agency selling. We then characterize the commission rate threshold below which the supplier chooses agency selling. Findings suggest that the threshold decreases with demand uncertainty when information is not shared. However, information sharing would cause a non-monotonic interaction between demand uncertainty and format choice. When demand uncertainty is relatively low or high, responsive pricing capability enhances the supplier’s preference for agency selling as demand uncertainty increases; otherwise, the supplier is decreasingly motivated to choose agency selling by increased demand uncertainty. Also notably, a high demand uncertainty might elicit the platform to withhold information. This study extends the existing literature by incorporating the quantity decision and responsive pricing capability. The findings provide insights to assist firms in deciding product quantity and offer valuable guidelines for suppliers’ selling format choice and platforms’ information-sharing strategy.

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Logistics
Lead Optimization
Market Psychology
Market Structure and Economic Design
Procurement
Sales and Distribution

Data Availability

All the data used in this work has been shown in the manuscript.

Notes

See https://rulechannel.tmall.com/?spm=a223k.28145804.0.0.4c3514fexOt5PL&type=detail&ruleId=20004150 &cId=379#/rule/detail?ruleId=20004150 &cId=379 and https://rule.jd.com/rule/ruleDetail.action?ruleId=638209647311982592&type=0, accessed Dec. 19th, 2024.
See https://gs.amazon.cn/sell?ref=as_cn_ags_hnav_sell_start#%23before_your_start, accessed on Dec. 19th, 2024.
Since $c < (1 - u) (1 - r)$ , the maximum of c can be 0.42 based on the values of u and r.
The letter I/N in Fig. 7 indicates the situation where the platform is equally profitable under information sharing and non-sharing.
The production cost needs to satisfy $c < \frac{(1 - r) (5 - r)}{5 + r} \leq 1 - r$ .

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Funding

This work was partly supported by the Startup Foundation for Introducing Talent of NUIST [grant number 2024r047], the Natural Science Foundation of Anhui Province [grant number 2308085QG241], the Scientific Research Foundation for High-level Talents of Anhui University of Science and Technology [grant number 2024yjrc198] and the National Natural Science Foundation of China [grant number 72001030].

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Authors and Affiliations

School of Business, Nanjing University of Information Science & Technology, 219 Ningliu Road, 210044 , Nanjing, China

Yanjun Wang
School of Economics and Management, Anhui University of Science & Technology, 168 Taifeng Road, 232001, Huainan, China

Yiming Fan

Contributions

Yanjun Wang: Conceptualization, Formal analysis, Methodology, Validation, Visualization, Writing – original draft. Yiming Fan: Project administration, Methodology, Validation, Visualization, Writing – original draft.

Corresponding author

Correspondence to Yiming Fan.

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Appendices

Appendix A Thresholds

Definition A1

The threshold ${\tilde{r}}^{N} (u, c)$ is presented as follows:

(i) When $0 \leq c \leq \frac{7 - 2 \sqrt{10}}{3}$ , then ${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + 8 c u + \sqrt{((1 + c)^{2} + 8 c u)^{2} - 16 c^{2}}}{4}$ .

(ii) When $\frac{7 - 2 \sqrt{10}}{3} < c \leq \frac{9 + 4 \sqrt{2}}{49}$ ,

(a) If $u \leq \frac{- 3 + 46 c - 3 c^{2} - 3 \sqrt{1 + 36 c - 74 c^{2} + 36 c^{3} + c^{4}}}{50 c}$ , then

${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + (1 - c) \sqrt{1 + 6 c + c^{2} - 8 c u} - 4 c u}{4 (1 - u)^{2}}$ ;

(b) If $\frac{- 3 + 46 c - 3 c^{2} - 3 \sqrt{1 + 36 c - 74 c^{2} + 36 c^{3} + c^{4}}}{50 c} < u \leq \frac{(1 - c)^{2}}{8 c}$ , then ${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + 8 c u + \sqrt{((1 + c)^{2} + 8 c u)^{2} - 16 c^{2}}}{4}$ ;

(iii) When $\frac{9 + 4 \sqrt{2}}{49} < c$ ,

(a) If $u \leq \frac{\sqrt{2} - 1}{\sqrt{2}} (1 - c)$ , then ${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + (1 - c) \sqrt{1 + 6 c + c^{2} - 8 c u} - 4 c u}{4 (1 - u)^{2}}$ ;

(b) If $\frac{\sqrt{2} - 1}{\sqrt{2}} (1 - c) < u$ , then ${\tilde{r}}^{N} (u, c) = 0$ .

Definition A2

The threshold ${\tilde{r}}^{I} (u, c)$ is presented as follows:

(i) If $u \leq \frac{(1 + c)^{2} - (1 - c) \sqrt{1 + 6 c + c^{2}}}{4 c}$ , then ${\tilde{r}}^{I} (u, c) = 1 - \frac{(1 + c)^{2} + (1 - c) \sqrt{1 + 6 c + c^{2}}}{4}$ .

(ii) If $\frac{(1 + c)^{2} - (1 - c) \sqrt{1 + 6 c + c^{2}}}{4 c} < u \leq \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , then

${\tilde{r}}^{I} (u, c) = 1 - \frac{(1 + c)^{2} + 4 c u + \sqrt{((1 + c)^{2} + 4 c u)^{2} - 32 c^{2} (1 + u^{2})}}{4 (1 + u^{2})}$ .

(iii) If $\frac{\sqrt{2} c + 1}{\sqrt{2} + 1} < u$ , then ${\tilde{r}}^{I} (u, c) = 1 -$ $\frac{(1 + u + 2 c)^{2} + \sqrt{(1 + u + 2 c)^{4} - 128 c^{2} (1 + u^{2})}}{8 (1 + u^{2})}$ .

It can be obtained that $lim_{c \to 0} \frac{(1 + c)^{2} - (1 - c) \sqrt{1 + 6 c + c^{2}}}{4 c} = 0$ by using the L’Hospital’s rule.

Appendix B Proofs

Proof of Lemma 1

Based on the realized random demand $\hat{ϵ}$ , the platform’s revenue function is $π_{P} (p | \hat{ϵ}, Q) = p min {\hat{ϵ} - p, Q}$ . Consider the following two cases to derive the responsive price.

Case 1: If $Q \leq \hat{ϵ} - p$ , the revenue function becomes $π_{P} (p | \hat{ϵ}, Q) = p Q$ . Since $π_{P}$ is increasing in p, the optimal price is $p^{*} = \hat{ϵ} - Q$ .

Case 2: If $Q > \hat{ϵ} - p$ , the revenue function becomes $π_{P} (p | \hat{ϵ}, Q) = p (\hat{ϵ} - p)$ , and the optimal price is $max {\frac{\hat{ϵ}}{2}, \hat{ϵ} - Q}$ . Therefore, $p^{*} = \frac{\hat{ϵ}}{2}$ and $π_{P} (Q | \hat{ϵ}) = π_{P} (p^{*} | \hat{ϵ}, Q) = \frac{{\hat{ϵ}}^{2}}{4}$ when $Q > \frac{\hat{ϵ}}{2}$ ; $p^{*} = \hat{ϵ} - Q$ and $π_{P} (Q | \hat{ϵ}) = π_{P} (p^{*} | \hat{ϵ}, Q) = (\hat{ϵ} - Q) Q$ when $Q \leq \frac{\hat{ϵ}}{2}$ . $◻$

Proof of Lemma 2

Substituting $p^{*}$ and the distribution of random demand $ϵ$ into the platform’s maximizing problem gives

Consider the following three cases to obtain the product quantity.

Case 1: If $Q > \frac{1 + u}{2}$ , the platform’s profit function becomes $π_{P} = - w Q + \frac{1}{2} [\frac{(1 + u)^{2}}{4} + \frac{(1 - u)^{2}}{4}]$ . Since $π_{P}$ is decreasing in Q, it is not optimal that $Q > \frac{1 + u}{2}$ .

Case 2: If $\frac{1 - u}{2} < Q \leq \frac{1 + u}{2}$ , the platform’s profit function becomes $π_{P} = - w Q + \frac{1}{2} [(1 + u - Q) Q + \frac{(1 - u)^{2}}{4}]$ . Therefore, the optimal quantity is $Q = \frac{1 + u - 2 w}{2}$ when $w < u$ , and $Q = \frac{1 - u}{2}$ when $w \geq u$ .

Case 3: If $Q \leq \frac{1 - u}{2}$ , the platform’s profit function becomes $π_{P} = - w Q + (1 - Q) Q$ . Therefore, the optimal quantity $Q = \frac{1 - w}{2}$ is when $w \geq u$ , and $Q = \frac{1 - u}{2}$ when $w < u$ .

Based on the above three cases, it can be concluded that: (i) When $w \geq u$ , we have $Q = \frac{1 - w}{2}$ , $π_{P} = \frac{(1 - w)^{2}}{4}$ and $π_{S} = (w - c) \frac{1 - w}{2}$ ; (ii) When $w < u$ , we have $Q = \frac{1 + u - 2 w}{2}$ , $π_{P} = \frac{(1 + u - 2 w)^{2}}{8} + \frac{(1 - u)^{2}}{8}$ and $π_{S} = (w - c) \frac{1 + u - 2 w}{2}$ .

Then, based on the obtained product quantity, we consider the following two cases to derive the supplier’s wholesale price.

Case 1: $w \geq u$ . In this case, the supplier’s profit is $π_{S} = (w - c) \frac{1 - w}{2}$ . Therefore, $w = \frac{1 + c}{2}$ when $u \leq \frac{1 + c}{2}$ ; $w = u$ when $u > \frac{1 + c}{2}$ .

Case 2: $w < u$ . In this case, the supplier’s profit is $π_{S} = (w - c) \frac{1 + u - 2 w}{2}$ . Therefore, $w = \frac{1 + u + 2 c}{4}$ when $u > \frac{1 + 2 c}{3}$ ; $w = u$ when $u \leq \frac{1 + 2 c}{3}$ .

From the above two cases, it is necessary to compare the supplier’s profits when $w = \frac{1 + c}{2}$ and $w = \frac{1 + u + 2 c}{4}$ . Then, it can be obtained that: (i) When $u \leq \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , we have $w = \frac{1 + c}{2}$ and $π_{S} = \frac{(1 - c)^{2}}{8}$ ; (ii) When $u > \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , we have $w = \frac{1 + u + 2 c}{4}$ and $π_{S} = \frac{(1 + u - 2 c)^{2}}{16}$ . $◻$

Proof of Lemma 3

The supplier’s revenue function based on demand realization is $π_{S} (p | \hat{ϵ}, Q) = p min {\hat{ϵ} - p, Q}$ . Consider the following two cases to derive the price decision.

Case 1: If $Q \leq \hat{ϵ} - p$ , then the revenue function becomes $π_{S} (p | \hat{ϵ}, Q) = p Q$ . Since $π_{S}$ is increasing in p, the optimal price is $p^{*} = \hat{ϵ} - Q$ .

Case 2: If $Q > \hat{ϵ} - p$ , then the revenue function becomes $π_{S} (p | \hat{ϵ}, Q) = p (\hat{ϵ} - p)$ , and the optimal price is $max {\frac{\hat{ϵ}}{2}, \hat{ϵ} - Q}$ . Therefore, $p^{*} = \frac{\hat{ϵ}}{2}$ and $π_{S} (p^{*} | \hat{ϵ}, Q) = \frac{{\hat{ϵ}}^{2}}{4}$ when $Q > \frac{\hat{ϵ}}{2}$ ; $p^{*} = \hat{ϵ} - Q$ and $π_{S} (p^{*} | \hat{ϵ}, Q) = (\hat{ϵ} - Q) Q$ when $Q \leq \frac{\hat{ϵ}}{2}$ .

Based on the expected retail price, we consider the following three cases to derive the supplier’s quantity decision before random demand is realized.

Case 1: If $Q \leq \frac{1 - u}{2}$ , the profit function becomes $π_{S} = (1 - r) (1 - Q) Q - c Q$ . Therefore, $Q = \frac{1}{2} - \frac{c}{2 (1 - r)}$ when $\frac{c}{1 - r} \geq u$ ; $Q = \frac{1 - u}{2}$ when $\frac{c}{1 - r} < u$ .

Case 2: If $\frac{1 - u}{2} < Q \leq \frac{1 + u}{2}$ , the profit function becomes $π_{S} = \frac{1 - r}{2} [(1 + u - Q) Q + \frac{(1 - u)^{2}}{4}] - c Q$ . Therefore, $Q = \frac{1 + u}{2} - \frac{c}{1 - r}$ when $\frac{c}{1 - r} < u$ ; $Q = \frac{1 - u}{2}$ when $\frac{c}{1 - r} \geq u$ .

Case 3: If $Q > \frac{1 + u}{2}$ , the profit function becomes $π_{S} = \frac{1 - r}{2} [\frac{(1 + u)^{2}}{4} + \frac{(1 - u)^{2}}{4}] - c Q$ . Since $π_{S}$ is decreasing in Q, it is not optimal that $Q > \frac{1 - u}{2}$ .

According to the above three cases, when $\frac{c}{1 - r} \geq u$ , the optimal quantity is $Q = \frac{1}{2} - \frac{c}{2 (1 - r)}$ , and when $\frac{c}{1 - r} < u$ , the optimal quantity is $Q = \frac{1 + u}{2} - \frac{c}{1 - r}$ . $◻$

Proof of Lemma 4

The supplier determines product quantity Q and retail price p simultaneously to maximize its profit based on random demand. By substituting the distribution of $ϵ$ , the supplier’s problem is re-expressed as

We argue that the $(\cdot)^{+}$ can be removed. First, p needs to be less than $1 + u$ , otherwise the profit is zero and cannot be optimized. Second, $p \leq \frac{1 + u}{2}$ must be satisfied when p is less than $1 - u$ , otherwise p will not be an optimal solution. Letting $1 - u > \frac{1 + u}{2}$ gives $u \leq 1 / 3$ , then the $(\cdot)^{+}$ operator is not active and can be removed in the following discussion.

For the supplier’s optimization problem, we first derive the retail price p by discussing different quantity levels. Consider the following four cases.

Case 1: $Q > 1 + u - p$ . In this case, the condition is $p > 1 + u - Q$ . The profit function becomes $π_{S} = - c Q + (1 - r) p (1 - p)$ , and it is concave in p, thus $p = \frac{1}{2}$ . To make the condition hold, Q has to be larger than $\frac{1 + 2 u}{2}$ . Therefore, it is optimal that $p = \frac{1}{2}$ and $π_{S} = \frac{1 - r}{4} - c Q$ when $Q > \frac{1 + 2 u}{2}$ .

Case 2: $Q = 1 + u - p$ . In this case, the condition is $p = 1 + u - Q$ . To make the condition hold, it requires

and

Therefore, it is optimal that $p = 1 + u - Q$ and $π_{S} = (1 - r) (1 + u - Q) (Q - u) - c Q$ when $\frac{1 + 3 u}{3} \leq Q \leq \frac{1 + 2 u}{2}$ .

Case 3: $1 - u - p < Q < 1 + u - p$ . In this case, the condition is $1 - u - Q < p < 1 + u - Q$ . The profit function becomes $π_{S} = - c Q + \frac{1 - r}{2} p (Q + 1 - u - p)$ , thus $p = \frac{1 - u + Q}{2}$ . To make the condition hold, it needs that $\frac{1 - u}{3} < Q < \frac{1 + 3 u}{3}$ . Therefore, it is optimal that $p = \frac{1 - u + Q}{2}$ and $π_{S} = \frac{(1 - r) (1 - u + Q)^{2}}{8} - c Q$ when $\frac{1 - u}{3} < Q < \frac{1 + 3 u}{3}$ .

Case 4: $Q \leq 1 - u - p$ . In this case, the condition is $p \leq 1 - u - Q$ . The profit function becomes $π_{S} = - c Q + (1 - r) p Q$ , thus $p = 1 - u - Q$ . To make the condition hold, it requires that

Therefore, it is optimal that $p = 1 - u - Q$ and $π_{S} = (1 - r) (1 - u - Q) Q - c Q$ when $Q \leq \frac{1 - u}{3}$ .

Then, we examine the following four cases to derive the supplier’s quantity decision.

Case 1: If $Q \leq \frac{1 - u}{3}$ , the profit function is $π_{S} = (1 - r) (1 - u - Q) Q - c Q$ . Therefore, if $\frac{c}{1 - r} \geq \frac{1 - u}{3}$ , then $Q = \frac{1 - u}{2} - \frac{c}{2 (1 - r)}$ ; if $\frac{c}{1 - r} < \frac{1 - u}{3}$ , then $Q = \frac{1 - u}{3}$ .

Case 2: If $\frac{1 - u}{3} < Q < \frac{1 + 3 u}{3}$ , the profit function is $π_{S} = \frac{(1 - r) (1 - u + Q)^{2}}{8} - c Q$ . Therefore, if $\frac{c}{1 - r} \geq \frac{2 - u}{6}$ , then $Q = \frac{1 - u}{3}$ ; if $\frac{c}{1 - r} < \frac{2 - u}{6}$ , then $Q = \frac{1 + 3 u}{3}$ .

Case 3: If $\frac{1 + 3 u}{3} \leq Q \leq \frac{1 + 2 u}{2}$ , the profit function is $π_{S} = (1 - r) (1 + u - Q) (Q - u) - c Q$ . Therefore, if $\frac{c}{1 - r} \leq \frac{1}{3}$ , then $Q = \frac{1 + 2 u}{2} - \frac{c}{2 (1 - r)}$ ; if $\frac{c}{1 - r} > \frac{1}{3}$ , then $Q = \frac{1 + 3 u}{3}$ .

Case 4: If $Q > \frac{1 + 2 u}{2}$ , the profit function is $π_{S} = \frac{1 - r}{4} - c Q$ and it is decreasing in Q, then the optimal result satisfies $Q \leq \frac{1 + 2 u}{2}$ .

From the above four cases, $π_{S}$ may be bimodal with respect to Q. Thus, it is necessary to compare the supplier’s profits when $Q = \frac{1 - u}{2} - \frac{c}{2 (1 - r)}$ and $Q = \frac{1 + 2 u}{2} - \frac{c}{2 (1 - r)}$ . Then, it can be obtained that: (i) When $\frac{c}{1 - r} \geq \frac{2 - u}{6}$ , we have $Q = \frac{1 - u}{2} - \frac{c}{2 (1 - r)}$ , $π_{P} = \frac{r [(1 - r) (1 - u) - c] [(1 - r) (1 - u) + c]}{4 (1 - r)^{2}}$ and $π_{S} = \frac{[(1 - r) (1 - u) - c]^{2}}{4 (1 - r)}$ ; (ii) When $\frac{c}{1 - r} < \frac{2 - u}{6}$ , we have $Q = \frac{1 + 2 u}{2} - \frac{c}{2 (1 - r)}$ , $π_{P} = \frac{r (1 - r - c) (1 - r + c)}{4 (1 - r)^{2}}$ and $π_{S} = \frac{(1 - r - c)^{2}}{4 (1 - r)} - c u$ . $◻$

Proof of Proposition 1

It shows that $Q^{A I}$ and $Q^{A N}$ are decreasing in r.

(i) If $u \leq \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , solving $Q^{A N} > Q^{R}$ with respect to r yields $r < 1 - min {\frac{2 c}{1 + c - 2 u}, \frac{6 c}{2 - u}}$ ; If $u > \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , solving $Q^{A N} > Q^{R}$ with respect to r yields $r < 1 - min {\frac{2 c}{1 + 2 c - 3 u}, \frac{6 c}{2 - u}}$ .

(ii) If $u \leq \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , solving $Q^{A I} > Q^{R}$ with respect to r yields $r < 1 - \frac{2 c}{1 + c}$ ; If $u > \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , solving $Q^{A I} > Q^{R}$ with respect to r yields $r < 1 - \frac{4 c}{1 + 2 c + u}$ .

(iii) Solving $Q^{A N} > Q^{A I}$ with respect to r yields $r < 1 - \frac{6 c}{2 - u} .$ $◻$

Proof of Proposition 2

Let $\bar{r} = 1 - r$ . The supplier’s profit under Scenario AN is

It can be seen that $π_{S}^{A N}$ is increasing in $\bar{r}$ . When $u \leq 1 / 3$ , the supplier’s profit under Scenario R is $π_{S}^{R} = \frac{(1 - c)^{2}}{8}$ , and it can be seen that $π_{S}^{R}$ remains unchanged with respect to $\bar{r}$ . By comparing several boundary points, the following results can be obtained:

(1) When $c < 5 / 18$ , then $0 < 1 / 3 < 2 - 6 c$ . Therefore, for $u \leq 1 / 3$ , we have $\frac{c}{1 - u} < \frac{6 c}{2 - u} < 1$ .

(2) When $5 / 18 \leq c < 1 / 3$ , then $0 < 2 - 6 c \leq 1 / 3$ . Therefore, for $u \leq 2 - 6 c$ , we have $\frac{c}{1 - u} < \frac{6 c}{2 - u} < 1$ , for $2 - 6 c < u \leq 1 / 3$ , we have $\frac{c}{1 - u} < 1 < \frac{6 c}{2 - u}$ .

(3) When $c \geq 1 / 3$ , then $2 - 6 c \leq 0 < 1 / 3$ . Therefore, for $u \leq 1 / 3$ , we have $\frac{c}{1 - u} < 1 < \frac{6 c}{2 - u}$ .

Consider the following cases to compare $π_{S}^{A N}$ and $π_{S}^{R}$ .

Case 1: When $\frac{c}{1 - u} < 1 < \frac{6 c}{2 - u}$ , for $\frac{c}{1 - u} < \bar{r} < 1$ , we have $π_{S}^{A N} = \frac{(\bar{r} (1 - u) - c)^{2}}{4 \bar{r}}$ and $π_{S}^{R} = \frac{(1 - c)^{2}}{8}$ .

(1.1) When $u \leq \frac{\sqrt{2} - 1}{\sqrt{2}} (1 - c)$ , let $\frac{(\bar{r} (1 - u) - c)^{2}}{4 \bar{r}} = \frac{(1 - c)^{2}}{8}$ , then it can be obtained that ${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + (1 - c) \sqrt{1 + 6 c + c^{2} - 8 c u} - 4 c u}{4 (1 - u)^{2}}$ .

(1.2) When $u > \frac{\sqrt{2} - 1}{\sqrt{2}} (1 - c)$ , it can be obtained that $\frac{(\bar{r} (1 - u) - c)^{2}}{4 \bar{r}} < \frac{(1 - c)^{2}}{8}$ and ${\tilde{r}}^{N} (u, c) = 0$ .

Case 2: When $\frac{c}{1 - u} < \frac{6 c}{2 - u} < 1$ , for $\frac{c}{1 - u} < \bar{r} < \frac{6 c}{2 - u}$ , we have $π_{S}^{A N} = \frac{(\bar{r} (1 - u) - c)^{2}}{4 \bar{r}}$ and $π_{S}^{R} = \frac{(1 - c)^{2}}{8}$ ; for $\frac{6 c}{2 - u} < \bar{r} < 1$ , we have $π_{S}^{A N} = \frac{(\bar{r} - c)^{2}}{4 \bar{r}} - c u$ and $π_{S}^{R} = \frac{(1 - c)^{2}}{8}$ .

(2.1) When $u \leq \frac{- 3 + 46 c - 3 c^{2} - 3 \sqrt{1 + 36 c - 74 c^{2} + 36 c^{3} + c^{4}}}{50 c}$ , let $\frac{(\bar{r} (1 - u) - c)^{2}}{4 \bar{r}} = \frac{(1 - c)^{2}}{8}$ , then it can be obtained that ${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + (1 - c) \sqrt{1 + 6 c + c^{2} - 8 c u} - 4 c u}{4 (1 - u)^{2}}$ .

(2.2) When $\frac{- 3 + 46 c - 3 c^{2} - 3 \sqrt{1 + 36 c - 74 c^{2} + 36 c^{3} + c^{4}}}{50 c} < u \leq \frac{(1 - c)^{2}}{8 c}$ , let $\frac{(\bar{r} - c)^{2}}{4 \bar{r}} - c u = \frac{(1 - c)^{2}}{8}$ , then it can be obtained that ${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + 8 c u}{4} - \frac{\sqrt{((1 + c)^{2} + 8 c u)^{2} - 16 c^{2}}}{4}$ .

(2.3) When $u > \frac{(1 - c)^{2}}{8 c}$ , it can be obtained that $\frac{(\bar{r} - c)^{2}}{4 \bar{r}} - c u < \frac{(1 - c)^{2}}{8}$ and ${\tilde{r}}^{N} (u, c) = 0$ .

Let $\hat{u} = \frac{- 3 + 46 c - 3 c^{2} - 3 \sqrt{1 + 36 c - 74 c^{2} + 36 c^{3} + c^{4}}}{50 c}$ , then we can obtain the results as follows:

(a) When $0 \leq c \leq \frac{7 - 2 \sqrt{10}}{3}$ , then $\hat{u} < 0 < 1 / 3 < \frac{(1 - c)^{2}}{8 c}$ , thus for $u \leq 1 / 3$ , the result in (2.2) can be obtained.

(b) When $\frac{7 - 2 \sqrt{10}}{3} < c \leq \frac{9 + 4 \sqrt{2}}{49}$ , then $0 < \hat{u} < \frac{\sqrt{2} - 1}{\sqrt{2}} (1 - c) < \frac{(1 - c)^{2}}{8 c} < min {2 - 6 c, 1 / 3}$ , the results in (2.1), (2.2) and (2.3) can be obtained.

(c) When $\frac{9 + 4 \sqrt{2}}{49} < c < 1 / 3$ , then $min {1 / 3, \hat{u}} > \frac{\sqrt{2} - 1}{\sqrt{2}} (1 - c) > \frac{(1 - c)^{2}}{8 c} > 2 - 6 c > 0$ . If $u \leq 2 - 6 c$ , then the results in (1.1) can be obtained; If $u > 2 - 6 c$ , then the results in (1.1) and (1.2) can be obtained. When $c \geq 1 / 3$ , then $min {1 / 3, \hat{u}} > \frac{\sqrt{2} - 1}{\sqrt{2}} (1 - c) > \frac{(1 - c)^{2}}{8 c} > 0 > 2 - 6 c$ , the results in (1.1) and (1.2) can be obtained.

Next, we take derivatives of the threshold with respect to u and c.

(i) When ${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + 8 c u + \sqrt{((1 + c)^{2} + 8 c u)^{2} - 16 c^{2}}}{4}$ , we have

$\frac{\partial {\tilde{r}}^{N} (u, c)}{\partial c} = \frac{1}{4} (\frac{32 c - 4 (c + 4 u + 1) (c^{2} + c (8 u + 2) + 1)}{2 \sqrt{(8 c u + (c + 1)^{2})^{2} - 16 c^{2}}} - 2 c - 8 u - 2) < 0$ , and

$\frac{\partial {\tilde{r}}^{N} (u, c)}{\partial u} = \frac{1}{4} (- \frac{8 c (8 c u + (c + 1)^{2})}{\sqrt{(8 c u + (c + 1)^{2})^{2} - 16 c^{2}}} - 8 c) < 0$ .

(ii) When ${\tilde{r}}^{N} (u, c) = 1 - \frac{(1 + c)^{2} + (1 - c) \sqrt{1 + 6 c + c^{2} - 8 c u} - 4 c u}{4 (1 - u)^{2}}$ , we have

$\frac{\partial {\tilde{r}}^{N} (u, c)}{\partial c} = - \frac{\frac{(1 - c) (c - 4 u + 3)}{\sqrt{c^{2} + c (6 - 8 u) + 1}} - \sqrt{c (c - 8 u + 6) + 1} + 2 c - 4 u + 2}{4 (1 - u)^{2}} < 0$ , and

$\frac{\partial {\tilde{r}}^{N} (u, c)}{\partial u} = - \frac{c^{2} - 4 c u - (c - 1) \sqrt{c (c - 8 u + 6) + 1} + 2 c + 1}{2 (1 - u)^{3}} -$ $\frac{\frac{4 (c - 1) c}{\sqrt{c (c - 8 u + 6) + 1}} - 4 c}{4 (1 - u)^{2}} < 0$ . $◻$

Proof of Proposition 3

Let $\bar{r} = 1 - r$ . The supplier’s profit under Scenario AI is

It can be seen that $π_{S}^{A I}$ is increasing in $\bar{r}$ .

The supplier’s profit under Scenario R is

It can be seen that $π_{S}^{R}$ remains unchanged with respect to $\bar{r}$ .

Consider the following cases to compare the supplier’s profits between these two scenarios.

Case 1: When $u \leq c$ , for $c < \bar{r} \leq 1$ , we have $π_{S}^{A I} = \frac{(\bar{r} - c)^{2}}{4 \bar{r}}$ and $π_{S}^{R} = \frac{(1 - c)^{2}}{8}$ . Let $\frac{(\bar{r} - c)^{2}}{4 \bar{r}} = \frac{(1 - c)^{2}}{8}$ , then it can be obtained that ${\tilde{r}}^{I} (u, c) = 1 - \frac{(1 + c)^{2} + (1 - c) \sqrt{1 + 6 c + c^{2}}}{4}$ .

Case 2: When $c < u \leq \frac{(1 + c)^{2} - (1 - c) \sqrt{1 + 6 c + c^{2}}}{4 c}$ , for $c < \bar{r} \leq \frac{c}{u}$ , we have $π_{S}^{A I} = \frac{(\bar{r} - c)^{2}}{4 \bar{r}}$ and $π_{S}^{R} = \frac{(1 - c)^{2}}{8}$ . Let $\frac{(\bar{r} - c)^{2}}{4 \bar{r}} = \frac{(1 - c)^{2}}{8}$ , then it can be obtained that ${\tilde{r}}^{I} (u, c) = 1 - \frac{(1 + c)^{2} + (1 - c) \sqrt{1 + 6 c + c^{2}}}{4}$ .

Case 3: When $\frac{(1 + c)^{2} - (1 - c) \sqrt{1 + 6 c + c^{2}}}{4 c} < u \leq \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , for $\frac{c}{u} < \bar{r}$ , we have $π_{S}^{A I} = \frac{(\bar{r} (1 + u) - 2 c)^{2}}{8 \bar{r}} + \frac{\bar{r} (1 - u)^{2}}{8}$ and $π_{S}^{R} = \frac{(1 - c)^{2}}{8}$ . Let $\frac{(\bar{r} (1 + u) - 2 c)^{2}}{8 \bar{r}} + \frac{\bar{r} (1 - u)^{2}}{8} = \frac{(1 - c)^{2}}{8}$ , then it can be obtained that ${\tilde{r}}^{I} (u, c) = 1 - \frac{(1 + c)^{2} + 4 c u + \sqrt{((1 + c)^{2} + 4 c u)^{2} - 32 c^{2} (1 + u^{2})}}{4 (1 + u^{2})}$ .

Case 4: When $\frac{\sqrt{2} c + 1}{\sqrt{2} + 1} < u$ , for $\frac{c}{u} < \bar{r}$ , we have $π_{S}^{A I} = \frac{(\bar{r} (1 + u) - 2 c)^{2}}{8 \bar{r}} + \frac{\bar{r} (1 - u)^{2}}{8}$ and $π_{S}^{R} = \frac{(1 + u - 2 c)^{2}}{16}$ . Let $\frac{(\bar{r} (1 + u) - 2 c)^{2}}{8 \bar{r}} + \frac{\bar{r} (1 - u)^{2}}{8} = \frac{(1 + u - 2 c)^{2}}{16}$ , then it can be obtained that ${\tilde{r}}^{I} (u, c) = 1 - \frac{(1 + u + 2 c)^{2} + \sqrt{(1 + u + 2 c)^{4} - 128 c^{2} (1 + u^{2})}}{8 (1 + u^{2})}$ .

Next, we take derivatives of the threshold with respect to u and c.

(i) When $u \leq \frac{(1 + c)^{2} - (1 - c) \sqrt{1 + 6 c + c^{2}}}{4 c}$ , we have $\frac{\partial {\tilde{r}}^{I} (u, c)}{\partial c} = \frac{1}{4} (- 2 - 2 c - \frac{(1 - c) (3 + c)}{\sqrt{1 + 6 c + c^{2}}} + \sqrt{1 + 6 c + c^{2}}) < 0$ and ${\tilde{r}}^{I} (u, c)$ is independent of u.

(ii) When $\frac{(1 + c)^{2} - (1 - c) \sqrt{1 + 6 c + c^{2}}}{4 c} < u \leq \frac{\sqrt{2} c + 1}{\sqrt{2} + 1}$ , we have $\frac{\partial {\tilde{r}}^{I} (u, c)}{\partial c} =$ $- \frac{1}{2 (1 + u^{2})} (1 + c + 2 u + \frac{1 + c^{3} + 2 u + c^{2} (3 + 6 u) + c (- 13 + 8 u - 8 u^{2})}{\sqrt{- 32 c^{2} (1 + u^{2}) + (1 + c^{2} + c (2 + 4 u))^{2}}}) < 0$ , and

(iii) When $\frac{\sqrt{2} c + 1}{\sqrt{2} + 1} < u$ , we have $\frac{\partial {\tilde{r}}^{I} (u, c)}{\partial c} = - \frac{1}{2 (1 + u^{2})} (1 + 2 c + u + \frac{(1 + 2 c + u)^{3} - 32 c (1 + u^{2})}{\sqrt{(1 + 2 c + u)^{4} - 128 c^{2} (1 + u^{2})}}) < 0$ , and

We can obtain that $\frac{\partial {\tilde{r}}^{I} (u, c)}{\partial u} < 0$ when u is small and $\frac{\partial {\tilde{r}}^{I} (u, c)}{\partial u} > 0$ when u is large. $◻$

Proof of Proposition 4

Note that $π_{S}^{A I} > π_{S}^{A N}$ . When $r = {\tilde{r}}^{I} (u, c)$ , we have $π_{S}^{A I} = π_{S}^{R} > π_{S}^{A N}$ . Since $π_{S}^{A N}$ is decreasing in r, then $π_{S}^{R} = π_{S}^{A N}$ if and only if $r < {\tilde{r}}^{I} (u, c)$ . Therefore, ${\tilde{r}}^{N} (u, c) < {\tilde{r}}^{I} (u, c)$ . $◻$

Proof of Proposition 5

When ${\tilde{r}}^{N} (u, c) < r < {\tilde{r}}^{I} (u, c)$ , the platform’s profit with information sharing is

and without information sharing is $π_{P}^{N} = π_{P}^{R} = \frac{(1 - c)^{2}}{16}$ .

If $c > u$ , letting $\frac{(1 - c)^{2}}{16} = \frac{r (1 - r + c) (1 - r - c)}{4 (1 - r)^{2}}$ gets $r_{1}$ . In this case, if ${\tilde{r}}^{N} (u, c) < r_{1} < {\tilde{r}}^{I} (u, c)$ , then the platform shares information when $r > r_{1}$ and does not share when $r < r_{1}$ ; otherwise, the platform provides information in the range of $r \in ({\tilde{r}}^{N} (u, c), {\tilde{r}}^{I} (u, c))$ .

If $c \leq u$ , letting $\frac{(1 - c)^{2}}{16} = \frac{r ((1 - r) (1 + u) + 2 c) ((1 - r) (1 + u) - 2 c)}{8 (1 - r)^{2}}$ $+ \frac{r (1 - u)^{2}}{8}$ gets $r_{2}$ . In this case, if ${\tilde{r}}^{N} (u, c) < r_{2} \leq 1 - \frac{c}{u}$ , then the platform shares information when $r > r_{2}$ and does not share when $r < r_{2}$ ; if $r_{2} > 1 - \frac{c}{u}$ and $1 - \frac{c}{u} < r_{1} < {\tilde{r}}^{I} (u, c)$ , then the platform discloses information when $r > r_{1}$ and conceal information when $r < r_{1}$ ; otherwise, the platform shares information in the range of $r \in ({\tilde{r}}^{N} (u, c), {\tilde{r}}^{I} (u, c))$ .

Finally, we show the value of ${\tilde{r}}^{o} (u, c) \in {r_{1}, r_{2}}$ . It can be obtained that $r_{1} = \frac{1}{12} (c^{2} - A + (- c^{4} + 4 c^{3} - 46 c^{2} - 12 c - 9) / A - 2 c + 9)$ , where $A = (- c^{6} + 6 c^{5} - 75 c^{4} + 116 c^{3} - 639 c^{2} + 24 \sqrt{3} \sqrt{- (c - 1)^{2} c^{2} (2 c^{3} - 9 c^{2} + 108 c + 27)} + 54 c + 27)^{1 / 3}$ , and $r_{2} = \frac{1}{12 (u^{2} + 1)} (c^{2} - B + (- c^{4} + 4 c^{3} - 2 c^{2} (44 u^{2} + 47) - 4 c (4 u^{2} + 3) - (4 u^{2} + 3)^{2}) / B - 2 c + 8 u^{2} + 9)$ , where $B = (- c^{6} + 6 c^{5} - 3 c^{4} (44 u^{2} + 49) + 20 c^{3} (12 u^{2} + 13) - 3 c^{2} (400 u^{4} + 824 u^{2} + 429) + 24 \sqrt{6} (- c^{2} (u^{2} + 1)^{2} (c^{6} - 2 c^{5} + 3 c^{4} (36 u^{2} + 35) + 4 c^{3} (68 u^{2} + 71) - c^{2} (304 u^{4} + 696 u^{2} + 405) + 6 c (4 u^{2} + 3)^{2} + (4 u^{2} + 3)^{3}))^{1 / 2} + 6 c (4 u^{2} + 3)^{2} + 64 u^{6} + 144 u^{4} + 108 u^{2} + 27)^{1 / 3}$ . $◻$

Proof of Lemma 5

Based on demand realization, the problem is $max_{p \geq 0} p min {1 + \hat{x} - p, Q}$ . Case 1: If $Q \leq 1 + \hat{x} - p$ , the optimal price is $p^{*} = 1 + \hat{x} - Q$ . Case 2: If $Q > 1 + \hat{x} - p$ , the optimal price is $max {\frac{1 + \hat{x}}{2}, 1 + \hat{x} - Q}$ . $◻$

Proof of Lemma 6

When $2 Q - 1 \geq u$ , we have that $π_{P} (Q)$ is decreasing in Q, thus $2 Q - 1 \geq u$ is not optimal. Taking derivatives of $π_{P} (Q)$ with respect to Q when $2 Q - 1 < u$ gives us

Thus, the optimal order quantity $Q^{R}$ is obtained by the first-order condition that satisfies

We work with $w (Q^{R})$ to solve the supplier’s problem, which is now reduced to

Taking derivatives of $π_{S}$ with respect to Q, we have

The first order conditions give us that $Q^{R} = Q^{*}$ or $Q^{R} = \frac{1 - c}{4}$ that achieves higher profit, where $Q^{*}$ is the solution of equation $(1 - 4 Q) \bar{F} (2 Q - 1) + \int_{2 Q - 1}^{u} x f (x) d x = c$ . $◻$

Proof of Lemma 7

Following a similar process of proving Lemma 6, we can get the outcomes in Lemma 7. $◻$

Proof of Lemma 8

Let $Q = 1 + z - p$ , then the problem becomes

We use the method introduced by Petruzzi and Dada (1999). We first derive p as a function of z. Taking derivatives of $π_{S} (z, p)$ with respect to p gives us

Then, the first order conditions give the optimal price $p^{*}$

Substituting $p^{*}$ into $π_{S} (z, p)$ , the optimization problem becomes a maximization over the single variable z. From the chain rule and taking derivatives of $π_{S} (z, p (z))$ with respect to z gives us

If $z \geq u$ , then $z^{*} = u$ ; If $z \leq - u$ , then $z^{*} = - u$ ; If $- u < z < u$ , searching over all values of z in the region $[- u, u]$ will determine $z^{*}$ that satisfies the first order condition $d π_{S} (z, p (z)) / d z = 0$ . $◻$

Proof of Proposition 6

It can be demonstrated that $Q^{A I} (r)$ is decreasing in r. For $r = 0$ , we have that $Q^{A I} (0)$ satisfies the equation $(1 - 2 Q) \bar{F} (2 Q - 1) + \int_{2 Q - 1}^{u} x f (x) d x = c$ . Since $Q^{R}$ satisfies the equation $(1 - 4 Q) \bar{F} (2 Q - 1) + \int_{2 Q - 1}^{u} x f (x) d x = c$ , it follows $Q^{A I} (0) > Q^{R}$ . By contrast, for $r = 1$ , we can numerically demonstrate that $Q^{A I} (1) < Q^{R}$ holds. Therefore, the equation $Q^{A I} (r) = Q^{R}$ has a unique solution $\hat{r}$ , and when $r < \hat{r}$ , we have $Q^{A I} (r) > Q^{R}$ . $◻$

Proof of Lemma 9

Substituting $p^{*}$ into the expected profit functions, and taking derivatives of $π_{S}$ with respect to $Q_{S}$ and $π_{P}$ with respect to $Q_{P}$ , respectively, gives

Solving the first-order conditions yields

Substituting $Q_{S}^{*} (w)$ and $Q_{P}^{*} (w)$ into the expected profit function $π_{S}$ and taking derivatives with respect to w gives us

Solving the first-order condition yields

By substituting, we get the optimal results in Lemma 9. $◻$

Proof of Proposition 7

Taking derivative of $π_{S}^{H}$ with respect to r give us

It is apparent that $\frac{d^{2} π_{S}^{H}}{d r^{2}} > 0$ , which means that $π_{S}^{H}$ is convex in r. First, we have $π_{S}^{H} > π_{S}^{A I} = \frac{(1 - r - c)^{2}}{4 (1 - r)}$ . Second, for $r = 0$ , we have $π_{S}^{H} (r) = \frac{(1 - c)^{2}}{4} > \frac{(1 - c)^{2}}{8} = π_{S}^{R}$ ; for $r = \frac{6 + c - \sqrt{c^{2} + 32 c + 16}}{2}$ , we have $π_{S}^{H} (r) = \frac{8 c (c^{3} - (\sqrt{c^{2} + 32 c + 16} - 24) c^{2} + (53 - 8 \sqrt{c^{2} + 32 c + 16}) c - 5 (\sqrt{c^{2} + 32 c + 16} - 4))}{(c + 4 - \sqrt{c^{2} + 32 c + 16})^{2} (4 - c + \sqrt{c^{2} + 32 c + 16})}$ $< π_{S}^{R}$ . Solving $π_{S}^{H} > π_{S}^{R}$ with respect to r yields $r < {\tilde{r}}_{1}$ , where $π_{S}^{H} ({\tilde{r}}_{1}) = π_{S}^{R}$ . $◻$

Proof of Proposition 8

It can be demonstrated numerically that $π_{P}^{A I} (r)$ is increasing in $r \in (0, {\tilde{r}}^{I} (u, c))$ . Also, we can know that $π_{P}^{A N} (r)$ is roughly increasing in $r \in (0, {\tilde{r}}^{N} (u, c))$ , except for a slight jump at $r = 1 - \frac{6 c}{2 - u}$ .

We first derive the platform’s commission rates for the information sharing and non-sharing cases, respectively. With information sharing, it can be obtained that $π_{P}^{A I} ({\tilde{r}}^{I}) > π_{P}^{R}$ , therefore the platform sets the optimal commission rate at ${\tilde{r}}^{I}$ . However, without information sharing, it can be obtained that $π_{P}^{A N} ({\tilde{r}}^{N}) > π_{P}^{R}$ when c is small, and $π_{P}^{A N} ({\tilde{r}}^{N}) < π_{P}^{R}$ when c is high. Therefore, the platform determines the optimal commission rate at ${\tilde{r}}^{N}$ when c is small; otherwise, when c is large, the commission rate is set too high for the supplier, which makes it opt for the reselling format.

Then, the platform’s profits are compared to derive the equilibrium. When c is small, we compare $π_{P}^{A I} ({\tilde{r}}^{I})$ and $π_{P}^{A N} ({\tilde{r}}^{N})$ . Since $π_{P}^{A I} (r) > π_{P}^{A N} (r)$ and ${\tilde{r}}^{I} > {\tilde{r}}^{N}$ , it follows that $π_{P}^{A I} ({\tilde{r}}^{I}) > π_{P}^{A N} ({\tilde{r}}^{N})$ . When c is large, we compare $π_{P}^{A I} ({\tilde{r}}^{I})$ and $π_{P}^{R}$ , which yields $π_{P}^{A I} ({\tilde{r}}^{I}) > π_{P}^{R}$ .

Therefore, the platform chooses to offer information and determines the optimal commission rate ${\tilde{r}}^{I}$ , and the supplier adopts the agency selling format eventually.

About this article

Cite this article

Wang, Y., Fan, Y. Product quantity and supplier format choice under platform information sharing. Oper Manag Res (2025). https://doi.org/10.1007/s12063-025-00555-y

Received 07 April 2024
Revised 13 May 2025
Accepted 06 June 2025
Published 10 July 2025
DOI https://doi.org/10.1007/s12063-025-00555-y

Keywords

Format choice
Information sharing
Product quantity
Pricing
Demand uncertainty