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Game theory applications in production research in the sharing and circular economy era

A local seller's app channel strategy concerning delivery

, &
Pages 220-255 | Received 20 Jul 2018, Accepted 29 Sep 2018, Published online: 23 Oct 2018
 

Abstract

When a local seller launches an app channel, it can either deliver online orders itself or use a third party service. Given this background, an interesting problem is whether and how a local seller's channel strategy will be affected by these delivery options. To investigate this problem, we focus on a local seller facing these two delivery options. For each option, besides setting the online and offline prices, the seller also determines its delivery service coverage. Assuming that consumers are evenly located along an infinite Hotelling line, we propose a joint pricing and delivery distance decision model and derive the local seller's optimal decisions in two subcases, i.e. the seller is a price taker or a price setter. By analyzing and comparing the seller's optimal decisions in these situations, we find that (i) the local seller's channel strategy will be changed dramatically by the delivery option – the seller will abandon the offline channel when it delivers itself; (ii) whether the seller is a price taker or price setter will also influence the seller's channel strategy; and (iii) unexpectedly, the highly developed just-in-time logistics is an important factor that helps the offline channel and app channel coexist in the Internet era.

Disclosure statement

No potential conflict of interest was reported by the authors.

Notes

1 The Robinson-Patman Act of 1936 is a United States federal law that prohibits anti competitive practises by producers, specifically price discrimination.

2 Let pT=k+pT(vpT)tx¯S. Solving this equation with respect to pT, we have the two solutions, i.e. 12(vtx¯S)12(vtx¯S)2+4ktx¯S and 12(vtx¯S)+12(vtx¯S)2+4ktx¯S. We can easily see that the former is negative and the latter is between 0 and v. Since pT>0, we see that the intersection point here must be the latter, i.e. pˆT=12(vtx¯S)+12(vtx¯S)2+4ktx¯S.

3 Letting pA=pT in pT=(pAk)(pAtx¯S)2pAvk and solving the equation with respect to pT, we obtain two solutions, i.e. 12(vtx¯S)12(vtx¯S)2+4ktx¯S and 12(vtx¯S)+12(vtx¯S)2+4ktx¯S. Similar to the previous footnote, examination shows that the former solution is negative and the latter is between 0 and v, meaning the latter is the proper solution.

4 Letting pA=v in pT=(pAk)(pAtx¯S)2pAvk results in pT=vtx¯S. Thus, the intersection is (vtx¯S,v).

5 Substituting pT=0 into pA=pˆA1 and solving the equation with respect to pA, we obtain pA=0.

Additional information

Funding

This work was supported by the National Natural Science Foundation of China (Grant no. 71671170, 71601176) and the Fundamental Research Funds for the Central Universities (Grant no. WK2040160026).

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