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Managing long queues for holiday sales shopping

Abstract

Long queues during holiday shopping events seem undesirable for both shoppers and retailers. However, the following article shows that, under some conditions, long queues benefit retailers for two reasons. First, long queues, by turning away high-time-cost shoppers, serve as a device of segmentation and targeting. Consequently, retailers deliver promotions only to low-time-cost shoppers. High-time-cost shoppers choose to purchase at a regular time (non-holiday shopping event) and pay the full price without having to make the wait. Second, longer queues prompt shoppers who stay in the line buy more products. In addition, the article shows that shoppers tend to wait longer when price discounts are greater. Accounting for the above findings, this article provides a numerical solution to jointly optimizing retailers’ promotional and operational decisions on holiday promotional sales.

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Notes

  1. We are aware that certain shoppers receive positive utility from the time spent on shopping even if they buy nothing. In this article we focus on those shoppers who deem such time-spending as a ‘loss’.

  2. It is possible that the per-unit shopping time might be smaller because of reduced crowdedness in regular-time shopping. Shoppers may self-adjust to spend a longer time purchasing when they become more relaxed. Yet the analysis gains little by adding an extra parameter at the expense of increased complexity.

  3. For high-time-cost shoppers, there may be some utility loss because of the delayed purchase. We assume the utility loss is minimal and do not consider it in the analysis for tractability.

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Authors and Affiliations

Authors

Corresponding author

Correspondence to Chun (Martin) Qiu.

Additional information

1received his PhD in marketing in 2008 from the University of Alberta, and is currently working as an Assistant Professor of Marketing at Wilfrid Laurier University, Canada. He is conducting research on retailing, competitive marketing strategies and mobile marketing.

Appendix

Appendix

Proof of Lemma 1

Differentiate w.r.t. , first-order condition The second-order condition is which is negative when is small, which is the necessary condition for shoppers to conduct holiday shopping.

We thus set d(U S )/dp 1 to 0 to solve for q S * with t 0=1 in the expression.

q R * is solved in a similar fashion by setting d=0 and w=0. □

Proof of Proposition 1

The sign is determined by the sign of the numerator of ∂q S */∂w. We need to show that

Since both sides are positive, we take the square of both sides of the above inequality. Now we need to show

which is satisfied since

 □Proof of Proposition 2

By definition of mean waiting time, we have (λ)/(2μ(μλ))≡w *, where Differentiate both sides w.r.t. to d, we have ((∂λ)/(∂d))/(2(μλ)2)=(∂* w)/(∂d). Note that is a function of both w * and d, that is, . Applying the chain rule, we have (∂λ)/(∂d)=

Substitute ∂λ/∂d back, and collect the items to simplify the expression, and we have

Everything else being equal, a bigger promotion will motivate more shoppers to join the queue, thus , and longer waiting will deter more shoppers from walking away, thus Therefore, both the numerator and the denominator of the right-hand side are positive. Hence, ∂* w/∂d>0.

The proof of ∂* w/∂K is similar. Note that in this case, μ is a function of K.

Since ∂μ/∂K>0, and that ; thus, the numerator of the right-hand side is negative. Also, we have so the denominator of the right-hand side is positive. Thus, ∂* w/∂K<0. □

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Qiu, C., Zhang, W. Managing long queues for holiday sales shopping. J Revenue Pricing Manag 15, 52–65 (2016). https://doi.org/10.1057/rpm.2015.46

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Keywords

  • shopping queues
  • segmentation
  • promotion
  • revenue management
  • retailing