DISCUSSION

Is Giant Eagle's Forced Self-Checkout CRM Tactic Smart?

Written by Guest contributor

Through a special arrangement, presented here for discussion is a summary of a current article from StorefrontBacktalk, a site tracking retail technology, e-commerce and mobile commerce.

Some Giant Eagle stores have started blocking access to self-checkout for anyone other than loyalty cardholders while they are using their cards. It's an interesting CRM move, in that it simultaneously discourages self-checkout usage but also gathers far more information about those who do opt to self-checkout using their cards.

For the last couple of years, retailers have had this intense love-hate relationship with self-checkout. Chains have touted "customer service" as their reason both for yanking the self-checkout units out and for adding more of them. Higher theft rates have been reported in some stores with self-checkouts and not others.

But the idea of requiring frequent shopper cards to self-checkout is a new twist. Chains — and that's triply true for grocery — have always struggled getting shoppers to use their loyalty cards. To be fair, many of those difficulties have been self-inflicted. We've seen many associates offering to scan generic loyalty cards to give the discounts to non-cardholders. Even worse, we've seen some associates stop customers trying to pull out the cards, saying, "That's not necessary. The discounts are being added automatically."

By forcing all self-checkout shoppers to use the cards, Giant Eagle could be testing how shoppers react to the cards when they are mandatory. Will they rebel or passively comply? If they comply, could requiring loyalty cards at other lanes be a possibility? Or is the thinking less severe, with merely the hope that by forcing people to use their cards in some parts of the store, they'll start using the cards in the entire store voluntarily?

In other words, once shoppers get used to the extra discounts, instant recall notices, and other loyalty card perks, will they stick with it always?

Discussion Thread0