My first thought is that you buy $40 items much more often than $2500 items; saving $20 on each $40 dollar purchase will save more money than the same savings on $2500 purchases. Especially given costs of driving around town and negotiating in high value purchases. I’ll agree that if you include these hidden costs and convert to a utility calculation people are probably not perfectly efficient, but doing the calculations themselves can be expensive.
In the example, the costs of making the savings are the same. Saving $20 on each $40 purchase will save you more money, but saving $20 on every purchase will save more still. If it’s worth going back to the store to save $20, it’s worth doing no matter how many times you do it.
Sure; on the other hand, I have one type of algorithm for dealing with $40 purchases; I do less than a day’s worth of research and think about whether I really want it or not. I make $20-$40 purchases at least once a week, maybe more. Changing my algorithm to save money on them will clearly give me a great increase in utility.
I have a separate algorithm for judging $2500 purchases. I think about them for days or weeks or maybe more, and don’t make the purchases until I’m certain that they’re sound. Some of the things I’ll need to decide on will be “will I use this” and “does it have the features I want,” which will be factors worth far more than $20. If I extend the set of things that I’m looking for to be “any factor which is worth at least $20” then I may end up spending too much time on analysis and end up not making a purchase when I could be using the the laptop already, or missing out on a sale and having the laptop end up costing $500 more. That last part has happened to me multiple times, quibbling over much more important factors than $20 in comparative cost.
Yes if you can go back to the store, return your laptop, and buy it again for $20 less, it’s probably worth it to do so. But this situation is much rarer and much less important than saving money on a $40 purchase.
This has been discussed before.
My first thought is that you buy $40 items much more often than $2500 items; saving $20 on each $40 dollar purchase will save more money than the same savings on $2500 purchases. Especially given costs of driving around town and negotiating in high value purchases. I’ll agree that if you include these hidden costs and convert to a utility calculation people are probably not perfectly efficient, but doing the calculations themselves can be expensive.
In the example, the costs of making the savings are the same. Saving $20 on each $40 purchase will save you more money, but saving $20 on every purchase will save more still. If it’s worth going back to the store to save $20, it’s worth doing no matter how many times you do it.
Sure; on the other hand, I have one type of algorithm for dealing with $40 purchases; I do less than a day’s worth of research and think about whether I really want it or not. I make $20-$40 purchases at least once a week, maybe more. Changing my algorithm to save money on them will clearly give me a great increase in utility.
I have a separate algorithm for judging $2500 purchases. I think about them for days or weeks or maybe more, and don’t make the purchases until I’m certain that they’re sound. Some of the things I’ll need to decide on will be “will I use this” and “does it have the features I want,” which will be factors worth far more than $20. If I extend the set of things that I’m looking for to be “any factor which is worth at least $20” then I may end up spending too much time on analysis and end up not making a purchase when I could be using the the laptop already, or missing out on a sale and having the laptop end up costing $500 more. That last part has happened to me multiple times, quibbling over much more important factors than $20 in comparative cost.
Yes if you can go back to the store, return your laptop, and buy it again for $20 less, it’s probably worth it to do so. But this situation is much rarer and much less important than saving money on a $40 purchase.