Purchasing a burger at a busy restaurant in a large city will not affect how many burgers they purchase from their distributor.
It will, by exactly 1 burger. More specifically, if their unit of buying is a box of 100 frozen burgers, and they use the surplus from each day to start the next, then in the long run they will have bought exactly 1 more burger than they would if you had not bought yours: one in 100 of the boxes they get through will have been purchased 1 day earlier than it would have been.
This is a common fallacy: saying that if a large change in X produces a large change in Y, then a small change in X will produce no change at all in Y. Stated like that it’s obviously absurd, but in concrete situations people apply the same wrong thinking as you have just done.
Compare the marketing parable (I don’t know if the exact scenario ever happened) of the manager at a burger chain who suggested putting just 5 sesame seeds less on every bun. No-one would notice and they’d save money over millions of buns. Repeat until they have no customers left.
Here’s another example. You are about to leave home to drive somewhere. There are many junctions with traffic lights on the way, and you will probably have to stop at some of them. If you are delayed by one second leaving home, by how much is your expected arrival time delayed?
Purchasing a burger at a busy restaurant in a large city will not affect how many burgers they purchase from their distributor.
It will, by exactly 1 burger.
What EricHerboso said wasn’t true in general but neither is that. I can well imagine that fast food places just buy a specific number of burgers periodically and discard the surplus. If there’s slack from this, buying 1 burger can have a far smaller effect upstream.
Might as well check this line of argument works with a toy example. Suppose the number of would-be burger buyers X at my local McDonald’s each day (discounting myself) is Poissonianly distributed with mean 80. The McDonald’s buys either 100 or 120 burgers per day: if it had >100 customers the previous day, it buys 120, otherwise just 100. Then, on average, it buys 100 P(X ≤ 100) + 120 P(X > 100) = 100.26 burgers. Now suppose I turn up and buy a burger. Then the expected number of burgers the restaurant buys the next day is 100 P(X+1 ≤ 100) + 120 P(X+1 > 100) = 100.34 burgers. My buying 1 burger makes the restaurant buy only 0.08 burgers more (on average).
This is a common fallacy: saying that if a large change in X produces a large change in Y, then a small change in X will produce no change at all in Y.
There is an analogous fallacy: assuming that if a large change in X causes a large change in Y, a small change in X causes a proportionally small change in Y.
Compare the marketing parable (I don’t know if the exact scenario ever happened) of the manager at a burger chain who suggested putting just 5 sesame seeds less on every bun. No-one would notice and they’d save money over millions of buns. Repeat until they have no customers left.
I hadn’t heard of that parable before, but I had heard the more upbeat business story of American Airlines saving $40,000 a year by putting one less olive in each salad it served in first class.
You are about to leave home to drive somewhere. There are many junctions with traffic lights on the way, and you will probably have to stop at some of them. If you are delayed by one second leaving home, by how much is your expected arrival time delayed?
Once, when I was younger, I found I could shave 5 minutes off my commute by leaving for the train station 5 minutes later in the morning!
Might as well check this line of argument works with a toy example.
The argument needs to look at the wider situation. How did the burger shop decide on their restocking algorithm? By looking at demand. They will continue to look at demand and review their algorithm from time to time. Buying one burger contributes to that, so the situation is that one more customer may result in them changing the numbers from 100 and 120 to 110 and 130. Ten extra burgers a day until they review the numbers again. The probability that your burger pushes them into ordering more is smaller than in the original example, but the number of extra burgers is proportionately larger. Slack in the chain doesn’t affect the mean effect, only the variance.
I hadn’t heard of that parable before, but I had heard the more upbeat business story of American Airlines saving $40,000 a year by putting one less olive in each salad it served in first class.
A crucial detail: “most passengers did not eat the olives in their salads”. At least, in that telling of the story. So there was reason to think that the olives weren’t earning their place in terms of passenger satisfaction.
Once, when I was younger, I found I could shave 5 minutes off my commute by leaving for the train station 5 minutes later in the morning!
The argument needs to look at the wider situation. How did the burger shop decide on their restocking algorithm? By looking at demand. They will continue to look at demand and review their algorithm from time to time. Buying one burger contributes to that, so the situation is that one more customer may result in them changing the numbers from 100 and 120 to 110 and 130. Ten extra burgers a day until they review the numbers again.
Incorporating this into the toy model shows this isn’t enough to guarantee proportionality either.
My hypothetical burger shop is now in Sphericalcowland, a land where every month is 30 days. It also has a new burger-buying policy. On day 1 of each month, it buys 100 burgers for that day, then uses a meta-decision rule to decide the burger-buying decision rule for the month’s remaining 29 days. Let Y be the number of customers in the previous month. If Y was no more than, say, 2500, the shop uses my earlier 100⁄120 decision rule for the remaining 29 days. But if Y > 2500, it uses your upgraded decision rule (buy 130 burgers if there were >100 customers the previous day, otherwise buy 110). X ~ Po(80) as before, so Y ~ Po(2400). (I’ve deliberately held constant the burgers bought for day 1 of each month to avoid applying a previous day-based decision rule for day 1 and causing inter-month dependencies.)
With the 100⁄120 decision rule, the shop buys an average of 3007.639 burgers a month. So with the 110⁄130 decision rule, it buys an average of 3297.639 a month.
If I don’t buy a burger, E(burgers bought next month) = (3007.639 × P(Y ≤ 2500)) + (3297.639 × P(Y > 2500)) = 3013.624 burgers, by my working.
If I buy a burger, E(burgers bought next month) = (3007.639 × P(Y+1 ≤ 2500)) + (3297.639 × P(Y+1 > 2500)) = 3013.920 burgers.
Hence in this example, the upstream marginal effect of my buying 1 burger is only 0.296 burgers. The presence of feedback doesn’t suffice to guarantee proportionality.
A crucial detail: “most passengers did not eat the olives in their salads”. At least, in that telling of the story. So there was reason to think that the olives weren’t earning their place in terms of passenger satisfaction.
For all I know, neither do sesame seeds on buns! In any case, the American Airlines story might be apocryphal in itself. I just bring it up to illustrate that there’s a countervailing anecdote to the parable.
Forgive me if I’m misunderstanding, but doesn’t the fallacy you bring up apply specifically to continuous functions only? For step-wise functions, a significantly small change in input will not correspond to a significantly small change in output.
Assuming the unit is a 100 burger box (100BB), then my purchase of a burger only affects their ordering choices if I bring the total burger sales over some threshold. I’m guesstimating, but I’d guess it’s around 1⁄3 of a box, or 33 burgers in a 100BB. So if I’m the 33rd additional customer, it might affect their decision to buy an extra box; but if I’m one of the first 32, it probably won’t. This puts a very large probability on the fact that my action will not have an effect.
Is my reasoning here flawed? I’ve gone over it again in my head as I wrote this comment, and it still seems to apply to me, even after reading your above comment. But perhaps I’m missing something?
This puts a very large probability on the fact that my action will not have an effect.
However, the complementary probability is the probability of a correspondingly large effect. The smallness of the probability and the largeness of the effect exactly cancel, giving an expected effect of 1 burger bought from the distributor for every burger bought by you.
The fact that the effect is nigh invisible due to the high level of stochastic noise does not mean it is not there.
To eat meat without it having been killed for your benefit, you should raid supermarket waste bins for the time-expired stuff they throw out.
When I first thought about this, I was fairly confident of my belief; after reading your first comment, I rethought my position but still felt reasonably confident; yet after reading this comment, you’ve completely changed my position on the issue. I had completely neglected to take into account the largeness of the effect.
You’re absolutely correct, and I retract my previous statements to the contrary. Thank you for pointing out my error. (c:
It will, by exactly 1 burger. More specifically, if their unit of buying is a box of 100 frozen burgers, and they use the surplus from each day to start the next, then in the long run they will have bought exactly 1 more burger than they would if you had not bought yours: one in 100 of the boxes they get through will have been purchased 1 day earlier than it would have been.
This is a common fallacy: saying that if a large change in X produces a large change in Y, then a small change in X will produce no change at all in Y. Stated like that it’s obviously absurd, but in concrete situations people apply the same wrong thinking as you have just done.
Compare the marketing parable (I don’t know if the exact scenario ever happened) of the manager at a burger chain who suggested putting just 5 sesame seeds less on every bun. No-one would notice and they’d save money over millions of buns. Repeat until they have no customers left.
Here’s another example. You are about to leave home to drive somewhere. There are many junctions with traffic lights on the way, and you will probably have to stop at some of them. If you are delayed by one second leaving home, by how much is your expected arrival time delayed?
What EricHerboso said wasn’t true in general but neither is that. I can well imagine that fast food places just buy a specific number of burgers periodically and discard the surplus. If there’s slack from this, buying 1 burger can have a far smaller effect upstream.
Might as well check this line of argument works with a toy example. Suppose the number of would-be burger buyers X at my local McDonald’s each day (discounting myself) is Poissonianly distributed with mean 80. The McDonald’s buys either 100 or 120 burgers per day: if it had >100 customers the previous day, it buys 120, otherwise just 100. Then, on average, it buys 100 P(X ≤ 100) + 120 P(X > 100) = 100.26 burgers. Now suppose I turn up and buy a burger. Then the expected number of burgers the restaurant buys the next day is 100 P(X+1 ≤ 100) + 120 P(X+1 > 100) = 100.34 burgers. My buying 1 burger makes the restaurant buy only 0.08 burgers more (on average).
There is an analogous fallacy: assuming that if a large change in X causes a large change in Y, a small change in X causes a proportionally small change in Y.
I hadn’t heard of that parable before, but I had heard the more upbeat business story of American Airlines saving $40,000 a year by putting one less olive in each salad it served in first class.
Once, when I was younger, I found I could shave 5 minutes off my commute by leaving for the train station 5 minutes later in the morning!
The argument needs to look at the wider situation. How did the burger shop decide on their restocking algorithm? By looking at demand. They will continue to look at demand and review their algorithm from time to time. Buying one burger contributes to that, so the situation is that one more customer may result in them changing the numbers from 100 and 120 to 110 and 130. Ten extra burgers a day until they review the numbers again. The probability that your burger pushes them into ordering more is smaller than in the original example, but the number of extra burgers is proportionately larger. Slack in the chain doesn’t affect the mean effect, only the variance.
A crucial detail: “most passengers did not eat the olives in their salads”. At least, in that telling of the story. So there was reason to think that the olives weren’t earning their place in terms of passenger satisfaction.
You knew the timetable and caught the same train?
Incorporating this into the toy model shows this isn’t enough to guarantee proportionality either.
My hypothetical burger shop is now in Sphericalcowland, a land where every month is 30 days. It also has a new burger-buying policy. On day 1 of each month, it buys 100 burgers for that day, then uses a meta-decision rule to decide the burger-buying decision rule for the month’s remaining 29 days. Let Y be the number of customers in the previous month. If Y was no more than, say, 2500, the shop uses my earlier 100⁄120 decision rule for the remaining 29 days. But if Y > 2500, it uses your upgraded decision rule (buy 130 burgers if there were >100 customers the previous day, otherwise buy 110). X ~ Po(80) as before, so Y ~ Po(2400). (I’ve deliberately held constant the burgers bought for day 1 of each month to avoid applying a previous day-based decision rule for day 1 and causing inter-month dependencies.)
With the 100⁄120 decision rule, the shop buys an average of 3007.639 burgers a month. So with the 110⁄130 decision rule, it buys an average of 3297.639 a month.
If I don’t buy a burger, E(burgers bought next month) = (3007.639 × P(Y ≤ 2500)) + (3297.639 × P(Y > 2500)) = 3013.624 burgers, by my working.
If I buy a burger, E(burgers bought next month) = (3007.639 × P(Y+1 ≤ 2500)) + (3297.639 × P(Y+1 > 2500)) = 3013.920 burgers.
Hence in this example, the upstream marginal effect of my buying 1 burger is only 0.296 burgers. The presence of feedback doesn’t suffice to guarantee proportionality.
For all I know, neither do sesame seeds on buns! In any case, the American Airlines story might be apocryphal in itself. I just bring it up to illustrate that there’s a countervailing anecdote to the parable.
Exactly.
Forgive me if I’m misunderstanding, but doesn’t the fallacy you bring up apply specifically to continuous functions only? For step-wise functions, a significantly small change in input will not correspond to a significantly small change in output.
Assuming the unit is a 100 burger box (100BB), then my purchase of a burger only affects their ordering choices if I bring the total burger sales over some threshold. I’m guesstimating, but I’d guess it’s around 1⁄3 of a box, or 33 burgers in a 100BB. So if I’m the 33rd additional customer, it might affect their decision to buy an extra box; but if I’m one of the first 32, it probably won’t. This puts a very large probability on the fact that my action will not have an effect.
Is my reasoning here flawed? I’ve gone over it again in my head as I wrote this comment, and it still seems to apply to me, even after reading your above comment. But perhaps I’m missing something?
However, the complementary probability is the probability of a correspondingly large effect. The smallness of the probability and the largeness of the effect exactly cancel, giving an expected effect of 1 burger bought from the distributor for every burger bought by you.
The fact that the effect is nigh invisible due to the high level of stochastic noise does not mean it is not there.
To eat meat without it having been killed for your benefit, you should raid supermarket waste bins for the time-expired stuff they throw out.
When I first thought about this, I was fairly confident of my belief; after reading your first comment, I rethought my position but still felt reasonably confident; yet after reading this comment, you’ve completely changed my position on the issue. I had completely neglected to take into account the largeness of the effect.
You’re absolutely correct, and I retract my previous statements to the contrary. Thank you for pointing out my error. (c: