Lambdia

A Hundred Fifty Thousand Gas Stations, and the Factor of Six Hiding in One Pump

Fourteen billion fill-ups a year divided by what a pump could do at full tilt gives 25,000 stations; divided by what a pump actually does it gives 149,829, inside the published range. The gap is exactly six, and the article proves that six is the ratio of the two throughput guesses alone, because the fleet, the fill-up frequency, the opening hours and the pumps per station all cancel. The utilisation of one sixth is Little's law read as 2.67 busy hours in a sixteen-hour day.

How many gas stations does the United States need? There is a chain that gets there from quantities you can nearly check by looking out of a car window, and it comes out at about 150,000, which is close to the published count. The same chain with one plausible-looking assumption changed gives 25,000, and the gap between the two is exactly a factor of six.

Which factor is worth more than the answer, because it turns out not to depend on any of the other guesses in the chain.

Fourteen billion fill-ups

Start with demand. Call the fleet 280 million cars and take fifty fill-ups a year each, which is roughly once a week and is a number you can test against your own habits. That gives the annual national demand:

F  =  2.8×108×50  =  1.4×1010 fill-ups a yearF \;=\; 2.8 \times 10^{8} \times 50 \;=\; 1.4 \times 10^{10}\ \text{fill-ups a year}
(1)

Both inputs are declared guesses and both are wrong in the third digit. Neither matters much, for a reason that shows up two sections down.

What a pump could do, and what it does

Now supply. A pump takes about five minutes per car, so it can serve twelve cars an hour, and a station open sixteen hours a day could push 192 cars per pump per day. Take eight pumps per station and divide the demand through:

Ncap  =  1.4×1010192×365×8  =  24,971N_{\text{cap}} \;=\; \frac{1.4 \times 10^{10}}{192 \times 365 \times 8} \;=\; 24{,}971
(2)

Twenty-five thousand stations for the whole country. Notice the direction of that error before going any further: assuming a pump does more work means the country needs fewer stations. An optimistic assumption about capacity undercounts, which is the opposite of what most people expect a generous assumption to do.

Now go and watch a real pump. Cars do not arrive every five minutes. Over a full sixteen-hour day the arrival rate at one pump is closer to one car every half hour, which is two an hour and thirty-two a day. The same division with that number:

N  =  1.4×101032×365×8  =  149,829N \;=\; \frac{1.4 \times 10^{10}}{32 \times 365 \times 8} \;=\; 149{,}829
(3)

which lands inside the published range for United States retail fuelling sites, between about 145,000 at the narrow definition and about 196,000 at the widest, and is 3.3 per cent off the narrow figure.

Fig. 1 — One assumption changed, one answer moved by a factor of six. The shaded band is the published range, which is itself 35 per cent wide.

Why the six survives every other guess

Write the chain with letters. Let λ\lambda be the cars a pump serves per hour, hh the hours a station is open, pp the pumps per station, and FF the annual demand from equation (1):

N  =  Fλh365pN \;=\; \frac{F}{\lambda\, h \cdot 365 \cdot p}
(4)

Take the ratio of the two answers and everything except the throughput cancels:

NobsNcap  =  F/(λobsh365p)F/(λcaph365p)  =  λcapλobs  =  122  =  6\frac{N_{\text{obs}}}{N_{\text{cap}}} \;=\; \frac{F / (\lambda_{\text{obs}}\, h \cdot 365 \cdot p)}{F / (\lambda_{\text{cap}}\, h \cdot 365 \cdot p)} \;=\; \frac{\lambda_{\text{cap}}}{\lambda_{\text{obs}}} \;=\; \frac{12}{2} \;=\; 6
(5)

The fleet, the fill-up frequency, the opening hours and the pumps per station all appear identically in the numerator and the denominator. Enumerated over 144 combinations of those four guesses, the ratio comes out at exactly six every time, with no deviation anywhere. So the size of the mistake is independent of every input you were unsure about. It depends only on the one input nobody thinks to question.

The number you can check at your own local station

Thirty-two fill-ups out of a possible 192 is a utilisation of one sixth, and stated as a time rather than a ratio it becomes something a person can verify: the pump sells fuel for under three hours of a sixteen-hour day.

Fig. 2 — The whole error, drawn. The trap prices the forecourt at the shaded width and then bills the country for the difference.
Utilisation

For a single server, the long-run fraction of time it is busy is the arrival rate times the service time, ρ=λobs/λcap\rho = \lambda_{\text{obs}} / \lambda_{\text{cap}}. The two throughput figures in this problem are not two guesses at the same quantity. Twelve an hour is the service rate and two an hour is the arrival rate, and the trap consists of putting one where the other belongs.

Once that is named, the idleness stops looking like waste. A forecourt is sized for its evening peak, because a station built for the average would have an hour-long queue at six o'clock. Capacity is bought for the worst twenty minutes of the day and paid for during the other fifteen and a half hours. Any chain that divides national demand by peak capacity is implicitly assuming a country with no rush hour.

Sweeping all four guesses at once

Equation (4) has four soft inputs, so the honest way to report the answer is a range. Sweeping all four across 81 combinations gives station counts from 68,493 up to 342,466, and the published narrow figure of 145,000 sits inside that interval. That is the same power of ten throughout, which is what this method buys, and it is worth stating rather than hiding behind the 3.3 per cent that the central combination happened to produce.

The spread is also a useful diagnostic. A factor of five between the extremes means the answer is not carried by any single assumption, which is a healthier state for an estimate than one input dominating.

Where the count stops meaning anything

The largest source of error here is not arithmetic, it is the definition. A gas station is not a well-defined object. Convenience stores selling fuel come to about 145,000; add truck stops, unattended card-only sites and dealer forecourts and the total runs to roughly 196,000. Those two numbers differ by 35 per cent, so an estimate accurate to three per cent against one of them is accurate to nothing in particular against the other. Any claim tighter than the width of the definition is noise.

The chain also assumes one representative station. Real ones range from a rural single-pump site doing a handful of cars a day to a motorway truck stop with twenty pumps running through the night. Dividing a national total by an average is fine for the total and tells you nothing about the distribution, so this method can never answer a question about the shape of the industry.

Finally, the model assumes fuel throughput is what determines whether a station exists. For a great many of them the forecourt is an accessory to a shop, and the economics that decide the count are the margin on coffee and cigarettes. When a chain like equation (4) lands close to the truth despite ignoring that, it is worth remembering which part was luck.

Sources and further reading

Nothing here asserts a true number of stations. What was checked, in exact rational arithmetic, is the arithmetic of both chains, the invariance of the factor six across all 144 combinations of the four declared guesses, the utilisation of one sixth read as 2.67 busy hours, and the 81-point sweep that spans 68,493 to 342,466 and contains the published figure.

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