11 ms·

Back of the envelope estimation hacks

234 points by bkudria 6y ago

MiguelVieira 6y ago
See also Street Fighting Mathematics https://ocw.mit.edu/courses/mathematics/18-098-street-fighting-mathematics-january-iap-2008/ https://ocw.mit.edu/courses/mathematics/18-098-street-fighti...
- borski 6y ago
  Came here to post this. I took his course as “The Art of Approximation in Science and Engineering” and absolutely loved it. So much so that it became the name of my CTF team for the next few years.
JadeNB 6y ago
> If you combine the rule of 72 with the powers of two, then you can quickly find out how long it will take for a quantity to increase by several orders of magnitudes. This seems to me like good but dangerous advice. I'm all for using back-of-the-envelope estimations as "small angle approximations" where they're not far off, but, in general, iterating the lossy approximation to get a large-scale answer is the wrong thing to do. It works well here, but only because we're dealing with logarithms, which genuinely do convert multiplication to addition, and not mentioning this point seems to tempt one to use it in situations where it's not applicable.
hoorayimhelping 6y ago
A post with some examples: http://highscalability.com/blog/2011/1/26/google-pro-tip-use-back-of-the-envelope-calculations-to-choo.html http://highscalability.com/blog/2011/1/26/google-pro-tip-use...
heipei 6y ago
Less sophisticated, but recently I realized that just plain remembering certain numbers like "average hours per month" (~730) or "seconds per day" (86400) is really helpful to make very quick estimates. A customer wants a quota of 100k queries a day? That averages out to a little more than one query per second assuming equal distribution.
- kolanos 6y ago
  Edit: I stand corrected. 365.25 / 12 * 24 = 730.5 Technically it is 731 if you round up.
  - mitjam 6y ago
    365 * 2 = 730 :)
    - deleted 6y ago
      [deleted]
    - pdonis 6y ago
      There aren't exactly 365 days in a year.
      - _1100 6y ago
        We understand nothing “exactly” in this universe, but this is a thread about estimation hacks so maybe cool it on the fun fact witch hunt.? Idk just a thought.
        pdonis 6y ago
        > We understand nothing “exactly” in this universe Apparently a lot of "we" don't, since a number of people were saying they did the "exact" calculation that kolanos did but got a different answer. > so maybe cool it on the fun fact witch hunt.? The "witch hunt" seems to me to be on the part of the people who downvoted a perfectly correct and legitimate post, and those who responded to it with a claimed "exact" calculation that was wrong.
  - g_sch 6y ago
    I did this exact calculation and it comes out to 730, exactly like GP said.
  - PureParadigm 6y ago
    No, that evaluates to 730. It's the same as 365*2=730.
noident 6y ago
I can't find a way to contact the author, so I'm just going to post this here: your RSS feed has broken links, it links to www.robertovitillo.com, but the www subdomain does not resolve to your site.
- redblackbit 6y ago
  Whoops, thanks!
DavidPeiffer 6y ago
I've found people hate giving numbers to things when estimating how a system would work, or what it would need. Ballparking can really help though. I've found success asking questions such as "Roughly how much will updating my abstract cost? I won't hold you to the number, but are we talking closer to $200 or $800?". I make sure the numbers are quite far apart, and use it to figure high/low ends of required performance, deciding if something that's not currently the bottleneck will become a bottleneck, etc. Also, converting between hourly and salaried, you can roughly use 40 hrs/week * 50 weeks/year = 2,000 hours/year.
- Aachen 6y ago
  I wonder what that number is anywhere else in the world. Most countries have a legal minimum paid time off higher than that (the USA is the exception when compared to the vast majority of countries, rich or poor), but most companies give a bit more than the minimum in our field, so what I'm wondering is whether the difference is large enough that rounding it up doesn't make sense. Personally I've got 6 paid weeks off every year (their default, I didn't negotiate that number) plus public holidays (iirc that adds up to another two weeks but maybe it's just one) with both my past and previous employer which were in different western European countries, but I don't know if that's average.
  - joe5150 6y ago
    I could be wrong, but I think 50 hours is the result of rounding down from 52, not the result of subtracting two weeks of unpaid vacation. Most people who work a full time hourly job in the US get paid for every week of the year if they only use paid time off.
- billme 6y ago
  >> “ Also, converting between hourly and salaried, you can roughly use 40 hrs/week * 50 weeks/year = 2,000 hours/year.” Even better, take just take an hourly rate, 2x it, add three zeros = annual salary; literally takes seconds to do once you know 2000 hours per year is the assumed hours worked per year.
qznc 6y ago
Rule of five: Take five samples. You can be 93% confident the median is between your five samples. Works for any distribution (not just normal).
- chrisseaton 6y ago
  > Works for any distribution (not just normal). How can this possibly be true? If I take five samples of the speed of my car, and I always take them while the car is just setting off, it's never going to be anywhere near the median speed over a twelve hour drive. I feel like there must be a huge list of extra constraints and caveats you aren't mentioning. (I'm really bad at stats - genuinely asking.)
  - qznc 6y ago
    From Douglas W. Hubbard, How to Measure Anything (3rd ed.) via https://lobste.rs/s/kk89vp/back_envelope_estimation_hacks#c_cffofh https://lobste.rs/s/kk89vp/back_envelope_estimation_hacks#c_... > There is a 93.75% chance that the median of a population is between the smallest and largest values in any random sample of five from that population. > It might seem impossible to be 93.75% certain about anything based on a random sample of just five, but it works. To understand why this method works, it is important to note that the Rule of Five estimates only the median of a population. Remember, the median is the point where half the population is above it and half is below it. If we randomly picked five values that were all above the median or all below it, then the median would be outside our range. But what is the chance of that, really? > The chance of randomly picking a value above the median is, by definition, 50%—the same as a coin flip resulting in “heads.” The chance of randomly selecting five values that happen to be all above the median is like flipping a coin and getting heads five times in a row. The chance of getting heads five times in a row in a random coin flip is 1 in 32, or 3.125%; the same is true with getting five tails in a row. The chance of not getting all heads or all tails is then 100% − 3.125% × 2, or 93.75%. Therefore, the chance of at least one out of a sample of five being above the median and at least one being below is 93.75% (round it down to 93% or even 90% if you want to be conservative).
    - 6y ago
stormdennis 6y ago
Working out loan repayments. If you borrow 100k over 20 years then roughly you'll be paying interest on 50k every year. If the rate is 8% then that's 4k in interest every year. You'll be paying back roughly 5k of the loan every year, so your annual repayment well be 9k a year or 750a month
- saagarjha 6y ago
  (Keep in mind that due to how compounding works, averaging interest in this way will underestimate it slightly.)
  - stormdennis 6y ago
    Yes but the point is that it's just a back of an envelope estimate. I'm amazed that it's been downvoted as it's actually a technique that's used in this way all the time, or at least was anyway. I don't know about now when everyone has a smartphone.
yongjik 6y ago
Order-of-magnitude is extremely important. Frequently I find myself asking "So how much money is it going to save us? $100/mo, $1k/mo, or $10k/mo?" It's basically a nicer (and more concrete) way to ask "Are we sure that this project is worth the money the company's paying us to do it?" (Unfortunately I don't always get a satisfying answer, but that's life...)
pulkitsh1234 6y ago
Can most people do these calculations in their head? Personally I find that I am unable to do these calculations/estimations unless I write things down. I get lost between powers of 2/MB/TB/nanoseconds and other unit conversions. I understand that this is elementary math, but still, I am just curious, can most people do similar calculations in their head (for instance during a meeting) ?
- physicles 6y ago
  I think it probably helps to know units and powers cold, so you can use your short term memory for the actual calculation. I was able to do the first calculation in the article because, as it says, it’s just summing exponents on powers of 10. It should be second nature that M=6, G=9, T=12, or that “nano” means -9. Powers of 2 are incredibly helpful too: I can estimate that the square root of 4000 is around 60 because 4000 is close to 2 to the 12th, for instance. As a coder it’s well worth your time to memorize powers of 2 up to at least 2 to the 16.
  - comradesmith 6y ago
    Further its good to think of your powers of 2 like: 10, kilo; 20, mega; 30, giga; And for 0-9 I have those memorized. So if I'm given the number 2^39 I unwrap that as 512 giga, or just over 500 billion (seems its ~550, but estimating with base 2 is still going to be closer than estimating with base 10)
- bo1024 6y ago
  Don't underestimate the effectiveness of practice!
saeranv 6y ago
Here's one that I learned from HN years ago (https://news.ycombinator.com/item?id=18463449 https://news.ycombinator.com/item?id=18463449) Rule of three: If you want to compare two things (i.e if code A is faster then code B), then measure each three times. If all three As are faster then the Bs, then there is a ~95% probability that is true. Derivation: We are trying to find the probability of selecting three As, from the set (ABABAB). We don't care about the order of the As, so this is a combinations without repetition problem: 6 choose 3 = n! / ((n-r)! r!)) = 6! / (3! 3!) = 120 / 6 = 20 There is only one state where all three As are chosen, so the probability of getting it by random chance (our null hypothesis) is: 1 / 20 = 5%. Hence 95% confidence.
- kevinwang 6y ago
  Uh oh... I think you're interpreting the probability incorrectly. Shouldn't this be "if we assume that a and b take the same amount of time, then there's a 5% chance that we would see this result?". Can't actually speak to the probability that a is faster than b without some kind of assumed prior distribution for a and b
  - saeranv 6y ago
    Yes, this is correct.
  - contravariant 6y ago
    It should also be pointed out that the one thing you're ruling out is that they have the same distribution. In particular it only provides weak evidence that the mean, mode, median of A is any faster than B's. In the worst case A happens to be a lot slower than B, but with very low probability.
  - bo1024 6y ago
    Yep. It gets at, what is the meaning of "faster?" Implicitly we are saying the average runtime of A is smaller than the average of B. But if for example, B always halts in 10 seconds, and A halts in 1 second 99.99% of the time but A runs for 10 years 0.01% of the time, then A is "slower" but you'd need 10,000 trials to notice.
deleted 6y ago
[deleted]
lpolovets 6y ago
It's been about 20 years since I read this, but for people interested in quick mental math, here's a great book on doing fast calculations in your head: Dead Reckoning: Calculating Without Instruments https://www.amazon.com/dp/B00BZE4916/ref=cm_sw_r_tw_apa_i_YuzYEbHK0KR6P https://www.amazon.com/dp/B00BZE4916/ref=cm_sw_r_tw_apa_i_Yu... IIRC it covers everything from multiplication to square roots to logarithms.
basementcat 6y ago
A large hamburger contains approximately one kilowatt hour worth of kilocalories.
gmcabrita 6y ago
Simon Eskildsen has a monthly newsletter called Napkin Math[0] entirely dedicated to practicing practical real-world Fermi problems like these. [0] - https://sirupsen.com/napkin/ https://sirupsen.com/napkin/
RcouF1uZ4gsC 6y ago
Another neat hack for estimating lower limits on latency is that light travels roughly 1 foot in a nanosecond. So if two servers are in a room 100 feet apart, the time it takes for them to communicate cannot be lower than 100 ns.
alexpetralia 6y ago
When tipping, I realized it's harder for me to multiply the bill (say $58) times 15% rather than times 30% / 2. In the latter case I simply do 60 * .3 / 2 which is $9. I thought about this after realizing, for example, that 18 * 5 is really 18 * 10 / 2. Edit: just realized this is more about mental math than about back-of-the-envelope estimations
- bcrosby95 6y ago
  I usually just do something with 10s. I usually tip 20% so i knock off a zero then double it. If I tipped 15%, I would probably knock off a zero, then add half to it.
londons_explore 6y ago
The quickest way to get good at this is practice. Just ask engineer friends random questions from time to time, and when there is no consensus, calculate the results properly. Eg. How much power could a solar powered fly collect? How many tons of batteries would I need to drive a train across the USA? How long can I power my laptop from a solar usb battery bank in the sun each day on a hiking trip? What's the load time of this webpage I just wrote on a satellite internet connection? How long would it take to boot up my OS if I could keep RAM intact during a reboot? How many kilos of coal does it take to cook a pizza?
kthejoker2 6y ago
Three other hacks which when combined have a lot of power .. Chebyshev's Theorem: 89% of all values in (almost every human-ish) distribution fall within 3 SDs of the mean. PERT estimation: base your estimates on the happiest (O for optimistic) unhappiest (P For Pessimistic), and most likely (ML) path - the recommended formula is (P + O + 4ML) / 6, but you can also just model them as a distribution. Secretary problem: when attempting to optimize a result, the effect of uncertainties can be ascertained in 1/e (~37%) of the total solution space. So putting this together .. If you have to give a time estimate for something, your best bet is to 1) get your 3 PERT numbers 2) calc the mean, SD, and 37% of mean + 3SD ("Assessment") - this is how much time it should take (based on your pessismism) to eliminate uncertainty for the remaining work estimate 3) if the mean is less than the Assessment time, there's too much uncertainty and you should offer to assess the effort instead of just doing the work 4) if the mean is more than the assessment, just offer to do the work with the PERT output from the formula above. I find most people are happy to hear you have a concrete plan for assessing something with the explicit purpose of providing precision.

234 points by bkudria 6y ago

See also Street Fighting Mathematics https://ocw.mit.edu/courses/mathematics/18-098-street-fighting-mathematics-january-iap-2008/ https://ocw.mit.edu/courses/mathematics/18-098-street-fighti...

borski 6y ago
Came here to post this. I took his course as “The Art of Approximation in Science and Engineering” and absolutely loved it. So much so that it became the name of my CTF team for the next few years.

JadeNB 6y ago

> If you combine the rule of 72 with the powers of two, then you can quickly find out how long it will take for a quantity to increase by several orders of magnitudes. This seems to me like good but dangerous advice. I'm all for using back-of-the-envelope estimations as "small angle approximations" where they're not far off, but, in general, iterating the lossy approximation to get a large-scale answer is the wrong thing to do. It works well here, but only because we're dealing with logarithms, which genuinely do convert multiplication to addition, and not mentioning this point seems to tempt one to use it in situations where it's not applicable.

hoorayimhelping 6y ago

A post with some examples: http://highscalability.com/blog/2011/1/26/google-pro-tip-use-back-of-the-envelope-calculations-to-choo.html http://highscalability.com/blog/2011/1/26/google-pro-tip-use...

heipei 6y ago

Less sophisticated, but recently I realized that just plain remembering certain numbers like "average hours per month" (~730) or "seconds per day" (86400) is really helpful to make very quick estimates. A customer wants a quota of 100k queries a day? That averages out to a little more than one query per second assuming equal distribution.

kolanos 6y ago
Edit: I stand corrected. 365.25 / 12 * 24 = 730.5 Technically it is 731 if you round up.
- mitjam 6y ago
  365 * 2 = 730 :)
  - deleted 6y ago
    [deleted]
  - pdonis 6y ago
    There aren't exactly 365 days in a year.
    - _1100 6y ago
      We understand nothing “exactly” in this universe, but this is a thread about estimation hacks so maybe cool it on the fun fact witch hunt.? Idk just a thought.
      - pdonis 6y ago
        > We understand nothing “exactly” in this universe Apparently a lot of "we" don't, since a number of people were saying they did the "exact" calculation that kolanos did but got a different answer. > so maybe cool it on the fun fact witch hunt.? The "witch hunt" seems to me to be on the part of the people who downvoted a perfectly correct and legitimate post, and those who responded to it with a claimed "exact" calculation that was wrong.
- g_sch 6y ago
  I did this exact calculation and it comes out to 730, exactly like GP said.
- PureParadigm 6y ago
  No, that evaluates to 730. It's the same as 365*2=730.

noident 6y ago

I can't find a way to contact the author, so I'm just going to post this here: your RSS feed has broken links, it links to www.robertovitillo.com, but the www subdomain does not resolve to your site.

redblackbit 6y ago
Whoops, thanks!

DavidPeiffer 6y ago

I've found people hate giving numbers to things when estimating how a system would work, or what it would need. Ballparking can really help though. I've found success asking questions such as "Roughly how much will updating my abstract cost? I won't hold you to the number, but are we talking closer to $200 or $800?". I make sure the numbers are quite far apart, and use it to figure high/low ends of required performance, deciding if something that's not currently the bottleneck will become a bottleneck, etc. Also, converting between hourly and salaried, you can roughly use 40 hrs/week * 50 weeks/year = 2,000 hours/year.

Aachen 6y ago
I wonder what that number is anywhere else in the world. Most countries have a legal minimum paid time off higher than that (the USA is the exception when compared to the vast majority of countries, rich or poor), but most companies give a bit more than the minimum in our field, so what I'm wondering is whether the difference is large enough that rounding it up doesn't make sense. Personally I've got 6 paid weeks off every year (their default, I didn't negotiate that number) plus public holidays (iirc that adds up to another two weeks but maybe it's just one) with both my past and previous employer which were in different western European countries, but I don't know if that's average.
- joe5150 6y ago
  I could be wrong, but I think 50 hours is the result of rounding down from 52, not the result of subtracting two weeks of unpaid vacation. Most people who work a full time hourly job in the US get paid for every week of the year if they only use paid time off.
billme 6y ago
>> “ Also, converting between hourly and salaried, you can roughly use 40 hrs/week * 50 weeks/year = 2,000 hours/year.” Even better, take just take an hourly rate, 2x it, add three zeros = annual salary; literally takes seconds to do once you know 2000 hours per year is the assumed hours worked per year.

qznc 6y ago

Rule of five: Take five samples. You can be 93% confident the median is between your five samples. Works for any distribution (not just normal).

chrisseaton 6y ago
> Works for any distribution (not just normal). How can this possibly be true? If I take five samples of the speed of my car, and I always take them while the car is just setting off, it's never going to be anywhere near the median speed over a twelve hour drive. I feel like there must be a huge list of extra constraints and caveats you aren't mentioning. (I'm really bad at stats - genuinely asking.)
- qznc 6y ago
  From Douglas W. Hubbard, How to Measure Anything (3rd ed.) via https://lobste.rs/s/kk89vp/back_envelope_estimation_hacks#c_cffofh https://lobste.rs/s/kk89vp/back_envelope_estimation_hacks#c_... > There is a 93.75% chance that the median of a population is between the smallest and largest values in any random sample of five from that population. > It might seem impossible to be 93.75% certain about anything based on a random sample of just five, but it works. To understand why this method works, it is important to note that the Rule of Five estimates only the median of a population. Remember, the median is the point where half the population is above it and half is below it. If we randomly picked five values that were all above the median or all below it, then the median would be outside our range. But what is the chance of that, really? > The chance of randomly picking a value above the median is, by definition, 50%—the same as a coin flip resulting in “heads.” The chance of randomly selecting five values that happen to be all above the median is like flipping a coin and getting heads five times in a row. The chance of getting heads five times in a row in a random coin flip is 1 in 32, or 3.125%; the same is true with getting five tails in a row. The chance of not getting all heads or all tails is then 100% − 3.125% × 2, or 93.75%. Therefore, the chance of at least one out of a sample of five being above the median and at least one being below is 93.75% (round it down to 93% or even 90% if you want to be conservative).
  - 6y ago

stormdennis 6y ago

Working out loan repayments. If you borrow 100k over 20 years then roughly you'll be paying interest on 50k every year. If the rate is 8% then that's 4k in interest every year. You'll be paying back roughly 5k of the loan every year, so your annual repayment well be 9k a year or 750a month

saagarjha 6y ago
(Keep in mind that due to how compounding works, averaging interest in this way will underestimate it slightly.)
- stormdennis 6y ago
  Yes but the point is that it's just a back of an envelope estimate. I'm amazed that it's been downvoted as it's actually a technique that's used in this way all the time, or at least was anyway. I don't know about now when everyone has a smartphone.

yongjik 6y ago

Order-of-magnitude is extremely important. Frequently I find myself asking "So how much money is it going to save us? $100/mo, $1k/mo, or $10k/mo?" It's basically a nicer (and more concrete) way to ask "Are we sure that this project is worth the money the company's paying us to do it?" (Unfortunately I don't always get a satisfying answer, but that's life...)

pulkitsh1234 6y ago

Can most people do these calculations in their head? Personally I find that I am unable to do these calculations/estimations unless I write things down. I get lost between powers of 2/MB/TB/nanoseconds and other unit conversions. I understand that this is elementary math, but still, I am just curious, can most people do similar calculations in their head (for instance during a meeting) ?

physicles 6y ago
I think it probably helps to know units and powers cold, so you can use your short term memory for the actual calculation. I was able to do the first calculation in the article because, as it says, it’s just summing exponents on powers of 10. It should be second nature that M=6, G=9, T=12, or that “nano” means -9. Powers of 2 are incredibly helpful too: I can estimate that the square root of 4000 is around 60 because 4000 is close to 2 to the 12th, for instance. As a coder it’s well worth your time to memorize powers of 2 up to at least 2 to the 16.
- comradesmith 6y ago
  Further its good to think of your powers of 2 like: 10, kilo; 20, mega; 30, giga; And for 0-9 I have those memorized. So if I'm given the number 2^39 I unwrap that as 512 giga, or just over 500 billion (seems its ~550, but estimating with base 2 is still going to be closer than estimating with base 10)
bo1024 6y ago
Don't underestimate the effectiveness of practice!

saeranv 6y ago

Here's one that I learned from HN years ago (https://news.ycombinator.com/item?id=18463449 https://news.ycombinator.com/item?id=18463449) Rule of three: If you want to compare two things (i.e if code A is faster then code B), then measure each three times. If all three As are faster then the Bs, then there is a ~95% probability that is true. Derivation: We are trying to find the probability of selecting three As, from the set (ABABAB). We don't care about the order of the As, so this is a combinations without repetition problem: 6 choose 3 = n! / ((n-r)! r!)) = 6! / (3! 3!) = 120 / 6 = 20 There is only one state where all three As are chosen, so the probability of getting it by random chance (our null hypothesis) is: 1 / 20 = 5%. Hence 95% confidence.

kevinwang 6y ago
Uh oh... I think you're interpreting the probability incorrectly. Shouldn't this be "if we assume that a and b take the same amount of time, then there's a 5% chance that we would see this result?". Can't actually speak to the probability that a is faster than b without some kind of assumed prior distribution for a and b
- saeranv 6y ago
  Yes, this is correct.
- contravariant 6y ago
  It should also be pointed out that the one thing you're ruling out is that they have the same distribution. In particular it only provides weak evidence that the mean, mode, median of A is any faster than B's. In the worst case A happens to be a lot slower than B, but with very low probability.
- bo1024 6y ago
  Yep. It gets at, what is the meaning of "faster?" Implicitly we are saying the average runtime of A is smaller than the average of B. But if for example, B always halts in 10 seconds, and A halts in 1 second 99.99% of the time but A runs for 10 years 0.01% of the time, then A is "slower" but you'd need 10,000 trials to notice.

deleted 6y ago

[deleted]

lpolovets 6y ago

It's been about 20 years since I read this, but for people interested in quick mental math, here's a great book on doing fast calculations in your head: Dead Reckoning: Calculating Without Instruments https://www.amazon.com/dp/B00BZE4916/ref=cm_sw_r_tw_apa_i_YuzYEbHK0KR6P https://www.amazon.com/dp/B00BZE4916/ref=cm_sw_r_tw_apa_i_Yu... IIRC it covers everything from multiplication to square roots to logarithms.

basementcat 6y ago

A large hamburger contains approximately one kilowatt hour worth of kilocalories.

gmcabrita 6y ago

Simon Eskildsen has a monthly newsletter called Napkin Math[0] entirely dedicated to practicing practical real-world Fermi problems like these. [0] - https://sirupsen.com/napkin/ https://sirupsen.com/napkin/

RcouF1uZ4gsC 6y ago

Another neat hack for estimating lower limits on latency is that light travels roughly 1 foot in a nanosecond. So if two servers are in a room 100 feet apart, the time it takes for them to communicate cannot be lower than 100 ns.

alexpetralia 6y ago

When tipping, I realized it's harder for me to multiply the bill (say $58) times 15% rather than times 30% / 2. In the latter case I simply do 60 * .3 / 2 which is $9. I thought about this after realizing, for example, that 18 * 5 is really 18 * 10 / 2. Edit: just realized this is more about mental math than about back-of-the-envelope estimations

bcrosby95 6y ago
I usually just do something with 10s. I usually tip 20% so i knock off a zero then double it. If I tipped 15%, I would probably knock off a zero, then add half to it.

londons_explore 6y ago

The quickest way to get good at this is practice. Just ask engineer friends random questions from time to time, and when there is no consensus, calculate the results properly. Eg. How much power could a solar powered fly collect? How many tons of batteries would I need to drive a train across the USA? How long can I power my laptop from a solar usb battery bank in the sun each day on a hiking trip? What's the load time of this webpage I just wrote on a satellite internet connection? How long would it take to boot up my OS if I could keep RAM intact during a reboot? How many kilos of coal does it take to cook a pizza?

kthejoker2 6y ago

Three other hacks which when combined have a lot of power .. Chebyshev's Theorem: 89% of all values in (almost every human-ish) distribution fall within 3 SDs of the mean. PERT estimation: base your estimates on the happiest (O for optimistic) unhappiest (P For Pessimistic), and most likely (ML) path - the recommended formula is (P + O + 4ML) / 6, but you can also just model them as a distribution. Secretary problem: when attempting to optimize a result, the effect of uncertainties can be ascertained in 1/e (~37%) of the total solution space. So putting this together .. If you have to give a time estimate for something, your best bet is to 1) get your 3 PERT numbers 2) calc the mean, SD, and 37% of mean + 3SD ("Assessment") - this is how much time it should take (based on your pessismism) to eliminate uncertainty for the remaining work estimate 3) if the mean is less than the Assessment time, there's too much uncertainty and you should offer to assess the effort instead of just doing the work 4) if the mean is more than the assessment, just offer to do the work with the PERT output from the formula above. I find most people are happy to hear you have a concrete plan for assessing something with the explicit purpose of providing precision.