Case Interview Example – Estimation Question and Answer
I was asked the following management consulting estimation question by a McKinsey interviewer many years ago:
“Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck”
Below you will see my answer to this estimation question and the process and rational I use to answer this specific question can be used as a template to practice answering other estimation questions as you prepare for case interviews.
The first thing to realize in an estimation question is that an acceptable answer MUST mention a specific number.
This question was how much time it takes to move an average mountain 1 mile (or something along those lines).
If the answer does not include a specific unit of time like X hours, Y days, Z years, then the answer is not acceptable.
By the way, I use the word “acceptable answer” instead of “correct answer” very deliberately. The interviewer’s evaluation in this type of question is in assessing the approach you took, not necessarily the specific answer you gave.
The next thing to the answer must include is that explicit assumptions must be made.
It is not possible to answer this question without making some assumptions. They key is to EXPLAIN to the interview that you are going to make some assumptions. Once you do and once you make a specific assumption, explain your rationale behind that assumption.
For example, when I was given this question. I knew that I needed to estimate the cubic volume of the mountain. And since the mountain loosely resembles a cone, I knew there was a geometric formula to calculate the volume of a cone–except I did not recall the specific formula off the top off my head.
So my interviewer suggested that I estimate the formula of a cone, which in turn I would use to estimate the volume of an average size mountain, which would then be part of a calculation to estimate the average time it would take to re-locate it.
Notice the estimate that is nested within the estimate here. This is very common. Most important thing is to not get mixed up and confused by your own work.
I find it is useful to just write out the formula that will produce the estimate FIRST, THEN go about making reasonable assumptions.
For the move the mountain case, the formula I wrote up on the white board during my interview was:
volume of mountain / volume of a truck * time per truck trip = total time to move a mountain
I would literally write that on the board. That is the amount of time it would take 1 truck to move an average size mountain 10 miles (the 1 truck is an assumption as well)
Then I went about estimating each of those 3 factors.
Assume the average size mountain is 1 mile tall, 1 mile wide, and the shape of a cone. That’s approximately 5,000 ft in height and base.
I forge the formula to calculate the volume of a cone, but if I eye ball it, it is probably a little more volume than half of a cube of similar size height and base.
The volume of a cube that’s 5,000 ft tall, 5,000 ft wide, and 5,000 ft deep is 125,000,000,000 cubic ft.
Since I’m trying to estimate a CONE, and not a CUBE, I’d then take 125,000,000,000 x 50% (my approximate guess as to how much smaller a cone is vs a cube of approximately the same height, and width and length at the base.
With some slight rounding, that gets us 60,000,000,000.
Then underneath my original formula, I would write the following:
60,000,000,000 cubic ft / volume of a truck * time per truck trip = total time to move mountain
Next, I would move on to estimate the volume of a truck.
The carrying capacity of a cargo truck is the width x length x heightof the cargo container.
I said, well I know those big trucks are a little wider than my car, but not by much since they still must be able to fit into a lane on the freeway. My car sits 3 people across, assuming 2 ft in shoulder width per person, that’s 6 ft of interior space. Let’s add on a little more and assume those big trucks are around 8 ft in width.
I know they are about double the length of most passenger sedans. And lets see if I were to lie down in the driver’s seat to take a nap, I cover most of the interior cabin space. And the hood and trunk of the car combined are about the same length as the interior cabin. I’m a little under 6ft tall, so that makes my car around 12 ft long. If I double that, I get the length of one of those trucks to be 24 ft long. I subtract out say 4 ft for the driver compartment, and that leaves me about 20 ft in length for the cargo area.
Last time I looked, I saw a worker standing in the back of one of the cargo areas, and the cargo area was taller than the person. I figure the cargo container is about 8 ft tall. And since most freeway bridges have signs that say “height 13 ft” and I know those trucks can go under those bridges, assuming an 8ft cargo section and a 4ft for the tires and chassis under the cargo area, that gives me 12 ft…which does seem to triangulate with the height of those underpasses. So I’ll say the cargo section is approximately 8 ft tall.
The volume of the cargo area of an earth moving truck is:
8 ft wide x 20 ft long x 8 ft tall = 1,280 cubic feet
For sake of simplicity, I’m going to round that down to 1,250 cubic feet and plug this number back into my original formula which now reads as follows:
60,000,000,000 cubic foot mountain / 1,250 cubic foot truck capacity * time for truck trip = total time to move a mountain
The only factor missing in our estimate is figuring out the round-trip time for a trip to move 10 miles, drop its load, and return the 10 miles. Let’s figure out the travel time first. Assume the truck travels on the freeway at 60 miles per hour.
For it to travel 10 miles, it does so in 1/6 and hour or 10 minutes. The drive time is 10 minutes to the new location, and 10 minutes returning to the old mountain for a total of 20 minutes. Assume that the off-loading process has been designed to be pretty quick. The load is just “dropped” and then repositioned while the truck is on its return trip (as opposed to being scooped out of the truck, one scoop at time which seems more time consuming).
That means each round trip takes 30 minutes or 0.5 hours.
Let’s go back to our formula again and update it.
60,000,000,000 cubic ft mountain / 1,250 cubic foot track capacity * 0.5 hours per truck trip = total time to move a mountain
Let me do the math now. For the first 2 components of the formula, that works out to about 50,000,000 (50 million truck loads).
50 million truck loads x 0.5 hours, thats 25 million hours to move a mountain.
If we assume a typical day has 25 hours (to make our math a little simpler), that’s 1 million days to move the mountain using only 1 truck. That works out to a bit under 3,000 years
That is the logic I just presented is a pretty good one that would most likely pass most estimation question interviews.
You will notice that for every little component I explain WHY I felt that was a reasonable assumption.
There is a big difference between making a wild assumption vs. a reasonable one. Your goal is to make as reasonable assumption as you can come up with. When you make such an assumption, it is very important you explain WHY you made the assumption you did.
The math is not that complicated (it’s math we all learned before high school) BUT communicating what you are doing is just as important.
It is also important that you do not make a math mistake. I wrote out this example quickly and hopefully I did not make a math mistake.
If I did make a math mistake, I would full expect to get rejected even if I got the logic and assumptions largely right.
That’s just the way it works. Practice your mental math. You DO use it a lot not just in interviews but with clients as well.
504 thoughts on “Estimation Question”
Assume radius of an average mountain – 1 km (which makes the circumference 6.2 km which is too big – but what the heck, how to define an ‘average size’ mountain)
Assume height of the mountain – 1/3 rd of its radius ~ 330m ~ 1/π km . Therefore volume of the mountain – 1/3 * π * 1*1*1/π = 0.33 km^3 ~ 300 million m^3
Assume the density of the mountain as a mountain and as disintegrated mud (rocks, shrubs included) in the truck are same
Assume the size of the truck container be 3m * 1.7 m * 1.7m (an average truck used in quarries are normally deep, but 1.7m may not be accurate). Hence the volume of the truck is approx 9 m3. Considering 1 m3 of sand can be heaped above the truck’s rectangular container , total transportable volume is 10m3
So in order to transport 300 million m3 of rock n sand with a 10m3 truck, we need to use the truck 30 million times.
Now we need to estimate how long it takes – Assuming a truck carrying a load of sand and rocks cannot travel more than 30 miles per hour. It takes 20 minutes to travel 10 miles and 20 minutes to travel back to the mountain. Take 20 minutes to load and unload (assuming we have some high performing loader) . So for 1 movement it takes 1 hour.
Hence for 30 million movements it would take 30 million hours. Assuming we work only 20 hours a day (4 hours for changing shift drivers, filling gas, stuck in traffic etc). we would take (30 million/20) days = 1.5 million days.
Assuming we work 350 days a year ( 15 days for Christmas holidays, Superbowl, Mitt Romney debate etc). It would take 1.5 million / 350 ~ 4200 years
2) Let’s assume it has a base area of 10 hectares
3) 2,000 meters tall*100,000 m2 base = 200,000,000 m3 (if square). Account for the cone shape, say ½, thus 100,000,000 m3 is the average volume of a mountain.
4) Assume an average truck bed is 2 meters x 3 meters and can be filled vertically to carry 10 m3
5) Thus, it would require 1,000,000 loads to move the mountain
6) Assume 30 minutes to cut and load 10m3 in a truck
7) Assume the truck travels at 60 mph X 10 miles
8) Round trip would be 20 minutes
9) Unloading time 10 minutes
10) Thus, we require 60 minutes to load and deposit 10m3 and return
11) 8 hour work day
12) Thus 8 loads / day
13) 252 work days in a year
14) Thus 2,016 loads transported / year (round down to 2,000)
15) 2,000, 10m3 loads per year would require 500 years to move 1,000,000 m3.
Step one did not paste in the initial post.
1) Let’s assume an average mountain is 2,000 meters tall
2) Let’s assume it has a base area of 10 hectares
3) 2,000 meters tall*100,000 m2 base = 200,000,000 m3 (if square). Account for the cone shape, say ½, thus 100,000,000 m3 is the average volume of a mountain.
4) Assume an average truck bed is 2 meters x 3 meters and can be filled vertically to carry 10 m3
5) Thus, it would require 1,000,000 loads to move the mountain
6) Assume 30 minutes to cut and load 10m3 in a truck
7) Assume the truck travels at 60 mph X 10 miles
8) Round trip would be 20 minutes
9) Unloading time 10 minutes
10) Thus, we require 60 minutes to load and deposit 10m3 and return
11) 8 hour work day
12) Thus 8 loads / day
13) 252 work days in a year
14) Thus 2,016 loads transported / year (round down to 2,000)
15) 2,000, 10m3 loads per year would require 500 years to move 1,000,000 m3.
First we have to figure the approximate square footage of material (mountain) to be moved. What is the height of an average mountain? This varies greatly. I took the tallest mountain in the world, Everest, which I estimated at 24,000 ft. But then, it doesn’t start at sea level. What is the base camp? I think I’ve heard something like 14,000ft. So the largest mountain we could guess has 10,000ft of vertical. From hiking I know that gaining 1000 ft to 2000 ft of vertical is a normal long hike, but I’m not sure I’d call that an average mountain. More a small mountain. From the range of possibilities from the hills in the center of America which they consider mountains, to the cascades or sierra nevadas or Olympics. I would estimate that the average vertical rise would be somewhere near 3,000ft.
Now that have this estimate, I want to estimate the volume of a mountain this large. If a mountain has proportions approximating an isoscolese triangle, then draw a line down from inside the tip of the triangle to the base labeled 3,000ft. The approximate proportions of long side to hypotenous in a right angle triangle put the hypotenous about 30% longer. this could be the measure of the length down the slope. This equals the other sides of the triangle, so the base is also the same. If our height is 3000ft then, we can guesstimate the base is 4000ft across. If we were to draw a cube around this basis of height, width and length, we would know how much volume was in that cube. 4,000×3,000×4,000= 48billion square feet. If we take that cube and draw our pyramid inside it, we can guestimate from looking at the proportions that our pyramid makes up about 1/3 of that cube. So take 1/3 of 48 billion and we have 16 billion square feet to move.
Now how long to move it? The question says average truck. It doesn’t specify what kind of truck. Dump truck? Pickup truck? I would say that the most “average” kind of truck is a consumer pickup. Some would argue that one would more likely use a dumptruck to move a mountain, but I would say that the question is already so absurd, perhaps best to go with the exact wording of “average”. So I choose pickup.
I estimate an average pickup to be 4 feet wide (can just fit a 4×8 piece of plywood, widthwise), and 7 feet long (I just fit with about 1 foot lee-way laying down in the back with the tailgate up.) The side height I’m estimating at 2 feet. So our cubic feet on one pickup load full is 4x7x2=56 square feet. A load can be piled up a bit in the middle so we’ll round up to 60. So how many loads do we need to move 16billion sq feet? 16billion/60= 266,666,666. We’ll round up to 270,000,000 total loads in the pickup truck.
How long will each load take?
10 miles each direction. Assume a normal highway and we can go 60mph. 10 minutes each direction. Add loading and unloading time. Let’s assume 5 minutes on each end for that. So one round trip will take about 30 minutes.
So now we can just calculate down our number of loads and load time. 270,000,000 loads x 30 minutes gives us total minutes, but we are about to divide that back down to get years, so let’s simplify and say 30 minutes is 1/48 of a day and divide 270,000,000 by 48. This equals about 5,400,000 days, divided by 365 gives us about 150,000 years.
This, of course, doesn’t take into account all kinds of variables like the trucks breaking down and driver issues, etc. etc. etc. (Which I imagine would SIGNIFICANTLY affect the final answer.)
Hypothesis:
1) The container of an Average size truck is long 7 meter x 3meter which is an Volume of 21 cube-meter. To semplify let’s use 20 cube-meter.
2)An average size mountain is 1000 meter long x 1000 meter high. His volume is 1,000,000 cube meter.
3) The average speed of the average truck is 80 km/h. Consequently, the truck will need 1/4 hour (15 minutes) to do one transport of 20 cube meter for 10 miles (or 20 km). Time needed:Distance/Speedrate–> Time needed: 20 km/ 80 km/h–> 1/4 h
Considering the following hypothesis the number of transportation-trip the truck will do in order to transport the whole mountain is : Volume of the mountain / Volume of the truck. Which is : 1,000,000 / 20. Which is 50,000.
Consequently, the time needed to transport the mountain for 10 miles (or 20 km) is equal to the time needed to do one transportation-trip times the total number of transportation-trip.
Total Number of Transportation-trip: 50,000
Time needed to do one transportation-trip: 1/4 h (15 min)
Total time needed to transport the mountain for 10 miles ( or 20 km) : 50,000 times 1/4 h : 12,500 h.
VMMCS
I should have divided it by 3. Busted
😀
I’ll assume the average mountain has the following dimensions:
Base area (ba): 10,000 m^2
Height (h): 100 m
For ease of calculation, I’ll use the formula to find a pyramid’s volume:
h.ba.1/3 = 100 . 10,000 . 1/3 = 1,000,000/3 ~ 330,000 m^3
I’ll also assume the truck travels at an average speed of 40 mph when it’s empty and at an average speed of 20 mph when it’s full and that it’s load capacity is of 200 m^3 (20m x 2m x 5m). It also takes 15 minutes to load and unload the truck.
Therefore, each time the truck travels to the mountain and back, it takes 1 hour and moves 200m^3 of it.
It’d take a total of 5,000 straight hours to move the mountain to somewhere 10 miles from it, but, considering it has only one driver, working 8 hours shift, it would take a total of 625 days, or 1 year 8 months 2 weeks and 6 days.
It is not specified whether or not the mountain is actually on the ground or not. Assuming it is floating in mid air (as this is an assumption based question), then it could take any amount of time varying from one minute or one hundred thousand days. The idea of a mountain of ‘average’ mass. There is no such thing as an average mountain as we cannot count the mass and size of every mountain in the world. We have no definition of a mountain, so a mountain could be as small as a grain of sand. The mountains we are aware of could be outliers in terms of what we know. An average size truck, by the same token, might not mean what we think it means. There is no time frame for such a question. This question could be as old as trucks are, or it could be a question meant for people in the future. Trucks come in a variety of sizes, and although there is a ‘common’ truck, there are trucks that could be quite small or quite large in comparison. This is all surrounded by the idea of an ‘average’. An average is all things considered and divided by the amount of all things. We cannot know ALL things, and therefore, we cannot judge. There is no wrong answer if the answer is in a time, so the time it would take is five minutes, simply because, until the answer is given, there is no way of getting it wrong because the answer is UNKNOWN knowledge. TOK’d.
You miss the point completely. If you are honestly trying to get a consulting job watch the videos on the main page to understand the whole point of asking a question like that. The goal is to demonstrate your reasoning and estimating process and skills. The exact numbers don’t matter. Just your process and skills.
Raghav,
George is absolutely correct. Although your answer is the most true, if you provide that answer in a case interview it is an automatic rejection.
Victor
Ok, first off, I’ll start with the size of the mountain and break that down. I’ve been to a few mountain ranges, and I know that typical peak elevation can be about 5,000 or 6,000 feet in altitude. Some can be a lot higher, but we’re saying average mountain, so let’s assume 5,000 feet. Also, the actual ground level in mountain ranges isn’t at sea level by any means – so let’s say the ground level is at 2,000 feet. Which means, the mountain is only 3,000 feet tall.
(5,000 – 2,000 = 3,000).
Next, the width of the mountain. I don’t really know how wide mountains are, but I think it’s safe to say they’re at least bigger than a football field. A football field is 100 yards, or 300 feet, so I guess a typical mountain would be about 500 feet.
Let’s assume the mountain is cone-shaped. Therefore the volume of the mountain is equal to the height of the mountain x half of the width squared x pi, or: 3,000 feet x 3.14 x (250 ft)^2
3,000 feet x 3.14 x 6,250 sq. feet
I’ll simply my numbers a bit to make the math easier:
3,000 ft x 3 x 6,000 sq. ft.
3×3 is 9, so that’s basically like saying 9×6, which is 54, and the two groups of thousands gives us another million, so
54 million cubic feet is the volume of our mountain.
Now, I need to know the volume of the truck. I would say the typical truck bed is about 9 feet long and 6 feet wide. That would give us a surface area of 54 squared feet (9×6), and we can probably load the dirt from the mountain about 2 feet high in the truck bed.
54 x 2 = 108.
So, I think we can feet 108 cubic feet of mountain into the truck at any one time.
with a 54 million cubic foot mountain and a truck bed of 108 cubic feet, we need 500 thousand truckloads in order to move the mountain.
10 miles isn’t too far a trip, but we also have to load and unload the truck bed each time. I’m going to assume each trip takes a total of one hour to load, move the mountain, unload and come back.
That means it takes us 500 thousand hours to move the mountain 10 miles.
If we assume there’s 25 hours in a day, then 500k/25 = 20k.
So, it takes us 20 thousand days.
1. Estimated size of an average size mountain:
Height: 30m
Width: 30m
Depth: 30m
Size in square meters: (30*30*30)/3=9000 square meters.
2. Estimated time for breaking the mountain using dynamite in order to enable transporting on the truck:
Estimation: 100 square meters of mountain exploded/ hour.
9000/100= 90 hours spent on breaking the mountain.
3. Estimated time for transport on truck:
Estimation: each truck load can take 100 square meters of stone. 9000/100= 90, meaning we need to make 90 trips to transport the whole mountain.
10 miles is approx. equal to 16 kilometers. If the truck has an estimated average speed of 80km/h it will takes us 80/16 = 1/5 of an hours in each direction, which is equal to 12 minutes (60/5= 12). Considering both directions we spend 12*2=24 minutes on the road .
4. Estimated time for on- and off loading of truck:
Off-loading: Assuming the truck has a platform that can be lifted allowing for the stones to slide off, off-loading is considered taking minimal time.
On-loading: 10 square meters of stone is estimated to take approx. 3.6 minutes to load. Therefore it will take 36 minutes to load one truck load of 100 square meters of stone ((100/10)*3.6=36).
5. Total time for transport:
On the road =24 minutes
Loading = 36 minutes
24 36=60 minutes or 1 hour.
From point 3: We need to make 90 trips. If each trip takes 1 hour it will take us 90 hours in total time for transport.
6. Total time for moving the average size mountain with an average size truck:
90h 90h=180 hours
or 180/24 = 7.5 days if we work 24 hours a day
or 180/8 = 22.5 days if we work 8 hours a day.
2400 yrs
Mountain Height=1000m
Mountain radius= 1000m
Volume = 1/3 * pi * 10^9 ~= 10^9 m^3
Each truck size = 2m (height) *2.5m (width) *10m (length) = 50 m^3
Number of truck loads= 2*10^7 truck loads
Speed of truck = 60 miles/hr
each way= 10 miles ( 10 minute)
number of loads per hour = 3 ( going back and forth)
hours of work per day = 8
number of loads per day = 24 ~= 25
Number of days requires= 2*10^7/25= 8* 10^5 days
Number of years= 8*10^5/365 ~= 8* 10^5/ (1000/3) = 2400 years