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”
Considering an average size truck needs 2 hours to drive 100 mile, in a flat road with very few trafic (no mountains in between, or curves). To transport the mountain, I would add an extra hour (i.e. it would take the truck 3 hours) as the truck driver also needs to pay attenion not to hurt anybody, or ruin anything. To accomplish this successfully the mountain would also have to have 4 wheals to make the transporation easier.
Average Size mountain:
1km diameter, round base, height 0.5km.
So ballpark estimation of Volume is 1/3*pai*0.5*0.5*0.5=0.1km3=1E8m3.
Once truck is loaded (5*2*2=20m3 much of material), it will take 1 hr to deliver and come back and refill (assuming 20 miles/hr).
So it will minimally consume 1E8/20=5E6 hrs=500 yrs
The rest will be compounding factors (in comparison to the 1 hr per round) that will make the time taken longer in a linear fashion for simplicity.
1. loading and unloading. 2X
2. truck driver shifts. Assuming 3 shifts of 8 hours and unlimited supply of workers. 1X
3. gas refill, mechanical breakdown, tire change (only 1 truck available), 2X.
4. Increment weather. 2X.
5. for the hundreds of years merely to transport, the time taken to break down the mountain (say by explosives in rounds of 5-10 seconds) can be neglected.
Considering the above, the estimated time taken will be :
500*2*2*1*2=4000 yrs.
My main issue here is the definition of average size..if the largest truck can take the largest mountian in one trip and the smalles one can take the smallest mountain in one trip then the average size truck will take one trip to locate the average sized mountain. So if the spead of the truck is 60 m/h then it will take about 10 minutes to relocate the mountian plus the load and unload time.
1.clarify what is the end-product the interviewer wants at the new location – still a mtn (hope not, may take very long) or pile of rocks
2. assume rocks are desired and only one truck; assume ave. mtn. weigh x tons and measure y cubic feet; assume an ave. truck capacity is m tons and n cubic feet; assume no need to build a road between the two sites; assume no need to dig a hole to dump the rocks
3. time to blow up the mtn, assume having super-dynamite, D minutes;
4. time to load&unload
A minutes per load, B mimutes per unload
N number of rounds = max(x/m, y/n)
total= (A B)*N
5. travel time
assume there is a road, but narrow&winding, C minutes one way
total =2*C*N
6. assume 12-hr per day operation(not safe to drive in the winding road at night), total days needed to move the mtn= (D (A B)*N 2*C*N)/60/12.
If all that I am given to move said average size mountain is an average sized truck… the problem is not solvable. The unknown or unallocated variables are significant:
1. What equipment do I have to load this truck? (shovel, backhoe, excavator, etc…)
2. What type of workforce to I have? (just myself, a team, an army)
3. What distance am I to move said mountain? (10 feet, 10 miles, cross country)
4. What class of truck are we talking about? (dual axle pick up truck, 2 ton or greater sized dump truck, heavy construction truck)
5. What is the scope of this mountain and its directive to move it? Are we taking it down to sea level from its current delineation of elevation… or simply until we might classify it as part of another ‘set’, such as a foothill?
Without additional information, any answer is arbitrary in nature and lacking necessary premises… therefore I submit my answer in similar fashion: 4
I pretty much got a process similar to what you guys had. But even before tackling the soil part, I took trees and cottages into consideration and assumed that it would take equal amount of hours of work to replicate the original condition in the new site as moving the soil. Now soil part, like some mentioned:
1)break
2)load
3)carry
4)unload
I assumed an average size mountain to be 1km high, 2.5 km in base and cone shaped which gives 1.3^2*pi (substitute with 3)*1(height)/3 = 1.7km^3 or 1.7billion m^3
an average truck may hold 4m(length)*1m(depth)*1(width) = 4m^3 and assume there are 100 trucks available for this project
4million travels needed altogether
now assume no traffic jam, no accidents, and trucks travel at 50km/hour, a mile is about 2km so it should take 0.3 appx one way and 1hr for both way.
and because 4million travels are needed, it takes them 4 million hours just by transportation
I’ve seen some construction site and it took them approximately 1 hr to load the truck full of soils and again assume that it takes equal amount of time to unload = 2hrs. additional labour such as breaking and rebuilding would take another 2hrs.
4*400million truck loading and unloading = 1600 million hours
added altogether for soil = 4million 1600 million = 1604 million appx 1600 million (gosh, transportation hours are non-material after all)
another 1600 million for trees and all other misc activities = 3.2 billion hours to move and replicate the mountain from one location to another.
Oops, forgot to turn the cuboid into a pyramid by /3…
@Jess:
Ah! I realise why this is so very small: mistake 1km cubed is not 1000m cubed, but 1000 x 1000 x 1000 = 1 bn m cubed! Would re-calculate to fit this
The following bits of information are the key ones I need to work out this answer:
– An estimation of the volume of the mountain; how many metres cubed of earth there is to shift
– An estimation of the volume of the ‘average truck’
– Use these to estimate how many times the truck must be filled to move all the earth
– Using this info about how many trips, work out how long it would take the truck to do the number of 20 mile return trips needed to shift the mountain.
This misses out any other time that would be needed for actually filling the truck, time needed for men to dig and move the earth: as the question seems to focus on the truck I would hope this is an ok assumption and check with the interviewer.
So, to start with the mountain volume.
I lived in Geneva last year, I was told 3000m is considered a high mountain, so I’d go for an ‘average’ height of 2000m.
For the base, I think 10km seems like a reasonable diameter.
As this is an estimation, I want to avoid pi, so instead of calculating a cone, I will calculate a pyramid. For the square base I’ll go for 10km x 10km, which gives me a base of 100km squared (hopefully not too far off what a diameter of 10km would give me). Times this by the height, and I get 200,000m cubed.
Now for the truck: estimating based on open trucks I have seen, I’d guess the depth of the loading area to be around 2m, the width around 3m and the length about 5m, giving a total volume of 30m cubed.
Dividing 200,000 by 30 gives 6666.66(recurring), which I will round up to 7000: this is the number of times the truck would need to be filled to transport all the earth.
So there need to be 7000 return trips. Each trip is 20 miles (10 there and back). So in total the truck will need to travel 7000 x 20 miles, 14000 miles.
A truck probably travels at about 50mph: they are usually slower than cars max speed, plus it is loaded, and may be travelling on rural roads.
A truck travelling 50mph will take 14000/50 = 280 hours.
So, in terms of truck travelling time, it would take 280 hours or about 12 days to shift the mountain.
Sanity check: this seems way too quick, and does not take into account the actual time of digging the earth, and loading the truck, nor reconstructing the mountain. I have oversimplified…
Edit:
Last EQ is missing /100:
• 2.500.000.000/(4*25) = 25.000.000 days -> 250000/365 = ~700 years
should be:
• 2.500.000.000/(4*25) = 25.000.000/100 days -> 250000/365 = ~700 years
Note: being Swedish I did my math in metric while keeping the distance in miles as stated.
I assume the mountain is to be moved piece by piece.
Assume data give represents bottlenecks – everything else runs as smoothly as possible i.e assemble and disassemble times are negligible.
—————————————————————————————–
You can either make the calculations based on volume or on weight being the limiting factor for the truck.
• {kg/(kg/day)}
• {m^3/(m^3/day)}
Density of rock is assumed to about 2,5 grams/cc = 2,5kg/l = 2,5 tonne/m^3
—————————————————————————————-
Determine What is an average truck:
• Assume Volume/weight capacity
o Volume = 20m*2,5m*2m=100m^3 -> 250 tons (seems too high in my opinion i.e weight is the bottle neck chosen here)
o Assumed Weight capacity = 20 tonnes (corrected to 25 to make the math easier)
o Motivation: I don’t know much about dump trucks or trucks in general, but I’ve seen big trucks transporting 10 or more cars and a car weighs ~1 ton. Assuming a dump truck is more specialized for its purpose I assume 20 tonnes
• Weight moved/day = (weight capacity) X (runs/day)
—————————————————————————————-
Runs/day
o Runs/day = (hours/day) / ({[(distance) X 2] / (speed)} [(loading t) X 2])
Runs/day = 10/({[(10) X 2] / (40)} [1 X 2] = 10/({20 / 40} 2) = 10/2,5 = 4
– Runs/day = 4
• Motivation and assumptions
o speed=distance/time -> t=d/s
o Runs= hours/(miles/[miles/hour])
o Assume average Speed = 40 mph
o Assume: Loading time = unloading time = 1 hour
o Assume work day = 10 h (we want to move it fast 🙂
—————————————————————————————-
Determine What is an average mountain:
• Assume Size (Volume)
o Shape = cone
EQ cone: (1/3)*pi*h*r^2 Define mountain
• Define mountain
o For me it is something you can climb or ski on, something bigger than a hill, in terms of abstract size it is something that takes you some effort to cross or climb – say more than an hour or so. By my definition it needs to be smaller than Mt. Everest (h = 9km) and bigger than a hill (highes mountain in Denmark is called a hill; h=0,3km).
– Assume h = 1 km
– Assume r = 1 km
o Volume = (1/3)*pi*1*1^2 = pi/3 km^3 ~ 1 km^3
– 1km^3 = 10^9 m^3
– Weight = 2,5 ton * 10^9 m^3 = 2.500.000.000 tonnes
—————————————————————————————-
Estimated time to relocate:
• (Mountain weight) / [(runs/day) X (weight moved/day)] {kg/(kg/day)}
• 2.500.000.000/(4*25) = 25.000.000 days -> 250000/365 = ~700 years
Conclusion: Except for taking a very long time and not being a remotely plausible project in real life, we can see that (Mountain weight) >> (other factors), leading to them being of very little importance. So the assumptions regarding the volume of the mountain and the density of the product moved are the central ones.