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”
Dump truck: 10 x 4m = 40m2
Mountain: 1000m x 1000m = 2000m2
Number of trips needed: 50
Total distance required: 500 miles
Average speed: 50 miles/hr
Time required: 100 hours
a mountain of 1000m high
the mountain is pyramidal with a square base
that volume is 1/6 of a cube
so the volume of the mountain is 1000*1000*1000/6 = 1.3*10^9 m^3
a truck has a volume of say 10*3*2=60m^3
#trips = mountain_volume/truck_volume = /- 2*10^7 m^3
10 miles = 16km
avg speed = 60km/h
so the truck takes 30min back and forth
add another 30min for loading and unloading
this results in 1h per round trip
time = 20.000.000 hours or approx 2.000 years
340,000 years
Estimations:
-size of mountain: two million tons
-truck can carry 100 tons a time
-distance travelled by truck for one go is 20 miles (10*2)
-average speed of trick is 60 miles an hour
2*10^6/100= 2*10^4
so the truck does 2*10^4 trips
total time= 2*10^4*20/60= 4*10^4 hours i.e 1667 days non-stop.
5,000 days
“Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck”
As no information is given as to how the truck is loaded, this time will be neglected.
average size mountain, 4000m, volume of mountain is around half of the 4000*4000*4000. assuming the mountain is symmetrical.
volume of mountain= 32000000000m^3
average size truck volume= 2m*8m*2m = 32m^3
1 billion trips.
truck drives 80km/h for 16km
each trip takes 24minutes (12 minutes there and back) or 0.4h
1 billion * 0.4h = 400 million hours
assuming working 10 hours a day, 400 million days. around 1 million years.
I will be calculating the volume of the mountain in order to correctly measure how I can move it using the truck. The formula to calculate the volume of the mountain (cone-shaped) is 1/3 x height x radius^2. To measure the radius, I start from calculating the roving of the bottom part. I would estimate that it will take 1 hour for a car with a speed of 60km/hr to fully circling through the mountain. Hence, with the roving formula: 60km = 3.14 x 2r. So the radius of the mountain is 10km.
The height of a mountain is probably similar to the height of stacking 100 5-meter-tall trees, which will give us 500 meters (0.5km).
From there, we can calculate the volume of the mountain, which is: 1/3 x 0.5 x 100 = 17km^3
The dimension of an average size truck is: 1.5 x 8 x 2 = 24m^3
The number of times a truck can carry each part of the mountain: (17 x 10^9) / 24 = 700 million times.
When empty, an average-sized truck’s normal speed is estimatedly 60miles/hr, but when fully-loaded, it might be reduced by 50%, which will give us the speed of 30miles/hr.
If the distance is 10miles, so it will take 20 minutes each trip or 40 minutes (2/3hr) for one round trip.
Total hours: 2/3 hours * 700 million times = 490 million hours
With 1 km height and 1 km radius, ie gradient 45degrees of the mountain, Volume calculated. Further, estimated a truck’s volume to be 12m^3. Dividing these, we obtain the no. of trips to be ((Pi/36)*10^9 trips )*2 (ie both coming and going back to fetch).
Further, estimating a speed of 48km/hr and the distance of 10 miles as 16km. So, 3 trips/hr or 216,000 trips/decade. (assumed only 60 trips/day and not 72, taking 4 hrs off for loading and unloading and rest to machinery,etc )
Dividing the above two no. with some round off we come to approx. 800 decades.
800 decades
my proxies are amount of rock or soil the average mountain contain, the speed of a truck, and the amount teh truck can realocate at full capacity (i decided to deal with volume and not with weight, so it was m^3 not kg).
1) 1 mile is 1.6 km, thus 10 miles = 16 km, and as the speed of a truck is 80 km per hour (assumption 1) it can cover 10 miles in 16/80 hours or 12 min. however it will take it 24 min to get to the point B and back to the mountian.
2) the average truck has a body with dimensions (assumption 2) of 4 m long, 2 m wide and 3 m high, which makes 4*2*3=24 m^3 of volume it can serve during one trip;
3) the mountain has a conic shape (assumption 3) and since it is the average mountain it is 1200 m heigh and it is base has a radius of 1000 m (assumption 4), so it contains (1/3)*3.14*1200*(1000)^2=1200 b m^3;
so it will take 1200 b m^3/24 m^3 = 50 b times to get all the rock or soil realocated;
which means 24 min*50 b=1200 b min = 20 b hours