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”
12 days and 18 hours 30 minutes (approx)
Assuming that an average mountain consist of 30,000 tons, an average truck can carry 10 tons per travel, and each travel will take 2 hrs (4 travels a day). Then we need 750 working days to move an average mountain 10 miles.
18 thousand years
First,considering the volume of an average size mountain. Assuming the mountain to be the size of a cone with base 8km and height 3km. It’s volume then is pie(r square)h, which turns out as 16kmcube! Now, assuming that the mountain can be broken into bricks of 10m, the number of total bricks will be 1600000000pie.
An average truck now, taking the volume of a cuboid, suppose can carry 5000 bricks in one trip.
Taking the speed of 50km/hour, one trip takes approx 0.212 hours, I.e. About 25 mins to the place of relocation and then coming back for another set of bricks makes it 50min for one set of 5000 bricks.
Total time then would be, 16000000pie minutes.
To solve the problem I would first calculate the volume of an average mountain. A mountain resembles a cone which has volume = pi*r^2*h/3. pi = 3.14, therefore I will round pi/3 = 1.
Volume = r^2*h. The height of mount Everest is around 8000 meters and the smallest knolls are around 1000 meters. So an avergage mountain is around 4000m.
Let’s say the radius of an average mountain is 200m.
Therefore volume of an avergage mountains = 200^2*4000
= 16*10^7 m3
Next, find the volume of a truck. A car is usually 8m long and 6m wide. An Average truck is about thrice as long as a car and 2 times wide. So the length and width of a truck (rounded off) is 25m and 10m. Lets say the height of a truck is 4 meters ( around 12 feet). Volume of truck is =h*b*l = 25*10*4 = 1000m3
Total time = No. of trips * Time per trip
No. of trips = vol of mount/vol of truck
= 160000
Time per trip = distance/speed unloading time
= 60mph(assume) * 20 miles( both ways) 1o mins( assume)
= 20mins 10 mins = 30 mins = 0.5 hours
Total time = 160000/2
= 80000 hrs ( only work 20 hours a day)
= 4000 days
= 10 years and 350 days or 11 years(rounding)
Assume that the average size of a mountain will be about 3x shorter than that of Mount Everest at the height of c.30,000ft. This gives us a height of c.10,000ft and assume the radius to be about 1,000 ft. Applying the formula of the volume of a cone of pi*r^2*(h/3). This gives us 3.142*1,000,000*10,000/3 that will be a proxy of 10,000,000,000ft^3. Assume an average truck with the dimensions of 5x2x4 = 40ft^3. The total number of trips required is 10,000 million / 40 = 2,500 million trips.
Now each trip will =travel at about 40mph given the heavy load (slightly slower than the 60mph), this means to travel 10mile distance, it requires 15mins. However, add the 5mins for loading and unloading, it requires 20mins which means 3 trips per hour. Thus, 2,500 million trips / 3 = c. 830 million hours. Proxy this to no. of days, I round up 830 million hours to 840 million hours for simpler calculations. 84/24 = 3.5*10 million = 350 million days.
I will solve the problem in two parts, firstly figure out the approximate weight of the average mountain and then I will find the time it takes to move it. Firstly, I will assume the mountain is in the shape of a cone with a diameter/base of 10000m. The height will be assumed to be 4000m as US’s tallest mountain is 6000m and the average height of a mountain would be lower than that. That’s the shape of the mountain. As for the composition, I will assume it to be a mixture of rock and soil compacted together. I will go ahead and make the assumption that a cube block of this mixture with a side length of 1m weighs around 100kg (1m^3). Following on from that, I will calculate the volume of the mountain as 50% of a cylinder with base of 5000m across and height of 4000m. This is to account for the mountain being in the shape of a cone that fits inside a cylinder.
V = ½ * pi(r^2)(h) = ½ * 3*(5000)^2 * 4000 = 150 billion m^3
Now to calculate the weight of the mountain, I will multiply the volume by 100kg giving 15000 billion kgs. I will assume that an excavating machine will be used to excavate the mountain for transportation by a truck; this machine will load the truck automatically at the rate of 100kg per min. I will assume the truck’s capacity to be 2000kg. Therefore it will take 20mins to load the truck each trip. I will also assume that the excavator is moved to desired locations while the truck is driving, so it doesn’t have to be added on to the driving time. Furthermore during transportation, it will drive at 30 miles per hour for safety reasons. One round trip would take 2*(10/30) hr round trip drive 1/3 hr loading time = 1 hr. So it takes 1 hour to transport 2000kg of the mountain. That would be about 7.5 billion hours to transport with continuous truck operation (15000 billion kgs/ 2000kg). If 1 man is working 5 days a week, 10 hr shifts, then the job will be done in 750 million working days (7.5 billion hrs/ 10hrs).
“Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck.”
I’ll assume that the average size truck is a large pick-up truck with a dump-lift, partly for fun, and partly because I don’t want to estimate this aspect of the question more precisely, which would require a further complex estimation.
I’ll assume that a crew supports the moving operation. So the pickup truck can be loaded and unloaded almost instantly. One front-loader fills up the bed instantly, and the lift dumps it quickly. So we’re looking at 1m in combined loading and unloading time.
I’ll assume that the truck is kept in constant operation, 24/7/365. When one driver tires, another replaces him. Refueling, tire changes, and repairs are made by a pit crew during the 1m while the truck is dumping its load. The truck does not otherwise break down. (If one wants to calculate the additional time serious repairs to the truck would require, that’s not difficult to add at the end.)
Finally, we assume that the mountain is broken down by a crew at one end, and when the truck arrives, it is ready to be instantly loaded. We also assume a crew at the other end to assemble the mountain while the truck is making its trips.
These assumptions reduce the remainder of the question to the time it takes the truck to transport the mountain 10m.
We start by estimating the volume of the mountain, the volume of the load the pick-up trick can handle, and the speed at which the pick-up truck can travel.
We assume a (rectangular) pyramidal shape for the mountain, and we estimate it’s height as 2m, or approximately 10,000 feet. We’ll assume the bases of the mountain are the same as its height (so 10,000 feet square). This gives a volume of:
333,333,333,333 CU FT
Our pick-up truck probably has a bed of about 8 FT X 4 FT x 2 FT. This loaded flat, it can handle 64 CU FT. Let’s pile it on. Call that an extra 20 CU FT (84 TL) and round up to 100 CU FT for easy math. It’s a large pick-up truck.
Total number of loads is now: 3,333,333,333.
We move to the time per trip. We’ll assume a paved road and a travel speed of 50 MPH. We assume no loss in the load at the speed on the paved road. (Perhaps the pit crew tosses in a tarp in the 1m load/unload time.) Since we’re heading 10m, that’s 12m each way, or 24m round trip. Given our loading/unloading time and other assumptions, we’re looking at 25/m round trip in total.
So now we have:
(3,333,333,333 * 25) / 60 / 24 / 365 = TL # of Years.
Or roughly, about 150,000 years.
Our pick-up truck’s load was 100 CU FT. A dump truck might be about 10x10x10, or 1000 CU FT. If using such a dump truck that can travel at 50 MPH and be similarly loaded very quickly (would need a very large loader, but I assume this is possible) we get 1,500 YRS.
Additional fudge factors can be now added as desired.
Time=((t1=how long it takes for a truck to drive 10miles) (t2=how long it takes to load and offload the cargo to the truck))*2*(n=volume of mountain/volume of truck))/(percentage of the time for driving a car in a week)
t1=10miles/truck speed=10miles/40miles/h=0.25h
t2=1h
n=volumn of mountain/volumn of truck=1/3*3.14*r*r*h/(5m*4m*1m)=1/3*3.14*1000m*1000m*800m/20m3=4*10 7
percentage of the time for driving a car in a week=10h/day*5/(24*7)
t=1.68*100million h
assume average speed of truck is 60 miles/h, to move one piece of rock 10 miles, it will take 1/6h.
Assume the mountain can be broken up into 1000 pieces, then it will take 1000 journeys to move the entire mountain. As there is only one car, the same car needs to return after completing the journey to take more rocks.
1/6h 1/6h * 2*999 = 1/6h 1/6h * 2*1000
estimates: 60h