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”
800,000 minutes:
answer = (mountain volume / truck volume) x round trip time
mountain volume: 800 m x 200 m = 160,000 m^2
truck volume: 4 m x 2 m = 8 m^2
round trip time: 30 mph -> 10 miles = 20 minutes each way x 2 = 40 minutes
(160,000 / 8) x 40 = 20,000 x 40 = 800,000 minutes
Considering a mountain of radius 1 km and height 2 kms. Volume will be around 2 cu. kms
Truck volume will be around 24 cu metres. the truck takes around 30mins to transfer and return.
Calculating it gives 35 years.
My answer, based on assumptions and calculation listed below is 2,559.8 hours.
Assumptions:
– The average mountain may be considered a mountain having average height of (maximum mountain in the world, which is approx. 8 km) and (lowest mountain in the world, which is 0 meters), which gives us a 4 km height
– The volume of the mountain for the simplicity may be calculated as volume of the cube, which gives us a 64 sq. km
– For simplicity reasons the 1000 ton weight of the mountain may be considered equal to 1 sq. km
– Average truck capacity may calculated as average of (maximum truck capacity, which is let’s say 40 tons) and minimum truck capacity (let’s say 0 tons) which gives us a 20 tons average truck capacity
– The mountain is located outside of the city (as most of mountains do), thus the country speed limits may be settled as in many countries, i.e. 90 km/h outside of the city
– Average full truck speed on the 10 miles road may be considered as 60 km/h maximum (given the time for reach this speed on this 10 miles road), average empty truck speed is 80 km/h
– Loading time may be considered as 15 min, unloading – 5 min (let’s suggest the truck has own mechanism for fast unload)
– The truck starts directly from the mountain
– All the arrangements to the reallocation are at hand (i.e. all the technics which will load the mountain mass into truck is at place, all the personnel is ready to work)
Based on these assumptions the average time of reallocating the mountain on a 10 miles distance may be calculated in the following way:
1) Number of tons in the mountain = 64,000 tons (i.e. 64 sq. km * 1000 tons per 1 sq. km);
2) Number of trucks required to reallocate the mountain = 3,200 trucks (i.e. 64,000 tons / 20 tons);
3) Delivering of first truck of mountain to the 10 miles destination = 0,6 hours (i.e. 16 km (which is 10 miles) / 60 km/h 15 min (load time) 5 min (unload time))
4) Number of rides left = 3,199 (i.e. 3,200 trucks – 1 truck);
5) Delivery of each next 20 tons will take 0.8 hours (i.e. time for road back of 16 km (which is 10 miles) / 80 km/h 0.6 (which is the same as time for delivering of first truck))
6) Time required for delivery of rest mountain = 2,559.2 hours (i.e. 3,199 trucks left * 0,8 hours)
7) Total time required for delivery of mountain = 2,559.8 hours (i.e. 0.6 hours (first time delivery) 2,559.2 hours (time required for delivery of rest of mountain))
To answer the question, I will try to estimate:
1) the mass of the mountain,
2) the overall time it will take an average truck to load, unload and drive 20 miles,
3) the load a truck could carry,
in order to estimate the time it will take to move the mountain 10 miles away.
To simplify my calculations I will use meters instead of miles and kg instead of lb.
Knowing that the world’s highest mountain, Mount Everest, rises over 8000 meters, I hypothesize that an average mountain’s height would be at 1000 meters and its base at around 2000 meters. I further assume that the shape of the mountain resembles a cone; inconsistencies around the shape are canceled out.
The volume of such a mountain is given by the formula (3.14)*(r^2)*(height/3). Since r and h are large numbers 3.14 is cancelled out with 3 and V=1,000,000,000 m^3.
Assuming that mountains avg density is 2, the mountain’s mass should be 2,000,000,000 kg (density=mass/volume).
Let’s assume that an avg truck can carry up to 5 tons that means it will take 400,000 truck rides to move the mountain, also assuming the volume of each load fits within the truck.
With an avg speed of 40 miles/hour it will take 15 min for the truck to complete half a ride and 30 mins for the whole ride (I use avg speed because the truck will return faster without the load). Adding 10 min to load and 5 to unload that means 45min in total for a truck to move 5 tons 10 miles away and get back.
Time to move the mountain 10 miles away= number of rides needed * minutes / ride= 400,000 rides * 45 min/ride= 18,000,000 min. Divided by 60 that is 300,000 hours. Divided by 24 this is 12,500 days. Divided by 365 that’s is approximate 34 yrs – assuming my calculations are correct.
Two things could change this result: 1) the truck’s speed, but I cannot argue that if the truck goes up to 70ml/hr that will change the result significantly 2) the load the truck could carry. Either one truck with double the size or an additional truck could take that time down to half.
time = # of travels * time of each travel
# of travels :
– the truck has 2 * 3 * 10 m3 of volume = 60 m3
– the mountain has a height of 4.000m and a base diameter of 4.000m. So it has a volume of 250.000.000m3.
time of each travel:
– average of 30 miles per hour
– so it is 20 minutes for going, 20 minutes for goind back and I considered 20 minutes for filling the truck. So it’s 1 hour per travel.
Finally, the time is 1 hour * 250.000.000 of travels, which is equal to 250.000.000 hours, or aproximately 10.000.000 days or 30.000 years.
The solution to this problem requires, the travel time and the number of trips required.
My first step was to establish a value for the volume of mountain. I used the equation for volume of a cone, this was in my opinion the best was to estimate the shape and size of an average mountain. I used a height of 6,000 ft (a value close to mountains near my hometown) and a radius of 5,000 ft (given mountains near me are fairly steep) to get an estimate of 150 billion cubic feet of mountain.
Next I assigned values of what an average truck bed volume would be. I used 4x7x10, a value I believe accounts for an average truck, bigger than an everyday pick up but smaller than a dump truck, and rounded up to 300 cubic ft.
Using the two volumes I calculated the required number of trips to be 500 million.
The next step was to establish a trip time. Given an average speed of 40 mph the 10 mile trip would take 15 minutes. Additionally it would take some time to load/unload the truck, I believe the load would take longer so I estimated 50 minutes for load and 35 to unload. This yields a total travel time of 100 minutes.
Finally multiplying total trips by total time, It would take approximately 50 billion minutes to move the average sized mountain 10 miles.
I approximated the mountain as a cone with the formula of pi*r^2*h/3 which I approximated to r^2*h. I assumed the average hight of a mountain is 800 meter and the radius to 500 meters. With that in mind, the volume of the mountain is 2 million m^3.
Assuming the standard truck have a loading space of 2*2*4 m^3 this create a need for approximatly 100 k rounds.
Assuming one round takes 1 h and 10 minutes (20 minutes to drive and 25*2 minutes to load) this will take about 100 00 hours. Here I assume that the mountain is worked on so that the driver can just load the parts.
one day is 24 hours which I round to 25 and one year consists of about 250 workdays, we get 100 k/25 = 4000. 4000/250 = 16.
Hence, my answer to this is 16 Years.
Estimated answer is 22 days.
100 miles is around 160km, given 100km/h, one cycle will be just short of 4 hrs, if load and dump time are included.
depending on the type of material of the mountain, we will assume it is consisting of a good solid manganese/iron ore which is quite heavy, around 100tons with an average size truck per cycle.
The mountain of average size, i compare with half the size of Table mountain (South Africa), should be around 500million tons, this depends heavily what your framework of an average size mountain is.
given that, 1million cycles ends up to be 4million hours, continuous operating (hot-seat change over) and with a 80% maintenance availability, million hours.
now what is 5 million hours in terms of days or months, it will take you 570 years minimum with one truck.
given that this most of the times are expected from mining companies all over the world, one can understand why they choose to invest billions of dollars into mega size equipment.
so economically scaled!
1. Size of the truck = 10 ft x 40 ft = 400 sqft
2. Size of the mountain = 4000 ft x 1000 ft = 4M sqft / 2 = 2M sqft
3. Truckload = 50,000 truckloads
4. Load time = 5 minutes x 2 = 10 minutes
5. Travel time = 10 minutes x 2 = 20 minutes
6. 30 min per load = 0.5 hours per load
7. 25,000 hours
8. 2000 hours per shift per year @ 3 shifts …roughly 4 years
Not accounting for maintenance, repair and gas stops. Maybe close to 5 years.