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”
Truck –
Capacity of the truck – 1000Kgs
Average time of the truck for one trip –
During Peak traffic – 2 hrs
During non-traffic time – 1 hr
Loading and unloading time total = 1 hr
Total time during Traffic for 1000Kgs – 3 hrs
Total Time during Non-traffic – 2 hrs
Assuming that the truck is refued while loading and unloading so no time is added for refueling
and the driver takes a break during the loading/unloading so no additional time there. The truck runs all the time till the process is completed
Mountain-
Weight of sand/gravel/other – 1000,000,000kgs
Road –
Peak time is there every 20 hrs- for every 3 hrs peak time there is 20 hrs non-peak traffic
No of trips = 1000,000,000/1000 = 1000,000
20hrs – 1000*10 = 10,000 kgs
3 hrs – 1000 kgs
23 hrs for 11000 kgs
how much for 1000,000,000 kgs
therefore 1000,000,000 * 23 /11000 = approx 2,090,000 hrs
Assume the average mountain is of a conical shape of 10km in base radius and 1km in height. That would give us the total volume of the mountain to be 1/3*pi*10km^2*1km=100km^3=1e11m^3. The average truck load volume can be estimated by 4m*2m*1m=8m^3, so it will take 1e11m^3/8m^3=1.25e10 runs for the truck to move the mountain away. Assume the truck drives at an average speed of 40mph, to make a round trip to and from 10 miles away would take half an hour. Therefore to move the entire mountain the total time would be 1.25e10*0.5hr=6.25e9 hrs. Each year has 365*24 hrs=8760 hrs. Hence 6.35e9 hrs equals approximately 7.1e5 years.
(1) Let’s first define the volume of averaged mountain:
Based on the experience, I am assuming the mountain occupies a square area (1 mile x 1 mile) and its height is 0.5 mile. To simplify the calculation, it appears as a cube. So the volume is: 1x1x0.5 = 0.5 mile^3 ~ 2 x 10^9 m^3.
(2) Now let’s consider the size of a averaged truck = 2x2x5 = 20 m^3
(3) Total number of transportations required: (2×10^9)/20 = 10^8 (times).
(4) In order to relocate, workers need time to mount the dirt, drive the car for 10 miles, and un-mount dirt. All these steps take time. Let’s assuming it takes 5 minutes to mount as well as unmount the dirt. And drivers drive at 30 miles/hr. so it takes 20 minutes total drive time for one way. Overall, a round trip costs 5 min (mount dirt) 20 min (drive to relocation spot) 5 min (unmount dirt) 20 min (drive to mountain) = 50 minutes.
(5) Now assuming workers can work 12 hours a day to relocate the mountain. Number of transportation within the 12 hrs is 12 (hr) x 60 (minutes/hr) / 50 (minutes) = 14 rounds (with 20 minutes remaining).
(6) As a result, the total working days required to relocate the mountain would be (10^8) / 14 = 7142857 working days ~ 19569 years.
“Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck”
An average mountain has a total amount of 1 million tons of earth. An average size truck can carry one ton of earth at once. It also takes 3 mins to load the truck and 1 mins to unload the truck. The truck drive at an average speed of 60 mile/hour which makes each round turn drive cost 20 mins. Therefore, for the truck to carry one ton of earth cost around 20 3 1 mins. So, to move an average mountain 10 miles, it takes about 24 million mins to do it
Moutain height = 3000m
Mountain basis = 500m
Moutain volume = 176 625 000 m3
Truck volume = 8 x 3 x 2 = 48m3
Loading time = unloading time = transport time = 20 min
Number of trucks needed = 176 625 000 / 48 = 3 679 688
Time needed for a truck to load, transport, unload and come back = 80 min (1,3 hours)
Time needed = 3 670 688 x 1,3 = 4 771 894, 4 hours (or 198829 days or 544,7 years)
How big is the mountain?
Having been to various mountains I would say that an average sized mountain is 2km in height and 4km in width. I will assume that the mountain takes a generic circular based cone shape. (i.e. radius is 2km)
Mass of mountain: (1/3)*pie*(radius^2)*height =~ (1/3)*3*4*2 = 8km^3 = 8,000,000,000 m^3
How much can the truck hold?
Assuming: length 6m; width/height 2.5m, the load of the truck for each journey is 6*2.5*2.5=37.5. Let’s round to 40 for simplicity.
Number of journeys required = 8bn/40 = 200 million
Time require = (# journeys)*(transport time load time unload time)
I assume an average speed of 40mph of truck (driving fully loaded at 30mph and returning unloaded at 50mph). Therefore transport time (to and fro) is 30min
Loading – I assume one can load the truck in 25 minutes. (at a rate =~ 1.5m^3/min)
Unloading – Much quicker because cargo is dropped out using hydraulic raisers and so unload time of 5 min.
Refueling – Ample time to do it when loading/unloading
Therefore, (in hours) Time required = 200m*(0.5 0.4 0.1)= 200 million
5,500 days.
First off, lets consider that we need to move all the mountain volume using the trucks loading capcity (i.e. the trucks volume).
Lets then consider an average size mountain to be a pyramid with a base that is a sqaure.
Then, lets say that this mountain has each side of its base equal to the length of 50 avg. sized trucks lined up one after the other and a height of 100 avg. sized trucks stacked up vertically.
As for the avg. sized truck, let us say that it is a rectangle with length of 4 meteres, base of 2 meters, and height of 2 meters.
Thus, each side of the mountain is 200 meters long (50*4) and has a height of 400 meters (100*4)
As the volume of a pyramid is calculated as: V=1/3*(b^2)*h, and b=200 meters and h=400 meters, the volume of the mountain is approximately 5,300,000 meters^3
The volume of a rectangle is calculated as: V = b*l*h, and b=2 meters, l=4 meters, and h=2 meters, the volume of the truck is 16 meters^3
Thus, we will need apprxoamitely 330,000 truck loads (5,300,000/16) to move the entire mountain
That means that the trucks with have to make 330,000 round trips which means that it will travel 10 miles 660,000 times. That makes for total travel of 6,660,000 miles.
If the truck travels at 50 miles per hour, our travel time is 132,000 hours which is equal to 5,500 days
For this problem, the principal underlying assumptions are the following:
-100% of the voulume of the truck can be used to move the mountain
-the truck is countiounsly moving at 50 miles per hour
-upon arrival to the mountain, the truck is instantly loaded at 100% of its capacity
-upon arrival at destination, the truck is instantly unloaded and instantely being reconstructed.
To estimate this question, I think we need to consider 5 factors:
1. Volume of an average mountain
2.Maximum volume that a truck can take
3.Driving speed to and from the 10 miles to dump the contents
4.Time it takes to upload the truck to full capacity
5.Time it takes to offload the truck contents
Starting from the first one. I would assume an average mountain is of the shape of a cone and thus has an average radius of 100 meters, a height of 1200 meters. Therefore, the volume of the mountain would be 1/3* pi * (100)^2*1200, i.e. 4*10^6*pi cubic meters.
2. Assume the container of an average truck is of 1 meter height, 4 meters long, and 2 meters wide. So, the volume of the container would be 1m*4m*2m=8 cubic meters.
3. Assume the driving speed to and from the dumping place is 20 miles/hour and thus the time it takes to and from the dumping place would be 2* (10/20)=1 hour
4. Time it takes to upload the mountain contents to the truck container. I assume there are two ppl using tools to haul the contents into the container and it takes them 3/4 hour to fill the container to full.
5. Assume it is much faster to offload the contents from the container to the dumping place, which takes only 1/4 hour.
Thus, overall time it takes to move an average mountain to 10 miles away would be (4*10^6*pi)/8*(1hour 3/4hour 1/4hour)=10^6*pi hours
I will assume that an average size mountain is about 1/4 of a mile high by 1/4 of a mile wide. To simplify, I will assume this mountain is a sphere. A mile is 5280 feet, so 1/4 of a mile is 1320. The volume of a sphere is (3.14 x d^3) / 6, so plugging my estimated average mountain into that equation, (3.14 x 1320^3) / 6 is about 230,000,000 sq ft. I will assume the average truck bed is 6x4x2 ft. With that assumption, the average truck can carry 48 sq ft of dirt. I will assume that each load will take 30 seconds to load into the bed of the truck and 30 seconds to unload. That means the actual loading and unloading will take about 4,800,000 minutes, which is 200,000 days or about 550 years. So, that’s just loading and unloading.
For the actual transportation, it takes 20 miles for the truck to drive to the depositing site and back. The truck will make about 4,800,000 of these trips, which is 96,000,000 miles of driving. Assuming the truck drives on average 30 mph, this will take 3,200,000 hours or about 350 years.
Adding the loading and unloading time with the driving time, it will take about 900 years for an average size truck to move an average size mountain 10 miles.
Mountain size assm: 5000′ high, 5000 radius. Vol = 1/3 * pi *r *h. Mountain volume = 125,000,000 cf
Truck size : 8’x8’x25′. Vol = LxWxH = 1600 cf.
Truckloads = Mtn Vol/ Truck Vol = 80,000 loads
Cycle : load drive unload return drive = 15 min 20 5 10 = 50 mins/cycle
Duration = 80,0000 loads x 50 mins/load = 4,000,000 mins ~ 9 yrs