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”
I would estimate 1.2 million days
Let’s compare the mountain with a cone with base a circle with a diameter of 900 m and a height of 1000 m. The volume of this cone is comparable to the volume of a cylinder with as a base a circle of 300 m. So we would have to move approximately 280000000 cubed metres. I guess you can about 20 cubed metres a time. So that would be 14000000 trips. Every trip will take about 2 hours (including loading and unloading the truck). So 28000000 hours, which about 1.2 million days.
Here I am assuming that we break the mountain into rocks like in a quarry and carry the load.
To calculate the time, we can say
Time taken = (No. of loads required to carry the whole mass * no. of trips per load * 10 miles/speed of truck) (Time to break the mountain mass) (unloading loading time ) * no. of loads
So, to calculate the number of loads needed,
One load is limited by the weight the truck can carry
Average mountain: say 200m high, 400 m in diameter and conical in shape (assumption)
volume of the mountain = 1/3 * 40000 * pi * 200 = 6 million cu. m (approx.)
Density of mountain mass
A brick which is approximately 15 cm by 5 cm by 5 cm weighs 1 kg. Since the mountain rock would have higher density, assuming twice that of a brick, we can say
density of mountain mass = 1 kg/ (15*5*5*2) cu.cm =1 kg per 750 cu. cm = 1000 kg/ cu. m (approx.)
Weight of the mountain =volume * density = 8 million*1000 =8 billion kg
Now, coming to the truck,
to get the amount of weight it could carry: atleast 20 ppl of 50kgs could be carried in the truck making it 1000 kg; taking an upside for concentrated mass, we could say 1200 kg per load per trip.
no. of loads needed = 8 billion kg/1200 kg
= 6 million trips (approx)
Loaded truck would travel at say 40 miles per hour and 60 miles per hr when empty so on an average say 50 miles per hr
So, our first term in the equation, will lead us to
= no.of loads *2* 10 miles/ speed of truck =( no. of loads*2*10/50 ) hrs = (0.4 * no. of loads) hrs
For the breaking of the rock and putting it back using cranes and other equipment, time taken will be negligible as it will be carried out simultaneously as well.
Now, time taken to load and unload
say unloading takes 10 mins
loading however takes, maybe 3 mins per 40 kgs in a crane, that would make it 1.25 hrs
In total for loading and unloading lets say it takes 1.5 hrs. to round off
Now calculating the time required,
= (0.4* no. of loads 1.5* no. of loads)
= 2(appx)*6 million = 12 million hours
Here I am assuming that we break the mountain into rocks like in a quarry and carry the load.
To calculate the time, we can say
Time taken = (No. of loads required to carry the whole mass * no. of trips per load * 10 miles/speed of truck) (Time to break the mountain mass) (unloading loading time)
So, to calculate the number of loads needed,
One load is limited by the weight the truck can carry
Average mountain: say 50m high, 20 m in diameter and conical in shape (assumption)
volume of the mountain = 1/3 * 50 * pi * 100 = 5000 cu. m
Density of mountain mass
A brick which is approximately 15 cm by 5 cm by 5 cm weighs 1 kg. Since the mountain rock would have higher density, assuming twice that of a brick, we can say
density of mountain mass = 1 kg/ (15*5*5*2) cu.cm =1 kg per 750 cu. cm = 1000 kg/ cu. m (approx.)
Weight of the mountain =volume * density
Time to break the mountain (Time to load time to move 10 miles*2 Time to unload)*No. of rounds
No. of rounds = volume of mountain/ volume space of truck
Volume of mountain= volume of square pyramid = 1/3 volume of cuboid
Side of base of average mountain = 200 mtrs
Height of average mountain = 75 mtrs
Volume of mountain = 200*200*75/3 = 1000000 cube mtrs
Volume space of truck = length (6 mtrs)* breadth (3 mtrs)* height (3 mtrs)= 54 cube mtrs ~ 50 cube mtrs
Time to break mountain = only time while breaking mountain to fill first truck is to be considered as other parts can be broken when truck is moving, so it will be negligible.
time to load (assuming loaded by crane) = (volume of truck/volume of crane loader)*time per loading = {50/(2*2*2)}*1 minute=~ 6 minutes= 0.1 hour
Time to move 10 miles = (10 miles/average speed i.e. 50 miles/hr.)*2 = 0.4 hour
Time to unload = 1 minutes (assuming truck has in-built crane unloader)
Total time = Negligible time to break {0.1 hr to load 0.4 hrs to move 1 minute to unload}*20,000 rounds = 10,000 hrs 20,000 minutes = 10,333 hrs
Time taken to move the mountain 10 miles =
Assuming that the mountain is already broken down into pieces which are ready to be put on the truck, the time taken to move the mountain can be written in the following form:
(Number of trips required to move the mountain)*(Time taken to make a round trip of 10 miles Time taken to load and unload the truck)
Taking each of the three required numbers one by one,
Number of Trips required to move the mountain:
The number of trips required to move the mountain are the trips that the truck needs to make to transfer the entire volume, when it is filled up to its capacity in each trip. Hence, it can be written as:
(Volume of the mountain/Volumetric Capacity of the truck)
Volume of the Mountain:
The mountain can be thought of as a pyramid. Since the volume of the pyramid as 1/3*A*h (A, being the area of the base and h being the height of the pyramid), I will now try to estimate both A and h
Assuming base is a square of side 100 m, A = 10000 m2
Also let’s assume the height of pyramid to be 100 m
Volume of the pyramid is 100*10000 = 1 mn cu.m.
Volumetric Capacity of Truck:
Assuming the interior of truck is a cubiod, volume of its interior can be taken as = x*y*z (x and y being the sides of the cross-section and z being the length). Now I will like to estimate the dimensions to calculate the volume.
An average truck is around 2 times the size of a human being in cross-section sides, hence I would estimate both x and y to be around 3 m (twice the average human size of 1.6 m)
From my experience, I feel that the length of the truck is slightly more than the twice of its cross-section side; hence I estimate it to be around 7 m.
Volume of the interior of the truck = 3*3*7 = 63 cu.m. Rounding it off to 60 cu.m for the ease of calculation.
The number of trips is hence = 1 mn. cu. m/60 cu. m. = 16,667 trips. Rounding it off to around 16,000 (since we reduced the volume of truck in previous rounding off)
Time taken to make a round trip:
Time taken to make a round trip = 20 miles/Average Speed
Assuming the average speed of around 40 miles per hour (average over loaded and unloaded condition) for the round trip.
Time taken to make a round trip = 30 minutes = 0.5 hours
Time taken to load and unload the truck
Though time taken to load could be more than the unloading time, I will take the average time of around 2 minutes to load and unload 1 cu. m. of mountain.
Time taken to load and unload the truck = 63 cu. m. * 2 minutes = 126 minutes = 120 minutes = 2 hours
Hence the total time comes out to be =
(Number of trips required to move the mountain)*(Time taken to make a round trip of 10 miles Time taken to load and unload the truck)
16000*(0.5 2) = 40,000 hours = Around 1700 days or around 5 years
One way would be to assume the average truck has a speed of 20 miles/hour , it can carry 10 tons and the mountain has 1 million rocks, each of them weighs 10 tons.
time=distance/speed= 2*10/20=1 hour/rock; 2 because the truck goes 10 miles and comes back 10 miles for another rock.
So it takes 1.000.000 hours.
Another way would be to assume that the mountain is a 10 tons rock, so the truck can move it in half an hour.
This is my first real estimation problem so I hope at least I get some points in favor. Thank you Victor!
1-Finding average truck capacity: I remember seeing average cargo trucks lift cars therefore I used the estimated weight my car has.
Truck capacity at 100%: 6 tons
2-Mountain area: I live in a big hilltop in Colombia. the height difference btn Bogota to it is from 1.700-2.400 mts (700 mts). The length of the mountain is approx 24 blocks, about 2 km.
Note: I took halft of this mountain into consideration due that the problem states average mountain.
(So height 700/2=35o mts and length from 2000 mts to 1000 mts.)
Mountain area in mts= 1000mts x 350 mts= 350.000 mts2
2,1-Converting mts2 to kg: I know that I litter is equal to 1 kg, therefore I estimated the area 1 litter milk carton.
Milk carton: 10 cm length x 20 cm width= 200 cm2 which equals to 2mts2.
*Mountain Area in tons: 350.000 mts2 / 2 mts2= 125.000 kg
or 12.500 tons.
3-Truck travel distance relocation: 10 miles
avg speed of truck at full capacity: 10 miles /hour (as I´ve recalled truck speed in my farm).
*Total distance time per load: 2 hours
4-Total working hours: 8 hours per day (mon-fri)
4,1-Lunch time per day: 1 hour
4,2-Hours taken to load and unload dirt per journey: 1 hour
*Total hours per journey: 3 hours
*Total available working hours: 7 hours
*Journeys per day: 7 hours / 3 hours = 2 journeys carrying 12 tons per day.
*Total number of days needed: 12.500 total tons / 12 tons per day= 1.040 days
Note: I want to take a wild hypothesis; 1 hour was “given” away per day, the problem could also solve which would be more cost efficient; hier this person for full time or per day.
I know the tallest mountain is 27k feet high. So I would estimate that the “avg” size mountain is about 8k feet high. Assuming a pyramid shape to the mountain, I would put the length and width of the mountain at about 2k feet each. I think we can probably cut the pyramid in the middle and put the two sides together to form a rectangular shape of volume 8k x 2k x 2k = 32b cubic feet.
If we assume average sand/rock carrying trucks, they seem to have about a one-car length x 5ft high x one-car width sized box in the back. One car length is about 12ft and width is about 6ft so total box volume is 360 cubic ft.
32/36 = 0.88; 1b/10 = 100m; 0.88 x 100m = 88m round trips. If each truck goes 30mph avg speed start to finish then 10 miles will take 20min to go there and 20min to come back = 40min, and let’s assume another 20min to load & unload = 1hr per round trip. So it will take 88m hours.
53 hours total
assumptions:
a truck can hold 100 kilos of soil and an average mountain has 10,000 kilos of soil
a truck will travel the same rate (time and distance) with or without the load from point A to point B. Let us assume that it takes 10hrs for the truck to travel
we would like to find out how many hrs it would take
Thus our equation should be
10,000 / 100 = number of trips
number of trips x (HRStotravelpoint1-2 (x2))
i multiplied it with x2 because it goes back and forth thus doubling the time