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”
In order to calculate the time we need to move an average size mountain along a 10 mile distance using the average size truck, we have to follow this formula:
Time = The volume of the mount/ The capacity of the truck x The time a truck complete the load, the unload, and the travel back and forth.
We will go to look at each the figure step by step.
First, start with the volume of an average mount. I reckon that a mount really resemble the shape of a cone with a square base. the volume of a cone is calculated using the following formula:
V = 1/3 x Height x Base area
Based on my observation, I estimate that the height of an average mount be 0.5 mile, and the base area is a rectangular with dimensions 2 mile x 3 mile. Applying these numbers into the above formula, we will get the volume figure to be 1 cubic mile for an average-sized mount.
The next step, we calculate the capacity of an average truck. The cargo of a truck is always a rectangular solid. Therefore, its volume is calculated using the following formula:
V = Height x Width x Length
I estimate that the length of a cargo is slightly more than 2 times the length of a normal sedan car, which is on average 9 feet long. So I would estimate that the length of a cargo is 20 feet. The width of a cargo is slightly bigger than that of a normal sedan car, which is on average 6 feet. SO I would estimate the width of a cargo to be 8 feet. Now, for the height of a cargo, from my observation, a cargo would be 1.5 times higher than a normal person standing on the base of the cargo. So I estimate the height of the cargo be 10 feet. SO using these estimation, we have the volume of a normal cargo to be 1600 cubic feet.
Now we calculate the last element of the big formula – that is, the time a truck need to complete all steps of moving the mount to a destination 10mile far away. First, assuming the road is in quite bad condition because this is a mountainous area, then the truck can only move with the constant speed of 20 mile per hour. Then, the time it needs to complete a round trip is 2 x 10 : 20 = 1 hr.
Assuming that the workers need 50 min to load the mount material on the cargo and the truck only need 10 min to unload, then we need 1 hour for the load and unload steps. In total, a truck would need 2 hr to complete a round trip to a 10 mile war away destination with the load and unload already involved.
So we use these numbers for the big formula above:
Time = 1 cubic mile : 1600 cubic feet x 2 hr = 5000^3 cubic feet : 1600 cubic feet x 2 hr = 125 x 10^9 : 16 : 10^2 x 2 = 2500 x 10^6 : 16 = (approximate) 150 x 10^6 hours = 150 x 10^6 : 864 years = (approximate) 170000 years
I would structure my solution in the following way.
To solve this, I need to know the time for each round trip with the averaged size truck, multiplied by the number of trips required to move the mountain.
If I assume a truck with a volume capacity of 20m cubed, and assume there are enough workers there to load and unload the earth into the truck in 10mins.
I then assume that a fully loaded truck can travel at 20mph while an empty truck can travel at 40mph. So The loaded journey will take 30mins and the return leg will take 15mins. The overall journey time is 10 (to load) 30 (to relocate) 10 ( to unload) 15 (to return) which gives 65mins.
If I assume our average mountain is roughly conical shaped with a base radius of a kilometer and a height of 2km, we can estimate the volume of earth contained to be 2*10^9 m cubed. This would result in about 10^8 trips required with a 20m cubed capacity truck. 10^8 * 65 mins gives 6.5*10^9 mins to complete the relocation process.
This assumes the work is done continuously, with no breaks in the process.
Using an average size truck, the mountain must be made of dirt
-Question is then how many truck-loads of dirt must be carried?
-How long does it take to move the truck 10 miles?
-At 30-35 miles an hour, it will take 20 minutes per trip
-An average size mountain is probably 3000 meters tall
-Let’s give it extra dimensions- 1000 meters long and 500 meters wide
-The volume is then 150, 000,000 cubic meters, or 150 million cubic meters
-An average truck is 5m *1m *1.5 m = 7.5 cubic meters
-150 million cubic meters of dirt, = 20 million truck loads
-20 million * 20 minutes = 400 million minutes
-5.75 million hours
-1 year = 8760 hours- roughly 10,000 hours
6,000,000/10,000
-Answer: 6 years, as a slight underestimation
In answering this question, I would first like to structure the problem such that it considers the relevant factors relating to the truck.
First off, the overarching formula to calculate the time taken to transport the average size load is a simple function of the speed/distance/time formula. From this, we know that:
Time = Distance/Speed
We have the distance (10 miles), therefore what we need to estimate is the speed the truck can travel at given its average size load. To estimate this, we can look at the maximum speed an average-size dump truck can travel and then adjust this downwards based on its load as well as driving adjustments. In my experience, the average shipping truck can travel at a maximum speed of around 60mp/h. However, a shipping truck is more mobile and less industrial than a dump truck so I’m revising that down to 50mp/h for your average dump truck.
Next we need to factor in how much it could be slowed down by an average size load. I would expect an average size load is close to (but not at) the maximum capacity of a dump truck because the average construction company or builder, for example, knows what capacity truck will be needed and will keep filling it up for each load, with a bit of room to spare. Therefore I’ll estimate an 80% load on the truck which we then just need use to work out how much this will slow the truck down.
Factors coming into play here include the driver wanting to be cautious and the weight slowing the truck. I’ll assume the driver applies a 10% margin off the maximum speed in his cautiousness and that the 80% load slows the truck by a factor of 40% of the maximum speed (considering gravity, engine strain and the assumption that the speed the truck travels at is inversely proportional to the load of the truck).
Therefore, the likely speed the truck will travel at over the 10 mile journey will be 50mp/h – 10%*50mp/h – 40%*50mp/h = 50 – 5 – 20 = 25mp/h.
We now just need to use the plausible speed we calculated in the original speed/distance/time formula stated above to arrive at an estimate of the time taken.
Time = Distance/Speed
Time for dump truck = 10/25 = 2/5 = 40% of an hour which yields 40%*60mins = 24 which I’ll round up to 25 minutes to make the journey.
Assumptions:
-An average track can move 4 cubic metres of mountain;
-Average mountine size 1000 meter height, 1000 meter in radius (base) making volume to be moved about 1 billion M3;
– Volume divided by truck volume gives .25 billion times needed
– One roundtrip will take .5 hours, and 8 hours working day, making around 20 trips a day;
Simple calculation gives 0.0125 million days needed.
“Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck”
Assume that a truck holds about 7 ft. x 30 ft. x 15 ft. The truck fills up the mountainous rock to the brim, so we have about 3150 ft^3.
Assume that the mountain is 3000 ft high, radius is 600 ft. Volume of a cone = pi * r^2 * h/3 = pi(360000)*1000 = 360000000pi ft^3.
Assume that the rate of filling up the tuck takes about 15 minutes and the dumping of the truck takes another 15 minutes, which is a total of 30 minutes per change in load.
The truck travels at a speed of 30 miles per hour, so it takes about 20 minutes to travel 10 miles to relocate the mountain.
So it will take approximately 360000000pi ft^3/ 3150 ft^3 = approximately 330,000 separate loads for the truck to go back and forth to move the mountain.
The truck has a gas tank that fills 20 gallons and it runs on 30 miles per gallon. Every 600 miles it needs to fill up gas which takes about 30 minutes.
The total amount of time it takes to move the mountain =
(time for change in load)*(# of loads) (# total time traveled)(# of loads) (# of loads*20 miles/600)*(time to get gas)
= 23111000 minutes
= 385000 hours
Assume there is only one truck with only one worker doing all the work. If the truck is average sized I assume it could carry about 4x6x4 (=96 –> round up to 100) cubic meters of clay per trip. The mountain is average sized, which I assume mean it consists of (200×100)/2=10.000 cubic meters. Each trip the driver could transport 100 cubic meters, which means it would take 100 trips.
One trip consist of driving both back and forth, as well as loading and unloading the truck. I assume the mountain is already broken down to transportable parts, and will therefore not calculate this in this estimation question.
10 miles equals 16 kmt. I assume he is driving on an industrialised road, which leads me to an other assumption that he has an average speed of 80 km/h for each trip, both ways. In total he needs to drive (16×2=) 32 km each trip, so each trip takes him (80/60 which is roughly 1,3 km/minute. 1,3*32 which is roughly 40 minutes. So each trip takes 40 minutes – only in driving. In addition to this I assume he spends 30 minutes on loading the truck, and 10 minutes unloading it. That means that each trip takes 80 minutes. As he has to take 100 such trips it would take him 800 minutes. 800/60 is about 13 hours.
The question asks how long (i.e. units of time) it will take to move an average sized mountain 10 miles away, using just one truck.
We’ll need to solve the following mathematical problem:
[time] = [number of truck trips] * [time of one roundtrip] – [time of return trip]
**As a note, the [time of return trip] is subtracted because once the final load is moved, the entire mountain has been moved. But this may be negligible in our calculation, and certainly too minute to consider in typical strategy level problems.
[number of truck trips] = [volume of mountain] / [volume of truck]
[time of one roundtrip] = [time to load the truck] [10 miles] / [average speed with full load] [time to unload the truck] [10 miles] / [average speed with no load]
To begin, let’s assume that or mountain takes the geometric form of a pyramid. Let’s also assume the average height of a mountain is 2000 m, with average base dimensions of 2000 m x 2000 m. Our volume is then (l * w * h)/3 = (2000 m * 2000 m * 2000 m)/3 = (8 billion m^3)/3 = 2.67 billion m^3. Let’s round up to 3 billion m^3 as our mountain volume.
Let’s assume that our truck can carry a load that fits into a cubic space of 3m x 2m x 5m, or 30 m^3.
So in this case, [number of truck trips] = [volume of mountain] / [volume of truck] = 3000 million / 30 = 100 million trips.
Next, we’ll estimate that it takes 0.5 hrs to load the truck, and the same amount of time to unload it. It will drive 40 mph with a filled load, and 60 mph empty.
In solving our predefined formula, [time of one roundtrip] = [time to load the truck] [10 miles] / [average speed with full load] [time to unload the truck] [10 miles] / [average speed with no load] = [0.5 hrs] [10 mi / 40 mph] [0.5 hrs] [10 mi / 60 mph] = [0.5 hrs] [.25 hrs] [0.5 hrs] [0.17 hrs] = 1.42 hrs. Let’s round this up to 1.5 hrs per roundtrip.
Because we are working with quite a large number, and to make it “relate-able” on the C-level, I will say that 1.5 hrs = 1/16 of a full day (yes, we will be working 24/7/365).
Finally, [time] = [number of truck trips] * [time of one roundtrip] = [100 million] * [1/16 day] = [100/16 million days] = 6.25 million days.
6,250,000 days / [365.25 days/yr] is roughly 17,000 yrs.
It will take about 17,000 years to move an average sized mountain 10 miles away using one truck.
Height of a typical mountain = 1000 m = 1 km
Base of a typical mountain = 1000 m = 1 km
So, volume of a typical mountain = 1/3 x 3.14 x 1 x 1 x 1 ~ 1 km cube
Volume of a typical truck = 10 m x 5 m x 5 m = 250 m cube
Avg. speed of a truck = 40 km/h
So, for covering 10 miles ~ 20 kms to and fro it would take 1 hours which means each 250 m cube relocation would take 1 hour
So, for relocating 1 km cube it would take = 1000 x 1000 x 1000 / 250 = 4 million hours
Time to relocate = {(i) (2 x truck travel time) (ii) (2 x load/unload time) } x (iii) #of trips made
(i) 2 times since truck would go back and forth to travel 10 miles
Assume average velocity = 50 miles/hr
= 2 x (10mil/50 mil/hr)
= 2 x 0.2 = 0.4 hours
(ii) 2 times since there will be load & unload times
= #of shoveling x time required per shovel x 2
# of shoveling = truck capacity (volume)/shovel capacity (volume)
Front row of truck can seat 4 people, assuming width of a person is 50 cm, then the truck width is 2m. Height is half of me if I stand up, around 0.75 cm. Length a little bit more than width but not twice assume 3m. Shovel volume assume as pyramid with base a size of A4 paper.
= 2m x 3m x 0.75 m / (0.3m x 0.2 m x 0.1m/2)
= 4.5 m3/ 0.003 m3
= 1500 times
Time per shovel 15 sec (assume man is standing near enough to truck).
= 1500 x 15 sec = 22,500 second/ truck = around 5 hours/truck x 2 = 10 hours/truck
(iii) no of trips made = truck volume/mountain volume
Truck volume = 4.5 m3
Mountain volume= 1/2 (phi x r2 x t) = 1/2 (3.14 x 1,000,000 x 1,000) = 15,000,000,000 m3
Average mountain height 2,000 m from sea level. Assume 1,000 m height, with 1,000 m radius since usually mountain are shaped like very short cone.
# of trips made = 3,300,000,000 trips
*****************************************************
Time to relocate = {(i) (2 x truck travel time) (ii) (2 x load/unload time) } x (iii) #of trips made
= 10.4 hours x 3,300,000,000 trips = 343,200,000,000 hours