# Estimation Question

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.

• Arthur Nov 11, 2010, 1:22 pm

I would take a pyramid 4km long, 3km wide, 2 km high to model my mountain. That’s 12km^3. Let’s assume we have 100 trucks to move rocks from the mountain. They can each move 3m^3 at once, and take one hour to go 10 miles away, dump the rock and come back. So that’s 300m^3 an hour. Assuming they work 8hrs/day, 300days/yr, that’s 720000m^3/yr, or 0.72 x 10^(-3) km^3.
12 / 0.72 x 10^(-3) = 1/6 x 10^6, so about 150 000 years. Better get started right now…

• Grigory Nov 11, 2010, 12:50 pm

The answer depends who (number of people & level of skill) is doing the mountain-moving and with what quarrying technology. If that’s the kind of technology they had in the stone ages, when they built the Stonehenge, – then very long indeed :). I will assume that we’re talking top-of-the-range modern excavators, drilling machines and trucks.

Independent variables needed to make a time estimation:
– Number and type of quarrying machines and their capacity

Dependent variables:
– The average time it takes to quarry 1 metric ton (drill, excavate, load on truck)
– The average time it takes to deliver 1 metric ton of mountain to 10 miles away

Constant:
– Volume of an average-sized mountain. (can be estimated by finding the volume of a cone with a radius of 1000 metres and a height of 1000 metres).

The formula:
Time it takes to quarry and transport 1 metric ton of mountain multiplied by the number of metric tons in the said mountain.

• GP Nov 11, 2010, 12:38 pm

(i am using SI units) let’s assume that an average mountian is 5000m high and that a average size mountain is a semisphere with a radius of 5000m. then the volume of it would be V=4/3xpixR^3×0.5. assuming that the mountainous material has density of 20, it’s mass is m=dxV.
since i am going to push it i assume i will exercise a constant force forever, resulting in a constant acceleration alpha = force/mass. i assume i can exercise force equal to my weight of 100kg, thus a force of 1000N, thus i know the acceleration alpha=1000/mass.
from physics we know that
distance=0.5xalphaxtime^2, distance in our case is 10 miles, or about 15km (which gives t=squareroot(2*distance/alpha)) which equals to roughly 25 years

• Sehar Nov 11, 2010, 12:38 pm

First some clarifying questions – where is the mountain, in the middle of a flat plain, in a mountain range, get a sense of the geography. Are we taking into account the time for equipment to help with the move to arrive?
Figure the mountain is attached at its base and must be removed from the base. We could either chop away at the bottom while having to make sure the mountain doesn’t topple over, perhaps having some sort of removable support devise put underneath, noting some of the mountain might be lost in the chopping, as in it crumbles away. This would take as much time as I assume it takes to drill a mine, which from my work with offshore drilling can be a few weeks. Or use explosive devices to try to break the mountain from its base which is a much faster technique that could be done in a matter of days, but we risk a possibly higher percentage of losing some of the mountain. Then we could either have the mountain rest on a plank and move it over land by pulling it with powerful automotive vehicles. It would probably take a few days to make sure the mountain was stable and then assuming we could get enough vehicle power while taking into account safety, we could move the mountain at a speed of 5 mph, or about in 20 hours. Or we could airlift the mountain by army grade helicopter devices, which I would assume would be able to move the mountain, taking into account safety, within a few hours. So the shortest amount of time to move the entire mountain including preparation for moving, could be a few days plus time to get equipment to the mountain to a few weeks. Physical movement from the mountain’s origin point to the point 1o miles away would range from a few hours to about 20 hours.

• Javier Nov 11, 2010, 12:37 pm

From the realistic point of view, the time that will take to relocate a mountain 10 miles, would the time the earth rotates from the first point where you measure the place till it reaches 10 miles. Saying that it will be different where the mountain is located..say that it is in the ecuator, so like that takes shorter time to do 10 miles..So, saying that, to rotate a mile the earth takes the whole distance of eacutor(perimeter)/24h..from this calculation we have miles/h..so then we will know how many min takes to move it
From the urealistic way of being able to move it with resorces, it will depend on how big it is, the path we have to cross to move it, etc..

• Medic Nov 11, 2010, 12:15 pm

revising my solution earlier, I should have split the mountain into three … upper third would take 30o hrs, mid 200 hrs, lower 100 hrs for 3.6m3 per man.

• all39 Nov 11, 2010, 12:07 pm

270 years / number of trucks

assumptions:

(1) earth movement by massive mining trucks
(2) decent road between the mountains
(3) sufficient earth moving equipment exists at the original mountain to dismantle it quickly enough such that trucks don’t have to queue
(4) sufficient earth moving equipment exists at new mountain site to assemble new mountain as quickly as the dirt comes in

• Erik G Nov 11, 2010, 11:53 am

So this is a question about a large scale industial project and
as such we will assume (cost) realistic amounts of available
labor and equtment and so forth.

Likely, the bottle neck in this process will be moving the large
amount of material 10 miles, and not the extracting of rock,
or unloading etc. Therefore we will estimate this by computing the
amount of material needing transport, and the rate at which we
could hope to move it.

A typical mountain (obviously ambiguous) we will assume to
be 1km in height, typically sloping at 45 degrees for simplicty,
and thus a radius 2km at the base. Since I am a physicist I
actually remember the formula for volume of a cone which
is (1/3)pi r^2 h, thus about 5 km^3= 5 * 10^9 m^3.

So I will assume a railway links the two sites, as this would
be a good investment if not already existing. A typical rail car has
dimension 5m*5m*30m=600 m^3. Assume 100 car
train, and there are two operating continuously. Realistically
it might take 1 hour to transport between the sites, so
assume a total of 10 loads per day.

Therefore 10x100x600=6*10^5 m^3/day. Dividing the
above figure we see that we need about 10^4 days to
transport the load which translates to 30 years.

Solution was typed out within a 5 minutes of seeing the
question.

• Surya S Nov 11, 2010, 11:41 am

I would start with a smile (!) and some clarifying questions:

1. Where is the mountain? On land or sea or some place else? (Say answer is land)
2. Does the mountain after move have to be in same shape and size? In other words, can it be broken and moved and the pieces just dumped 10 miles away? (Say answer is it can be broken)
3. Is there a direction preference and are there any obstacles to consider in that direction? (Say answer is no preference and no obstacles)
4. Is there any indication on what an ‘average size mountain is? (Say answer is 4000 meters in height, 2000 meters wide and 2000 meters long and assume it’s a pyramid)

Ok. With these inputs, the total time would be the sum of time of:

1. Breaking
3. Moving

Assumptions:
1. All tools and eqiupment needed are readily available and there is no lead time involved in procuring or setting up any of these.
2. We use dynamite to break the mountain and trucks to move the mountain.
3. One dynamite explosion breaks one truckload worth of mountain.
4. Truck carrying capacity is 20mx10mx10m = 2000m3.
5. To optimize time and labor, loading starts every 4 explosions and it takes 1 hour to load trucks (8 of them) and it takes 1 hour to setup the next set of 4 explosions.
6. Unloading is just dumping the truckload 10 miles away and on mountain regions truck move slowly so assuming speed to be 30 mph
1. Breaking Time
One explosion breaks 2000m3 of mountain. To break complete mountain, it takes approx 700K sets of 4 explosions. With each set taking 1 hour, it will take 700K hours to break the mountain.

Each loading starts after 4 explosions and takes 1 hour each, so to load 700K sets will take 700K hours but as loading happens in parrallel, only last load will take additional time of 1 hour.

Moving Time
3. As trucks will move in a mountain region slowly, let’s assume they take 20 mins one way to cover 10 miles (30 miles per hour speed) so will make a round trip in 40 mins and let’s assume it takes only 20 mins to unload the trucks as unloading is simple dumping of the mountain. So by the time one load is ready for pickup, the trucks are available and so no additional time involved in moving as the move also happens in parallel.

Additional Moving Time = 20 mins (one way for last load)

Assumed that unloading is just dumping the load and it takes 20 mins and this happens in parralel to breaking.

Total Time = 700K + 1 + 20/60 + 20/60 = 702K hours = 80 years (approx)

• Jakobicek Nov 11, 2010, 11:17 am

Oki that is a good one… lets see…

Lets try to approach it as follows we are going to move the mountain by moving material using trucks. The rest of the work is negligible. So we need to see how much material there is in an average mountain and how much can one truck pick up per load.

First, we need to somehow get in grasp with what is an average mountain… this is quite a vague description, but let us start with modeling an average mountain as follows – 500m high, square with a side of 10km

Now lets count how much material that is…

1/3*500*10000*10000 m^3

Fine next step is determining what an average size truck looks like…
yeah a truck is nothing more than a box that is 10m long, 4m wide and 2m high.

so now we have:

Now lets determine how many trips per hours you can make. 10 miles is about 16km so a truck that goes 80km/h should be able to get about 2 trips in per hour.( do not forget to count for return trip) with a work day of 10 hours we get to 20 trips per day.

500*10000*10000/(10*3*2*4*20)days

Oki now let us assume we have 100 trucks for the job

500*10000*10000/(100*10*3*2*4*20)days

Now lets get this mess sorted
500*10000*10000/(100*10*3*2*4*20)days = 500*100*100/(4*2*3*2)days ~ 500*2*100 days = 100000 days for 100 trucks ~ 300 years

• Himanshu Ahuja Mar 27, 2017, 11:45 am

Why have you taken *3 while calculating the load a truck can carry because the formula should be L*B*H i.e. 10*4*2 instead of 10*4*2*3