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It started with a lunch conversation that slowly turned into a coffee time mini-debate. You see, I moved into a new job recently at the Applied Cloud Computing Lab in HP Labs Singapore. My first task was to hire engineers to staff the lab and while the final numbers were not absolute, the number 70 was bandied around for the population of the facilities. We have half a floor at Fusionopolis, plenty of space really but the bone of contention was more delicately focused on more sanitary facilities i.e. the toilets.

The problem was that the whole floor shares a single pair of toilets (one for male and another for female of course). I was totally convinced that with 70 people in the office, we will reach bladder or bowel apocalypse, a disaster in the waiting. My other colleagues however were less worried — the other floors seem to be doing fine. It wasn’t as if we could do anything about it, no one’s going to magically add in new toilets for us no matter how long the toilet queue was; but I had to know. I was also curious of course — what is the algorithm used to determine the number of toilets per square meter of occupancy?

Looking up building regulations found nothing (at least not online) in Singapore but UK has some regulations covered by the Health and Safety Executive.

Looking at facilities used by men only:

Well, the toilets seem to almost fit into the UK regulations, but these are just regulations. How did these numbers come about and how realistic are they? To better model the usage scenarios, I did a Monte Carlo simulation on the usage pattern of the toilets, based on a simple set of assertions.

Taking an educated guess of 1 in 7 of engineers to be female, that leaves us with 60 male engineers sharing a single toilet facility. The toilet facility has 3 urinals and 2 stalls, one of which is a flush toilet, the other is a squat toilet (which in general no one uses any more).

An important piece of information is to find out how many times an average person need to urinate a day. It turns out that our bladder will signal us to make a trip to the toilet when it fills up to about 200ml. Combine with the information that an average urine output of an adult is 1.5 liters a day, this means that we roughly need to go about 8 times a day. This in turn means 8 times in 24 hours or 1 time in 3 hours on an average. The normal working hours in HP are 8:30am – 5:30pm, which is 9 hours, so this means on an average a person makes a beeline to the toilet 3 times during the course of the working day in the office.

I modeled the scenario discretely, per minute. This means a person in the office would make a decision to go to the loo or not every minute within the 9 hours in the office, or 540 minutes. Coupled with an average of 3 times in 540 minutes, this means the probability that a person would go to the toilet every count of the minute is 3 out of 540. In the case of urinating, my estimate is that a person will be able to complete the necessary job within 2 minute on an average and vacate the facility.

Armed with this model, let’s look at the simulation code:

require 'rubygems'
require 'faster_csv'

# population of 60 people on the same floor
population = 60
# period of time from 8:30am -> 5:30pm is 9 hours
period = 9
# 8 times a day -> 8 times in 24 hours -> 1 time every 3 hours
# within a period of 9 hours -> 3 times
# probability in every min -> 3 out of 9 * 60
times_per_period = 3
# rate of servicing
mu = 2
# number of people in the queue
queue = 0
# number of stalls available
stalls = 3
# number of occupied stalls
occupied = 0

# total of all queue sizes for calculating average queue size
queue_total = 0
# max queue size for the day
max_queue_size = 0

FasterCSV.open('queue.csv', 'w') do |csv|
  csv << %w(t queue occupied)
  (period * 60).times do |t|
   # at every rate of service interval, vacate any and all people from the toilet and replace them with people in the queue
    if t % mu == 0
      occupied = 0
      while occupied < stalls and queue > 0
        occupied += 1
        queue -= 1
      end
    end
    population.times do
      # does the engineer want to relieve himself?
      if (rand(period * 60) + 1) <= times_per_period
        if occupied < stalls
          occupied += 1
        else
          queue += 1
        end
      end
    end

    csv << [t, queue, occupied]
    max_queue_size = queue if queue > max_queue_size
    queue_total = queue_total + queue
  end
end

puts "** Max queue size : #{max_queue_size}"
puts "** Total mins in queue by all : #{queue_total} mins"

Most of the code are quite self explanatory. However notice that mu is 3 (minutes) instead of 2 as we have discussed earlier. The reason for this is that in keeping the simulation code simple, I’ve not factored in that if we clear the toilets every 2 minutes, people who entered during the 2nd minute (according to the simulation) will be cleared in just 1 minute. To compensate for this, I use a mu variable that is 1.5 times the service rate.

The simulation writes to queue.csv, which contains the queue size and toilet occupancy at every interval. This are the results:

** Max queue size : 3
** Total mins in queue by all : 17 mins

I opened the CSV up and used the data to create a simple histogram:

From the results it’s pretty obvious that the architect and engineers who built Fusionopolis knew their stuff. These were the quantitative conclusions:

1. Queues were formed less than 3% of the time of the day
2. The maximum number of people in the queue were 3 people and that happened for only less than 2 minutes in a day
3. The total time spent by everyone in the office, in a queue to the toilet was 17 mins

This is for the men’s urinals, let’s look at the stalls now. The same simulation script can be used, just changing the parameters a bit. Instead of mu = 3, we use mu = 8. Instead of 3 urinals, we have 2 stalls and instead of 3 times over the period, we use 1 time over the working period.

These are the results:

** Max queue size : 3
** Total mins in queue by all : 81 mins

The problem seems a bit more serious now though not significantly so, though the chances of being in a a queue are higher now (10%).

In conclusion, I was wrong but I’m glad to be wrong in this case. Waiting in a toilet queue is seriously not a fun thing to do.

In my previous blog post, I came to the conclusion that

If you’re a hiring manager with a 100 candidates, just interview a sample pool of 7 candidates, choose the best and then interview the rest one at a time until you reach one that better than the best in the sample pool. You will have 90% chance of choosing someone in the top 25% of those 100 candidates.

I mentioned that there are 2 worst case scenarios for this hiring strategy. Firstly, if coincidentally we selected the 7 worst candidates out of 100, then the next one we choose will be taken regardless how good or bad it is. Secondly, if the best candidate is already in the sample pool, then we’ll be going through the rest of the population without finding the best. Let’s talk about these scenarios now.

The first is easily resolved. The probabilities are already considered and calculated in our simulation so we don’t need to worry about that any more. The second is a bit tricky. To find out the failure probability, we tweak the simulation a bit again. This time round instead of getting every sample pool size, we focus on the best results i.e. sample pool size of 7.

require 'rubygems'
require 'faster_csv'

population_size = 100
sample_size = 0..population_size-1
iteration_size = 100000
top = (population_size-25)..(population_size-1)
size = 7
population = []
frequencies = []

iteration_size.times do |iteration|
 population = (0..population_size-1).to_a.sort_by {rand}
 sample = population.slice(0..size-1)
 rest_of_population = population[size..population_size-1]
 best_sample = sample.sort.last
 rest_of_population.each_with_index do |p, i|
 if p > best_sample
 frequencies << i
 break
 end
 end
end

puts "Success probability : #{frequencies.size.to_f*100.0/iteration_size.to_f}%"

Notice we’re essentially counting the times we succeed over the iterations we made. The result for this is a success probability of 93%!

Let’s look at some basic laws of probability now. Let’s say A is the event that the strategy works (the sample pool does not contain the best candidate), so the probability of A is P(A) and this value is 0.93. Let’s say B is the event that the the candidate we find out if this strategy is within the best quartile(regardless if A succeeds or not), so the probability of B is P(B). We don’t know the value of P(B) and we can’t really find P(B) since A and B are not independent. What we really want is  . What we do know however is P(B|A) which is the probability of B given A happens. This value is 0.9063 (from the previous post),

Using the rule of multiplication:

We get the answer:

which is 84.3% probability that the sample pool does not contain the best candidate and we get the top quartile of the total candidate population.

Next, we want to find out, given that we have a high probability of getting the correct candidates in, how many candidates will we have to interview before we find one that is better than the best in the sample pool? Naturally if we have to interview a lot of candidates before finding one, the strategy isn’t practical. Let’s tweak the simulation again, this time to count the successful ones and get a histogram of the number of interviews before we hit jackpot.

What kind of histogram do you think we will get? An initial thought was that it would be a normal distribution, which is not great news for this strategy, because a histogram peaks around the mean, which means the number of candidates we need to interview is around 40+.

require 'rubygems'
require 'faster_csv'

population_size = 100
sample_size = 0..population_size-1
iteration_size = 100000
top = (population_size-25)..(population_size-1)
size = 7
population = []
frequencies = []

iteration_size.times do |iteration|
 population = (0..population_size-1).to_a.sort_by {rand}
 sample = population.slice(0..size-1)
 rest_of_population = population[size..population_size-1]
 best_sample = sample.sort.last
 rest_of_population.each_with_index do |p, i|
 if p > best_sample
 frequencies << i
 break
 end
 end
end

FasterCSV.open('optimal_duration.csv', 'w') do |csv|
 rest_of_population = population[size..population_size-1]
 rest_of_population.size.times do |i|
 count_array = frequencies.find_all{|f| f == i}
 csv << [i+1, count_array.size.to_f/frequencies.size.to_f]
 end
end

The output is a CSV file called optimal_duration.csv. Charting the data as a histogram, surprisingly we get power law graph like this:

This is good news for or strategy! Looking at our dataset, the probability of the first candidate in the rest of the candidate population to be the one is 13.58% while the probability of getting our man (or woman) in the first 10 candidates we interview is a whopping 63.25%!

So have we nailed the question if the strategy is good one? No, there is still one last question unanswered and this is the breaker. Stay tuned for part 3! (For those who are reading this, an exercise — tell me what you think is the breaker question?)

When I was working in Yahoo at Suntec City, I used to take the MRT to work every day and my route is completely sheltered right to the doorsteps of the office so the weather didn’t really have much effect on me. I still take the MRT to the office now that I am at Garena, but there is a short distance from the MRT station to the office, about 5 minutes of walking, where it is subject to the elements.

Now I don’t take any umbrellas along with me, so with the recent change in weather (it has been raining cats and dogs and all sorts of mammals in between) I have been caught a few times in some unpleasant wetness. It is during one of these wet commutes from home to office I was thinking of ways of making sure that will not happen again.

The idea was like this. I will place some umbrellas at the office and also at home. Whenever I leave the house or the office and if it is raining, I will take the umbrella to the office or back home respectively. Naturally if it’s not raining then I wouldn’t want to look like a dork and carry one.

Therein lies the problem. Let’s say I keep 1 umbrella at the office and another at home. If it rains when I’m leaving the office, I will carry that one home, leaving 2 at home and none at the office. If it rains the next day when I leave home it’s ok, I will just take an umbrella to the office, making it 1 – 1 again. If it doesn’t rain and therefore I don’t take any umbrellas to the office, I will be left with 2 – 0. If it happens to rain when I’m leaving the office I’m stuck with getting wet again.

What if I have 2 umbrellas at the office and 2 at home? In this case I will be caught wet only if the same occurrence (rain when leaving the office followed by no rain when leaving the house) happens 2 days in a row, which probability is lower than if it happens just once. Still, that can happen and I want to get assurance that the risk of getting wet is really minimal. In that case, the question becomes, how many umbrellas should I leave at the office and similarly at home such that the probability of getting wet is negligible? Note that it doesn’t really matter if one place has more umbrellas than the other since it will be the same as the lower number of umbrellas in the long run.

To find out, I used a popular (and really pragmatic and therefore very ‘engineering’) method of finding probabilities — the Monte Carlo simulation method. The name sounds fancy but it’s really just an algorithm that uses repeated random samples to find the answer. In our case, what we want to do is to find out, over large sample, the average number of trips I will make before I run the risk of getting wet.

To find this, I will need to inspect the probability that it will rain at any given day. For easy of calculations, I use a probability between 1% chance that it will rain and 99% chance that it will rain. Why don’t I include 0% or 100%? This is because if it is 0% it means it will never rain and so I will never need to use an umbrella, and if it 100% it will always rain and therefore I will always bring an umbrella back or forth.

On top of that, I will run the simulation for the case where I have 2 umbrellas at home and at the office, up to 50 umbrellas at home and at the office. To give a large enough representation, I run this over 1000 iterations (more will be better but it will take too long to run). This is the Ruby code for the simulation.

require 'rubygems'
require 'faster_csv'

umbrellas_range = 2..50
probability_range = 1..99
@location = :home

total_trips = {}
iterations = 1000

def walks_to(loc)
 location = loc
 @num_of_trips += 1
end

def office?()  @location == :office; end
def home?() @location == :home; end
def raining?(probability) rand(99) < probability; end

FasterCSV.open('umbrellas.csv', 'w') do |csv|
 csv << [''] + probability_range.to_a
 umbrellas_range.each do |umbrellas|
 probability_range.each do |probability|
 total_trips[probability] = 0
 iterations.times do
 @num_of_trips = 0
 wet = false
 home = umbrellas
 office = umbrellas

 while not wet
 if home?
   walks_to :office
   if raining?(probability)
     if home > 0
       home -= 1
       office += 1
     else
       wet = true
     end
   end

 elsif office?
   walks_to :home
   if raining?(probability)
     if office > 0
       office -= 1
       home += 1
     else
       wet = true
     end
   end
 end
 end

 total_trips[probability] += @num_of_trips
 end
 end

 row = [umbrellas]
 total_trips.sort.each { |pair| row << pair[1]/iterations }
 csv << row
 end
end

The Ruby program writes to a simple CSV file called umbrellas.csv with the first row being the probability, and the first column the number of umbrellas at each location. If you chart each row with the y-axis being the number of trips and the x-axis being the probability of rain from 1% to 99%, you will find a chart like this:

Now that I have the average number of trips that I will make before I get wet, I want to actually see what that means for Singapore. To do this, I went to the National Environment Agency’s website and dug out the weather statistics for Singapore.  There is a statistic that gives the average number of rainy days in any given month for the past 118 years. I take that for each month and divide that with the number of days in that month to derive the monthly probability of rain as in the following table:

Finally I match that with the Monte Carlo simulation I made earlier. The average number of trips for each month, must be between 56 to 62 days (each trip is one way only, so every day is 2 trips). For example, the statistics shows that the wettest months are in November and December which is 63% and 61% probability of rain respectively. This means I will need to have 38 umbrellas at each location for each of these 2 months (please check your umbrellas.csv file; open it up with Excel). In February where the probability of rain is the lowest (39%), I will only need 23 umbrellas at each location to almost guarantee that I will not get wet (the whole deal assumes at the end of each month, I replenish each location with the necessary number of umbrellas, of course).

That evening when I proudly told my wife my findings at home, she stared at me for while (frostily if I might add) then gave me a small folding umbrella, which I now carry in my laptop bag.

And that is the reason why engineers should get married.

(I have been inspired by this book — Digital Dice : Computational Solutions to Practical Probability Problems, by Paul J. Nahin)