Yes, women take longer, yes, women go more often. Thus for an equal number of men and women at a venue there should be 2 female stations for one male.

The British Toilet Association is campaigning – too quietly, perhaps – for public facilities to uphold a 2:1 ratio of women\’s to men\’s loos. They haven\’t triumphed yet.

However, there\’s something at the back of my mind that I recall from the dim and distant past that I think I erad upon the intertubes. Something I couldn\’t find when I wrote this.

At the back of my mind there is something from queueing theory: queues do not grow as a direct multiple of the time taken, they obey a square law. That is, the wait to use something will grow as the square of the difference between usage times, not a simple multiple. Which means that if women take twice as long as men then to equalize waiting times we need to have four times (2 times 2) as many women’s as men’s. Or maybe it’s an inverse square law meaning we need 1 plus the square root of two times as many…..does anyone actually know?

Anyone?

Hmmn. Not sure your queueing theory is right. Maybe for an M/M/1 queue, but it’s usually one queue for several booths (making it an M/M/n queue). And I’m not sure the arrival rates are uniform, either.

How could Wembley Stadium possibly provide an ideal number of toilets for men and women when the sex ratio for different events at the venue varies so wildly?

The correct proportion for a U2 concert might well be 2 women’s to 1 mens but if they had the same ratio for the Rugby League game that they hosted at the weekend the appropriate ratio might be 1:3 rather than 2:1.

If women went in took a piss and got out instead of fannying around chatting and checking their hair then there would be no problem.

Woman + deadline = Disaster

“If women went in took a piss and got out instead of fannying around”

http://www.shewee.com/

Solution: unisex toilets as provided in some bars in Sweden. But as Tim would say there are trade-offs.

Little’s Law says the number of entities in a system is equal to the product of the rate at which entities enter (equivalently leave) and the time they spend in the system. The offered load (no, that really is the technical term) thus scales linearly with duration of visit and with arrival rate which is 1/frequency of arrival.

Little’s Law

Sorry, arrival rate = frequency of arrival

David,

Isn’t that for infinite queues?

Suppose there is a fixed limit to the time people can cross their legs, and suppose you want to keep the probability of someone having a “little accident” below some acceptable bound. Then it’s not the average behaviour that matters, so much as the outliers of the distribution. You will need more capacity than the average rate dictates to cope with peaks in traffic.

If the time taken to do one’s business is non-deterministic, then while n cubicles has n times the average throughput, the standard deviation will increase proportional to sqrt(n). The rate of service will be more stable, and long runs of people with the runs will have less impact.

What do you think?

No, Little’s Law works for all systems with objects entering the system and persisting in it for a finite period. If inflow = outflow then the number of objects in the system attains a steady, finite, state. If λ is the average arrival rate and t is the time an average object spends in the system, then the offered load L = λ t. The great thing is it is

independent of the underlying probability distributions. It’s a very useful little intuitive gizmo to have in your mental toolbox. But it is a long-term average.In a real world queuing system, e.g. a packet-switched network, arrival times are usually modeled, as one would expect, as a Poisson process with a given mean. You can get pretty good bounds on contention ratios and tune the throughput to have an acceptable number of dropped packets (soiled undies). I don’t know if architects and human factors engineers use the same techniques.

I believe the student pubs in Manchester solved this problem: when busy, the women use the mens’ toilets and the men use the sinks.

Never mind the queues, I was just bowled 🙂 over by the thought that there is a British Toilet Association. But it seems most nations have one eg we have an Australian Toilet Association. I assume the head and deputy of each are known as No. 1 and No.2.

But the kicker is that the world body is called the World Toilet Organisation. The WTO !! Puns galore.

With a bit of imagination erlang C queing theory used forsizing call centres can be applied to this problem.

http://www.erlang.com/calculator/erlc/

Let:

number of agents = number of cubicles

calls per hour = number of visits expected. This

can be calculated from knowing the nnumber of women in the crowd and how often they go

call duration = length of visit

average delay = how long women are prepared to spend in a queue

David,

So what is there in Little’s Law that prevents the queue growing too long?

Let’s say that the time to do one’s business has a Bernoulli distribution of 1% taking 49.5 minutes and 99% taking 0.5 minutes. The mean is 0.99 minutes.

So let’s say the ladies arrive at the single cubicle on average 1 per minute, which means in the long run we can cope. If they arrive at

exactlyone per minute, and every 100th lady is slow, then you’ll get a 49.5 minute wait, then 49.5 minutes of rapid action, the stall will be empty for a minute, then another 49.5 minute pause, and so on. In each cycle, 100 ladies are served in 100 minutes, and we have a minute’s worth of slack time. The longest wait can be no longer than 49.5 minutes. Of course, 49.5 minutes is a long time to hold it, but let’s suppose it’s just possible.But now suppose things are not quite so deterministic, and purely by chance, 3 slow ones come along all at once. (It’s a million to one shot, but it could just happen.) Now this is of course balanced by intervals in which you get several cycles in which everyone is fast, so the queue will get processed eventually, but in the meantime, the person who arrives just after them will be waiting for a bladder-bursting 148.5 minutes – nearly two and a half hours!

You see, if you insist that the queue

must notexceed your set limit, you require more capacity than Little’s Law stipulates. Little’s Law quite correctly states that any queues will get dealt witheventually, with probability 1, but it sets no limit on how long it might take.If the limit were 50 minutes, and the arrival of slow ladies Poisson, say, then what proportion of ladies would be in trouble? I haven’t worked it out, but I think it’s quite large.