In the previous installment, we discussed the dynamics of a (very) simple network of queues, and showed how much extra capacity was required to accommodate the build-up of population inside the queue, based on two factors: the rate at which people enter and leave the queue.

Today, we will look at a related question. Last time we determined the expected queue size at equilibrium, given the flow of people into the queue. This time, we want to consider the reverse problem: if you knew how many people are in the queue at equilibrium, what population breakdown would you expect between the two queues?

The question may sound theoretical – it isn’t. If you knew the total size of a market, the relative preferences of consumers between the products, and how long it takes them to replace their product, then determining how many consumers would be using each product at any given time is equivalent to the question we are considering.

Let’s illustrate on a fictional example. Imagine there is a disease, which can be treated two ways – using a blue pill, or a red pill. Doctors prescribe the blue pill to 25% of the patients, and the red one to 75%. The blue pill treatment takes 5 weeks, and the red pill treatment 8 (which we convert to average rates of exit of 0.2 and 0.125 per week). Suppose you knew that currently, 1000 people were under treatment: how many patients would you expect to be treated with a blue pill?

(picture from www.hackthematrix.org)

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I was recently inspired by an article in OR/MS mag to write 2 posts on using Excel data tables to resolve optimization problems. As it turns out, these puzzles are not only published in the magazine: they also have their online home at Puzzlor.com. So if you like optimization and math problems in general, and like puzzles, check it out!

I found a bug in my code the other day. It happens to everybody - apparently I am not the only one to write bugs – but the bug itself surprised me. In my experience, once you know a piece of code is buggy, it’s usually not too difficult to figure out what the origin of the problem might be (fixing it might). This bug surprised me, because I knew exactly the 10 lines of code where it was taking place, and yet I had no idea what was going on – I just couldn’t see the bug, even though it was floating in plain sight (hint: the pun is intended).

Here is the context. The code reads a double and converts it into a year and a quarter, based on the following convention: the input is of the form yyyy.q, for instance, 2010.2 represents the second quarter of 2010. Anything after the 2nd decimal is ignored, 2010.0 is “rounded up” to 1st quarter, and 2010.5 and above rounded down to 4th quarter.

Here is my original code:

public class DateConverter
{
    public static int ExtractYear(double dateAsDouble)
    {
        int year = (int)dateAsDouble;
        return year;
    }

    public static int ExtractQuarter(double dateAsDouble)
    {
        int year = ExtractYear(dateAsDouble);
        int quarter = (int)(10 * (Math.Round(dateAsDouble, 1) - (double)year));
        if (quarter < 1)
        {
            quarter = 1;
        }
        if (quarter > 4)
        {
            quarter = 4;
        }
        return quarter;
    }
}

Can you spot the bug?

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In my last post, I illustrated how to quickly to pick the best value from a selection to get the optimal result, by using Excel Data Tables. This time, we will see how to pick the best possible pair of values.

We are trying to figure out which 2 bridges we should build, in order to minimize the overall travel time for the inhabitants of the island.

I worked out the math for one bridge last time. We will start we a similar setup, but adjust our spreadsheet so that for each islander, we compute the travelling distance for 2 bridges, and select the shortest route.

The ranges B1 and B2 are named Bridge1 and Bridge2. Column I now contains the formula computing the shortest route for each islander. For row 5 for instance, the formula is

=MIN(D5+E5-H5,F5+G5-H5)

Cell I10 is the total of the vertical distances travelled by each individual.

We can select from 4 bridge locations: 2, 4, 7 and 12. What we need is to find out which 2 numbers give us the lowest total travel. Let’s build our data table, this time using 2 bridge positions.

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Today I came across a solution to Euler Problem 205 on The Daily Dose of Excel. The problem is stated as follows:

Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2, 3, 4.
Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4, 5, 6.

Peter and Colin roll their dice and compare totals: the highest total wins. The result is a draw if the totals are equal.

What is the probability that Pyramidal Pete beats Cubic Colin? Give your answer rounded to seven decimal places in the form 0.abcdefg

I thought it was a pretty cool problem; I love probability problems, and had never come across something similar, so it piqued my interest. The solution presented in The Daily Dose was essentially a pretty efficient brute-force enumeration, and I wondered if it was possible to go a bit faster that 6 minutes follow a different approach – using my language of predilection, C#. [Edited August 9. Note to self: before commenting on other people’s blog posts, I should make sure I read them properly. Especially when discussing their code’s performance. Otherwise, I will look foolish].

The probability that Pete wins can be written as:

Refreshing a bit my memory in probability through Wikipedia, “the exact probability distribution F_s,i of a sum of i s-sided dice can be calculated as the repeated convolution of the single-die probability distribution with itself” as follows:

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