Asynchronous Versus Synchronous: A Performance Perspective

I just had my coffee from my favorite coffee shop down the corner. I’m a regular at that coffee shop that the baristas know me and I’m able to sit down after I order and do something else while they make my coffee and serve to me when ready.  If I order from other branches of this coffee shop, I would have to wait for my coffee in a corner and not being able to do anything productive.  So while I was seated comfortable waiting for my coffee, a thought came to me. This is a great example of synchronous versus asynchronous service invocation.

A synchronous service invocation is just like buying your coffee and having to wait in the corner for the baristas to complete your coffee before you can even sit down and do something useful. Asynchronous service invocation is having to get seated and do something else while waiting for your coffee to be served.  My instinct tells me that asynchronous invocation is better than synchronous in terms of performance especially when it takes a long for the service to complete and the waiting time is much better used doing something else. But how can we show that this is the case?

We can model the synchronous/asynchronous invocation as a Markov Chain and compute a few metrics from our model. If you’re not familiar with this approach, you can refer to this article MODELING QUEUING SYSTEMS USING MARKOV CHAINS.

First, we model the asynchronous invocation. We can identify each state of our system by  2 slots each of which contains a number. The first slot is the number of responses waiting for the client to process and the second slot is the number of requests waiting for the server to process.  To simplify the modeling, we assume the following:

  1. The maximum number of requests that a server can handle at any given time is 2. Which means that a client cannot anymore send a request if the second number is equal to 2.
  2. When the server responds, the first number increments by 1. Once a client receives a response, it will stop whatever it is doing and process that response. When the client finishes processing the response, the first number goes down to zero. As a consequence, this assumption says that the maximum value of slot number 1 is 1 at any given time.

With these assumptions, we can draw the Markov Diagram of the asynchronous system to come up with the below diagram:

Explanation of the Diagram

The system initially starts at 00 where there are no requests and no responses. It will then transition to state 01 when the client will send a  request which increments the server’s pending requests to 1. From this state, the system can transition to either state 02 or state 10. State 02 occurs when the client sends another request without waiting for a response. State 10 is when the server is able to process the request and return a response, thereby decrementing it’s pending items and incrementing the client’s number of responses to process.

At state 02, the client cannot anymore send requests since the maximum requests that the server can handle is 2. At this state, it can only transition to state 11, where the server has processed one of the requests (thus decrementing the second slot from 2 to 1 and incrementing the first slot from 0 to 1).

State 11 can only transition to 01 since the client will stop whatever it is doing to process the single response it has received (thus decrementing the first slot from 1 to 0).

The numbers a,b,c are the rates at which the transitions will occur. For example, the rate at which the system will transition from state 00 to state 01 is b.

In the remainder of this article, we assume that the client can request at the rate of 60 requests per unit time and can process the response at 60 per unit time. We also assume that the server can process requests at 30 per unit time. We therefore have the following:

a=60,  b=60,  c=30

Computing the Probabilities

At steady state, the flow going into each state is equal to the flow out of that state. For example, for state 00, the following is true:

\displaystyle bP_{00} = aP_{10}

Doing this for all states, we get the following balance equations:

\begin{array}{rlr}  \displaystyle  bP_{00} &= aP_{10} & (1)\\  bP_{00} + a P_{11} &= bP_{01} + cP_{01} & (2)\\  bP_{01} &= cP_{02} & (3)\\  aP_{10} &= cP_{01} & (4)\\  aP_{11} &= cP_{02} & (5)  \end{array}

Since the sum of the probabilities should be equal to 1, we have our last equation:

P_{00} + P_{01} + P_{02} + P_{10} + P_{11} = 1

We have 6 equations in 5 unknowns, one of the equations above is actually redundant. In fact, you can show that equation 2 above is redundant.

We can form the matrix equation of the system of equations above and solve for the probabilities:

\begin{bmatrix}  -b & 0 & 0 & a & 0 \\  0 & b & -c & 0 & 0 \\  0 & -c & 0 & a & 0 \\  0 & 0 & -c & 0 & a \\  1 & 1 & 1 & 1 & 1  \end{bmatrix}  \begin{bmatrix}  P_{00}\\  P_{01}\\  P_{02}\\  P_{10}\\  P_{11}  \end{bmatrix}  =  \begin{bmatrix}  0\\  0\\  0\\  0\\  1  \end{bmatrix}

Solving this matrix equation when a=60, b=60 and c=30, we can find the probabilities:

\begin{bmatrix}  P_{00}\\  P_{01}\\  P_{02}\\  P_{10}\\  P_{11}  \end{bmatrix}  =  \displaystyle  \frac{1}{ab^2+abc+ac^2+b^2c+bc^2}  \begin{bmatrix}  ac^2\\  abc\\  ab^2\\  bc^2\\  b^2c  \end{bmatrix}  =  \begin{bmatrix}  1/10 \\ 1/5 \\2/5\\ 1/10\\ 1/5  \end{bmatrix}

Utilization and Throughput: Asynchronous Case

The client utilization is defined to be the probability that the client is busy. In the same way, the server utilization is the probability that the server is busy. Looking at the diagram, the client is busy sending requests at state 00 and state 01. It is busy processing responses at state 10 and 11.  On the other hand, the server is busy processing requests at state 01 and 02.

Therefore, the client utilization is  equal to

\begin{array}{rl}  U_{\text{client}} &= P_{00} + P_{01} + P_{10} + P_{11}\\  &= 1/10 + 1/5 + 1/10 + 1/5\\  &= 3/5\\  &= 60\%  \end{array}

The server utilization is equal to

\begin{array}{rl}  U_{\text{server}} &= P_{01} + P_{02}\\  &= 1/5 + 2/5 \\  &= 3/5 \\  &= 60\%  \end{array}

The system througput is the number of requests the client is able to submit and is equal to

\begin{array}{rl}  X &= 60P_{00} + 60 P_{01} \\  &= 60*1/10 + 60 * 1/5 \\  &= 18  \end{array}

Comparison with Synchronous Invocation

For the synchronous invocation, the client will submit a request at state 00. The server will then receive this request and process it immediately at state 01. The client will get the response and process it at state 10 and do the loop again at state 00. Please see the Markov diagram below describing this process.

We can solve the probabitlies of this Markov Chain by solving the balance equation

\begin{bmatrix}  b & 0 & -a \\  0 & c & -a \\  1 & 1 & 1  \end{bmatrix}  \begin{bmatrix}  P_{00}\\  P_{01}\\  P_{10}  \end{bmatrix}  =  \begin{bmatrix}  0\\0\\1  \end{bmatrix}

Solving for the probabilities, we get

\begin{bmatrix}  P_{00}\\  P_{01}\\  P_{10}  \end{bmatrix}  =  \displaystyle  \frac{1}{ab + ac + bc}  \begin{bmatrix}  ac\\  ab\\  bc  \end{bmatrix}  =  \begin{bmatrix}  1/4\\ 1/2\\ 1/4  \end{bmatrix}

At state 00, the client is busy sending requests at the rate of 60 requests per unit time, therefore the system throughput is

\begin{array}{rl}  X &= 60 P_{00} \\  &= 60/4 \\  &= 15  \end{array}

which is less than the Asynchronous case. In addition, the client utilization is

\begin{array}{rl}  U_{\text{client}} &= P_{00} + P_{10}\\  &= 1/4 + 1/4\\  &= 1/2  \end{array}

This is lower than the Asynchronous case where the client utilization is 60%.

Therefore, the asynchronous service invocation is indeed better compared to the synchronous case. It has a much higher throughput and a much higher utilization compared to the synchronous case.

P.S. I used Mathics to solve the matrix equations. Mathics is an open source alternative to Mathematica. It features Mathematica-compatible syntax and functions so if you’re a user of Mathematica you can run basic commands in Mathics. If you’re a Mathics user, you can easily switch to Mathematica as well. Please visit to know more about Mathics.

When Average Is Not Enough: Thoughts on Designing for Capacity

Designing a system from scratch to handle a workload you don’t know is a challenge. If you put to much hardware, you might be wasting money. You put little, then your users will complain of how slow the system is.

If you’re given only a rate, like 6000 hits/hour, you don’t know how these are distributed in a minute by minute or per second interval. We can make a guess and say that there are about 100 hits per minute or 1.67 hits/sec. If hits come uniformly at that rate, then we can design a system that can handle 2 hits/sec and all users will be happy since all requests will be served quickly and no queueing of requests. But we know it’s not going to happen. There will be some interval where the number of hits is less than 3 and some more than 3.

Theoretically, requests to our server come randomly. Let’s imagine 60 bins represented by seconds in one minute. We also imagine that requests are like balls we throw into the bins. Each bin is equally likely to be landed by a ball. It’s possible that all balls land on only one bin!


After throwing the balls into bins, let’s see what we have.


As you can see, some bins have more than 2 balls (which is the average number of balls in a bin). Therefore if we design our system based on the average, 50% of our users will have a great experience while the other 50% will have a bad experience. Therefore we need to find how many requests per second our server needs to handle so that our users will have a good experience (without overspending).

To determine how many requests per second we need to support, we need to get the probability of getting 4, 5, 6 or more request per second. We will compute the probability starting from 3 requests per second and increment by one until we can get a low enough probability. If we design the system for a rate that has a low probability, we are going to spend money for something that rarely occurs.

Computing the Probability Distribution

We can view the distribution of balls into bins in another way. Imagine labeling each ball with a number from 1 to 60. Each number has an equal chance to be picked. The meaning of this labeling is this: the number that was assigned to the ball is the bin (time bucket) it belongs to. After labeling all balls, what you have is a distribution of balls into bins.

Since each ball can be labeled in 60 different ways and there are 100 balls, the number of ways we can label 100 different balls is therefore

\displaystyle 60^{100}

Pick a number from 1-60. Say number 1. Assume 2 balls out of 100 are labeled with number 1. In how many ways can you do this ? Choose the first ball to label. There are 100 ways to choose the ball. Choose the second ball. Now there are 99 ways to choose the second ball. We therefore have 990 ways to select 2 balls and label them 1. Since we don’t really care in what order we picked the ball, we divide 990 with the number of possible arrangements of ball 1 and ball 2, which is 2! (where the exclamation mark stands for “factorial”). So far, the number of ways to label 2 balls with the same number is

\displaystyle \frac{100 \times 99}{2!}

Since these are the only balls with label 1, the third ball can be labeled anything except number 1. In that case, there are 59 ways to label ball 3. In the same way, there are 59 ways to label ball 4. Continuing this reasoning until ball 100, the total ways we can label 2 balls with number 1 and the rest with anything else is therefore:

\displaystyle \frac{100 \times 99}{2!} \times 59^{98}

Notice that the exponent of 59 is 98 since there are 98 balls starting from ball 3 to ball 100.

Therefore, the probability of having two balls in the same bin is

\displaystyle \frac{100 \times 99}{2!} \times \frac{59^{98}}{60^{100}} = 0.2648

We can also write this as

\displaystyle \frac{100!}{2! \times 98!} \times \frac{(60-1)^{98}}{60^{100}} = \binom{100}{2} \frac{(60-1)^{98}}{60^{100}}

In general, if m is the number of balls, n the number of bins and k the number of balls with the same label, then the probability of having k balls within the same bin is given by

\displaystyle \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}}



\displaystyle \binom{m}{k} = \frac{m!}{k!(m-k)!}

is the binomial coefficient.

It turns out that this is a probability distribution since the sum of all probabilities from k=0 to k=m is equal to 1. that is

\displaystyle \sum_{k=0}^{n} \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}} = 1

To see this, recall from the Binomial Theorem that

\displaystyle \big( x + y \big)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k

If we let x=n-1 and y=1, we can write the above equation as

\displaystyle  \begin{array}{ll}  \displaystyle \sum_{k=0}^{m} \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}} &= \displaystyle \sum_{k=0}^{m} \binom{m}{k} \frac{(n-1)^{m-k}\cdot 1^k}{n^{m}}\\  &= \displaystyle\frac{(n-1+1)^m}{n^{m}}\\  &= \displaystyle\frac{n^m}{n^m}\\  &= \displaystyle 1  \end{array}

Here is a graph of this probability distribution.


Here’s the plot data:

1 0.315663315854
2 0.264836171776
3 0.146632456689
4 0.060268424995
5 0.019612775592
6 0.005263315484
7 0.001197945897
8 0.000236035950
9 0.000040895118
10 0.000006307552

We can see that for k=9, the probability of it occurring is .004%. Anything beyond that we can call rare and no need to spend money with.

Just For Fun

What’s the probability that a given bin is empty, that is, there are no balls in it?

Other Probability Distributions

Our computation above was based on a uniform probability distribution. However, there are other distributions that are more suitable for arrival of requests. One of the most widely used is called the Poisson Distribution where you can read from here.

R Code

The R code to generate the simulation:

  for(i in 1:100){
  plot(tabulate(x),type="h", ylab="tx", xlab="secs")

for(i in 1:16){

The R code to generate the probability distribution:


for(i in 1:10){
plot(tt,type="h",xlab="Number of Balls",ylab="Probability")


This post is dedicated to my friend Ernesto Adorio, a mathematician. He loves combinatorial mathematics.

Rest in peace my friend! I miss bouncing ideas with you.

Divided We Compute, United We Reduce

Once upon a time, in a far away village lay a dying old man. He called his sons to his deathbed and spoke to them one last time. He said “Sons, see that bundle of sticks? Each one of you try to break it. The one who can break it will inherit all my riches”. Each son, being greedy, wanted all the riches for himself. So each one of them tried to break the bundle of sticks but none of them succeeded. The old man asked his servant to untie the bundle and said to his sons, “Each one of you now get one stick and break it”. Without any effort, each son was able to break the stick. The old man said “You see, when you unite, no task will be difficult. The riches that I told you was a lie. We are broke. When i’m dead, make sure  you unite so that you can survive.”

Fast forward to modern times. You can think of the bundle of sticks to be a complex problem that is itself composed of smaller problems. The sons are the processors of your computer. When each processor was given the task to solve the complex problem, it fails to solve it in a reasonable amount of time. When the complex problem is decomposed into smaller problems and given to each processor, each processor is now able to solve the smaller problems quickly thereby solving the big problem quickly as well.

The process of decomposing a problem into smaller problems and solving them in separate processors is called Parallel Computing. In this article, we will compute how fast a certain algorithm will run when parallelized. The problem we want to investigate is sorting an array of a million (2^{20}) integers.

Efficient Sorting

Suppose you have an array \{ a_1, a_2, a_3, ..., a_n \} that you want to sort based on pairwise comparison.  The sorted array is just one of the many permutations of the array \{a_1, a_2, a_3,\ldots, a_n\}. In fact, if you have n different objects to sort, then there are exactly n! ways to arrange these objects, and one of them is the sorted state. You can imagine the sorting process as a decision tree. Say, for example we have array A={ a,b,c }. To sort this, we first compare a with b and there are 2 ways this can go. Either a \le b or a > b. If a \le b, we then compare b and c. This also give either b \le c or b > c. As you can see from the diagram below, this is nothing but a decision tree.

Since the height of this binary tree is lg(n!), then we have

\lg(n!) = \lg\Big[ n \cdot (n - 1) \cdot (n-2) \cdots 1\Big] \le \lg n + \lg (n-1) \cdots \lg1 \le \underbrace{\lg n \cdots \lg n}_\text{ n times}
\lg(n!) \le n\cdot \lg n

There are efficient algorithms that are able to sort of this complexity. For example, the merge sort has this complexity. Therefore, if you have an array of 2^{20} elements, then the complexity is

2^{20} \cdot \lg(2^{20}) = 2^{20} \cdot (20) = 20971520

that is, it takes about 20 million comparisons to sort an array of 1 million. Could we do any better than this? We can either upgrade the cpu of the machine doing the sorting or use two or more machines to divide the work among those machines. In this article, we are going to investigate the impact of dividing the work into smaller chunks and farming it to other processors.

Divide and Conquer

Assume we have an array n=2^{20} elements that we need to sort and suppose we have two identical processors we can use. Divide the array into 2 equal sized arrays. Give the first array to the first processor and the other half to the second processor. Apply an efficient sorting algorithm to the subarrays to produce a sorted array for each processor. We then combine the result of processor 1 and processor 2 to one big array by merging the two sorted arrays. The diagram below illustrates the process of computation:

This is also known as the MapReduce algorithm. Mapping is the process of assigning subsets of the input data to processors where each processor computes the partial result. Reducing is the process of aggregating the results of each processor to the final solution of the problem.

The process of merging is straightforward. Given two sorted arrays, begin by comparing the first element of each array. The smaller of the two will then occupy the first slot in the big array. The second element of the array from which we took the smallest element will now become the first element of that array. Repeat the process until all elements of both arrays have already occupied slots in the big array. The diagram below illustrates the algorithm of merging.

If you count the total number of comparisons that you need to merge two sorted arrays, you will find that it takes n-1 comparisons. Therefore, the complexity of the merging process is O(n).

Since each processor has n/2 sized subarrays, the sorting complexity is therefore n/p \lg (n/p). Furthermore, since the merging process takes O(n) comparisons, the total complexity of the parallel sorting process is therefore

\displaystyle n/p \lg(n/p) + n

In our example, C=2^{20}/2 \lg(2^{20}/2) + 2^{20}=  11010048 comparisons compared to 2^{20} \lg(2^{20}) = 20971520 when run sequentially. For large values of n, n/p \lg(n/p) dominates n, therefore the complexity of the parallel algorithm is O(n/p \lg(n/p)).

Can we do any better?

For a given value of n, what do you think is the value of p that reduces the running time to O(n)? If we take n=2^{20} and plot complexity against p = \{ 2, 4, 8, 16, 32\} we get the diagram below.

In this diagram, we also plotted the horizontal line y=2^{20}. The intersection of this line with the plot of \displaystyle f(p) = \frac{n}{p} \lg(\frac{n}{p}) gives us the value of p such that the total comparisons is already linear, that is,

\displaystyle f( p ) = n
\displaystyle \frac{n}{p} \lg(\frac{n}{p})  = n

To get the value of p numerically, we have to solve the root of the equation

\displaystyle g( p ) = \frac{n}{p} \lg(\frac{n}{p}) - n = 0


\displaystyle \frac{1}{p} \lg(\frac{n}{p}) - 1 = 0
\displaystyle \lg(\frac{n}{p}) = p
\displaystyle \frac{n}{p} = 2^p
\displaystyle p2^p - n = 0

Since this is a non-linear equation, we can solve this using the Newton's method. It is a method to compute the roots by approximation given an initial value of the solution. Starting from a guess solution p_1, the root can be approximated using the recursive formula

\displaystyle p_{n+1} = p_n - \frac{g( p_n)}{g\prime ( p_n)}

where g\prime ( p ) is the first derivative of g( p ) . Applying the rules of derivatives, we get

\displaystyle g\prime ( p ) = p\cdot 2^p \ln 2 + 2^p

Substituting this to the formula for Newton's method, we get

\displaystyle p_{n+1} = p_n - \frac{p2^p - n}{p2^p \ln 2 - 2^p}

Below is an R code using newton’s method to compute the root of the equation g(p).

	p* 2^p - n

	p*2^p *log(2) - 2^p

	tmp = p
	for(i in 1:iter){
		if(abs(p-tmp)< 0.0001){


Running this code, we get the value of p = 16:

> newton(15,n,100)
[1] 16.80905
[1] 16.08829
[1] 15.98603
[1] 16.00286
[1] 15.99944
[1] 16.00011
[1] 15.99998
[1] 16

Ignoring network latency, by distributing the input evenly into 16 processors, we get a running time of O(n) time complexity for n=2^{20} array of items. Therefore, instead of doing 20 million comparisons, you only need 1 million comparisons to sort 1 million objects.

In this age of multicore processors, parallel computing is fast becoming the norm than the exception. Learning to harness the power of multicores is becoming an extremely handy skill to have.

Performance of Shortest Queue Discipline

We computed the performance of random line queuing discipline in the previous article. However, in real life, we don’t really throw a coin in order to pick which of the two lines we take. Most of the time, we go to the shortest line for service. Using the techniques we have learned in modeling, we will compute the performance of this queue and compare the performance to the other queueing discplines we have computed so far.

Markov Diagram


As usual, the arrival rate of customers is 2 customers per minute. Each customer is given a service at an average of 20 secs, which means that an average of 3 customers are being serviced per minute.

Continue reading Performance of Shortest Queue Discipline

Computing Performance of Symmetric Multi-Processing Scheduling

You might have noticed that when you do a transaction in a bank, you usually see a single line being served by 2 or more tellers. You must have perceived that this technique of lining up customers is better than when each teller has its own line. This will lessen the chance of you being stuck for a long time when the guy before you takes a long time to complete transaction.

In computer systems, you can view the customers as tasks to be executed by 2 or more processors. Each processor will pick from a common line, a task that it will execute. In this article, we will analyze the performance of symmetric multiprocessing using continuous markov chain analysis. We will also be using the tool available from to help us compute the probabilities.

Continue reading Computing Performance of Symmetric Multi-Processing Scheduling

Computing Continuous Markov Chains Using Extreme Optimal Solver

In the last article, we computed the probabilities associated with the states in a continuous markov chain model using balance equations. The computation, although elementary, was tedious and distracts us from the real problem, which is understanding the performance of a queueing model. Fortunately, there is an online tool we can use in order to speed up the computation of continuous markov chains. This tool can be found in the site This site contains numerous solvers that can be of interest to the computational scientists. However, the tool we are interested in is the Continuous Markov Chain solver which can be accessed in this link.

Continue reading Computing Continuous Markov Chains Using Extreme Optimal Solver

Modeling Queuing Systems Using Markov Chains

Have you ever noticed how various establishments make people line in a certain way? Have you gone to a bank and noticed a number of tellers servicing each client but there is only one line? Wouldn’t it be more natural to let people line up before each teller like you see in a grocery? The way customers are lined up in a server or system of servers is called Queueing Discipline.
In this article, we will learn how various queuing disciplines will affect the values of the performance metrics. We will use continuous markov chain analysis that we learned in the last article.

Motivating Scenario

Assume that the counter of a certain fast food store is arrange in a 2-stage way. The customers first place their order in counter 1, pays the cashier and proceeds to counter 2 to wait for their order to be picked up. Let a be the rate at which customers arrive, b the rate at which customers places their order and also the rate at which they leave the second counter. Let us draw the markov chain diagram of this system.

Each state is described by two numbers. The first number refers to the number of customers in the first counter. The second number refers to the number of customers waiting at the second counter. For example, state 00 mean that there are no customers being serviced. State “12” mean that one customer is in counter 1 and 2 customers in counter 2. The blue arrows represent arrivals at rate a and red arrows are represent the rate of customers going to counter 2 and those leaving the counter 2.


Let us examine the four states in the diagram below. There is a blue arrow from state “00” to state “10” because when the first customer arrives, the first station will have 1 customer but no one in line yet in counter 2. This means that when the first customer arrives, the system will be in state “10”. There is an arrow from state “10” to “01”. The means that the first customer is already in counter 2 to pickup his/her order but no new customer has arrived yet. There is a red arrow from state “01” to “00”. This means that the single customer has already picked up his/her food and has left leaving the system with no customers at both stations. Notice that there is no arrow from state “10” to “00” because the customer who is in counter 1 cannot just leave the system but has to go to counter 2 before leaving.

Balance Equations
Now that we have understood why the arrows point from one state to the next, we now setup the balance equations. At steady state, the sum of flows going to one state must equal the sum of flows going out of that state. By definition, the flow along an arrow is equal to the rate along the arrow multiplied by the probability of the system being in the current state.Let us do the balance equation for state “00”. There is one arrow going to state “00” and one arrow going out of state “00”. Therefore the flow going to state “00” is equal to the rate going from state “01” to “00” multiplied by the probability of the system being in state “01” : bP_{01}. This is equal to the flow going out of state “00” to state “10”: a*P00. Therefore,we have,

\displaystyle a P_{00} = bP_{01}

Let us now compute the balance equation for state “01”. As you can see, there are 2 arrows out of state “01” and one arrow going into state “01”. The balance equation is therefore:

a P_{01} + b P_{01} = b P_{10}

Doing this for all states, we get the following balance equations:

\displaystyle aP_{02} + bP_{02} = b P_{11} + b P_{03}
\displaystyle b P_{03} = bP_{12}
\displaystyle aP_{00} + bP_{11} = bP_{10} + aP_{10}
\displaystyle aP_{01} + bP_{12} + bP_{20} = bP_{11} + bP_{11} + aP_{11}
\displaystyle bP_{12} + bP_{12} = aP_{02} + bP_{21}
\displaystyle bP_{20} + aP_{20} = aP_{10} + bP_{21}
\displaystyle bP_{21} + bP_{21} = aP_{11} + bP_{30}
\displaystyle b P_{30} = a P_{20}

We add to these equations the condition that the sum of all probabilities is equal to 1:

\displaystyle P_{00} + P_{01} + P_{02} + P_{03} + P_{10} + P_{11} + P_{12} + P_{20} + P_{21} + P_{30} = 1

Solving for the probabilities using elementary algebra, we get the following:

\displaystyle P_{01} =\frac{ab^2}{b^3+2ab^2+3a^2b + 4a^3}
\displaystyle P_{02} =\frac{a^2b}{b^3+2ab^2+3a^2b + 4a^3}
\displaystyle P_{03} =\frac{a^3}{b^3+2ab^2+3a^2b + 4a^3}
\displaystyle P_{10}  =P_{01}
\displaystyle P_{11} =P_{02}
\displaystyle P_{12} =P_{03}
\displaystyle P_{20} =P_{02}
\displaystyle P_{21} =P_{03}
\displaystyle P_{03} =P_{30}
\displaystyle P_{00} =(b/a)P_{01}

A simple Example

Suppose that customers arrive on the average of 1 customer per minute and can place and pay for their order in 2 minutes. After that they wait on the average of 2 minutes at the second counter to take out their order. Using the formula above, we have a = 1 and b=2. We can now solve for the probabilities by substituting these values to the above formula.

\displaystyle P_{01} = \frac{ab^2}{b^3+2ab^2+3a^2b + 4a^3} = \frac{1\cdot 2^2}{1^3+ 2\cdot 1\cdot 2^2 + 3\cdot 1^2\cdot 2 + 4\cdot 1^3}
\displaystyle =\frac{2}{13}

Computing the rest of the values, we get the following table:

P00 4/13
P01,P10 2/13
P02,P11,P20 1/13
P03,P12,P21,P30 1/26

We can compute for the Utilization of the first counter by using it’s idle time. Notice that the first counter does not have anything to do when there are no customers lining in front of it. This happens in states P00, P01, P02 and P03. Notice that these states have the first number equal to 0 (which means: no customer in that counter). Therefore, the utilization is equal to 1 minus the idle time of counter 1, i.e,

\displaystyle U_\text{counter 1} = 1 - P_{00} - P_{01} - P_{02} - P_{03}
\displaystyle =\frac{11}{26}

Using the Utilization law we learned in a previous article, we can compute the throughput of counter 1 to be:

\displaystyle X_\text{counter 1} = \frac{U_\text{counter 1}}{S_\text{counter 1}}
\displaystyle =\frac{11/26}{2}
\displaystyle =\frac{11}{52}

This happens to be the system throughput also since each customer will visit a counter only once per transaction.

Now that we know how to model using markov chains, in the next article, we will analyze other queueing disciplines and compare them according to throughput and response time.