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Little Law of Queuing Theory and How It Impacts Load Testers

First published on MSDN on Oct 27, 2017

Authored by Christopher Witte

Load testing is all about queuing, and servicing the queues. The main goals in our tests are parts of the formula itself. For Example: the response times for a test is equivalent to service times of a queue, load balancing with multiple servers is the same as queue concurrency. Even when we look at how we are designing a test we see relationships to this simple theory.

I will walk you through a few examples and hopefully open up this idea for you to use in amazing ways.

How to determine concurrency during a load test? (Customer Question What was the average number of concurrent requests during the test? What about for each server?)

Little's Law -> Adjusted to fit load testing terminology

L = λW

L is the average concurrency

λ is the request rate

W is the Average Response Time

For Example: We have a test that ran for 1 hour, the average response times were 1.2 seconds. The average requests sent to the server by the test per second was 16.

L = 16 * 1.2

L = 19.2

Average number of concurrent request processing on the servers was 19.2 averaged per second.

Let’s say we have perfect load balancing, of 5 servers. M = Number of Servers/Nodes

L = Concurrent requests per Server/Node

L = 19.2/5

So, each server is processing ~3.84 concurrent requests per second on average

This explains why, when we reach our maximum concurrency on a single process, response times begin to grow and throughput levels off. Thus, we need to optimize the service times, or increase the concurrency of processing. This can be achieved by either scaling out or up depending on how the application is architected.

Also, this is not all entirely true, there will be time spent in different parts of the system, some of which have little to no concurrency, two bits cannot exist concurrently on the same wire electrically, as of writing this article.

Can we achieve the scenario objectives given a known average response time hypothesis?

For tests that use “per user per hour pacing”, we can determine the minimum number of users necessary to achieve the target load given a known average response time under light load. You can obtain the average response times from previous tests or you can run a smoke test with light load to get it.

W = Average Response Times in seconds for the test case you are calculating

3600 seconds in an hour

Number of users needed U = 3600/W

G = Throughput Time Goal Let’s say we have an average response time of 3.6 seconds obtained from a shakeout test, and our goal is to run this test case 2600 times per hour. What would be the minimum number of users to achieve this?

U = ceil(2600/(3600/3.6))

U=3

Pacing = U / G

Pacing = 866

So the scenario would be 3 users and the test case would be set to execute 866 times per user per hour. Personally I like to add 20% additional users to account for response time growth, just in case the system slows down as the load is increased. For example: I would run with 4 vUsers at 650 Tests per User per Hour, or 5 vUsers at 520 Tests per User per Hour.

There are endless possibilities to using this formula to understand and predict behavior of the system, post some examples of how you would use queuing theory in the comment section below.

Have a great day and happy testing!