Take the 2-minute tour ×
Server Fault is a question and answer site for professional system and network administrators. It's 100% free, no registration required.

If I put a reverse proxy in front of a load balancer Then does it take thrice the bandwidth to do requests?

E.g. request 100mb video from http://site.com/

site.com gets request from proxy <- load balancer <- server

  • server returns data to load balancer (100 mb sent)
  • load balancer passes to proxy (another 100 mb sent)
  • proxy to client (final 100mb sent)

is that how it works?

share|improve this question
    
It depends on at what point you measure the bandwidth usage. In many cases the only place worth measuring is where it leaves your internal network. –  sciurus Mar 14 '11 at 22:42

1 Answer 1

Your question is a bit vague, but I'll try my best to explain.

Load balancers, depending on the implementation, generally just forward requests to the servers. Sometimes they act as a proxy, but that's not how they're usually intended. Think of normal load balancers as a traffic cop that divides the requests between multiple servers.

Proxies do increase the overall number of packets sent across a network for a single unique request, yes. Generally, a proxy is put in a place where it can consolidate requests so that 100 people asking for a specific file do not all have to download it from the original server. The real benefit in this is making a proxy keep a copy of a dynamic page for a set amount of time. Wikipedia has a very good implementation of this, which is partially described on Wikipedia's Meta site

So, in your example, a 100Mb file would generate ~200Mb of total network traffic plus overhead. 100 users, on the other hand, would generate only 10100Mb (100Mb * 101) of traffic, and only 100Mb of it would ever reach the original server.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.