Network Throughput in a 2-Link Network

Networkrouterthroughput

I was reading "Computer Networking; A Top-Down Approach" by "Kurose Ross" that I bumped in a question. Consider a 2 Link Network with a router as the following picture:

by definition:

The instantaneous throughput at any instant of time is the rate
(in bits/sec) at which Host B is receiving the file

and

the average throughput of the file transfer is F/T bits/sec.

Consider we have file with size XL in which X is the number of packets and L is the size of packet.
So as we see in the picture, there is Rs bits/sec Link bottleneck, so (X-1) packets will be transferred to client after the first packet gets to the router. the first packet needs L/Rs seconds to get to the router. It means transferring the whole file needed XL/Rc + L/Rs seconds, Which by definition gives the following throughput:

but in book, the answer of the same question is:

Having determined the throughput, we
can now approximate the time it takes to transfer a large file of F bits from server to client as F/min{Rs, Rc}. For a specific example, suppose you are downloading an MP3 file of F = 32 million bits, the server has a transmission rate of Rs = 2 Mbps, and you have an access link of Rc = 1 Mbps. The time needed to transfer the file is then 32 seconds.

I can not understand why the book doesn't care about the last packet. What is the reason?

Best Answer

I think the reason is :

Having determined the throughput, we can now approximate the time it takes...

The last packet is received in about a ms, and this doesn't change much the calculated value.

In other word, it is negligible.

Note that, anyway, actual communication in a network use various protocols, each of which has some kind of overhead that will impact the actual throughput.

Related Solutions

Cisco – Output Drops on Serial interface: Better queueing or Output queue size

You're right, you wouldn't really see the burstiness easily on SNMP. 1GE can send 1.48Mpps, so it takes very very little time to congest the the 45Mbps, which can handle less than 75kpps.

If your ingress is 1GE and egress is 45Mbps, then obviously the congestion point of 45Mbps will need to drop packets. This is normal and expected. If you increase buffers you'll introduce more delay.
1GE takes 0.45ms to send 40 1500B IP frames, which is right now the amount of burst you can handle. However dequeueing them on the 45Mbps already takes 10ms.

If you don't have any acute problem, I would probably not do anything about it. But if some traffic is more eligible for dropping than other, then you should replace FIFO with class-based queueing. Say maybe you want to prioritize so that more ftp is dropped and less voip.
Then it'll also make more sense to add more buffering on the ftp traffic, as it's not really sensitive to delay.

If you want to try your luck with deeper buffers, something like this should suffice:

policy-map WAN-OUT
 class class-default
    fair-queue
    queue-limit 200 packets
!
interface Serial1/0
  service-policy output WAN-OUT

This would cause 50ms buffers on the Serial1 and would allow you to handle up-to 2.25ms burst from single Gige interface.

Switch – Typical throughput on switched network

With throughput defined as the amount of data transferred from point A to point B in a period of time. The significant variables in throughput are: Latency, Packet Size, and Retransmissions (quality). This was proven with the Mathis Equation.

http://www.slac.stanford.edu/comp/net/wan-mon/thru-vs-loss.html

TCP Basics: TCP connections establish a session with a SYN, the receiving machine sends and ACK, Then data flows. When the window (setting in windows saying how much data to collect before it asks for validation/acknowledgement of receipt) fills up another set of acknowledgements are sent to start the next flow of data.

If your environment has the default MTU of 1500, and you introduce a device that has an MTU of 1460, this will slow down your network because all packets flowing through the device will be fragmented; when the 1500 packet hits the device with MTU of 1460 it will fragment the packet into two packets and transmit it.

If you have Jumbo frames enabled and your MTU is 9220 you will get much higher throughput, 5 times higher. Each payload is larger even though the packet has the same latency.

In short: Throughput is conditional and depended on both host and network device settings. Use the Mathis Equation as a guide for what you could expect with the given data points for your network. Validate your findings and verify units; Megabit/sec and Megabyte or Kilobit and Kilobyte are not the same.

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