- Jun 21, 2014
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Suppose that 100kBytes of user data has to be sent from host A to host B passing through 4 switches employing packet switching
- Thread starter Devi Pagavan
- Start date
- Jun 21, 2014
- 4
- 0
- 4,510
I found it
Answer Description
There are 5 hops between Host A and Host B. Let p bits be the payload in a packet. The header bits are 408= 320. Time to send this packet is given by
hop-count * packet-tx-time + (Number of packets-1) * packet-txt-time
The first part calculates the time it takes for the first packet to reach host B. After this, rest of the packets follow separated by transmission time, which is captured by the second part.
5 * (p+320)/110^6 + [(1000008/p)-1] * (p+320)/110^6
This expression needs to be minimized with respect to p. Differentiating and setting to zero yields p = 1000bytes.
Answer Description
There are 5 hops between Host A and Host B. Let p bits be the payload in a packet. The header bits are 408= 320. Time to send this packet is given by
hop-count * packet-tx-time + (Number of packets-1) * packet-txt-time
The first part calculates the time it takes for the first packet to reach host B. After this, rest of the packets follow separated by transmission time, which is captured by the second part.
5 * (p+320)/110^6 + [(1000008/p)-1] * (p+320)/110^6
This expression needs to be minimized with respect to p. Differentiating and setting to zero yields p = 1000bytes.
Your formula completely ignores real life.
The hop count means very little actually. All a extra hop adds is the tiny amount of time it takes to clock the data into a buffer on the switch and then reclock it out the other port. This depends on the port speed which your formula must take into account. For a 10m port it might be fairly high but for 1g or even 10g ports it is such a tiny fraction of a millisecond that it will be hidden by other delays. The much more important number is the distance between the devices. This delay is pretty much some multiple of the speed of light depending on what type of media it is traveling over.
Then you can not assume some fixed header sizes. It greatly depends on the protocols being used. This is why true video steaming uses UDP rather than TCP. On top of all this you need to know the minimum MTU in the path. You can get hugely different through put if you were to compare a actual lan running something like jumbo frames compared to a VPN connection that has huge overhead and only allows 1100 bytes of payload maximum.
The hop count means very little actually. All a extra hop adds is the tiny amount of time it takes to clock the data into a buffer on the switch and then reclock it out the other port. This depends on the port speed which your formula must take into account. For a 10m port it might be fairly high but for 1g or even 10g ports it is such a tiny fraction of a millisecond that it will be hidden by other delays. The much more important number is the distance between the devices. This delay is pretty much some multiple of the speed of light depending on what type of media it is traveling over.
Then you can not assume some fixed header sizes. It greatly depends on the protocols being used. This is why true video steaming uses UDP rather than TCP. On top of all this you need to know the minimum MTU in the path. You can get hugely different through put if you were to compare a actual lan running something like jumbo frames compared to a VPN connection that has huge overhead and only allows 1100 bytes of payload maximum.
WyomingKnott
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