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Consider transmitting a 10,000 bit message over two links, from source to destination hosts via a router (see figure below). The router uses store and forward packet switching. The link between the source host and the router is 1,000 meters long, and the link between the router and the destination host is also 1,000 meters long. Both links have a transmission rate of 1 Mbps. Assume a propagation speed of 2*10^8 m/sec. Assume that the only delays are transmission delays and propagation delays.

a.Suppose the message is sent as one 10,000 bit packet. What is total delay in sending the message from source to destination?

b.Now suppose that the message is broken into four packets, each 2,500 bits. What is the total delay in sending the message from source to destination?

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Hi, I'm new to networking and need some help on my assignment. I think the first one might be dprop+dtrans = 1000/(2*10^8) + 10000/(10^6), but not sure. Can anyone give some help?

Layla
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2 Answers2

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Q1. The time taken for the first bit to reach first hop: 1000/2*10^8s. Time taken for transmitting 10000 bits : 10000/1M Secs

So, by the end of `10000/1M + 1000/2*10^8 sec`. the whole packet will reach the router.

As you said in your question - assuming there is no processing delay in the router, at the end of the before said time, the router starts transmitting the packet. And from the router to destination takes the same amount of time as source to router. So at the end of 2* before said time, destination gets the whole packet.

Q2. The only difference between first scenario and second scenario is based on the store and forward logic you mentioned in the question. That means, the router is able to transmit only after the whole packet is received. so it has to wait the previously calculated amount of time to start transmitting. But in the second scenario since the packet size if 1/4 of original, it can start transmit the first pkt at the end of 1/4 of original time. But in both the scenarios the time at which last bit is received on the destination will be the same.

Does it answer your question?

joe
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  • why 10000/1Mbps = 10sec ？ – Layla Oct 28 '13 at 01:50
  • ahh. sorry, I kept reading it as Kbps. My mistake. thank you for pointing this out. btw, in networking 1K is normally taken as 1000. I corrected that. otherwise, the answer holds good, doesn't it? Sorry for the delay in response. – joe Nov 02 '13 at 02:09
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Q1: In this case there is only 1 packet of 10000bits So here transmission delay will be 2(10000/1mbps) And propogation delay will be 2(1000/2×10^8) And the total delay will be prop delay + transmission delay.

Q2: the packet is divided into packets the propogation delay is not affected by this as it is independent of the number of packets or bits in each packet but this changes the transmission delay it will be 2(2500) /1mbps for the first packet and 1(2500)/1mbps for other 3 packets so total transmission delay becomes 2(2500)/1mbps+3(1(2500))/1mbps

Umer Farooq
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