please explain step by step. I do not undestand anything. In modern packet- swit
ID: 3800684 • Letter: P
Question
please explain step by step. I do not undestand anything.
In modern packet- switched networks, including the Internet, the source host segments long, applicationlayer messages (for example, an image or a music file) into smaller packets and sends the packets into the network. The receiver then reassembles the packets back into the original message. We refer to this process as message segmentation. Figure 1.27 illustrates the end- to- end transport of a message with and without message segmentation. Consider a message that is 8×106 bits long that is to be sent from source to destination in Figure 1.27. Suppose each link in the figure is 2 Mbps. Ignore propagation, queuing, and processing delays. a. Consider sending the message from source to destination without message segmentation. How long does it take to move the message from the source host to the first packet switch? Keeping in mind that each switch uses store- and- forward packet switching, what is the total time to move the message from source host to destination host? b. Now suppose that the message is segmented into 800 packets, with each packet being 10,000 bits long. How long does it take to move the first packet from source host to the first switch? When the first packet is being sent from the first switch to the second
a. Source b. Source Message Packet switch Packet switch Destination Packet Packet switch Packet switch Destination End-to-end message transport: (a) without message segmentation; (b) with message segmentationExplanation / Answer
1)
tm to send msg frm sorce host to 1st pckt swtch=8*106/2*106 sec=4sec
uses store- and- forward packet switching,
the total time to move the message from source host to destination host=4sec*3hops=12sec
2)
Tme 2 snd 1st packt from srce host to 1st packt switch =1*104/2*106=5msec
Tme wch 2nd packt received at the 1st swch = tme wch 1st packet is received at the 2nd swch = 2*5msec=10msec
3)
tme wch 1st packt recved at destnatin host=5m sec*3 hops=15m sec
when aftr 5 m sec tme at which last (800th) packet is received =15m sec +799*5m sec=4.01sec
when msge segmentation is used there cn be a less delay
4)
With msg segemtation,If ter is single bit,the whole msg get transmitted and huge packt are not sent into the network.
Drawback:
a)At destination packt should be in sequence
b)msg segmntatin results many small pckts,the totall amt of header byte is more because the header size is usual same for all packts with their sizes also.
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