(a) A router RI receives traffic from two Local Area Networks LAN 1 and LAN 2, a
ID: 3874343 • Letter: #
Question
(a) A router RI receives traffic from two Local Area Networks LAN 1 and LAN 2, and forwards traffic on a link at 8 Mbits/sec towards a router R2. Router R2 receives also traffic from another Local Area Network LAN 3, and forwards packets on a link towards a LAN 4 (see figure Q4) The average arrival rate of the packets into R1 from LAN 1 is 400 packets/sec, and LAN 3 sends packets into R2 on average every 6 msec. Assume all arrival processes into the routers are Poisson processes, and the size of packets is exponentially distributed with an average size of 1250 bytes. 4. Figure Q4 (i) Consider Rl as an MM/1 queue, and assume LAN 2 sends no packets. Calculate the average delay experienced by packets in R1. [6 marks] (ii) Assume R1 and R2 operate both as MM/I queues. Assume LAN 2 sends no packets. Calculate what the forwarding rate on the outgoing link from router R2 should be in bits/sec to have an average delay per packet from LAN 1 to LAN 4 of 10 msec. (Consider the propagation delays negligible.) (6 marks] (ii) Assume R1 operates as an M/M/1/N queue, and R2 operates as an MM/1 queue with the forwarding rate found i Now LAN 2 sends 200 packets/sec. What should be the buffer size in router R1 so that its throughput is limited to a value that does not lead to an unstable system in R2 (i.e. service rate in R2 should remain greater than arrival rate)? Give your result as an integer number. 6 marks] (iv) Assuming for RI the finite capacity calculated ini, establish what the number of dropped packets per second in R1 will be when LAN 2 stops sending packets. 6 marks] Note: The probability P of having n packets in a queuing system accommodating at most N packets (M/M/1/N) is: p where is the traffic intensityExplanation / Answer
In packet switched networks, including the Internet, the source host segments long, application layer messages i.e. an image or a music file, into lighter packets and sends the packets into the network. The receiver then resembles the packets back into the original message. This process is known as message segmentation.
Consider the message that is (1250 bytes) 8*10^6 bits long that is to be sent from a source to a destination which are separated by two routers in between. Each of the links in this path is 8 Mbps. Ignoring propagation, queuing and processing delays.
Now each the packet is segmented into 400 packets with each packet being 10000bits long.
Time to send message from source to first packet switch is
= (8* 10^6) /( 2 * 10^6 ) = 4 sec.
With store and forward to move message from source to destination= 4*3 = 12 sec.
Time to send 1st packet from source host to first packet switch
= (1*10^4) / (2*10^6) = 5 m sec.
Time at which second packet is received at the first switch = time at which first packet is received at the second switch = 2 × 5m sec = 10 m sec
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