Consider a 3-hop communication path characterized as follows: a) All links opera
ID: 3649138 • Letter: C
Question
Consider a 3-hop communication path characterized as follows:a) All links operate at 48000 bps
b) Propagation delay is 20 msec per link
c) Connection set-up time is made of 23 sec for dialing and 100 msec to send the
request and receive the confirmation
d) Queueing as well as processing delays are negligible'
1. Using circuit switching, compute the total delay to transfer a 10,000 bytes
long message.
2. Assume that the message is divided into datagrams, each carrying 50 bytes
of header information and P bytes of data information. Consecutive
datagrams are submitted to the network 20 msec apart of each other. What
is the value of P that guarantees better delay for datagram service over
circuit switching?
You dont need to perform the actual calculations. You only need to set up the equations
and say what needs to be done to find P. Find the actual value of P only if you want to do
it
Explanation / Answer
Given:
bitrate=48,000 bps
Propagation delay=20msec per link
Connection setup time=23 s(for dialing) + 100 msec(send Req +Rcv confirmation)
=23.1 s
1) Using circuit switching :
Message length=10,000 bytes
No of hops=M=3
Queueing and processing delay=0
Link propagation delay=L=20msec=0.02 s
Transmission speed=W bps=48,000 bps
Message size=10,000 bytes=10,000*8=80,000 bps
Total delay=3ML+B/W+(M-1)P
=3*3*0.02+(80,000/48,000)+2*0
=0.18+1.666+0
=1.8466
We add up connectup setup dealy,
Total delay=1.8466+23.1=24.9466s
2)Using datagram packet switching,
Delay=ML+NT+(M-1)T+(M-1)P
No of packets=N
Per hop processing delay,P=20msec=0.02sec
Link popagatiion delay,L=20 msec=0.02sec
Packet size=50 bytes+P bytes
=(50*8+P*8)bits
=(400+8P) bits
Transmission speed=20msec=0.02sec
Packet transmission delay,T=Packetssize/Transmission speed
=(400+8P)/0.02
Delay=ML+NT+(M-1)T+(M-1)P
=3*0.02+N*(400+8P/0.02)+2*0.02
=0.06+(400+8P/0.02)+0.04
=0.1+(400+8P/0.02)
OBSERVATIONS:
Circuit switching delay=24.9466 s
Datagram packet switching delay=0.1+(400+8P/0.02) where P is bytes of data transferred.
With the advancement in transmission speeds, the total delays are dominated by the propagation delays.
Typically, the packet size in datagram packet switcing network is 1000 bytes.Header contains information about the order of packets and the checksum.
N=10,000/1000=10 datagram packets can be sent over the nework.
Hence, 50+P=1000
P=1000-50=950 bytes of data.
We should try to minimise the store and forward delays,packet transmission delays and increase the transmission speeds of the links in order to guarantee better delay for over circuit switching.
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