Packet switching

Question

Q 30). Consider a route in a store and forward networking going through $9$ intermediate nodes.The packet contains $1100$ bits and are transmitted at $64 \  Kpbs $ . Assume propogation delay over the links are negligible.As packet travels along the route, it encounters an average of $5$ packets when it arrives at each node. How long does it take for the packets to get to the receiver if the nodes transmit on a " first come first served" basis (in ms) ?

Comments

is the answer, 945 ms??

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Yes,how u got it please explain

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Pls refer Network performance measures: Example 2 Page 21

http://csiweb.ucd.ie/Staff/jmurphy/networks/comp30040_2-basics.pdf

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i think answer should be 2.5 ms.

As for 1pkt total time is only transmission time as propogation delay is nil. and so transmission is 8*t as 9 hops are there

so, total time=5*8*4/(64*1000)s

Correct me if i m wrong.

Answer 1

It is given that at each router we encounter 5 packets. That means at each router we will have to wait for 5*Tt+self Tt = 6Tt

now there are 9 routers. So total delay = 6*9*Tt = 54Tt

now we also have to add Tt for sender to transmit that packet

so total delay now becomes 54Tt + Tt = 55*Tt

now Tt=1100/64 = 17.1875 msec

total delay = 55 * 17.1875 = 945.3125 msec

Comments

Plz explain that part self Tt .

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Plz explain

1)  here every router encounter 5 packets ok. So 5*Tt  but why add self Tt.

2) another add one more Tt for sender to receiver

@Digvijaysingh Gautam

@rahul Sharma 5

 

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@ Hradesh patel

@Retina Ray

 But why add self Tt ?

because after the router transmits those 5 packets it will now transmit the packet sent by the host 

so total 5tt+1tt

another adds one more Tt for the sender to receiver?

draw a sketch for 9 routers and 2 hosts you will get that