How long does it take for the packet to get to the receiver?

Question

Consider a route in a store and forward network going through 9 intermediate nodes. The packet contains 1100 bits and are transmitted at 64 Kbps. Assume propagation delay over the links are negligible. As a 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 packet to get to the receiver if the nodes transmit on a "first come first served" basis (in ms) ?

Comments

is the answer, 945 ms??

Answer 1

945.3125 msec (Ans) 

Here,  packet goes like this -  sender > 9 intermediate nodes > receiver

 Transmission Time for one packet  (T.T.) =  $\frac{Frame\ Size}{Bandwidth}$ 

                                                                   =  $\frac{1100\ bits}{64\ Kbps} = 17.1875\ msec$

whenever a packet is transmitted by a node , it eventually get received by

next node immediately ,   as propagation delay is negligible.

After one T.T. it is received by next node , when it encounters 5 more

 packets which are to be transmitted before it (FCFS used).

So, for this packet to be transmitted for first intermediate node = 6 * T.T.  

       so, it gets received immediately by 2nd intermediate node (t_p = 0)

So, Total time = T.T.  +  9 * (6 T.T.)      { At Sender  1 T.T.  & at each

                                                                     Intermediate node 6 T.T.}

                       = 55 T.T. = 55 * 17.1875

                       = 945.3125 msec (Ans)                 

     

Comments

if "FCFS" not given then what should be the answar?Plz someone explain this.....