Suppose there is exactly one packet switch between a sending

Suppose there is exactly one packet switch between a sending host and a receiving host. Assume each link has a distance 1, 250 Km with propagation speed 2.5 times 10^8 mps. Link 1 has a transmission rate 10 Mbps and Link 2 has a transmission rate 20 Mbps. Assuming that the switch uses store-and-forward packet switching, what is the total end-to-end delay to send a 2KB packet? (Ignore queuing and processing delay.

Solution

Dear Student,

Here is the answer...

This question is from Computer Networking. And to solve this problem simply use time and distance formula.

We know that   

time = distance/velocity

So here

D1 = D2 = 1250                  ( D1 and D2 is the link distance)

Propagation speed P = 2.5 X 10⁸ mps

Time to travel between sending host and recieving host = distance/speed = 1250/2.5x10⁸ = 0.005 seconds.

Link 1 transmission rate t1 = 10 Mbps

Link 2 transmission rate t2 = 20 Mbps

Packet Length = 2KB

# Suppose at time t₀ the transmitting of the the packet occur.

# At time t1 the sending host will begin to transmit the packet

so

time t1 = 2KB/10Mbps = 2 x 2¹⁰ x 8 / 10 x 2 ²⁰ = 16 / 10 x 2¹⁰ = 0.0015625 seconds

Because switch uses store and forward mechanism so packet will be first stored and then transmitted so total time = t1+t2.

# Now At time t2 the the switch will completes transmission.

So time t2 = t1 + 2 x 8 x 2¹⁰ / 20 x 2²⁰ = 0.0015625 + 0.00078125 = 0.00234375 seconds.

Hence the end to end delay is = t1+t2 = 0.00234375 seconds.           Ans..

Kindly Check and Verify Thanks....!!!