Answer:
260us, 520us and 166us
Explanation:
a) Given,
Transmission time
= 12000bits/100MBPS
= 12000/100×10 raised to power 6
= 120us
Packet reaches at switch
=120+10
=130us
Then from switch to destination =130
Therefore, total latency,
= 130+ 130
= 260us
Please kindly go to attachment for the step by step solution of b and c.
Given Information:
Network speed = 100 Mbits/s
packet size = 12 kbits
Propagation delay = 10 µs
Answer:
(a) Latency = 260 µs
(b) Latency = 520 µs
(c) Latency = 142 µs
Explanation:
Latency is the time taken for the data to travel to its destination over some network.
The total time taken is the sum of transmission delay and propagation delay
Latency = Ttrans + Tprop
Latency = (packet size)/(Network speed) + Tprop
(a) 100-Mbps Ethernet with a single store-and-forward switch in the path and a packet size of 12,000 bits.
Since single switch has two links, each link will take a time of
Latency = (12x10³ bits)/(100x10⁶ bits/s) + 10 µs
Latency = 120 µs + 10 µs
Latency = 130 µs
For two links,
Latency = 130*2
Latency = 260 µs
(b) Same as (a) but with three switches.
Three switches corresponds to a total of 4 links and each link takes a time of 130 µs
Latency = 130*4
Latency = 520 µs
(c) Same as (a), but assume the switch implements "cutthrough" switching; it is able to begin retransmitting the packet after the first 200 bits have been received.
The Latency in this case would be
Latency = [(200)/(100x10⁶) + 10 µs] + [(12x10³)/(100x10⁶) + 10 µs]
Latency = [ 2 µs + 10 µs] + [ 120 µs + 10 µs]
Latency = [ 12 µs ] + [ 130 µs]
Latency = 142 µs
