Question

What is the total delay latency for a frame of size 5 million bits that is being sent

Answer 1

It seems like the question is not complete. The question goes something like:

Question: What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 2 μs and a processing time of 1 μs. The length of the link is 2000 Km. The speed of light inside the link is $2 \times 10^{8} m/s$. The link has a bandwidth of 5 Mbps.

Answer: 1.01000030 s

Step-by-step explanation:

Given,

Frame of size = 5 million bits = 5 × $10^{6}$ bits = 5000000 bits

Queuing time = 2 μs =  2 × $10^{-6}$ sec = 0.000020 sec

Processing time = 1 μs = 1 × $10^{-6}$ sec = 0.00001 sec

Length of link, distance d = 2000 Km = 2000 × $10^{3}$ m

The speed of light (Propagation speed) = $2 \times 10^{8} m/s$

Bandwidth , B = 5 Mbps = 5 × $10^{6}$ bps

For delay (latency), use the following formula

Delay (latency) = Propagation time + Transmission time + Queuing time + Processing time ...........(i)

Propagation time = $\frac{\text{Distance}}{\text{Propagation speed}}$

                             = $\frac{2000 \times 10^{3}}{2 \times 10^{8}}$

                             = 0.01 sec

Transmission time = $\frac{\text{Frame of size}}{\text{Bandwidth}}$

                              = $\frac{5000000 \times 10^{3}}{5 \times 10^{6}}$

                              = 1 sec

Now, from equation (i) we get,

Delay (latency) = 0.01 + 1 + 0.000020 + 0.000010

                         = 1.010030 sec

Answer 2

Answer:

Propagation time = distance / propagation speed

= 2000 Km / 2 x 108 m/s = 10 ms

Transmission time = Message size / Bandwidth

= 5 x 106 bits/ 5 Mbps = 1 s

Queuing time = 10 routers * 2 us = 20 us

Processing Delay = 10 routers * 1 us = 10 us

Total delay (latency) = 10 ms + 1 s + 20 us + 10 us

= 1010.03 ms = 1.01003 s  1 s