Transmission and Propagation Delays: Suppose s = 2.5 × 10^8 meters/sec, L = 3000 bytes, and R = 10 Mbps. Find the distance m so that dprop = dtrans.

Question

Suppose s = 2.5 × 10^8 meters/sec, L = 3000 bytes, and R = 10 Mbps. Find the distance m so that dprop = dtrans. Explain how I can do step by step. Thanks!

Answer 1

corrected ,,,was 2.4*10^-4, corrected to 2.4*10^-3

propagation time is (3000 bytes)(8bits/byte)/(10⁷ bits/second)=2.4*10^-3 seconds,,,,,

if the transmission time of first bit = 2.4*10^-4seconds, then the transmission distance of

of the first bit is:

(2.5*10⁸ meters/second)(2.4*10^-3 seconds) meters=(2.4)(2.5)*10⁵ meters=600km

Answer 2

Here is the procedure step-by-step! Let me know if you understand it or are confused about something :) Also, I would appreciate it if anyone could check it as well!

Suppose S = 2.5×10⁸ m/s, L = 3000 bytes and R = 10 Mbps

Find the distance m so that d_prop=d_trans

d_prop= propagation delay

d_trans= transmission delay

Mbps = Megabytes per second = 10⁶ per second

m/s = meters per second

d_{end−to−end} = (m/S + L/R)

m/S = L/R

m = L/R ∗ S

So, m = L/R ∗ S = (3000 bytes)/(10 ×10⁶ bps) ∗ 2.5×10⁸ m/s = 75000 m = 75 km

Answer: m = 75 km