Interactive end-of-chapter exercises


Computing end-end delay (transmission and propagation delay)

Consider the figure below, with three links, each with the specified transmission rate and link length.




Assume the length of a packet is 16000 bits. The speed of light propagation delay on each link is 3x10^8 m/sec

Round your answer to two decimals after leading zeros



Question List


1. What is the transmission delay of link 1?

2. What is the propogation delay of link 1?

3. What is the total delay of link 1?

4. What is the transmission delay of link 2?

5. What is the propogation delay of link 2?

6. What is the total delay of link 2?

7. What is the transmission delay of link 3?

8. What is the propogation delay of link 3?

9. What is the total delay of link 3?

10. What is the total delay?




Solution


Link 1 transmission delay = L/R = 16000 bits / 100 Mbps = 0.00016 seconds

Link 1 propagation delay = d/s = ()2 Km) * 1000 / 3*10^8 m/sec = 6.67E-6 seconds

Link 1 total delay = d_t + d_p = 0.00016 seconds + 6.67E-6 seconds = 0.00017 seconds

Link 2 transmission delay = L/R = 16000 bits / 1000 Mbps = 1.60E-5 seconds

Link 2 propagation delay = d/s = ()1000 Km) * 1000 / 3*10^8 m/sec = 0.0033 seconds

Link 2 total delay = d_t + d_p = 1.60E-5 seconds + 0.0033 seconds = 0.0033 seconds

Link 3 transmission delay = L/R = 16000 bits / 100 Mbps = 0.00016 seconds

Link 3 propagation delay = d/s = ()3 Km) * 1000 / 3*10^8 m/sec = 1.00E-5 seconds

Link 3 total delay = d_t + d_p = 0.00016 seconds + 1.00E-5 seconds = 0.00017 seconds

The total delay = d_L1 + d_L2 + d_L3 = 0.00017 seconds + 0.0033 seconds + 0.00017 seconds = 0.0037 seconds



That's incorrect

That's correct

The answer was: 0.00016

Question 1 of 10

The answer was: 6.67E-6

Question 2 of 10

The answer was: 0.00017

Question 3 of 10

The answer was: 1.60E-5

Question 4 of 10

The answer was: 0.0033

Question 5 of 10

The answer was: 0.0033

Question 6 of 10

The answer was: 0.00016

Question 7 of 10

The answer was: 1.00E-5

Question 8 of 10

The answer was: 0.00017

Question 9 of 10

The answer was: 0.0037

Question 10 of 10

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We greatly appreciate the work of John Broderick (UMass '21) in helping to develop these interactive problems.

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