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 8000 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 = 8000 bits / 1000 Mbps = 8.00E-6 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 = 8.00E-6 seconds + 6.67E-6 seconds = 1.47E-5 seconds

Link 2 transmission delay = L/R = 8000 bits / 10 Mbps = 0.0008 seconds

Link 2 propagation delay = d/s = ()500 Km) * 1000 / 3*10^8 m/sec = 0.0017 seconds

Link 2 total delay = d_t + d_p = 0.0008 seconds + 0.0017 seconds = 0.0025 seconds

Link 3 transmission delay = L/R = 8000 bits / 1000 Mbps = 8.00E-6 seconds

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

Link 3 total delay = d_t + d_p = 8.00E-6 seconds + 6.67E-6 seconds = 1.47E-5 seconds

The total delay = d_L1 + d_L2 + d_L3 = 1.47E-5 seconds + 0.0025 seconds + 1.47E-5 seconds = 0.0025 seconds



That's incorrect

That's correct

The answer was: 8.00E-6

Question 1 of 10

The answer was: 6.67E-6

Question 2 of 10

The answer was: 1.47E-5

Question 3 of 10

The answer was: 0.0008

Question 4 of 10

The answer was: 0.0017

Question 5 of 10

The answer was: 0.0025

Question 6 of 10

The answer was: 8.00E-6

Question 7 of 10

The answer was: 6.67E-6

Question 8 of 10

The answer was: 1.47E-5

Question 9 of 10

The answer was: 0.0025

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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