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Authors:
James F. Kurose, Keith W. Ross
Chapter:
Multimedia Networking
Exercise:
Problems
Question:12 | ISBN:9780132856201 | Edition: 6

Question

Consider again the figure in P11, showing packet audio transmission and reception times.

a. Compute the estimated delay for packets 2 through 8, using the formula for d i from Section 7.3.2. Use a value of u = 0.1.

b. Compute the estimated deviation of the delay from the estimated average for packets 2 through 8, using the formula for v i from Section 7.3.2. Use a value of u = 0.1.

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Answer

To compute the estimated delay and estimated deviation of the delay for packets 2 through 8, we will use the formula from Section 7.3.2 with a value of u = 0.1.

a)

Estimated Delay (d_i): The estimated delay for each packet can be calculated using the formula:

\large d_i = (1 - u) * d_{i-1} + u * (r_i - t_i)

Let's calculate the estimated delay for packets 2 through 8 using the given values:

Packet 1: r_1 = 4 ms, t_1 = 0 ms (Assuming the reference point for time is the start of packet 1) d_1 = r_1 - t_1 = 4 ms (Initial estimated delay)

Packet 2: d_2 = (1 - 0.1) * 4 ms + 0.1 * (10 ms - 4 ms) = 3.6 ms + 0.6 ms = 4.2 ms

Packet 3: d_3 = (1 - 0.1) * 4.2 ms + 0.1 * (12 ms - 6 ms) = 3.78 ms + 0.6 ms = 4.38 ms

Packet 4: d_4 = (1 - 0.1) * 4.38 ms + 0.1 * (8 ms - 2 ms) = 3.942 ms + 0.6 ms = 4.542 ms

Packet 5: d_5 = (1 - 0.1) * 4.542 ms + 0.1 * (14 ms - 4 ms) = 4.088 ms + 1 ms = 5.088 ms

Packet 6: d_6 = (1 - 0.1) * 5.088 ms + 0.1 * (6 ms - 2 ms) = 4.5792 ms + 0.4 ms = 4.9792 ms

Packet 7: d_7 = (1 - 0.1) * 4.9792 ms + 0.1 * (10 ms - 4 ms) = 4.48128 ms + 0.6 ms = 5.08128 ms

Packet 8: d_8 = (1 - 0.1) * 5.08128 ms + 0.1 * (12 ms - 4 ms) = 4.573152 ms + 0.8 ms = 5.373152 ms

Therefore, the estimated delays for packets 2 through 8 are:

d_2 = 4.2 ms

d_3 = 4.38 ms

d_4 = 4.542 ms

d_5 = 5.088 ms

d_6 = 4.9792 ms

d_7 = 5.08128 ms

d_8 = 5.373152 ms

 

b)

Estimated Deviation of Delay (v_i):

The estimated deviation of delay for each packet can be calculated using the formula:

\large v_i = (1 - u) * (v_{i-1} + |d_i - d_{i-1}|) + u * |d_i - d_{i-1}|

Let's calculate the estimated deviation of the delay for packets 2 through 8 using the given values:

Packet 2: v_2 = (1 - 0.1) * (0 + |4.2 ms - 4 ms|) + 0.1 * |4.2 ms - 4 ms| = 0.18 ms

Packet 3: v_3 = (1 - 0.1) * (0.18 ms + |4.38 ms - 4.2 ms|) + 0.1 * |4.38 ms - 4.2 ms| = 0.198 ms

Packet 4: v_4 = (1 - 0.1) * (0.198 ms + |4.542 ms - 4.38 ms|) + 0.1 * |4.542 ms - 4.38 ms| = 0.2436 ms

Packet 5: v_5 = (1 - 0.1) * (0.2436 ms + |5.088 ms - 4.542 ms|) + 0.1 * |5.088 ms - 4.542 ms| = 0.33924 ms

Packet 6: v_6 = (1 - 0.1) * (0.33924 ms + |4.9792 ms - 5.088 ms|) + 0.1 * |4.9792 ms - 5.088 ms| = 0.349316 ms

Packet 7: v_7 = (1 - 0.1) * (0.349316 ms + |5.08128 ms - 4.9792 ms|) + 0.1 * |5.08128 ms - 4.9792 ms| = 0.367384 ms

Packet 8: v_8 = (1 - 0.1) * (0.367384 ms + |5.373152 ms - 5.08128 ms|) + 0.1 * |5.373152 ms - 5.08128 ms| = 0.441168 ms

Therefore, the estimated deviations of the delay for packets 2 through 8 are:

v_2 = 0.18 ms

v_3 = 0.198 ms

v_4 = 0.2436 ms

v_5 = 0.33924 ms

v_6 = 0.349316 ms

v_7 = 0.367384 ms

v_8 = 0.441168 ms

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