This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: d 2 /s 2 . Similarly, we can find the delay caused by the second switch and the third link: L/R 3 , d proc , and d 3 /s 3 . Adding these five delays gives d end-end = L/R 1 + L/R 2 + L/R 3 + d 1 /s 1 + d 2 /s 2 + d 3 /s 3 + d proc + d proc To answer the second question, we simply plug the values into the equation to get 6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec. Problem 13 It takes seconds to transmit the packets. Thus, the buffer is empty when a batch of packets arrive. The first of the packets has no queuing delay. The 2nd packet has a queuing delay of seconds. The th packet has a delay of seconds. The average delay is . Problem 26 a) 80,000,000 bits b) 800,000 bits, this is because that the maximum number of bits that will be in the link at any given time = min(bandwidth delay product, packet size) = 800,000 bits. c) .25 meters Problem 28 Recall geostationary satellite is 36,000 kilometers away from earth surface. a) 150 msec b) 1,500,000 bits c) 600,000,000 bits...
View Full Document
This note was uploaded on 12/11/2009 for the course TELECOM 2310 taught by Professor Rezgui during the Fall '09 term at Pittsburgh.
- Fall '09