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HW1 Solutions

# HW1 Solutions - Problem 11 The arriving packet must first...

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HW1 Solutions: Problem 5 a) s m d prop / seconds. b) R L d trans / seconds. c) ) / / ( R L s m d end to end seconds. d) The bit is just leaving Host A. e) The first bit is in the link and has not reached Host B. f) The first bit has reached Host B. g) Want 893 10 5 . 2 10 28 100 8 3 S R L m km. Problem 6 Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires 3 10 64 8 48 sec=6msec. The time required to transmit the packet is 6 10 1 8 48 sec= 384 sec. Propagation delay = 2 msec. The delay until decoding is 6msec + 384 sec + 2msec = 8.384msec A similar analysis shows that all bits experience a delay of 8.384 msec. Problem 8 a) 10,000 b) M N n n M n p p n M 1 1

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c) Problem 10 Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Thus, d end-end = L/R + d 1 /s 1 + d 2 /s 2 For the values in Problem 9, we get 8 + 16 + 4 = 28 msec.
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Unformatted text preview: Problem 11 The arriving packet must first wait for the link to transmit 3,500 bytes or 28,000 bits. Since these bits are transmitted at 1 Mbps, the queuing delay is 28 msec. Generally, the queuing delay is [nL + (L - x)]/R . Problem 14 a) The transmission delay is R L / . The total delay is I R L R L I R IL 1 / ) 1 ( b) Let R L x / . Total delay = ax x 1 Problem 15 a) There are Q nodes (the source host and the 1 N routers). Let q proc d denote the processing delay at the q th node. Let q R be the transmission rate of the q th link and let q q trans R L d / . Let q prop d be the propagation delay across the q th link. Then Q q q prop q trans q proc end to end d d d d 1 . b) Let q queue d denote the average queueing delay at node q . Then Q q q queue q prop q trans q proc end to end d d d d d 1 ....
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HW1 Solutions - Problem 11 The arriving packet must first...

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