Hw1 solution PDF

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Summary

solution to written hw 1...

Description

Message Switching (Store and Forward)

Packet Switching (Store and Forward)

1. a)

1. b) Datagram Network R bps R bits/sec

Time

bits

1 2 3

1 2

M

M M

◆

“Message switching” on N links » N * 8(F+h)/R s total time to send file from A to B

◆

N*8(P+h)/R

...

...

M Packets, P+h Bytes M

8(F+h)

1 2 3 4

...

8(F+h)/R sec

...

8(F+h)/R sec

...

F Bytes

8(F+h)/R sec

1 2 3 4

N*8(P+h)/R + (M-1)*8(P+h)/R

Packet-switching: » Total time to send file = (M+N-1)*8(P+h)/R

1

2

Packet Switching (Store and Forward)

Circuit Switching

1. c) Virtual-Circuit Network

1. d)

R bps R bits/sec

Time

1 2

...

M

TS + 8(F+h/2)/R sec 8(F+h/2) bits

M M

◆

N*8(P+h/2)/R

...

1 2 3

...

...

M Packets, P+h/2 Bytes M

1 2 3 4

...

F Bytes

1 2 3 4

N*8(P+h/2)/R + (M-1)*8(P+h/2)/R

◆

Total time to send file from A to B: » TS + 8(F+h/2)/R

Packet-switching: » Total time to send file = (M+N-1)*8(P+h/2)/R + TS 3

4

Packet Switching vs. Circuit Switching

Bandwidth-Delay Product

1. e)

2. a), 2. b), 2. c) R = 1.5 Mbps

R bits/sec

d = 5,000 km

◆

Tpropagation = 5,000 km / 2.5 x 108 m/s = 0.02 s

◆

One-way Bandwidth-delay Product = 1.5 Mbps * 0.02 s

F Bytes

◆

Circuit Switching Time > Packet Switching Time

= 30,000 bits TS + 8(F+h/2)/R > (M+N-1)*8(P+h)/R … TS > 8/R * ((N-1)(P+h) + Mh – h/2) 5

◆

Max # of bits on the link at any one time = 30,000

◆

BDP represents the storage capacity of the network 6

Total Delay: Message Switching

Total Delay: One UnACKed Packet Only

2. d)

2. e) R = 1.5 Mbps

d = 5,000 km

◆

1 2 3 4

1

50

2

R = 1.5 Mbps

d = 5,000 km

Total delay for 450,000 bits = processing + queueing + transmission + propagation delays

3

= 0 + 0 + 450,000/(1.5 Mbps) + 0.02 = 0.32 s ◆

Total delay for 50 packets, 9,000 bits each = 50*(end-to-end frame delay + end-to-end ACK delay) – (end- to-end ACK delay for 50th ACK) = 50 * ( (9,000 b)/(1.5 Mbps) + 0.02 + 0.02 ) – 0.02 = 2.28 s

7

8

Total Delay: Upto m UnACKed Packets

Total Delay: m Outstanding Packets

2. f)

2. f)

1 2 3 4

1 2 3

50

4 5 6

R = 1.5 Mbps

R = 1.5 Mbps

d = 5,000 km

d = 5,000 km

◆

50 pkts, 9,000 bits each, m pkts pipelined Time for 1st ACK to return

◆

To keep pipeline full: m*0.006 > 0.046

◆

Sender is in middle of transmitting 8th packet when 1st ACK arrives

◆

= (9,000 b)/(1.5 Mbps) + 0.02 + 0.02 = 0.006 + 0.02 + 0.02 = 0.046

7

◆

=> m = 8 packets

Up to m packets can be sent before waiting for an ACK e.g., m = 3

»

9th packet will be transmitted, starting at 8*0.006 = 0.048 s

Frame 2m will be sent starting at: (2m-1)*0.006 = 0.09 s ◆ Total end-to-end delay = 50*0.006 + 0.02 = 0.32 s

◆ 9

10

Total Delay: m Outstanding Packets

Geostationary Satellite

3.

4. R = 600 Mbps

d = 5,000 km

◆

d = 3.6 * 107 m

◆

R = 10 Mbps Image size = 37*106

◆ ◆

a) 12 Mb

◆

b) 450 Kbits d) 20.75 ms

One un-acknowledged image at a time a) Images transmitted per minute: Total time per image = transmissionDelay + 2*propagationDelay = 37 * 106/107 + 2* (3.6 * 107)/(2.4 * 108) = 4 seconds Images / minute = 60/4 = 15

e) 1.98075 s f) m = 2,668 packets

◆

b) Desired throughput = one image / second Total time per image = X/107 + 2*propagationDelay...