CSC458 Computer Networks Problem Set #1 Tutorial

CSC458 Com
Problem Set #

Lilin Zhang
Fall 2015

mputer Networks
#1 Tutorial

1

Chapter 1 – 1.

Calculate the total time requir
following cases, assuming an R
initial 2*RTT of “handshaking”

(a) The bandwidth is
continuously.

(b) The bandwidth is
each data packet we must wai

(c) The link allows inf
such that only 20 packets can

(d) Zero transmit tim
can send one packet, during th
during the second RTT we can
can send four (2^3-1) packets,

.4

red to transfer a 1.5MB file in the
RTT=80ms, a packet size of 1KB, and an
” before data is sent.

10 Mbps, and data packet can be sent

10 Mbps, but after we finish sending
it one RTT before sending the next.

finitely fast transmit, but limits bandwidth
be sent per RTT.

me as in (c), but during the first RTT we
he first RTT we can send one packet,
n send two packets, during the third we
, etc.

2

Chapter 1 – 1.

Calculate the total time requir
following cases, assuming an R
initial 2*RTT of “handshaking”
a)(a) The bandwidth is 10 Mbps
continuously.

Continu

A starts
After T
While b

.4

red to transfer a 1.5MB file in the
RTT=80ms, a packet size of 1KB, and an
” before data is sent.
s, and data packet can be sent

uously: transmit a whole bunch of bits.
s to send the first bit at 10 Mbps,
T_{propagation} time, the bit arrives at B,
bit#1 is propagating, bit #2 starts transmission.

3

Chapter 1 – 1.4

(ba) ) The bandwidth is 10 Mbps
packet we must wait one RTT

After receiving each packet, the reception of
bits is only paused for RTT/2: the propagation
the last bit is concurrent with the first half of
pause at the transmitter A. Hence, we have

4

s, but after we finish sending each data
T before sending the next.

packet
n of
f RTT

4

Chapter 1 – 1

(c) The link allows infinitely fa
that only 20 packets can be se

RTT = 80 ms
Tt = 0 ms
Tp = 40 ms

Total time = 2xRTT + 76xRTT
= 160 + 6080 + 40 ms
= 6.28 s

1.4

ast transmit, but limits bandwidth such
ent per RTT.

T + Tp request
RTT reply

confirm
Ack

RTT

...

5

Chapter 1 – 1

(d) Zero transmit time as in (c
one packet, during the first RT
second RTT we can send two
four (2^3-1) packets, etc.

RTT = 80 ms
Tt = 0 ms
Tp = 40 ms
# of packets = 1536
# of waits (1+2+…2n = 2^(n+1) -1)

Since 2^11 -1 =2047 packets,
At n = 10, we are able to send all 1
packets.

Total time = 2xRTT + 10xRTT + Tp
= 160 + 800 + 40 ms
= 1 s

1.4

c), but during the first RTT we can send
TT we can send one packet, during the

packets, during the third we can send

A B
RTT
request
RTT
reply
confirm
Ack

1536 ...

6

Chapter 1 – 1

Calculate the latency (from fir

(a) 1-Gbps Ethernet with a s
path and a packet size of
introduces a propagation
retransmitting immediate
packet.

(b) (b) Same as (a) but with t

(c) Same as (b), but assume th
switching; it is able to begin re
128 bits have been received.

1.17

rst bit sent to last bit received) for:
single store-and-forward switch in the
5000 bits.Assume that each link

delay of 10 and that the switch begins
ely after it has finished receiving the

three switches.
he switch implements “cut-through”
etransmitting the packet after the first

7

Chapter 1 – 1

Calculate the latency (from fir
(a) 1-Gbps Ethernet with a sin
and a packet size of 5000 bits.
propagation delay of 10 and
immediately after it has finishe

A S
1st bit: time 0
Last bit: 5 μs

Tp = 10 μs

Last bit rec: 15μs
Last bit sent: 20μs

Last bit rec: 30μs

1.17

rst bit sent to last bit received) for:
ngle store-and-forward switch in the path
.Assume that each link introduces a
d that the switch begins retransmitting
ed receiving the packet.

B

A->S
S ->B

8

Chapter 1 – 1

(b) Same as (a) but with three
4 links equal to 4 Tp delay
4 transmissions equal to 4
Total: 4Tp + 4Tt = 60 μs
(c) Same as (b), but assume th
switching; it is able to begin re
128 bits have been received.

Tot

A S1 S2 S3 B
1st bit: time 0
128th bit: 0.128 μs 10.

last bit: 5 μs 20

3

1.17

e switches.

4 Tt delay

he switch implements “cut-through”
etransmitting the packet after the first

tal time = (128*10^{-9} + 10)*3
+ 5000*10^{-9} +10

= 10.128 * 10^{-6}
+ 5 * 10^{-6} + 10 * 10^{-6}

= 45.384 μs

.128 μs
.256 μs

30.384 μs: 1st bit sent from S3-> B

9

Chapter 2 – 2.

Show the 4B/5B encoding, and the
for the following bit sequence:
1101 1110 1010 1101 1011 1110 1

According to the 4B/5B encoding (
11011 11100 10110 11011 10111 1
NRZI:
1 – a transition from the current si

.3

resulting NRZI signal,
1110 1111
(p82 in textbook), the encoded sequence is:
11100 11100 11101
ignal; 0 – stay at the current signal.

10

Chapter 2 – 2.

Show that two-dimensional parity
information to correct any 1-bit er
only 1 bit is bad), but not any 2-bit

For example, a frame contains
0101001 1
1101001 0

1000000 1 2
----------------------
1 bit corrupted

0100001 1
1101001 0

1000000 1

.13

provides the receiver enough
rror (assuming the receiver knows
t error.
s 2 bytes of data.:

2 bits corrupted – unable to correct
0100001 1
1100001 0
1000000 1

11

Chapter 3 – 3.

Given the extended LANs shown i
suffers catastrophic failure. Indicate
spanning tree algorithm after the r
has been formed.

Spanning tree: a subgraph that covers all the
but contains no cycles.

Spanning tree algorithm (p196):
The bridge of the smallest ID is chosen as r
All ports of the root bridge are selected.

Each bridge computes the shortest path to
The ports on the shortest paths are selecte

as preferred path to the root.

Each LAN elects a designated bridge, the on
being the closest to the root.

.14

in the figure, assume that bridge B1
e which ports are not selected by the
recovery process and a new tree

e vertices,

root.
root.
ed
ne

12

Chapter 3 – 3.

After B1 fails, B2 is selected as the roo

All ports of B2 is selected – B2: A, B, D

The shortest paths to B2: Design
B3->B2: 2 LAN A:

B4->B2: 4 LAN B:

B5->B2: 2 LAN C

B6->B2: 4 LAN D

B7->B2: 2 LAN E:

.14

ot bridge.

D.

nated bridges: Blocked ports:
: B2 LAN F: B3 B5: idle
B6: I
: B2 LAN G: B3

C: B7 LAN H: B3

D: B2 LAN I: B4

: B3 LAN J: B6

13

References

—José María Foces Morán, lecturer
note on computer networks,
http://paloalto.unileon.es/cn/Comp
—Loukas Lazos, University of Arizo
Networks, www2.engr.arizona.ed

r. University of León, Complementary lecture
plNotesCN.Ch1.pdf
ona, Lecture #3 Fundamentals of Computer
du/~ece578/lectures/03_Performance.ppt

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