Computer Networks - Solved Worked out Problems

**WORKED OUT PROBLEMS**

1.Calculate the total time required to transfer a 1.5MB file in the following cases,assuming a RTT of 80 ms, a packet size of 1 KB data, and an initial 2×RTT of“handshaking” before data is sent.

(a)The
bandwidth is 10Mbps, and data packets can be sent continuously.

The bandwidth is 10Mbps, but after we finish
sending each data packet we must wait one RTT before sending the next.

The link allows infinitely fast transmit, but
limits bandwidth such that only 20 packets can be sent per RTT.

Zero transmit time as in (c), but during the first
RTT we can send one packet, during the second RTT we can send two packets,
during the third we can send four = 23−1, and so on.

__SOLUTION__

We will
count the transfer as completed when the last data bit arrives at its
destination.

1*.*5 MB =
12*,*582*,*912 bits. 2 initial RTT’s *(*160
ms*)* + 12*,*582*,*912/10*,*000*,*000
bps (transmit)+

RTT/2
(propagation) ≈ 1*.*458 seconds.

Number of packets required = 1*.*5 MB/1 KB = 1*,*536. To the
above we add the time for 1,535 RTTs (the number of RTTs between when packet 1
arrives and packet 1,536 arrives), for a total of 1*.*458+122*.*8 = 124*.*258 seconds.

Dividing the 1,536 packets by 20 gives 76.8. This
will take 76.5 RTTs (half an RTT for the first batch to arrive, plus 76 RTTs
between the first batch and the 77th partial batch), plus the initial 2 RTTs,
for 6.28 seconds.

Right after the handshaking is done we send one
packet. One RTT after the handshaking we send two packets. At *n* RTTs past the initial handshaking we
have sent 1+2+4+・ ・ ・+2*n* = 2*n*+1 −1 packets. At *n* = 10 we have thus been able to send all 1,536 packets; the last
batch arrives 0.5 RTT later. Total time is 2 +10*.*5 RTTs, or 1 second.

2.Consider
a point-to-point link 50 km in length. At what bandwidth would propagation
delay (at a speed of 2 × 108 m/sec) equal transmit delay for 100- byte packets?
What about 512-byte packets?

__SOLUTION :__

Propagation
delay is 50 × 103 m/*(*2 × 108 m/sec*)* = 250 *μ*s 800 bits/250 *μ*s is 3.2
Mbits/sec. For 512-byte packets, this rises to 16.4 Mbit/sec.

3.Suppose
a 128-Kbps point-to-point link is set up between Earth and a rover on Mars. The
distance from Earth to Mars (when they are closest together) is approximately
55 Gm, and data travels over the link at the speed of light—3×108 m/sec.

Calculate the minimum RTT for the link.

Calculate the delay × bandwidth product for the
link.

A camera
on the rover takes pictures of its surroundings and sends these to Earth. How
quickly after a picture is taken can it reach Mission Control on Earth? Assume
that each image is 5 MB in size.

__SOLUTION :__

(a)
Propagation delay on the link is *(*55
× 109*)/(*3 × 108*)* = 184 seconds. Thus the RTT is 368 seconds.

The delay × bandwidth product for the link is = 184
× 128 × 103 = 2*.*81 MB.

After a picture is taken it must be transmitted on
the link, and be completely propagated before Mission Control can interpret it.
Transmit delay

for 5 MB
of data is 41,943,040 bits/128×103 = 328 seconds. Thus, the total time required
is transmit delay + propagation delay = 328 + 184 = 512 seconds.

4.Calculate
the latency (from first bit sent to last bit received) for:

· A 1-Gbps
Ethernet with a single store-and-forward switch in the path, and a packet size
of 5,000 bits. Assume that each link introduces a propagation delay of 10 *μ*s and that the switch begins
retransmitting immediately after it has finished receiving the packet

Same as (a) but with three switches.

Same as (b) but assume the switch implements
cut-through switching: it is able to begin retransmitting the packet after the
first 128 bits have been received.

__SOLUTION :__

· For each
link, it takes 1 Gbps/5 kb = 5 *μ*s to
transmit the packet on the link, after which it takes an additional 10 *μ*s for the last bit to propagate across
the link. Thus, for a LAN with only one switch that starts forwarding only
after receiving the whole packet, the total transfer delay is two transmit
delays + two propagation delays = 30 *μ*s.

For three switched and thus four links, the total
delay is four transmit delays + four propagation delays = 60 *μ*s.

For “cut-through,” a switch need only decode the
first 128 bits before beginning to forward.

This
takes 128 ns. This delay replaces the switch transmit delays in the previous
answer for a total delay of one Transmit delay + three cut-through decoding
delays + four propagation delays =45*.*384
*μ*s.

**5. **For the following, as in the
previous problem, assume that no data compression is done.** **Calculate the bandwidth necessary for transmitting in real time:

HDTV high-definition video at a resolution of 1*,*920 × 1*,*080, 24 bits/pixel, 30 frames/sec.

Plain old telephone service (POTS) voice audio of
8-bit samples at 8 KHz.

GSM mobile voice audio of 260-bit samples at 50 Hz.

HDCD high-definition audio of 24-bit samples at
88.2 kHz.

__SOLUTION :__

· 1*,*920×1*,*080×24×30 = 1*,*492*,*992*,*000
≈ 1*.*5 Gbps.

8×8*,*000 =
64 Kbps.

260 ×50 = 13 Kbps.

24×88*,*200 = 216*,*800 ≈ 2*.*1 Mbps.

6. Show
the 4B/5B encoding, and the resulting NRZI signal, for the following bit
sequence:

1101 1110
1010 1101 1011 1110 1110 1111

__SOLUTION :__

The 4B/5B
encoding of the given bit sequence is the following. 11011 11100 10110 11011
10111 11100 11100 11101

**7 **Suppose the following sequence of
bits arrive over a link:** **011010111110101001111111011001111110

Show the
resulting frame after any stuffed bits have been removed. Indicate any errors
that might have been introduced into the frame.

__SOLUTION :__

Let ∧ mark each position where a
stuffed 0 bit was removed. There was one error where the sever consecutive 1s
are detected (** err**). At the end of the bit sequence, the end of frame was
detected (

01101011111∧101001111111*err*0 110 01111110*eof*

*
*Suppose we want to transmit the message 1011 0010
0100 1011 and protect it from errors using the CRC8 polynomial *x*8 +*x*2
+*x*1 + 1.

(a) Use
polynomial long division to determine the message that should be transmitted.

(b)
Suppose the leftmost bit of the message is inverted due to noise on the
transmission link.

What is
the result of the receiver’s CRC calculation? How does the receiver know that
an error has occurred?

__SOLUTION :__

· We take
the message 1011 0010 0100 1011, append 8 zeros and divide by 1 0000 0111 (*x*8 + *x*2
+* x*1 + 1). The remainder is 1001
0011. We transmit the original message with this remainder* *appended, resulting in 1011 0010 0100 0011 1001 0011.

(b)
Inverting the first bit gives 0011 0010 0100 1011 1001 0011. Dividing by 1 0000
0111 (*x*8 +*x*2 +*x*1 + 1) gives a a
remainder of 1011 0110.

Suppose you are designing a sliding window protocol
for a 1-Mbps pointto-point link to the stationary satellite evolving around
Earth at 3 × 104 km altitude. Assuming that each frame carries 1 KB of data,
what is the minimum number of bits you need for the sequence number in the
following cases? Assume

the speed
of light is 3×108 meters per second.

(a)
RWS=1.

(b)
RWS=SWS.

__SOLUTION :__

One-way
latency of the link is 100 msec. (Bandwidth) × (roundtrip delay) is about 125
pps×0*.*2 sec, or 25 packets. SWS
should be this large.

If RWS = 1, the necessary sequence number space is
26. Therefore, 5 bitsare needed.

IfRWS =
SWS, the sequence number space must cover twice the SWS, or up to 50.
Therefore, 6 bits are needed.

10.Given
the extended LAN shown in Figure 3.34, assume that bridge B1 suffers
catastrophic failure. Indicate which ports are not selected by the spanning
tree algorithm after the recovery process and a new tree has been formed.

__SOLUTION :__

The
following list shows the mapping between LANs and their designated bridges. B1
dead[B7]

B2 A,B,D

B3
E,F,G,H

B4 I

B5 idle

B6 J

B7 C

11.Suppose
we have the forwarding tables shown in Table 4.13 for nodes A and F, in a
network where all links have cost 1. Give a diagram of the smallest network
consistent with these tables.

__SOLUTION :__

The
following is an example network topology.

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