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CEN444-Computer Networks

Tutorial for Chapter 4

Pure and Slotted ALOHA problems and Solutions

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Tutorial for Chapter 4 Problem 1

CEN444 – Computer Networks of 48

Problem 1 Frames arrive randomly at a 100-Mbps channel for transmission. If the channel is

busy when a frame arrives, it waits its turn in a queue. Frame length is exponentially

distributed with a mean of 10,000 bits/frame. For each of the following frame arrival

rates, give the delay experienced by the average frame, including both queueing time

and transmission time.

(a) 90 frames/sec.

(b) 900 frames/sec.

(c) 9000 frames/sec.



Solution The delay is calculated using the formula: 𝑇 = 1/(𝜇𝐶 − 𝜆).

Data rate 𝐶 = 100 × 106 = 108 bits/sec.

Mean frame length 1

𝜇= 10000 bits/frame. Thus 𝜇 =

1

10000= 10−4 frames/bit.

Mean service rate 𝜇𝐶 = 10−4 frames/bit × 108 bits/sec = 10000 frames/sec

(a) Arrival rate 𝜆 = 90 frames/sec. 𝑇 =1

𝜇𝐶−𝜆=

1

10000−90= 0.0001009 ≈ 0.1 msec

(b) Arrival rate 𝜆 = 900 frames/sec. 𝑇 =1

𝜇𝐶−𝜆=

1

10000−900= 0.000109 ≈ 0.11 msec

(c) Arrival rate 𝜆 = 9000 frames/sec. 𝑇 =1

𝜇𝐶−𝜆=

1

10000−9000= 0.001 = 1 msec



Problem 2 A group of 𝑁 stations share a 56-kbps pure ALOHA channel. Each station outputs a

1000-bit frame on average once every 100 sec, even if the previous one has not yet

been sent (e.g., the stations can buffer outgoing frames). What is the maximum value

of 𝑁?



Solution In pure ALOHA, the maximum throughput is 0.184.

Maximum usable bandwidth = 0.184 × 56000 bits/sec = 10300 bits/sec.

Each station requires 1000 bits

100 sec= 10 bits/sec.

So, 𝑁 =10300

10= 1030 stations.



Problem 3 A large population of ALOHA users manages to generate 50 requests/sec, including

both originals and retransmissions. Time is slotted in units of 40 msec.

(a) What is the chance of success on the first attempt?

(b) What is the probability of exactly 5 collisions and then a success?

(c) What is the expected number of transmission attempts needed?



Solution (a) 𝐺 = number of transmission + retransmissions per frame (slot) time

Slot time = 40 msec = 0.04 sec/slot. Number of slots per sec =1

0.04= 25 slots/sec.

Frames transmissions = 50 frames/sec. 𝐺 =50 frames/sec

25 slots/sec= 2 frames/slot.

Chance of success on first attempt 𝑃0 = 𝑒−𝐺 = 𝑒−2 = 0.135.

(b) The probability of a transmission requiring 𝑘 − 1 collisions followed by success is:

𝑃𝑘 = 𝑒−𝐺(1 − 𝑒−𝐺)𝑘 −1

In this case, 𝑘 − 1 = 5

𝑃6 = 𝑒−2(1 − 𝑒−2)5 = 0.0654

(c) The expected number of attempts = 𝑒𝐺 = 𝑒2 = 7.389



Problem 4 What is the length of a contention slot in CSMA/CD for:

(a) a 2-km cable (signal propagation speed is 82% of the speed in vacuum)?

(b) a 40-km multimode fiber optic cable (speed is 65% speed in vacuum)?



Solution (a) Signal propagation speed = 0.82 × 3 × 108 = 2.46 × 108m/sec.

Propagation time = 𝜏 =Distance

Speed=

2000 m

2.46×108 m/sec= 8.13 × 10−6 sec.

Length of contention slot = 2𝜏 = 2 × 8.13 × 10−6 = 16.26 × 10−6 sec.

(b) Signal propagation speed = 0.65 × 3 × 108 = 1.95 × 108m/sec.

Propagation time = 𝜏 =Distance

Speed=

40000 m

1.95×108 m/sec= 205.128 × 10−6 sec.

Length of contention slot = 2𝜏 = 2 × 205.128 × 10−6 = 410.256 × 10−6 sec.



Problem 5 Consider the following the token passing protocol: When a host receives the token, the

host may transmit for at most 1 ms duration at the rate of 100 Mbps, and then pass the

token to the next host. Suppose that the time required to pass the token between

adjacent hosts is 0.05 ms. Assume that the token ring consists of 10 hosts.

a. Estimate the maximum aggregate (إجمالي) throughput achievable using the above

protocol.

b. Suppose that only one of the hosts on the token ring has a backlog (عمل متراكم), and

no other host has any data to transmit. Determine the maximum throughput

achieved by the backlogged host.



Solution a. Time for token to circulate around the ring is:

Token hold time × no. of stations with data + Station transmit time × no. of stations

= 1 ms × 10 + 0.05 ms × 10 = 10.5 ms

To maximize the aggregate throughput, each host should maximally utilize 1 ms for

transmission.

The aggregate number of bits that can be transmitted during this 10.5 ms period is:

1 ms × 100Mbps × 10 = 106bits.

The aggregate throughput = total number of bits / total transmission time

= 106 bits / 10.5 ms = 95.23 Mbps.



b. Since only one host is backlogged, only one host has data to transmit.

Time for token to circulate around the ring is:

Token hold time × no. of stations with data + Station transmit time × no. of stations

= 1 ms × 1 + 0.05 ms × 10 = 1.5 ms

The aggregate number of bits that can be transmitted during this period is:

1 ms × 100Mbps = 105bits.

The aggregate throughput = total number of bits / total transmission time

= 105 bits / 1.5 ms = 66.67Mbps.



Problem 6 Consider five wireless stations, A, B, C, D, and E. Station A can communicate with all

other stations. B can communicate with A, C and E. C can communicate with A, B and

D. D can communicate with A, C and E. E can communicate A, D and B.

Show the steps of the MACA protocol until the frame is successfully transmitted.

Indicate which nodes must be silent during each step.

(a) When A sends a frame to B.

(b) When B sends a frame to A.

(c) When B sends a frame to C.



Solution (a)

1. A sends RTS to B.

All stations hear the RTS, so they must stay silent until CTS

comes back.

A B C D E

A x x x x

B x x x

C x x x

D x x x

E x x x

2. B sends CTS to A.

C and E will hear the CTS, so they must remain silent during the transmission of the

upcoming data frame.

D doesn’t hear the CTS.

3. A sends data frame to B.

C and E are silent. D can transmit.



(b)

1. B sends RTS to A.

C and E hear the RTS, so they must remain silent until CTS comes back.

2. A sends CTS to B.

All stations hear the CTS, so they must remain silent during the transmission of the


3. B sends data frame to A.

All stations are silent.



(c)

1. B sends RTS to C.

A and E will hear the RTS, so they must remain silent until CTS comes back.

2. C sends CTS to B.

A and D hear the CTS, so they must remain silent during the transmission of the


E doesn’t hear the CTS.

3. B sends data frame to C.

A and D are silent. E can transmit.



Problem 7 Six stations, A through F, communicate using the MACA protocol. Is it possible for

two transmissions to take place simultaneously? Explain your answer.



Solution

If they are in a straight line and that each station can reach only its nearest neighbors.

Then A can send to B while E is sending to F. So, the answer is Yes.

A B E C D F



Problem 8 A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of 200

m/μsec. Repeaters are not allowed in this system. Data frames are 256 bits long,

including 32 bits of header, checksum, and other overhead. The first bit slot after a

successful transmission is reserved for the receiver to capture the channel in order to

send a 32-bit acknowledgement frame. What is the effective data rate, excluding

overhead, assuming that there are no collisions?



Solution

The effective data rate is calculated as: data length without headers

total time required to complete transmission

The one-way propagation delay 𝜏 =cable length

signal speed=

1000 m

200 m/10−6sec= 5 × 10−6sec

The round-trip propagation time of the cable =2𝜏

A complete transmission has six phases:

1. Transmitter seizes cable:

Required time is 2𝜏 = 10 × 10−6 sec

2. Transmit data:

Required time is frame length

data rate=

256

10×106 = 25.6 × 10−6 sec



3. Delay for last bit to get to the end:

Required time is 𝜏 = 5 × 10−6 sec

4. Receiver seizes cable:

Required time is 2𝜏 = 10 × 10−6 μsec

5. Acknowledgement sent:

Required time is ack length

data rate=

32

10×106 = 3.2 × 10−6 sec

6. Delay for last bit to get to the end:

Required time is 𝜏 = 5 × 10−6 sec

Total time required = (10 + 25.6 + 5 + 10 + 3.2 + 5) × 10−6 = 58.8 × 10−6 sec

Effective data rate =256−32

58.8×10−6 = 3862068.96 ≈ 3.8 × 106 bps



Problem 9 A switch designed for use with fast Ethernet has a backplane that can move 1 Gbps.

How many frames/sec can it handle in the worst case?



Solution The worst case is an endless stream of 64-byte (512bit) frames.

The backplane can handle 109 bps

The number of frames it can handle is:

109

512= 1953125 frames/sec



Problem 10 In the following figure, four stations, A, B, C, and D, are shown. Which of the last two

stations do you think is closest to A and why?



Solution Station C heard the RTS and responded to it by asserting its NAV signal.

D did not respond.

Thus, C is the closest to A.

D must be outside A’s range.



Problem 11 Suppose that an 11-Mbps 802.11b LAN is transmitting 64-byte frames back-to-back

over a radio channel with a bit error rate of 10−7. How many frames per second will

be damaged on average?



Solution Let error probability per bit 𝑝 = 10−7

Probability of an 𝑛-bit frame arriving correctly is 𝐴 = (1 − 𝑝)𝑛

Thus, probability of a frame arriving correctly is (1 − 10−7)64×8 = 0.9999488

Probability of frame being damaged is (1 − 𝐴) = (1 − 0.9999488) = 5.12 × 10−5

Data rate of 802.11b is 11 × 106bits/sec

Number of frames transmitted per second is 11×106 bits/sec

64×8 bits/frame= 21484.375 frames/sec

No. of damaged frames/second = No. of frames/sec × probability of frame damaged

= 21484.375 × 5.11986918 × 10−5 = 1.0999 ≈ 1.1 frames/sec



Problem 12 Consider an 802.11 wireless LAN with the following parameters:

Physical layer data rate = 54Mbps MAC layer data payload = 1452 bytes

MAC header = 28 bytes ACK Frame size = 14 bytes

RTS length = 20 bytes CTS length = 14 bytes

DIFS time = 34μs SIFS Time = 16μs

MAC layer throughput is defined as the number of bits sent by the MAC layer in a

given period of time. Assuming that there are two stations exchanging a data frame

using 802.11 DCF and that the two stations are using RTS/CTS transaction.



a. Draw a timeline diagram describing this communication.

b. Calculate the required time for this communication.

c. Calculate the MAC layer throughput.



Solution a.



b.

Total frame size = data size+ header size = 1452+28 = 1480 bytes

Data frame transmission time =1480×8 bits

54 Mbps= 219.25 μs

ACK frame transmission time = 14×8 bits

54 Mbps= 2.07 μs

RTS frame transmission time = 20×8

54= 2.96 μs

CTS frame transmission time = 14×8

54= 2.07 μs

Total time required for transmission of one frame is:

Time of DIFS + RTS + SIFS + CTS +SIFS + Data + SIFS + ACK

34 + 2.96 + 16 + 2.07 + 16 + 219.25 + 16 + 2.07 = 308.35 μs



c.

We are using 308.35 μs to transmit 1452 bytes

MAC layer throughput is:

1452 × 8 bits

308.35 μs= 37.67 Mbps



Problem 13 In the previous problem, suppose that RTS/CTS transaction is not used.

a. Draw a timeline diagram describing this communication.

b. Calculate the required time for this communication.

c. Calculate the MAC layer throughput.



Solution a.



b.

Total frame size = data size+ header size = 1452+28 = 1480 bytes

Data frame transmission time =1480×8 bits

54 Mbps= 219.25 μs

ACK frame transmission time = 14×8 bits

54 Mbps= 2.07 μs

Total time required for transmission of one frame is:

Time of DIFS + Data + SIFS + ACK

34 + 219.25 + 16 + 2.07 = 271.32 μs



c.

We are using 308.35 μs to transmit 1452 bytes

MAC layer throughput is:

1452 × 8 bits

271.32 μs= 42.81 Mbps


M. Dahshan CEN444 – Computer Networks of 48

Problem 14 Consider an 802.11 wireless LAN using DCF operation with the following parameters:

SIFS = 1 slot, DIFS = 3 slots, ACK = 1 slot, CWmin = 8, All Data frames = 4 slots.

There are four stations, S1, S2, S3 and S4.

At T = 0, S2 starts transmitting a Data frame to S3.

At T = 2, S1 has a Data frame to transmit to S4.



Assume backoff values, of S1 = 6, S3 = 3 and S4 = 1.

Draw a timeline diagram describing this communication. RTS/CTS are not used.



Solution



Problem 15 Store-and-forward switches have an advantage over cut-through switches with respect

to damaged frames. Explain what it is.



Solution Store-and-forward switches store entire frames before forwarding them.

After a frame comes in, checksum can be verified.

If the frame is damaged, it is discarded immediately.

With cut-through, damaged frames cannot be discarded by the switch because by the

time the error is detected, the frame is already gone.



Problem 16 Consider the extended LAN connected using bridges B1 and B2 in the figure. Suppose

the hash tables in the two bridges are empty. List all ports on which a packet will be

forwarded and hash table updates for the following sequence of data transmissions:

(a) A sends a packet to C. (b) E sends a packet to F.

(c) F sends a packet to E. (d) G sends a packet to E.

(e) D sends a packet to A. (f) B sends a packet to F.



Solution Initially, the hash tables are empty.

Switches don’t know which address is associated with which port.

B1 B2

Address Port Address Port

- - - -



(a) B1 will forward this packet on ports 2, 3, and 4. B2 will forward it on 1, 2 and 3.

Switch B1 B2

From To Port 1 2 3 4 1 2 3 4

A C X X X X X X

Updated hash tables:

B1 B2


A 1 A 4



(b) B2 will forward this packet on ports 1, 3, and 4. B1 will forward it on 1, 2 and 3.

Switch B1 B2

From To Port 1 2 3 4 1 2 3 4

E F X X X X X X

Hash tables are updated as follows:

B1 B2


A 1 A 4

E 4 E 2



(c) B2 will not forward this packet on any of its ports, and B1 will not see it.

Switch B1 B2

From To Port 1 2 3 4 1 2 3 4

F E


B1 B2


A 1 A 4

E 4 E 2

F 2



(d) B2 will forward this packet on port 2. B1 will not see it.

Switch B1 B2

From To Port 1 2 3 4 1 2 3 4

G E X


B1 B2


A 1 A 4

E 4 E 2

F 2

G 3



(e) B2 will forward this packet on port 4 and B1 will forward it on port 1.

Switch B1 B2

From To Port 1 2 3 4 1 2 3 4

D A X X


B1 B2


A 1 A 4

E 4 E 2

D 4 F 2

G 3

D 1


of 48

(f) B1 will forward this packet on ports 1, 3 and 4. B2 will forward it on port 2.

Switch B1 B2

From To Port 1 2 3 4 1 2 3 4

B F X X X X


B1 B2


A 1 A 4

E 4 E 2

D 4 F 2

B 2 G 3

D 1

B 2

CEN444 –Computer Networks

tutorial for chapter 4 - wordpress.com · 2018. 8. 17. · tutorial for chapter 4 problem 3 cen444...

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