Tutorial 04

Tutorial for Chapter 4 CEN 444 – 444 – Computer Computer Networks Dr. Mostafa Dahshan Department of Computer Engineering College of Computer and Information Sciences King Saud University [email protected]

Last update: 30/12/2015

Tutorial for Chapter 4

Problem 1

Problem 1 Frames arrive randomly randomly at a 100-Mbps channel for transmission. transmission. If the channel is busy when a frame frame arrives, it waits its its turn in a queue. queue. Frame length is exponentially distributed with a mean of 10,000 10,000 bits/frame. For each of the following frame 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.

M. Dahshan

CEN444 – CEN444 – Computer Computer Networks

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Problem 1

Problem 1 Frames arrive randomly randomly at a 100-Mbps channel for transmission. transmission. If the channel is busy when a frame frame arrives, it waits its its turn in a queue. queue. Frame length is exponentially distributed with a mean of 10,000 10,000 bits/frame. For each of the following frame 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.

M. Dahshan


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Problem 1

Solution The delay is calculated using the formula:  = 1/(  ) . Data rate  = 100 100 × 10 = 10  bits/sec. 







Mean frame length = 10000 bits/frame. Thus  =

= 10− frames/bit.

Mean service rate  = 10− frames/ frames/bit bit × 10 bits/sec = 10000 frames/sec (a) Arrival rate  = 90 frames/sec.  =

 −

(b) Arrival rate  = 900 frames/sec.  =

 −

(c) Arrival rate  = 9000 frames/sec.  =

M. Dahshan

=

 −

=

 −

= 0.0001009 ≈ 0.1 msec

 −

=

= 0.000109 ≈ 0.11 msec

 −


= 0.001 = 1 msec

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Problem 2

Problem 2 A group of  stations share a 56-kbps pure ALOHA ALOHA channel. Each station outputs 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 buffer outgoing frames). frames). What is the maximum maximum value of ?

M. Dahshan


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Problem 2

Solution In pure ALOHA, the maximum throughput is 0.184. Maximum usable bandwidth = 0.184 × 56000 bits/sec = 10300 bits/sec. Each station requires So,  =

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 

 bits  sec

= 10 bits/sec.

= 1030 stations.

CEN444 – Computer Networks

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Problem 3

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?

M. Dahshan


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Problem 3

Solution (a)  = number of transmission + retransmissions per frame (slot) time Slot time = 40 msec = 0.04 sec/slot. Number of slots per sec = Frames transmissions = 50 frames/sec.  =

 frames/sec  slots/sec

 .

= 25 slots/sec.

= 2 frames/slot.

Chance of success on first attempt  =  − =  − = 0.135. (b) The probability of a transmission requiring   1 collisions followed by success is:

 =  − (1   − ) − In this case,   1 = 5

 =  − (1   − ) = 0.0654 (c) The expected number of attempts =   =   = 7.389

M. Dahshan


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Problem 4

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)?

M. Dahshan


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Problem 4

Solution (a) Signal propagation speed = 0.82 × 3 × 10 = 2.46 × 10 m/sec. Propagation time =  =

Distance Speed

=

 m .× m/sec

= 8.13 × 10− sec.

Length of contention slot = 2 = 2 × 8.13 × 10− = 16.26 × 10− sec.

(b) Signal propagation speed = 0.65 × 3 × 10 = 1.95 × 10 m/sec. Propagation time =  =

Distance Speed

=

 m .× m/sec

= 205.128 × 10− sec.

Length of contention slot = 2 = 2 × 205.128 × 10− = 410.256 × 10− sec.

M. Dahshan


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Problem 5

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 (

) throughput achievable using the above

protocol. b. Suppose that only one of the hosts on the token ring has a backlog (

), and

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

M. Dahshan


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Problem 5

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 = 10 6 bits. The aggregate throughput = total number of bits / total transmission time = 106 bits / 10.5 ms = 95.23 Mbps.

M. Dahshan


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Problem 5

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 = 10 5 bits.

The aggregate throughput = total number of bits / total transmission time = 105 bits / 1.5 ms = 66.67Mbps.

M. Dahshan


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Problem 6

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.

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Problem 6

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 B x x x x x x x

C D E x x x x x x 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.

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Problem 6

(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 upcoming data frame.

3. B sends data frame to A. All stations are silent.

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Problem 6

(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 upcoming data frame. E doesn’t hear the CTS.

3. B sends data frame to C. A and D are silent. E can transmit.

M. Dahshan


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Problem 7

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.

M. Dahshan


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Problem 7

Solution

A

B

C

D

E

F

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.

M. Dahshan


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Problem 8

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 b its 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?

M. Dahshan


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Problem 8

Solution The effective data rate is calculated as: The one-way propagation delay  =

data length without headers total time required to complete transmission

cable length signal speed

=

 m  m/sec

= 5 × 10− sec

The round-trip propagation time of the cable = 2 A complete transmission has six phases: 1. Transmitter seizes cable: Required time is 2 = 10 × 10− sec 2. Transmit data: Required time is

M. Dahshan

frame length data rate

=

 ×

= 25.6 × 10− sec


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Problem 8

3. Delay for last bit to get to the end: Required time is  = 5 × 10− sec 4. Receiver seizes cable: Required time is 2 = 10 × 10− μsec 5. Acknowledgement sent: Required time is

ack length data rate

=

 ×

= 3.2 × 10− sec

6. Delay for last bit to get to the end: Required time is  = 5 × 10− sec Total time required = (10 + 25.6 + 5 + 10 + 3.2 + 5) × 10− = 58.8 × 10− sec Effective data rate =

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− .×

= 3862068.96 ≈ 3.8 × 10 bps


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Problem 9

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?

M. Dahshan


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Problem 9

Solution The worst case is an endless stream of 64-byte (512bit) frames. The backplane can handle 10 bps The number of frames it can handle is:

10 512

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= 1953125 frames/sec


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Problem 10

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?

M. Dahshan


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Problem 10

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.

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Problem 11

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− . How many frames per second will be damaged on average?

M. Dahshan


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Problem 11

Solution Let error probability per bit  = 10− Probability of an -bit frame arriving correctly is  = (1  ) Thus, probability of a frame arriving correctly is (1  10− )× = 0.9999488 Probability of frame being damaged is (1  ) = (1  0.9999488) = 5.12 × 10− Data rate of 802.11b is 11 × 10 bits/sec Number of frames transmitted per second is

× bits/sec × bits/frame

= 21484.375 frames/sec

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

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

M. Dahshan


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Problem 12

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.

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Problem 12

a. Draw a timeline diagram describing this communication. b. Calculate the required time for this communication. c. Calculate the MAC layer throughput.

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Problem 12

Solution a.

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Problem 12

b. Total frame size = data size+ header size = 1452+28 = 1480 bytes Data frame transmission time =

× bits

ACK frame transmission time = RTS frame transmission time = CTS frame transmission time =

 Mbps

= 219.25 μs

× bits  Mbps ×  × 

= 2.07 μs

= 2.96 μs = 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

M. Dahshan


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Problem 12

c. We are using 308.35 μs to transmit 1452 bytes MAC layer throughput is:

1452 × 8 bits 308.35 μs

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= 37.67 Mbps


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Problem 13

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.

M. Dahshan


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Problem 13

Solution a.

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Problem 13

b. Total frame size = data size+ header size = 1452+28 = 1480 bytes Data frame transmission time =

× bits

ACK frame transmission time =

 Mbps

= 219.25 μs

× bits  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

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Problem 13

c. We are using 308.35 μs to transmit 1452 bytes MAC layer throughput is:

1452 × 8 bits 271.32 μs

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= 42.81 Mbps


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Problem 14

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. At T = 3, S3 has a Data frame to transmit to S2. At T = 14, S4 has a Data frame to transmit to S2. Assume backoff values, of S1 = 6, S3 = 3 and S4 = 1. Draw a timeline diagram describing this communication. RTS/CTS are not used.

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Problem 14

Solution

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Problem 15

Problem 15 Store-and-forward switches have an advantage over cut-through switches with respect to damaged frames. Explain what it is.

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Problem 15

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.

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Problem 16

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.

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Problem 16

Solution Initially, the hash tables are empty. Switches don’t know which address is associated with which port. B1

B2

Address

Port

Address

Port

-

-

-

-

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Problem 16

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

To

A

C

Port

B1 1

B2

2

3

4

1

2

3

X

X

X

X

X

X

4

Updated hash tables: B1

B2

Address

Port

Address

Port

A

1

A

4

M. Dahshan


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Problem 16

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

To

E

F

Port

B1

B2

1

2

3

X

X

X

4

1

2

X

3

4

X

X

Hash tables are updated as follows: B1

B2

Address

Port

Address

Port

A

1

A

4

E

4

E

2

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Problem 16

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

To

F

E

Port

B1 1

2

B2 3

4

1

2

3

4

Hash tables are updated as follows: B1

B2

Address

Port

Address

Port

A

1

A

4

E

4

E

2

F

2

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Problem 16

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

To

G

E

Port

B1 1

2

B2 3

4

1

2

3

4

X

Hash tables are updated as follows: B1 Address A E

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B2 Port 1 4

Address A E F G


Port 4 2 2 3

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Problem 16

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

To

D

A

Port

B1 1

2

B2 3

4

1

2

3

4

X

X

Hash tables are updated as follows: B1 Address A E D

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B2 Port 1 4 4

Address A E F G D


Port 4 2 2 3 1

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Tutorial 04

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