Data Transmission Delay: Propagation, Emission, Packetization, and CRC - Prof. Malathi Vee, Assignments of Computer Systems Networking and Telecommunications

Solutions to various problems related to data transmission delays, including propagation and emission delays, packetization delays, and crc codes. It covers topics such as image file size, emission delay, propagation delay, packetization for different codecs, and crc calculations.

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CS/ECE 757 Fall 2007
Homework 1 Solution
Instructions:
Be sure to write your name on your submission.
Show all your steps and state your assumptions.
Complete the homework individually.
Pledge: On my honor as a student I have neither given nor received aid on this exam/assignment.
(Write out and sign).
The only help that you can get is from your class notes and the reference textbooks.
Be neat. Unreadable solutions will not be graded.
Number of problems: 7
Total points: 30 points (will be scaled down for the final grade)
Due Date: Oct. 2, 2007 (submit in class, at the start of class)
Problem 1. [2 points]
Assume that the SNR of a received signal is 20dB, and that the bandwidth of the communication chan-
nel is 2.4kHz. What is Shannon’s channel capacity? [Do not forget the unit.]
Answer:
SNR is 20dB. Since , is 100, since . Given that W=2.4kHz,
we compute Shannon’s channel capacity as
Problem 2. (4 points):
A scanner has a resolution of pixels per square inch and creates 8 bits of data per pixel. This
scanner is used to scan an image into a data file. How long does it take to transfer this data file
across a 10km link? Assume that the transmission rate of the link is 100Mbps, and that the dielectric con-
stant of the medium is 1.7. Your answer should consider delays due to emission and propagation. Assume
no errors occur during the transfer, and there are no flow-control effects.
Answer:
Since the question asks for the time to send the data file across the link, we need to include both prop-
agation delay and emission (transmission) delay. The emission delay (time to emit the data file on to the
channel) is pipelined with the time to propagate all the bits except for the last bit. Hence we need to add
SNR dB() 10 SN()
10
log=SN() 100
10
log 2=
CH 1S
N
----
+
⎝⎠
⎛⎞
2
log 2400 1 100+()
2
log 2400 101()
10
log
2
10
log
-------------------------- 15980bits/sec== = =
600 600×
810×
pf3
pf4
pf5

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Download Data Transmission Delay: Propagation, Emission, Packetization, and CRC - Prof. Malathi Vee and more Assignments Computer Systems Networking and Telecommunications in PDF only on Docsity!

CS/ECE 757 Fall 2007

Homework 1 Solution

Instructions:

  • Be sure to write your name on your submission.
  • Show all your steps and state your assumptions.
  • Complete the homework individually.
  • Pledge: On my honor as a student I have neither given nor received aid on this exam/assignment.

(Write out and sign).

  • The only help that you can get is from your class notes and the reference textbooks.
  • Be neat. Unreadable solutions will not be graded.
  • Number of problems: 7
  • Total points: 30 points (will be scaled down for the final grade)

Due Date: Oct. 2, 2007 (submit in class, at the start of class)

Problem 1. [2 points]

Assume that the SNR of a received signal is 20dB, and that the bandwidth of the communication chan-

nel is 2.4kHz. What is Shannon’s channel capacity? [Do not forget the unit.]

Answer:

SNR is 20dB. Since , is 100, since. Given that W=2.4kHz,

we compute Shannon’s channel capacity as

Problem 2. (4 points):

A scanner has a resolution of pixels per square inch and creates 8 bits of data per pixel. This

scanner is used to scan an image into a data file. How long does it take to transfer this data file

across a 10km link? Assume that the transmission rate of the link is 100Mbps, and that the dielectric con-

stant of the medium is 1.7. Your answer should consider delays due to emission and propagation. Assume

no errors occur during the transfer, and there are no flow-control effects.

Answer:

Since the question asks for the time to send the data file across the link, we need to include both prop-

agation delay and emission (transmission) delay. The emission delay (time to emit the data file on to the

channel) is pipelined with the time to propagate all the bits except for the last bit. Hence we need to add

SNR dB ( ) = 10 log 10 ( SN ) ( SN ) log 10100 = 2

C H 1

S

N

⎝⎛ +^ ----⎠⎞

2

log 2400 log 2 ( 1 + 100 ) 2400

log 10 ( 101 ) log 102

= = = -------------------------- =15980bits/sec

600 × 600

8 ″ × 10 ″

the propagation delay for the last bit to the emission delay to get the total delay to send the file over the

link.

To compute propagation delay, compute the speed of light in the medium using , where the

dielectric constant is given to be 1.7.

The propagation delay is given by:

To compute the emission (transmission) delay, we need to first compute the size of the image data file:

Therefore the emission (transmission) delay is

The total delay is transmission delay + propagation delay. Since the propagation delay is much smaller,

the total delay is dominated by the transmission delay.

Problem 3. (6 points):

Compute the packetization delays incurred in creating a 40-byte voice packet using a 8.5kbps Truespeech

codec versus a 160-byte packet using a PCM codec. [Hint: see the Digital representation of information

slides to find the rate of a PCM codec].

a. Which packetization delay is lower?

b. Which of these two packets will experience a lower emission delay if the link transmission

rate is 64kbps?

c. Provide a formula describing the relationship between the packet sizes of the two types of

packets if we want to make the total delay (packetization plus emission) the same in both

cases. (Denote the two packet sizes, and , and express one in terms of the other.)

Answers:

a. Convert bytes to bits by multiplying by 8. It takes ms to fill the

40-byte voice packet with the Truespeech codec. The PCM codec rate is 64kbps. It takes

ms. So the packetization delay is lower for the PCM case.

b. The emission delay for the Truespeech packet will be ms. The emission delay for

the PCM packet will be ms. So the emission delay is lower for the Truespeech

packet.

v c ε

=^ ------

v c ε

8 ×

= = -----------------^ =2.3 × 10 8 ms

propagation delay distance speed of light in medium

------------------------------------------------------------ 10 km^ ×^1000 (^ m^ ⁄ km ) 2.3 10 8 × m/s

= = ---------------------------------------------------- =43.5μ s

image size ( 600 × 600 ) pixels sq. inch

------------------ (^) ( 8 × 10 )sq. inch × 8 bits pixel = × ------------ =2.304 × 10 8 bits

emission delay image size bit rate

------------------------- 2.304^10

8 × bits 100 × 10 6 b/s

= = -------------------------------------^ =2.304sec

x y

( ( 40 × 8 )bits) ⁄ ( 8.5kbps)=37.

( ( 160 × 8 )bits) ⁄ (64kbps ) = 20

40 ×8bits 64kbps

160 ×8bits 64kbps

The received bit string, 110110111, can be expressed as.

c. Answer: 111; The degree of the generator polynomial is 3, which means there are three check

bits. The last three bits of the bit string 110110111 are the check bits.

Problem 5. (6 points):

Consider a data link layer protocol that uses a sliding window Go-back-N ARQ scheme and sliding-win-

dow flow-control scheme. Assume that the sending window size, , is 6 frames, and that the receiver

buffer can hold a maximum of 7 frames. Assume that the number carried in the ACK indicates the

sequence number of the next expected frame. Frame numbering starts at 0.

Make the following assumptions:

  • Each user data frame is 1500 bytes long.
  • The link rate is 1.5Mbps.
  • The retransmission timer value is 32 ms. Assume that the retransmission timer starts as soon as

the last bit of a frame is emitted. If an ACK is not received within 32ms, the sender will initiate

retransmission of the frame as soon as possible after the retransmission timer expires.

  • The round-trip propagation delay is 20 ms.
  • ACK generation delays (processing delays) and ACK emission delays are negligible.
  • The maximum receiver buffer size has been communicated to the sender at the start of the session.

The figure shows that user data frames are sent from A to B, and ACKs with flow-window values are sent

from B to A. See the figure and answer the following questions:

a. What is the time value when Frame 0 is fully emitted on to the link?

b. What is the time value in the figure?

c. What is the time value in the figure?

d. Is the sender allowed to send any new frames after it completes retransmission of Frame 6? If

so, how many frames can it send without waiting for an ACK?

b x ( ) x 8 x 7 x 5 x 4 x 2 = + + + + + x + 1

Remainder

x 8 x 7 x 5 x 4 x 2 x + + + + + x + 1 (^3) + x (^2) + 1

x 4

x^8 + x^7 + x^5

x 5

  • x + 1

x^4 + x^3 + x

x 3

  • x^2 + 1 x^3 + x^2 + 1 0

W (^) s

t 1

t 2

t 3

5. During the time-out interval for frame 6, if new frames were delivered to the data-link layer

by the higher layer, is the DLL sender allowed to send any frames? If so, how many can it

send, and at what point in time can it start sending frames?

Answers:

a. Each frame takes 8ms to emit ( ). Therefore.

b. Time is the round-trip propagation delay, since ACK processing delays and emission delays

are negligible. It is 20ms, as specified in the problem.

c. Time is the retransmission time-out delay. As specified in the problem, this number is

32ms.

d. Since the flow-window is 7, , and there can be a maximum of ( , flow window), i.e.,

6 frames outstanding at any instant in time. There is already one outstanding frame, i.e., the

retransmitted frame 6. Therefore, after this retransmission, the sender can send 5 more frames

without awaiting an ACK.

e. During the time-out, the sender cannot send any new frames until it receives the ACK 6, flow-

window = 7 control signal because until that point the last report from the receiver said that

flow-window was only 1 and Frame 6 was still outstanding (before this point, flow window is

a constraint). However, when this ACK 6, flow-window = 7 control signal arrives, the sender

can immediately start sending new frames, such as 7, 8, 9, 10, 11, if the higher-layer had

passed down data for transmission to the DLL. In other words, it could have sent 5 more

frames since and frame 6 is still outstanding.

Entity A Entity B

Frame 0

Frame 1

Frame 2

Frame 3

Frame 4

Frame 5

Frame 6

ACK 2, flow window = 5

ACK 4, flow window = 3

Time-out for frame 6

ACK 7, flow window = 6

Frame 6

t =

t =t (^1)

Start emission at Complete emission of frame 0 at

t =t 1 +t 2

ACK 1, flow window = 6

ACK 3, flow window = 4

ACK 5, flow window = 2

ACK 6, flow window = 1

start emitting Frame 6 (retransmission)

( t 3 ) ACK 6, flow window = 7

Receiver’s buffer

has been depleted

by the higher-layer

allowing it to

notify the sender

that it has more

space available now.

Dashed arrows carry ACK and flow-window values

Complete emission of frame 6

1500 B

1.5 Mbps

---------------------- (^) = 8 ms t 1 = 8 ms

t 2

t 3

W (^) s = 6 W (^) s

W (^) s = 6

10 snvang-losang.abilene.ucaid.edu (198.32.8.95) 78.058 ms 78.148 ms 84.177 ms

11 110.pos-1-0.core0.eug.oregon-gigapop.net (198.32.163.17) 90.303 ms 91.659 ms 90.299 ms

12 vl-110.uonet8-gw.eug.oregon-gigapop.net (198.32.163.131) 90.547 ms 90.784 ms 90.558 ms

13 ge-5-2.uonet2-gw.uoregon.edu (128.223.3.2) 149.257 ms 244.681 ms 90.675 ms

14 www.uoregon.edu (128.223.142.89) 90.670 ms 90.537 ms 90.550 ms

Abilene backbone routers on the path include ATLA (Atlanta), HSTN (Houston), LOSA (Los Angeles),

SNVA (Sunnyvale). It is unclear whether the Washington IP router of Abilene is on the path or not.