About the exercises of Data Communications/Networking 4e

exercises of data and signals

Hi, I am a recruit and have some questions while resolving the exercises of Mr. Forouzan's "data communications and networking 4th international edition".

The exercises that I confused are listed below for help.

Added after 32 minutes:

-------- Chapter 3 Data and Signals -------

40. A signal with 200 milliwatts power passes through 10 devices, each with an average noise of 2 microwatts. What is the SNR? What is the SNRdb?

My solution:

a)SNR = the SNR of each devices = 200 milliwatts / 2 microwatts = 200000 microwatts / 2 microwatts = 100000

b)SNRdb = the sum of the SNR of all the 10 devices = 10 * log10(SNR) = 10 log10(200000) = 10 (log10(2) + 5) = 10*5.301 = 53.01 db

Is this correct?

Added after 56 minutes:

48. What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 2µs and a processing time of 1µs. The length of the link is 2000Km. The speed of light inside the link is 2X100,000,000 m/s. The link has a bandwidth of 5Mbps. Which component of the total delay in dominant? Which one is negligible?

a) Queuing time + Processing time = 10*(2µ+1µ)X5000000 = 30s
Should I multiply a frame ( 10(2µ+1µ)*1 ) instead of ( 10(2µ+1µ)*5000000 )?

b) Propagation time = 2000X1000 / 2X100,000,000 = 0.01s

c) Transmission time = 5000,000/5000,000= 1s

transmit 100 digitized voice channels

-------- Chapter 5 Analog Transmission --------

About the students' odd-numbered solutions

17. What is the required bandwidth for the following cases if we need to send 4000 bps?
a. ASK
b. FSK with 2Δf= 4KHz
c. QPSK
d. 16-QAM

I think one of the solutions released from official web site has 3 errors.

a. r=1 , B = (1+1) X (1/1) X 4Kbps = 8Kbps
b. r=1 , B = (1+1) X (1/1) X 4Kbps + 2Δf = 8Kbps + 4Kbps = 12Kbps other than 8Kbps
c. r=2 , B= (1+1) X (1/2) X 4Kbps = 4Kbps other than 2Kbps
d. r=4 , B= (1+1) X (1/4) X 4Kbps = 2Kbps other than 1Kbps

Is the official solution really incorrect ?

data communication exercises

-------- Chapter 6 Multiplexing and Spreading --------

14. We need to transmit 100 digitized voice channels using a pass-band channel of 20KHz. What should be the ratio of bits/Hz if we use no guard band?

I guess that I couldn't resolve this exercise because of the absence of some conditions which should be given.

Could anyone help me, please?

data communication + example exercises

-------- Chapter 7 Transmission Media --------
8. Name the advantages of optical fiber over twisted-pair and coaxial cable.

It's so easy that I should not post it there to make you laugh.

The real question is how to remember the advantages and disadvantages of fiber-optic cable if you have a exam and unfortunately you are requested to define these with no permission of opening textbook.

My idea listed below:

Advantages : BEST CoW
Bandwidth: higher
Electromagnetic interference : immunity
Signal attenuation : less
Tapping : great immunity
Corrosive materials : resistance
Weight : light

Disadvantages : Im Uc (I am ..... you see?)
Installation & Maintenance
Unidirectional light prpagation
Cost

The method listed above seems not funny enough.......

Does anybody have more funny methods helpful for remembering these ?
Thanks anyway.

solutions of exercises of data and signals

-------- Chapter 10 Error Detection and Correction --------

(Review Questions) 10. Can the value of a checksum be all 0s (in binary)? Can the value be all 1s (in binary)? Define your answer.

My answers are both "yes" because I use the figure 10.24 in section 10.5 and set sender's checksum all 0s or 1s, then derive the "possible" wrapped sum and sum of datacodes from this checksum. It seems that the wrapped sum and the sum are valid.

Besides, the solution (requested checksum) of exercise 33 (this problem shows a special case in checksum handling. A sender has two datacodes to send: 0x4567 and 0xBA98. What's the value of the checksum? ) is all 0s .

Is something wrong about my solution? :roll:

PS:
In page 295 of the international 4th edition, the blue block "If a generator cannot divide x^t + 1 (t between 0 and n-1), then all isolated double errors can be detected." has a little mistake. t's range is between 2 and n-1 instead of 0 and n-1.

I really enjoy reading Mr. Forouzan's "Data communications and networking".

csma/cd, forouzan fourth edition

-------- Chapter 12 Multiple Access -------

Before resolving exercises in this chapter, I have found a strange figure in text.

Figure 12.17 at page 379 shows flow diagram for CSMA/CA.

This figure seems mixed with something about CSMA/CD in figure 12.14 at page 376. According to the definition of contention window at page 378, the time slot is selected binary-exponentially, not at random (between 0 and 2^k-1) as showed in figure 12.17.

Do I mistake anything?

forouzan exercises

-------- Chapter 12 Multiple Access -------

After reading the text about contention window at page 378 carefully and repeatedly, I'm sure that I have mistaked something.

"The number of slots in the window changes according to the binary exponential back-off strategy. This means that it's set to one slot the 1st time and then doubles each time the station cannot detect an idle channel after the IFS time."

According the text listed above, I assumed a station which suffered 3 times back-off, then listed all of the time slots it has occupied below:

At 1st back-off, the station taked 1 time slot of contetion window.
At 2nd back-off, the station taked 2 time slots of contention window.
At 3rd back-off, the station taked 4 time slots of contention window.

But after reading the text flowing the previous one, I found it says: "This is very similar to (the original CSMA's) p-persistent method except that a random outcome defines the number of slots taken by the waiting station."

So in figure 12.17 the block of "choice a random number R between 0 and (2^k)-1" is undoubted exact.

Once again, the example I prefered must be changed below:

At 1st back-off, the station taked a random number of time slots of contetion window with size 0, which is derived from the expression (2^k)-1 where K=0. In other words, the station's 1st sending didn't wait. As the flow in the figure 12.17 showed, it didn't receive ACK and increased k by 1, where k = 1 and less than 15.

At 2nd back-off, the station taked a random number of time slots of contetion window with size 1, which is derived from the express (2^k)-1 where K=1. It's obviously the number of time slots is neither 0 nor 1. No more one another to "be randomly choose". The station still failed to sending data and k was increased to 2.

At 3rd back-off, the station taked a random number of time slots of contetion window with size 3, which is derived from the express (2^k)-1 where K=2.

The station would have 0 time slot to wait and send immediately.
Or It would try to send after waiting for 1 time slot.
Maybe it tried to send after waiting for 2 or 3 time slots.

The number of time slots to wait before try to sending data is between 0 and 3 where the size of contention window is 3. I think that the binary exponential back-off strategy means the growth of the size of contention window, not the "random" waiting time.

I should read this good book more carefully.

data communication exercise

-------- Chapter 20 Network Layer: Internet Protocol -------

There may be a error in solution 23.

Flags and fragmentation = 0x5850 = 010.1100001010000 not 0x0000
So D=1, which means "don't fragment", and M=0, which means "no more fragments".
The datagram is not fragmented.

Checksum = 0x0000 not 0x5850, which means the packet is not corrupted.

The solution interchanges the values of checksum and flags/fragmentation.

data communication exercise

-------- Chapter 21 Network Layer: Address Mapping, Error Reporting, and Multicasting -------

Review question 12.
Why is there no need for the ICMPv4 message to travel outside its own network?

I wonder why not a ICMPv4 travels outside its own network.
An ICMPv4 packet is encapsulated in an IP datagram. Why can't it travel through the gateway to a destination host in another network?

exercise about data communication

-------- Chapter 23 Process-To-Process Delivery: UDP, TCP, and SCTP -------

Exercise 16.
A client has a packet of 68,000 bytes. Show how this packet can be transferred by using only one UDP user datagram.

The size of a UDP datagram is 0 to 65,535 bytes.
According to the text at page 711, there is a descritption: "Howerver, the total length needs to be much less because a UDP user datagram is stored in an IP datagram with a total length of 65,535."

How can we tranffer this big packet by using only one UDP user datagram?