ANSWERS FOR MISCELLANEOUS HOMEWORK-1 374...1 ANSWERS FOR MISCELLANEOUS HOMEWORK-1 1. There are two...

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ANSWERS FOR MISCELLANEOUS HOMEWORK-1 1. There are two networks N and NN interconnected through SS7. First network N has 2 SSP (named as SSP-N1 and SSP-N2), 2 STP (named as STP-

N1 and STP-N2) and 2 SCP (named as SCP-N1 and SCP-N2). Second network NN also has 2 SSP (named as SSP-NN1 and SSP-NN2), 2 STP

(named as STP-NN1 and STP-NN2) and 2 SCP (named as SCP-NN1 and SCP-NN2). Assume STPs are at the same hierarchical levels. a. Draw the overall SS7 network, clearly indicating the types of all the signaling links,

voice trunks and the subscriber lines between the relevant network elements. In your drawing show one telephone subscriber as connected to each SSP.

b. The subscriber connected to SSP-N1 wants to reach a service provided by lines starting with 800 (e.g. car rental) whose software is in SCP-NN1 and SCP-NN2 so the subscriber dials the car rental company's advertised 800 number.

Write the steps that occur until the SSP-N1 picks a trunk to the relevant destination and generates an initial address message (IAM) and proceeds. - The subscriber connected to SSP-N1 dials the company's advertised 800

number - When the subscriber has finished dialing, switch SSP-N1 recognizes that this is

an 800 call and that it requires assistance to handle it properly.

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- Switch SSP-N1 formulates an 800 query message including the calling and called number and forwards it to STP-N1 (or STP-N2) over the A link between SSP-N1 and STP-N1 (or between SSP-N1 and STP-N2).

- STP-N1 determines that the received query is an 800 query and selects the

800 database in SCP-NN1 which is connected to STP-NN1.

- STP-N1 forwards the query to STP-NN1 over the B link between STP-N1 and STP-NN1.

- STP-NN1 forwards the query to SCP-NN1 over the A link between STP-NN1

and SCP-NN1.

- SCP-NN1 receives the query, extracts the passed information and (based on its stored records) selects either a real telephone number or a network (or both) to which the call should be routed

- SCP-NN1 formulates a response message with the information necessary to

properly process the call, addresses it to switch SSP-N1, picks an STP, e.g. STP-NN2 and routes the response to STP-NN2 through the A link between SCP-NN1 and STP-NN2.

- STP-NN2 receives the response message, recognizes that it is addressed to

switch SSP-N1 which is connected to STP-N2 and routes the response message to STP-N2 over the B link between STP-NN2 and STP-N2.

- STP-N2 receives the response message, recognizes that it is addressed to

switch SSP-N1 and routes the response message to SSP-N1 over the A link between STP-N2 and SSP-N1.

- Switch SSP-N1 receives the response and uses the information to determine

where the call should be routed

- Switch SSP-N1 then picks a trunk to that destination, generates an initial address message (IAM) and proceeds to set up the call.

2. a. The bit sequence in one T1 frame is given as:

100011001101000111010111000000111000100001111101001110000110001001000000001111010111100001111000101000001000100000100100001010001001000100101000111000011110010111100110101110000110011001010001

Write the signalling sequence in this frame. This is Channel Associated Signaling or Call Associated Signaling (CAS). Taking each 8th bit in each channel of total 24 channels, signalling sequence is 011101000100000010110001

b. What are the rates of voice and signalling in part a?

7 bits for voice and 1bit for signalling in one 64 Kbps channel. Thus each channel has 56 Kbps of voice and 8 Kbps of signaling.

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c. The bit sequence in one E1 frame is given as:

1000110011010001110101110000001110001000011111010011100001100010010000000011110101111000011110001010000010001000001001000010100010010001001010001110000111100101111001101011100001100110010100011010001101100110011001100000111110011001111110011001100011110000

Write the signalling sequence in this frame. This is Common Channel Signaling (CCS). Taking the 16th channel bits where it starts with the 0th channel, signalling sequence is 10010001

d. What are the rates of voice and signalling in part c?

Slot 16 is allotted for signaling only, thus signalling and 30 voice channels each has 64 Kbps.

e. Is the voice quality in part a or in part c better? Why? Voice quality in part c is better because voice uses the full 64 Kbps in all of the 30 voice channels. However, in part a, voice uses only 56 Kbps in each of the 24 channels.

3. a. You have options of subscribing to X.25, ATM and Frame Relay services. Which

one do you prefer for the following traffic types? i. Data at 155 Mbps. ATM ii. Data at 512 Kbps. Frame Relay iii. Audio. ATM iv. Data at 19200 bps. X.25 v. Video. ATM.

b. Among PSTN, SDH, DWDM, IP, ATM and Frame Relay systems, which ones set

up the connection before information transfer occurs? PSTN, SDH, DWDM, ATM, Frame Relay

c. Write separately for each of the the below listed network elements whether it is

definitely needed in introducing Advanced Intelligent Network (AIN) services in a PSTN infrastructure. i. IP/ATM Network. No. ii. Local Exchange. Yes. iii. PSTN Subscriber. Yes. iv. SS7. Yes. v. VoIP. No

4. In mapping ATM in STM-1 payload area, 258 columns by 9 rows of bytes in STM-1 payload area is used only for the transport of ATM packets. ATM cells can be split between the two consecutive rows in the STM-1 payload area.

a. Find the maximum integer number of AAL1- ATM cells that an STM-1 frame can carry.

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There are total space for 258 x 9 = 2322 bytes in the payload area. AAL1- ATM cell size is fixed and 53 bytes

Since AAL1- ATM cells can be split between the two consecutive rows in the payload area, the maximum integer number of AAL1- ATM cells that an STM-1 frame can carry = 2322 / 53 = 43 AAL1- ATM cells.

b. Find the number of AAL1- ATM payload bits (carrying information) in one STM-1

frame when AAL1- ATM are mapped as in Part a.

In one AAL1- ATM cell, there are 47 bytes in the AAL1- ATM payload. İ.e. 47 payload bytes. Thus, in one AAL1- ATM cell, there are 47 bytes x 8 bits / byte = 376 bits in the ATM payload. İ.e. 376 payload bits. In part a 43 ATM cells are found in one STM-1 frame. İ.e. there are 376 x 43 = 16,168 bits of AAL1- ATM payload (carrying information) in one STM-1 frame

c. Find the rate of the information found in Part b if one STM-1 frame is transmitted at

every 0.000125 seconds (1/8000th of a second). One STM-1 frame is transmitted at every 0.000125 seconds (1/8000th of a second) In Part b it is found that the information in one STM-1 frame is 16,168 bits Thus rate of the information found in Part b is 16,168 bits in 0.000125 seconds. İ.e 16,168 bits x (1/0.000125 sec) = 129.344 Mbps

d. Find the rate of AAL1- ATM found in Part a if one STM-1 frame is transmitted at

every 0.000125 seconds (1/8000th of a second).

In one STM-1 frame, 43 AAL1- ATM cells are found in Part a. 43 AAL1- ATM cells are 43 AAL1- ATM cells x 53 bytes/ AAL1- ATM cell x 8 bits / byte =18,232 AAL1- ATM bits in one STM-1 frame. Thus rate of ATM found in Part a is 18,232 bits in 0.000125 seconds. İ.e 18,232 bits x (1/0.000125 sec) = 145.856 Mbps

5. Length of the raw data sequence to be communicated between two nodes is L bits. In a communication system using OSI Reference Model, headers with lengths of A bits, P bits, S bits, T bits, N bits and D bits are added at the Application, Presentation, Session, Transport, Network and Data Layers, respectively. At the Data Layer, additionally a trailer with length of Tr bits is also added.

a. Find the total number of transmitted bits and the percentage of the raw data in

the transmission. Total number of transmitted bits = (L+A+P+S+T+N+D+Tr) bits.

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Percentage of the raw data in transmission = (L / (L+A+P+S+T+N+D+Tr)).100% b. Fill in the appropriate layers in the below table.

Layer Function or Network Component

File Transfer

Multiplexer

Defines the methods used to transmit and receive data on the network

ATM

Flow control of data

Encryption

Name recognition (identification) so only the designated parties can participate

6. a. Prepare your own question which you think would best cover (in general) the topics

studied in this course. b. Answer your question in part a.


Application File Transfer

Physical Multiplexer

Data Defines the methods used to transmit and receive data on the network

Network ATM

Transport Flow control of data

Presentation Encryption

Session Name recognition (identification) so only the designated parties can participate

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ANSWERS FOR MISCELLANEOUS HOMEWORK-2

1. Network I1 has one SSP, one STP and one SCP. Network I2 also has one SSP, one

STP and one SCP. I1 and I2 are interconnected by No.7. Write the total number of: a. A – links, B – links, 4,1 b. C – links, F – links, 0,0 c. Trunks, unspecified

If a second SCP is added to Network I1, find the additional number of d. A – links, B – links, 1, 0 e. C – links, F – links, 0, 0

2. a. The bit sequence in one E1 frame is given as:

1001100011010100111001110001000110001000010101010001101001101010010101000011010101101010011111001010110010011000001001100000100010011001001110101110100111000101110001101011110001101110010110011011001101100010011101100010111110111001110010011001110011010010

Write the signalling sequence in this frame. This is Common Channel Signaling (CCS). Taking the 16th channel bits where it starts with the 0th channel, signalling sequence is 10011001

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b. What are the rates of voice and signalling in each channel in part a?

Slot 16 is allotted for signaling only, thus signalling and 30 voice channels each has 64 Kbps.

c. The bit sequence in one T1 frame is given as:

101011011101000100010111000101011000100001111101001110000110001001000000001111010111100001011000001000001000100000100100001010001001000100101000110101011110010111111110101100000110001001010100

Write the signalling sequence in this frame. This is Channel Associated Signaling or Call Associated Signaling (CAS). Taking each 8th bit in each channel of total 24 channels, signalling sequence is

111101000100000010110000

d. What are the rates of voice and signalling in each channel in part c?

7 bits for voice and 1bit for signalling in one 64 Kbps channel. Thus each channel has 56 Kbps of voice and 8 Kbps of signaling.

e. Does part a or part c have better voice quality? Why? Voice quality in part a is better because voice uses the full 64 Kbps in all of the 30 voice channels. However, in part c, voice uses only 56 Kbps in each of the 24 channels.

3. a. Do you prefer X.25, ATM or Frame Relay subscription for the following applications?

i. Data at constant 512 Kbps. Frame Relay ii. Data at 622 Mbps. ATM iii. Data at 19200 bps. X.25 iv. 140 Mbps video. ATM v. 64 kbps audio. ATM.

b. Indicate whether the following network concepts are related to the circuit switched

or packet switched networks. i. PVC. Packet switched networks ii. SVC. Packet switched networks iii. CIR. Packet switched networks iv. T1. Circuit switched networks v. ABR. Packet switched networks

c. Do you need the following in order to introduce Intelligent Network (IN) services in

a PSTN infrastructure. i. SS7. Yes. ii. Database. Yes. iii. IP Network. No. iv. PSTN Subscriber. Yes.

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iv. ATM Network. No. d. Write whether the following telecommunication systems can be considered as WAN

(Wide Area Network) or not: i. VPN. Yes. ii. PSTN. Yes. iii. Internet. Yes. iv. Intelligent Network. No. v. LAN. No.

4. An ethernet frame with 1414 bytes of payload is being carried by ATM cells. a. Find the number of ATM cells needed to carry this ethernet frame.

There are total of (8+6+6+2+1414+4)=1440 bytes in the total ethernet frame. In ATM we can replace 48 bytes in each cell 1440 bytes / 48 bytes = 30 cells

b. Find the ratio of ethernet payload bits to the total number of bits transmitted in the ATM network in part a.

No. of ethernet payload bits=1414 x 8 = 11312 bits Total number of bits transmitted in the ATM network in part a = 53 bytes / (ATM cell) x 30 (ATM cells) x 8 bits / byte = 12720 bits Ratio of ethernet payload bits to the total number of bits transmitted in the ATM

network in part a = 11312 bits / 12720 bits = 0.889

5. Information data sequence to be communicated between two nodes is Q bits. In a communication system using OSI Reference Model, headers with lengths of HA bits, HP bits, HS bits, HT bits, HN bits and HD bits are added at the Application, Presentation, Session, Transport, Network and Data Layers, respectively. At the Data Layer, additionally a trailer with length of TD bits is also added.

a. In the transmission using this OSI Reference Model scheme, find the total

number of transmitted bits and the percentage of the information data in the transmission.

Total number of transmitted bits = (Q + HA + HP + HS + HT + HN + HD +TD) bits. Percentage of the information data in transmission =[Q/(Q+HA+HP+HS+HT+HN+HD+TD)].100% b. If the total number of bits in part a are transmitted in 1 second, find the rate of

the flow of sequence in the physical layer. (Q + HA + HP + HS + HT + HN + HD +TD) bits / sec. c. If the total number of bits in part a are transmitted in 1 second, find the rate of

the information data. Q bits / sec. d. In the following table, fill in the layers that suits most the function and network

component.


Mail forwarding and storage

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Ethernet coaxial cable

Framing, addressing and checksumming of Ethernet packets

ATM

Defines if and how retransmissions will be used to ensure data delivery

Translation of different control codes

Ensures that the information exchange is completed appropriately


Application Mail forwarding and storage

Physical Ethernet coaxial cable

Data Framing, addressing and checksumming of Ethernet packets

Network ATM

Transport Defines if and how retransmissions will be used to ensure data delivery

Presentation Translation of different control codes

Session Ensures that the information exchange is completed appropriately

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1. Network NN has two signal switching points, two signal tranfer points and one signal

control point. Network N also has two signal switching points, two signal tranfer points and one signal control point. N1 and N2 are interconnected by No.7. For this SS7 a. Draw this SS7 configuration Solution:

b. Find the total number of A – links Solution: Total number of A – links = 12 c. Find the total number of B – links Solution: Total number of B – links = 4 d. Find the total number of C – links Solution: Total number of C – links = 2 e. How many trunk channels do you suggest if all the signal switching points have 5000 subscribers each. Explain. Solution: Around 480 or 500 (%10 of the subscriber number) channel trunks can reasonably be connected between each signal switching point.

2. Considering two types of ethernet frame formats, the first type having 46 bytes in the payload area and the second type having 1500 bytes in the payload area. Rate of ethernet is 2 Mbps. The actual information data to be carried is 69 kbytes. a. Find the total number of ethernet frame format bits needed to carry the 69 kbytes of actual information data when the first type of ethernet frame format is used. Solution: We need total of 69 kbytes / 46 bytes = 1500 frames to carry the 69 kbytes of actual information data In one frame of the first type there are 8+6+6+2+46+4=72 bytes = 72 x 8 = 576 bits Thus the total number of ethernet frame format bits needed to carry the 69 kbytes of actual information data when the first type of ethernet frame format is used = 1500 frames x 576 bits / frame = 864 kbits b. Repeat part a when the second type of ethernet frame format is used. Solution: We need total of 69 kbytes / 1500 bytes = 46 frames to carry the 69 kbytes of actual information data In one frame of the second type there are 8+6+6+2+1500+4=1526 bytes = 1526 x 8 = 12208 bits.

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Thus the total number of ethernet frame format bits needed to carry the 69 kbytes of actual information data when the second type of ethernet frame format is used = 46 frames x 12208 bits / frame = 561568 bits c. Find the rate of the actual information data when the first type of ethernet frame format is used. Solution: For the first type of ethernet frame, the ratio of the actual information data bits to the total number of bits in one ethernet frame = 46 / (8+6+6+2+46+4) = 46 / 72 = 0.63889 So, the rate of the actual information data when the first type of ethernet frame format is used = 2 Mbps x 0.63889 = 1.28 Mbps d. Repeat part c when the second type of ethernet frame format is used. Solution: For the second type of ethernet frame, the ratio of the actual information Data bits to the total number of bits in one ethernet frame = 1500 / (8+6+6+2+1500+4) = 1500 / 1526 = 0.98296 So, the rate of the actual information data when the second type of ethernet frame format is used = 2 Mbps x 0.98296= 1.97 Mbps e. Find the rate of the non-actual information data in parts c and d. Solution: In part c, the rate of the non-actual information data = (2 - 1.28) Mbps = 0.72 Mbps = 720 kbps In part d, the rate of the non-actual information data = (2 - 1.97) Mbps = 0.03 Mbps = 30 kbps 3. a. Write 2 similarities and 3 differences between X.25 and Frame Relay. Solution: Similarities: Both are packet switched networks Both are meant mainly for data Differences: Rates are different. X.25 is for much lower rates. Packet structure and sizes are different. Class of services they provide are different. b. In an OSI Reference Model which layer covers the following functionality? i) Encryption Solution: Presentation Layer ii) Framing of ethernet packets Solution: Data Link Layer iii) Electrical specifications Solution: Physical Layer iv) Routing the packets Solution: Network Layer v) File transfer Solution: Application c. Indicate 3 points that are needed to adapt an IN to the existing network? Also write 2 features of AIN that makes it different when compared to IN. Solution: 3 points that are needed to adapt an IN to the existing network: Software- defined hooks or triggers, Database at the SCP, Management system to support the SCP providing the application 2 features of AIN that makes it different when compared to IN: There is service-independent software in the SSP (Switch), and SCP service logic and the service management system are service-independent, not service specific. d. Write 5 examples for Wide Area Networks Solution: PSTN, Internet, VPN, Frame Relay, ATM

e. Explain PVC and SVC and indicate their counterparts in circuit switching

applications. Compare the rates used for PVC and SVC. Solution: PVC is virtual circuit formed in a packet switched network on a permanent basis.

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SVC is virtual circuit formed in a packet switched network on a switched basis, i.e the packet switching network forms a virtual circuit upon demand. Counterpart of PVC in circuit switched networks is leased line or dedicated circuits Counterpart of SVC in circuit switched networks is the dial up application There exists no rate distinction between PVC and SVC. 4. a. Find the maximum number of digital voice channels that a 80-λ DWDM system can

carry if each carrier is modulated at 10 Gbps. Maximum number of digital voice channels = 10 Gbps x 80 / 64 kbps = 80 x 10 10 / 64 x 10 3 = 12,500,000 voice channels. b. Write whether the following telecommunication systems can be considered as

WAN (Wide Area Network) or not: i. VPN. Yes. ii. PSTN. Yes. iii. Internet. Yes. iv. Intelligent Network. No. v. LAN. No.

d. Find the minimum and maximum percentage of data that can be carried in an

ethernet frame format. Total number of non data bytes in one ethernet frame format is 8+6+6+2+4=26

bytes Minimum number of data bytes in one ethernet frame format = 46 bytes

Thus minimum percentage of data that can be carried in an ethernet frame format = [46 / (26 + 46)] x 100 = 63.89 %.

Maximum number of data bytes in one ethernet frame format = 1500 bytes


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1. a. Is E-1 related to i. Multiplexing? Yes. ii. Voice? Yes. iii. Packet switching? No. iv. VPN? No. v. Leased Line? Yes.

2. a. An ethernet frame format has a structure with minimum number of data bits. How

many such ethernet frames are needed to carry a 69000 bytes of data? Minimum number of data bytes in one ethernet frame format = 46 bytes 69000 bytes / 46 bytes = 1500 such ethernet frames are needed to carry a

69000 bytes of data

b. For part a, find the total number of non data bytes to carry the total 69000 bytes. Total number of non data bytes in one ethernet frame format is 8+6+6+2+4=26

bytes In part a, 1500 ethernet frames are found.

Thus, total number of non data bytes to carry the total 69000 bytes is 1500 frames x 26 bytes / frame = 39000 bytes

c. An ethernet frame format has a structure with maximum number of data bits.

How many such ethernet frames are needed to carry a 69000 bytes of data? Maximum number of data bytes in one ethernet frame format = 1500 bytes 69000 bytes / 1500 bytes = 46 such ethernet frames are needed to carry a

69000 bytes of data d. For part c, find the total number of non data bytes to carry the total 69000 bytes.

Total number of non data bytes in one ethernet frame format is 8+6+6+2+4=26

bytes In part c, 46 ethernet frames are found.


e. Find the percentages of data that can be carried in the ethernet frame formats specified in part a and in part c.

Total number of non data bytes in both ethernet frames in part a and in part c

are 8+6+6+2+4=26 bytes

In part a, percentage of data that can be carried in the ethernet frame = [46 / (26 + 46)] x 100 = 63.89 %.

In part c, percentage of data that can be carried in the ethernet frame = [1500 /

(26 + 1500)] x 100 = 98.3 %.

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ANSWERS FOR MISCELLANEOUS HOMEWORK-5 1. One ethernet frame is sent in 10-4 second.

a. Find the minimum ethernet rate in bps. Ans: Minimum size ethernet is 8+6+6+2+46+4 = 72 bytes =72 bytes x 8 bits/byte = 576 bits Minimum ethernet rate = 576 bits / 10-4 = 5.76 Mbps b. Find the maximum ethernet rate in bps. Ans: Maximum size ethernet is 8+6+6+2+1500+4 = 1526 bytes = 1526 bytes x 8 bits/byte = 12208 bits Maximum ethernet rate = 12208 bits / 10-4 = 122.08 Mbps

c. How many bits of information is carried in one second in the payload of the ethernet

frames described in part a? Ans: In part a, in one ethernet frame there are 46 bytes = 46 bytes / frame x 8 bits/byte = 368 bits / frame. One ethernet frame is sent in 10-4 seconds so 104 frames are sent in one second. Thus, the number of bits of information carried in one second in the payload of the ethernet frames described in part a = 368 bits / frame x 104 frames = 368 x 104 bits = 3.68 Mbits d. In how many seconds can 600 Mbits of payload be sent with the ethernet frames

described in part b? Ans: In part b, in 10-4 second, one ethernet frame is sent i.e., in 10-4 second, 1500 bytes = 1500 bytes x 8 bits/byte = 12000 bits of payload are sent

Thus, the number of seconds in which 600 Mbits of payload be sent with the ethernet frames described in part b = 600 Mbits x 10-4 second / 12000 bits = 5 sec

e. Write one similarity and one difference between token and carrier sense in local

area networks. Ans: Both are messages used to organize the message flow in local area

networks. Token is used in ring topology, carrier sense is used in ethernet. 2. Consider 3 networks interconnected by the North American SS7 signalling architecture.

Each network has one signal switching point, two signal control point and two signal transfer points. a. Find the maximum number of A – links needed. Answer: For one network: 2 A - links from SSP to STPs,

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2 A - links from the first SCP to STPs, 2 A - links from the second SCP to STPs, i.e total of 6 A - links for one network Since there are 3 networks, the maximum number of A – links needed = 6 A – links / network x 3 networks =18 A – links. b. Find the maximum number of B – links needed. Answer: In the connection of the first network to the second network: 4 B - links from STPs to STPs, In the connection of the first network to the third network: 4 B - links from STPs to STPs, In the connection of the second network to the third network: 4 B - links from STPs to STPs, Thus, the maximum number of B – links needed = 4 B – links / network x 3 networks =12 B – links. c. Find the maximum number of C – links needed. Answer: C – links are used between the mated STPs. There is maximum one C – link per network. Thus, in 3 networks, the maximum number of C – links needed = 1 C – link / network x 3 networks = 3 C – links. d. Find the minimum number of E – links needed. Answer: Since it is not compulsory to use E links, the minimum number of E – links needed = 0.

e. Describe signalling and SS7.

Answer: Signalling is the exchange of information between call components required to provide and maintain service.

Signalling System 7 (SS7) is a standart architecture for performing out-of- band signalling to support call-establishment, billing, routing and information- exchange functions of the public switched telephone network (PSTN). 3. a. Among X.25, Frame Relay, ATM and E hierarchy, write the ones which i) Provide the highest rate Answer: ATM and E –7. ii) Have packet structure Answer: X.25, Frame Relay and ATM. iii) Support voice transmission Answer: ATM and E hierarchy iv) Support data transmission Answer: X.25, Frame Relay, ATM and E hierarchy

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v) Have packet size greater than or equal to 53 bytes. Answer: X.25, Frame Relay and ATM b) Among the OSI layers, which ones use i) Mainly hardware Answer: Physical, Data, Network ii) Mainly software Answer: Session, Presentation, Application iii) Fiber optics cable Answer: Physical iv) Modulation Answer: Physical v) Network support Answer: Physical, Data, Network c) Write one similarity and one difference between leased line and PVC connection. Answer: Both appear on permanent basis, leased line is done through physical circuit, PVC is formed by virtual circuit. d) Write two differences between leased line and SVC connection. Answer: Leased line is on permanent basis, SVC is switched connection Leased line is done through physical circuit, PVC is formed by virtual circuit. e) Write one similarity and one difference between SVC and PVC connection. Answer: Both are virtual circuits, PVC is permanent, SVC is switched connection 4. a. Among IN, AIN, CAS and CCS, which ones i) Are related with signalling? Answer: IN, AIN, CAS and CCS ii) Use in-band? Answer: CAS iii) Are used to inject a single new service in a PSTN infrastructure?

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Answer: IN iv) Can be used to provide “800” and “900” services? Answer: AIN v) Are related to PSTN infrastructure? Answer: IN, AIN, CAS and CCS b. Among PSTN, X.25, LAN, Frame Relay and ATM, which ones i) Are considered as WAN? Answer: PSTN, X.25, Frame Relay and ATM ii) Can carry ethernet packets? Answer: LAN, X.25, Frame Relay and ATM iii) Use circuit switching? Answer: PSTN iv) Are used for data communication? Answer: PSTN, X.25, LAN, Frame Relay and ATM v) Form virtual circuits in WAN? Answer: X.25, Frame Relay and ATM

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ANSWERS FOR MISCELLANEOUS HOMEWORK-6 1. Three networks N, NN and NNN are interconnected in the North American SS7

signaling architecture. Each of N, NN and NNN has one signal switching point, one signal transfer point, each of N, NN has one signal control point and NNN has no signal control point.

a. Sketch this SS7 connections showing all the relevant components and links. Answer:

b. Find the total number of A links Answer: Total number of A links = 5 c. Find the total number of B and D links Answer: Total number of B and D links = 3 d. Find the total number of C links Answer: Total number of C links = 0 e. Is it absolutely needed to have trunk connections between the signal switching

points of N and NN, between the signal switching points of N and NNN and between the signal switching points of NN and NNN altogether or is it still operational if one of these trunk connections does not exist? Explain.

Answer: It is not absolutely needed to have trunk connections between the signal

switching points of N and NN, between the signal switching points of N and NNN and between the signal switching points of NN and NNN altogether. It is still operational if one of these trunk connections does not exist, i.e., if SSP-N is connected SSP-NN and SSP-NN is connected SSP-NNN, it is not necessary to connect SSP-N and SSP-NNN because the trunks between SSP-N and SSP-NNN can still be established by using the existing trunks between SSP-N, SSP-NN and between SSP-NN, SSP-NNN.

2. Each ethernet frame has 46 bytes in the payload area. Ethernet rate is 72 Mbps. a. How many Mbytes of actual information data can be carried in 1 minute? Answer: In one second total (actual data+control) 72 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second, 72 Mbits x 46 / (46+26) = 72 Mbits x 46 / 72 = 46 Mbits of

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actual data is carried.

Thus in one minute, 46 Mbits x 60 of actual data bits ⇒ 46 Mbits x 60 / 8 = 345 Mbytes of actual data bytes are carried. b. Find the rate of the actual information data. Answer: In one second total (actual data+control) 72 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second 72 Mbits x 46 / (46+26) = 72 Mbits x 46 / 72 = 46 Mbits of actual data is carried. Thus the rate of the actual information data is 46 Mbps. c. Find the number of ethernet packages used in one second. Answer: In one second, total (actual data+control) 72 Mbits are carried. However, in one frame, there are 8+6+6+2+4+46=72 bytes, i.e., in one second, 72 Mbits / 72 bytes = 72 Mbits / (72 x 8) bits = 125,000 ethernet packages are used d. How many Mbytes of actual information data can be carried in 1 minute if each ethernet frame has 694 bytes in the payload area? Ethernet rate is again 72 Mbps. Answer: In one second total (actual data+control) 72 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second, 72 Mbits x 694 / (694+26) = 72 Mbits x 694 / 720 = 69.4 Mbits of actual data is carried.

Thus in one minute, 69.4 Mbits x 60 of actual data bits ⇒ 69.4 Mbits x 60 / 8 = 520.5 Mbytes of actual data bytes are carried. e. Compare the results found in parts a and d. Comment on the comparison.

Answer: Number of Mbytes of actual information data that can be carried within unit time and under the same rate is larger in part d as compared to part a, because in part d we use much larger payload than in part a.

3. a. Write the corresponding words used in packet switching for the following circuit

switching words: i) Physical circuit Answer: Virtual circuit ii) Leased line Answer: PVC (Permanent Virtual Circuit) iii) Switched line Answer: SVC (Switched Virtual Circuit) iv) Non efficient use of the network Answer: Efficient use of the network v) High rate data transmission Answer: The same, i.e., high rate data transmission b. In OSI Reference Model, write the related layer for the following: i) X.25 packet formation Answer: Data Link Layer ii) Fiber optics cable Answer: Physical Layer iii) Control code translation Answer: Presentation Layer iv) Re-transmission of packets Answer: Transport v) Re-routing of ethernet frames Answer: Network Layer

20

c. What is achieved by introducing intelligent network in an existing telecommunications infrastructure? Your answer should be one sentence only. Answer: New features can be introduced in the existing old network. d. What is the other alternative to be done in the network to arrive at the same goal of establishing an intelligent network? Your answer should be one sentence only. Answer: To upgrade the sofware of all the existing exchange equipments to introduce the new feature. e. Write the Wide Area Network used mainly for

i) Voice transmission Answer: PSTN ii) Data transmission Answer: Internet iii) Connection of two local area networks Answer: Frame relay iv) Very low rate data communications Answer: X.25 v) Multimedia signal transmission Answer: Internet 4. a. In OSI, what would have happened if the number of layers used were much more than 7? Your answer should be maximum 2 sentences. Answer: The functionalities would be unnecessarily split into very small pieces. b. In OSI, what would have happened if the number of layers used were much less than 7? Your answer should be maximum 2 sentences.

Answer: The functionalities would be all combined in a few layers which would result in layers with so many functionalities that probably would not be used totally In all the applications. This means inefficient software and hardware designs.

c. In OSI, which layer(s) carry the actual information data. Answer: All the 7 layers d. The bit sequence in one T1 frame is given as:

101011001101000000010110000101001000100101111100001110010110001101000001001111000111100101011001001000011000100100100101001010011001000000101001110101001110010011111110110110001011000110101010

Write the signaling bit sequence in this frame. Answer: This is Channel Associated Signaling or Call Associated Signaling (CAS). Taking each 8th bit in each channel of total 24 channels, signaling sequence is

000010111011111101000010

21

e. If all the channels in part d carry voice signals, find the total number of voice signal bits in the T1 frame given in part d?

Answer: 7 bits for voice in one channel and there are 24 channels, so the total

number of voice signal bits in the T1 frame given in part d is 7 bits/channel x 24 channels = 168 bits

22

ANSWERS FOR MISCELLANEOUS HOMEWORK-7 1. a. Is the following a WAN or not? i. PSTN Answer: WAN ii. Token ring Answer: Not iii. Frame Relay Answer: WAN iv. CSMA Answer:Not v. ATM Answer:WAN b. Write 5 ATM services in the order of the lowest priority, i.e., write the lowest priority

service first and the highest priority service last. Answer: UBR, ABR, nrt_VBR, rt-VBR, CBR c. Can a frame relay system CIR 2 Mbps + EIR 1 Mbps subscriber i. Transmit 3 Mbps data? Answer: Yes ii. Receive 2 Mbps data for sure? Answer: Yes iii. Send 2 Mbps voice? Answer: No iv. Transmit 3 Mbps video? Answer: No v. Connect 2 LANs at 0.5 Mbps data rate? Answer: Yes d. Among ATM, frame relay and X.25, which one i. Can transmit at 10 Gbps? Answer: ATM ii. Is the first introduced in the infrastructure? Answer:X.25 iii. Is used to carry video? Answer: ATM iv. Is used to carry audio? Answer: ATM v. Is currently used to connect LANs? Answer: ATM, Frame Relay e. Can an ethernet at 1 Mbps be carried by i. CIR? Answer: Yes ii. CBR? Answer: Yes iii. X.25? Answer: No iv. rt-VBR? Answer: Yes v. UBR? Answer: Yes 2. Network N1 has one SSP, two STP and one SCP. Network N2 has one SSP, one STP

and two SCP. N1 and N2 are interconnected by No.7. Write the total number of: Answer:

a. A – links, B – links: Answer: 7,2

23

b. C – links, F – links: Answer: 1,0 c. Voice trunks if the SSP of Network N1 is connected to the SSP of Network N2 by E2. Answer: 120 If a second STP is added to Network N2, find the maximum number of additional Answer:

d. A – links, B – links, C – links, F – links, E-links Answer: 3, 2, 1, 0, 1 (Additional E-

links is 1, since before adding the second STP to Network N2, there was 3 E-links). e. Voice trunks Answer: 0

3 a. Are the following correct or wrong? i. The ATM packet with CLP bit 1 has higher priority than the ATM packet with CLP

bit 0. Answer: Wrong ii. Intelligent network is the only solution of introducing new services in a classical

network. Answer: Wrong iii. Every telecommunication device must have all the 7 layers of OSI. Answer:

Wrong iv. Errors in ATM payload can be detected by HEC in the ATM header. Answer:

Wrong v. Addressing in ATM is done in one step. Answer: Wrong b. How many AAL1 ATM packets are required to carry one minimum sized ethernet

packet? Answer: Minimum sized ethernet packet has 26+46=72 bytes. AAL1 ATM packet has 48-1=47 bytes of payload, Thus 2 AAL1 ATM packets are required to carry one minimum sized ethernet packet c. Write the corresponding words used in circuit switching for the following packet

switching words: i) Switched Virtual Circuit Answer: Switched line (dial up) ii) Packetized digital bit stream Answer: Continuous bit stream iii) Virtual circuit Answer: Physical circuit iv) Efficient use of the network Answer: Non efficient use of the network v) Permanent Virtual Circuit Answer: Leased line d. In OSI Reference Model, write the related layer for the following: i) ATM Answer: Data Link Layer (2) ii) FTP (File Transfer Protocol) Answer: Application Layer (7) iii) Encryption Answer: Presentation Layer (6)

24

iv) Addressing of internet Answer: Network (3) v) Connector Answer: Physical Layer (1) e. How many bits are there in the payloads and the VPI, VCI of UNI-ATM and NNI-

ATM headers? Answer:In UNI-ATM header: 48x8=384 bits of payload, 8 bits of VPI, 16 bits of VCI In NNI-ATM header: 48x8=384 bits of payload, 12 bits of VPI, 16 bits of VCI 4. The rate of ethernet is 236 Mbps and in each ethernet frame there are 92 bytes

payload. a. How many Gbytes of the actual information data is carried in an hour? Answer: In one second total (actual data+control) 236 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second, 236 Mbits x 92 / (92+26) = 236 Mbits x 92 / 118 = 184 Mbits of actual data is carried.

Thus in one hour, 184 Mbits x 60 x 60 of actual data bits ⇒ 184 Mbits x 60 x 60 / 8 = 82800 Mbytes = 82.8 Gbytes of actual data bytes are carried.

b. How many ethernet packages are needed in one minute? Answer: In one second, total (actual data+control) 236 Mbits are carried. However, in one frame, there are 8+6+6+2+4+92=118 bytes, i.e., in one second, 236 Mbits / 118 bytes = 236 Mbits / (118 x 8) bits = 250,000 ethernet packages are used Thus, in one minute 250,000 x 60 = 15 Million ethernet packages are used c. Ethernet rate is unchanged but each ethernet frame has 210 bytes in the payload

area. Find the number of actual information data in Mbytes that can be carried in 1 second.

Answer: In one second total (actual data+control) 236 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second, 236 Mbits x 210 / (210+26) = 236 Mbits x 210 / 236 = 210 Mbits = 210/8 Mbytes = 26.25 Mbytes of actual data is carried. d. What is the rate of the actual information data in part c? Answer: In one second total (actual data+control) 236 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second 236 Mbits x 210 / (210+26) = 236 Mbits x 210 / 236 = 210 Mbits

of actual data is carried. Thus the rate of the actual information data is 210 Mbps. e. In part c, what percentage of the total ethernet packet is the header?

Answer: 26 header bytes / (210 payload bytes + 26 header bytes) = 26 / 236 = 0.11

⇒ 11 %

25


1. a. Write the OSI layer(s) that i. communicate peer-to-peer. Ans: Layers 4-7 ii. communicate through the connections of intermediate nodes. Ans: Layers 1-3 iii. is responsible from encryption. Ans: Presentation, i.e. layer 6 iv. provides the technical specification of the BNC connector. Ans: Physical, i.e.,

layer 1 v. contains the whole bit stream. Ans: Physical, i.e., layer 1 b. Find the number of AAL1 ATM packets that are required to carry one maximum

sized ethernet packet? Answer: Maximum sized ethernet packet has 26+1500=1526 bytes. AAL1 ATM packet has 48-1=47 bytes of payload. Thus 1526 / 47 = 32.46809, i.e., 33 AAL1 ATM packets are required to carry one maximum sized ethernet packet. c. Write the number of bits in VPI and VCI of UNI-ATM and NNI-ATM headers. How

many VPI adresses can be formed in UNI-ATM? Answer:In UNI-ATM header: 8 bits of VPI, 16 bits of VCI In NNI-ATM header: 12 bits of VPI, 16 bits of VCI Number of VPI adresses that can be formed in UNI-ATM is 28 = 256 d. Can a frame relay system providing CIR 1 Mbps + EIR 512 kbps i. transmit 2 Mbps video? Answer: No ii. connect 2 LANs at 512 kbps data rate? Answer: Yes iii. receive 1.5 Mbps data? Answer: Yes iv. receive 1.5 Mbps data for sure? Answer: No v. transmit 5 Mbps data? Answer: No e. For each item given below, write the one which is unrelated to the others. i. CBR, VPI, CIR, VBR, EIR. Answer: VPI ii. FDDI, PVC, CSMA, Exponential Backoff, Collision. Answer: PVC iii. PVC, Leased Line, Dial Up, SVC, E-1. Answer: E-1. iv. AIN, IN, SVC, SS7, SCP. Answer: SVC v. CAS, Out-of-Band Signalling, In-Band Signalling, VCI, CCS. Answer: VCI 2. An ethernet frame contains 114 bytes payload and the rate of ethernet is 140 Mbps. a. Find the total actual information data (in Mbps) carried in one second. Answer: In one second total (actual data+control) 140 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second, 140 Mbits x 114 / (114+26) = 140 Mbits x 114 / 140 = 114 Mbits of actual data is carried. b. Find the number of ethernet packages needed in one second. Answer: In one second, total (actual data+control) 140 Mbits are carried. However, in one frame, there are 8+6+6+2+4+114=140 bytes,

26

i.e., in one second, 140 Mbits / 140 bytes = 140 Mbits / (140 x 8) bits = 125,000 ethernet packages are used. c. If the ethernet rate is kept the same but the payload of each ethernet frame has

254 bytes, how many Gbytes of actual information data can be carried in one hour? Answer: In one second total (actual data+control) 140 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second, 140 Mbits x 254 / (254+26) = 140 Mbits x 254 / 280 = 127 Mbits = 127/8 Mbytes = 15.875 Mbytes of actual data is carried. Thus in one minute, 15.875 x 60 x 60 /1000 =57.15 Gbytes of actual data is carried in one hour. d. For part c, find the rate of the actual information data. Answer: In one second total (actual data+control) 140 Mbits are carried. However, in one frame, there are 8+6+6+2+4=26 control bytes, i.e., in one second 140 Mbits x 254 / (254+26) = 140 Mbits x 254 / 280 = 127 Mbits

of actual data is carried. Thus the rate of the actual information data is 127 Mbps. e. Find the percentages of the header bits for a maximum and a minimum sized ethernet frame. Answer: For a maximum sized ethernet frame, 26 header bytes / (1500 payload

bytes + 26 header bytes) = 26 / 1526 = 0.01704 ⇒ 1.7 % For a minimum sized ethernet frame, 26 header bytes / (46 payload

bytes + 26 header bytes) = 26 / 72 = 0.36111 ⇒ 36.11 % 3. Network N1 has one SSP, one STP and two SCP. Network N2 has one SSP, two STP

and one SCP. If N1 and N2 are interconnected by No.7, write the total number of: Answer:

a. A – links and B – links. Answer: 7 and 2 b. C – links and F – links. Answer: 1 and 0 c. Voice trunks if the SSP of Network N1 is connected to the SSP of Network N2 by

E1. Answer: 30 A second STP is added to Network N1. In this new configuration, find the maximum number of Answer:

27

d. A – links, B – links, C – links, F – links, E-links. Answer: 10, 4, 2, 0, 4. e. Voice trunks. Answer: 30.

4. a. Considering ATM, frame relay and X.25, which can be used i. to transmit video? Answer: ATM ii. currently to connect LANs? Answer: ATM, Frame Relay iii. to communicate with a fixed packet size? Answer: ATM iv. to transmit audio? Answer: ATM v. for data communication at 622 Mbps? Answer: ATM b. Can 140 Mbps ethernet be carried by i. VBR? Answer: Yes ii. X.25? Answer: No iii. EIR? Answer: No iv. CIR? Answer: No v. CBR? Answer: Yes c. For each of the following items, write whether the statement is correct or wrong. i. Errors in ATM payload cannot be detected by HEC in the ATM header. Answer:

Correct ii. Two step adressing is used in ATM. Answer: Correct. iii. It is not possible to introduce a new service feature into an existing telecommunications infrastructure without employing an intelligent network.

Answer: Wrong iv. There are no bits in the ATM packet that are somehow related to priority.

Answer: Wrong v. It is necessary that each equipment used in the telecommunications

infrastructure should include all the layers of OSI. Answer: Wrong d. Is the following a WAN or not? i. ATM. Answer:WAN ii. Multiplexer. Answer: Not iii. AIN. Answer: Not iv. CAS. Answer: Not v. Local Area Network. Answer: Not e. Are PAD (Packet Assembler / Disassembler) and SAR (Segmentation and Reassembly) the same? Why?

28

Ans: No, because in PAD (Packet Assembler/Disassembler) packet assembly for outgoing data and packet disassembly for incoming packets are performed in X.25. However, SAR (Segmentation and Reassembly) layer in ATM accepts a 47-byte from CS and adds a one byte header.

29


1. a. Can VPN be a replacement for

i. SVC? Yes. ii. PVC? Yes. iii. PSTN Dial-up? Yes. iv. Leased Line? Yes. v. Dedicated Line? Yes.

b. Write separately for each of the below listed network elements whether it is definitely needed in introducing Next Generation Network services in a PSTN infrastructure. i. SS7. No. ii. Local Exchange. Yes. iii. VoIP/Media Gateway. Yes. iv. IP/ATM Network. Yes v. PSTN Subscriber. Yes.

c. In each of the following items (i, ii, iii, iv, v) 5 systems are named. For each item,

write the name of the system which is unrelated to the other 4 systems. i. OADM, 140 Mbps/565 Mbps Mux., MEM, All Optical Networks, DWDM. 140 Mbps/565 Mbps Mux. ii. Fiber Access Network, E3, STM-1, STM-3, Satellite, Copper Network. Satellite iii. Wireless, FDMA, ATM, Frequency Hopping, CDMA. ATM iv. ATM, PPTP, VPN, SS7, T-3, L2TP. ATM v. FDMA, XDSL, Access Network, BRA, Local Loop. FDMA

d. Give very short (around one line) description for the following:

i. VPN.

A private network that uses a internet to connect remote sites or users together. ii. Q0S.

Provides service differentiation and performance assurance for internet applications

iii. Fiber Access. Provides high rate services (like xDSL) to existing copper subscribers by using

fiber network. iv. All Optical Network. Network with all elements operating in optical domain, thus utilizing the high

rate advantage of optics all over the network. v. Next Generation Network.

Provides internetworking among PSTN, AIN and Next Generation IP based Services.

e. Explain the three multiple access methods, emphasizing their difference.

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In TDMA (Time Division Multiple Access) simultaneous calls can be made in the same bandwith, by dividing the total time frame into separate time slots.

In FDMA (Frequency Division Multiple Access) simultaneous calls can be made at

the same time, by dividing the total frequency band into smaller frequency bands. In CDMA (Code Division Multiple Access) simultaneous calls can be made in the

same spread bandwith and at the same time, by giving different codes (PN) to different users.

2. Three spreading codes used in a CDMA system are given below where C(1) is the

spreading code of user-1, C(2) is the spreading code of user-2, C(3) is the spreading code of user-3. Chip period is 100 nsec. Each code signal has 10 chips, information rate is 1 Mbit/sec. In your plots clearly show all the relevant scales with appropriate numbers incorporated.

C(1)

1

0

Time

-1

C(2)

1

0

Time

-1

C(3)

1

0

Time

-1

a. Find the spreading factor. Spreading factor = Gsf = Tb / Tc Tb = 1 / (1 Mbit/sec) = 10 -6 sec Tc = 100 nsec = 100 x 10 -9 sec = 10 –7 sec Spreading factor = 10 -6 sec / 10 –7 sec = 10

31

b. For C(1), plot the spread signal from t=0 to t=1 µsec., if the amplitude of the information bit symbol is 1 (representing 1).

c. For C(2), plot the spread signal from t=0 to t=1 µsec., if the amplitude of the

information bit symbol is -1 (representing 0).

m(2)(t)

1

0

1 µsec Time

-1

d. ( )

ci

n and ( )

cj

n can be used as spreading codes if

9( ) ( )

0

c c 0i j

n n

n=

=∑ . Using this

condition, find which of the 3 spreading codes are appropriate for CDMA.

9(1) (2)

0

c c 0n n

n=

=∑ ,

9(1) (3)

0

c c 2n n

n=

=∑ ,

9(2) (3)

0

c c 4n n

n=

=∑

Thus (1)

cn and

(2)c

n are appropriate for CDMA.

(3)c

n is not appropriate because it

gives

9( ) (3)

0

c c 0i

n n

n=

≠∑ .

e. To which CDMA system can the information rate specified in this problem belong to?

WCDMA or cdma2000. 3. For the transport of ATM packets in STM-1, 258 columns by 9 rows of bytes in STM-1

payload area is used. ATM cells can be split between the two consecutive rows in the STM-1 payload area. IP with 24 bytes of header and 504 bytes of total packet size is carried by AAL1- ATM cells.

a. Find the maximum integer number of AAL1- ATM cells that an STM-1 frame can

carry.

There are total space for 258 x 9 = 2322 bytes in the payload area.

m(1)(t)

1

0

1 µsec Time

-1

32

AAL1- ATM cell size is fixed and 53 bytes

Since AAL1- ATM cells can be split between the two consecutive rows in the payload area, the maximum integer number of AAL1- ATM cells that an STM-1 frame can carry = 2322 / 53 = 43 AAL1- ATM cells.

b. Find the total number of AAL1- ATM payload bytes (carrying information) in one

STM-1 frame when AAL1- ATM are mapped as in Part a.

In one AAL1- ATM cell, there are 47 bytes in the AAL1- ATM payload. İ.e. 47 payload bytes. Thus, in one AAL1- ATM cell, there are 47 bytes x 8 bits / byte = 376 bits in the ATM payload. İ.e. 376 payload bits. In part a 43 ATM cells are found in one STM-1 frame. i.e., there are 376 x 43 = 16,168 total bits of AAL1- ATM payload (carrying information) in one STM-1 frame i.e., there are 16,168 / 8 = 2,021 total bytes of AAL1- ATM payload (carrying information) in one STM-1 frame

c. Find the maximum integer number of IP packets that the total AAL1- ATM payload

in one STM-1 frame as found in Part b can carry. The maximum integer number of IP packets that the total AAL1- ATM payload in

one STM-1 frame as found in Part b can carry = 2,021 total bytes of AAL1- ATM payload / 504 bytes of total IP packet size = 4 IP packets.

d. If the data portions of IP packets in Part c are fully occupied by TCP, find the total

number TCP bytes in one STM-1 frame. There are 504 bytes total IP packet size - 24 bytes of IP header = 480 bytes of

data in one IP packet. Since the data portions of IP packets are fully occupied by TCP, the total number TCP bytes in one IP packet = 480 bytes.

Thus total number TCP bytes in one STM-1 frame = 4 x 480 bytes = 1920 bytes. e. If the data portions of TCP packets are fully occupied by HTTP and the header of

TCP is 24 bytes, find the total number HTTP bytes in one STM-1 frame. There are total 1920 TCP bytes in one STM-1 frame. There are total 4 TCP packets x 24 bytes TCP header = total 96 bytes TCP header

in one STM-1 frame. Since the data portions of TCP packets are fully occupied by HTTP, the total

number HTTP bytes in one STM-1 frame = 1920 bytes – 96 bytes = 1824 bytes. 4. a. Find the maximum number of digital voice channels that a 80-λ DWDM system can

carry if each carrier is modulated at 10 Gbps. Maximum number of digital voice channels = 10 Gbps x 80 / 64 kbps

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= 80 x 10 10 / 64 x 10 3 = 12,500,000 voice channels. b. Write whether the following telecommunication systems can be considered as

WAN (Wide Area Network) or not: i. VPN. Yes. ii. PSTN. Yes. iii. Internet. Yes. iv. Intelligent Network. No. v. LAN. No.

c. Considering QoS (Quality of Service):

i. Is 500 bits error in one billion bits acceptable for conversation? Why? Yes because bit error rate ≤ 10-6 is acceptable for conversation . ii. Is end-to-end delay of 600 ms acceptable for telephone speech? Why?

No because around 250 ms end-to-end delay is acceptable for telephone speech.

iii. Is loss of 1000 packets within a total of one million packets acceptable for uncompressed video? Why?

Yes because packet loss ≤ 10-2 is acceptable for uncompressed video. iv. Is loss of 1000 packets within a total of one million packets acceptable for

compressed video? Why? No because packet loss ≤ 10-11 is acceptable for compressed video. v. Is the rate 1 Mbps acceptable for HDTV uncompressed quality? Why?

No because > 1 Gbps is acceptable for HDTV uncompressed quality. d. Find the minimum and maximum percentage of data that can be carried in an

ethernet frame format. Total number of non data bytes in one ethernet frame format is 8+6+6+2+4=26

bytes Minimum number of data bytes in one ethernet frame format = 46 bytes


Maximum number of data bytes in one ethernet frame format = 1500 bytes


e. Write five points which you think would be the most important in telecommunication

networks in the year 2020.

• Extremely high rates in the backbone networks and very high rates in the access networks,

• All optical networks,

• Almost all digital networks,

• Networks being able to follow multiprotocol, i.e., one huge network worldwide,

• Intelligent programmable network: From anywhere in the world, any service or feature will be accessible irrespective of the connected network provider or network platform.

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1. a. Is E-1 related to i. Multiplexing? Yes. ii. Voice? Yes. iii. Packet switching? No. iv. VPN? No. v. Leased Line? Yes.

b. Are the following pair of concepts or systems related to each other or not? i. MEMS and Exchange. Related. ii. Intelligent Network and DWDM. Not. iii. SS7 and IP. Not. iv. VOIP and Internet Protocol. Related. v. All Optical Network and E-4. No.


write the name of the system which is unrelated to the other 4 systems. i. Fiber Access Network, MEMS, Optical Switch, Wavelength Multiplexing, DWDM.

Ans: Fiber Access Network ii. CDMA, GSM, FDMA, ATM, Frequency Hopping. Ans: ATM . iii. HTTP, PPP, PPTP, IP, TCP. Ans: PPTP iv. Local Loop, Access Network, FSO, Last Mile, VPN. Ans: VPN v. ATM, IP, Intelligent Network, Next Generation Network, VOIP. Ans: Intelligent

Network d. Write 5 important elements appearing in VPN. Ans: Layer Two Tunneling Protocol (L2TP), Point-to-Point Tunneling Protocol

(PPTP), Authentication, Encryption, Generic Routing Encapsulation (GRE) e. Write 5 important elements appearing in Multiple Access Mobile Ssytems. Ans: TDMA, FDMA, CDMA, Frequency Hopping, Spread Spectrum. 2. a. An ethernet frame format has a structure with minimum number of data bits. How

many such ethernet frames are needed to carry a 69000 bytes of data? Minimum number of data bytes in one ethernet frame format = 46 bytes 69000 bytes / 46 bytes = 1500 such ethernet frames are needed to carry a

69000 bytes of data

b. For part a, find the total number of non data bytes to carry the total 69000 bytes. Total number of non data bytes in one ethernet frame format is 8+6+6+2+4=26

bytes In part a, 1500 ethernet frames are found.


35

c. An ethernet frame format has a structure with maximum number of data bits. How many such ethernet frames are needed to carry a 69000 bytes of data?

Maximum number of data bytes in one ethernet frame format = 1500 bytes 69000 bytes / 1500 bytes = 46 such ethernet frames are needed to carry a

69000 bytes of data d. For part c, find the total number of non data bytes to carry the total 69000 bytes.

Total number of non data bytes in one ethernet frame format is 8+6+6+2+4=26

bytes In part c, 46 ethernet frames are found.


e. Find the percentages of data that can be carried in the ethernet frame formats specified in part a and in part c.

Total number of non data bytes in both ethernet frames in part a and in part c

are 8+6+6+2+4=26 bytes

In part a, percentage of data that can be carried in the ethernet frame = [46 / (26 + 46)] x 100 = 63.89 %.

In part c, percentage of data that can be carried in the ethernet frame = [1500 /

(26 + 1500)] x 100 = 98.3 %. 3. An IP packet frame is shown below.

a. Write the IP address of the sender in dotted decimal notation. Ans: 255.127.48.3 b. Write the IP address of the receiver in dotted decimal notation. Ans: 127.128.17.48

c. Write the number of digits used for padding.

36

Ans: <10 d. Find the number of data bytes.

Ans: 1024 total bytes - 28 header bytes = 996 data bytes e. Find the total number of TCP + HTTP bits carried by this IP packet frame.

Ans: TCP + HTTP bits carried by this IP packet frame occupy the data part which

is found in part d as 1024 total bytes - 28 header bytes = 996 data bytes which is equal to 996 x 8 = 7968 bits.

f. Find the number of ATM-AAL1 cells needed to carry this IP packet frame

Ans: There are 1024 total bytes (header + data) in this IP packet frame. In an ATM-AAL1, 47 bytes of data can be replaced in one cell. Thus, the number of ATM-AAL1 cells needed to carry this IP packet frame is 1024 bytes / 47 bytes, i.e. > 21, i.e., 22 cells.

g. Re-write the possible first 8 digits of the 3rd row in this IP packet frame when this

IP packet reaches the next hop in its travel from source to destination. Ans: Will probably be one less than the current Time-to-Live value where one less

is 00011110

4. a. Why all optical networks are required? Ans: To achieve very high bit rate transmissions in the network without going

through electo-optic and opto-electronic conversions. b. How can you compare the rates in all optical networks with the rates in 3G Mobile

Systems. Ans: In all optical networks, rates of 10 15 bits / sec, 10 18 bits / sec or even infinite

rates are being pronounced whereas in the 3G Mobile Systems, rate is only 2 Mbit / sec.

c. For QoS (Quality of Service), are the following values acceptable or not? Why?

i. For conversation, 300 bits of error in one billion bits. Yes because bit error rate ≤ 10-6 is acceptable for conversation . ii. For telephone speech, 500 ms end-to-end delay.

No because around 250 ms end-to-end delay is acceptable for telephone speech.

iii. For uncompressed video, loss of 500 packets within a total of 500,000 packets.

Yes because packet loss ≤ 10-2 is acceptable for uncompressed video. iv. For compressed video, loss of 2000 packets within a total of one million

packets. No because packet loss ≤ 10-11 is acceptable for compressed video. v. For HDTV uncompressed quality, rate of 1 Mbps.

No because > 1 Gbps is acceptable for HDTV uncompressed quality.

37

ANSWERS FOR MISCELLANEOUS HOMEWORK-11 1. a. List the following telecommunication systems in the order of lowest to highest rate

transmission. (Write the lowest rate first, highest rate last). (i) Fiber optics, (ii) E-4 hierarchy, (iii) All optical networks, (iv) GSM, (v) Third generation mobile networks Ans: iv, v, ii, i, iii

b. Compare wavelength switching with packet switching. Ans: In packet switching, switching is performed by routing the information flow

packet by packet. However, in wavelength switching, wavelengths each carying all the relevant multiplexed information are routed on wavelength basis.


write the name of the system which is unrelated to the other 4 systems. i. VPN, Wavelength Switching, Holographic Switch, All Optical Networks, DWDM. Ans: VPN. ii. Packet Switching, ATM, CBR, VBR, CIR. Ans: CIR iii. EIR, CIR, Frame Relay, X.25, 2.048 Mbps. Ans: X.25 iv. GRE, VPN, PPTP, SS7, L2TP. Ans: SS7 v. FDMA, CDMA, TDMA, Multiple Access Systems, Fiber Optics. Ans:Fiber Optics

d. Name 5 type of systems used in multiple access technology. Ans: TDMA, FDMA, CDMA, Frequency Hopping, Spread Spectrum.

e. To your opinion, which aspects of telecommunication networks will become more important in future? Indicate 5 points.

• Extremely high rates in the backbone networks and very high rates in the access networks,

• All optical networks,

• Almost all digital networks,

• Networks being able to follow multiprotocol, i.e., one huge network worldwide,

• Intelligent programmable network: From anywhere in the world, any service or feature will be accessible irrespective of the connected network provider or network platform.

2. Below structure shows an IP packet frame.

38

a. Write the IP address of the sender in dotted decimal notation. Ans: 252.124.60.15 b. Write the IP address of the receiver in dotted decimal notation. Ans: 124.131.19.51

c. Write the number of digits used for padding.

Ans: 0 d. Find the number of data bytes.

Ans: 988 total bytes - 28 header bytes = 960 data bytes e. Find the total number of TCP + HTTP bits carried by this IP packet frame.

Ans: TCP + HTTP bits carried by this IP packet frame occupy the data part which

is found in part d as 960 data bytes which is equal to 960 x 8 = 7680 bits. f. Find the number of ATM-AAL1 cells needed to carry this IP packet frame

Ans: There are 988 total bytes (header + data) in this IP packet frame. In an ATM-AAL1, 47 bytes of data can be replaced in one cell. Thus, the number of ATM-AAL1 cells needed to carry this IP packet frame is 988 bytes / 47 bytes, i.e. > 21, i.e., 22 cells.

g. Re-write the possible first 8 digits of the 3rd row in this IP packet frame when this

IP packet reaches the next hop in its travel from source to destination. Ans: Will probably be one less than the current Time-to-Live value where one less

is 00011011 3. Three spreading codes used in a CDMA system are given below where C(1) is the

spreading code of user-1, C(2) is the spreading code of user-2, C(3) is the spreading code of user-3. Chip period is 1 microsec. Each code signal has 10 chips, information rate is 100 kbit/sec. In your plots clearly show all the relevant scales with appropriate numbers incorporated.

39

a. Find the spreading factor. Solution: Spreading factor = Gsf = Tb / Tc Tb = 1 / (100 kbit/sec) = 10 -5 sec Tc = 1 microsec = 1 x 10 -6 sec = 10 –6 sec Spreading factor = 10 -5 sec / 10 –6 sec = 10 b. For C(1), plot the spread signal from t=0 to t=10 µsec., if the amplitude of the

information bit symbol is 1 (representing 1). Solution:

c. For C(2), plot the spread signal from t=0 to t=10 µsec., if the amplitude of the

information bit symbol is 1 (representing 1). Solution:

40

d. ( )

ci

n and ( )

cj

n can be used as spreading codes if

9( ) ( )

0

c c 0i j

n n

n=

=∑ . Using this

condition, find which of the 3 spreading codes are appropriate for CDMA.

9(1) (2)

0

c c 0n n

n=

=∑ ,

9(1) (3)

0

c c 2n n

n=

=∑ ,

9(2) (3)

0

c c 4n n

n=

=∑

Thus (1)

cn and

(2)c

n are appropriate for CDMA.

(3)c

n is not appropriate because it

gives

9( ) (3)

0

c c 0i

n n

n=

≠∑ .

e. Write two points in CDMA that seems to be contradicting with classical telecommunication principles.

Solution: - Information signal is intentionally spread in spectrum. - Clean information signal is intentionally multiplied by a noise like PN sequence 4. a. Name 10 designs that can be implemented in all optical networks. Ans: DWDM, Narrowband Lasers, EDFA, Optical Add/Drop Multiplexers, Optical

Switches, Thin Film Substrates, MEMS, Buble Switches, Thermo-optical Switches, Tunable Lasers

b. Are the following comments correct or wrong?

i. HTTP is in the payload of TCP Ans: Correct ii. TCP header is in the payload of IP Ans: Correct iii. HTTP is in the payload of IP Ans: Correct iv. PPP header is in the payload of IP Ans: Wrong v. IP header is in the payload of TCP

Ans: Wrong

c. When QoS (Quality of Service) is considered, are the following specifications acceptable or not? i. 1.1 Gbps for HDTV uncompressed quality Solution: Acceptable because > 1 Gbps is acceptable for HDTV uncompressed

quality. ii. 1 bit error in one billion bits for conversation Solution: Acceptable because bit error rate ≤ 10-6 is acceptable for conversation iii. Loss of 1 packet within a total of one thousand packets for compressed video Solution: Not acceptable because packet loss ≤ 10-11 is acceptable for

compressed video. iv. Loss of 10 packets within a total of hundred packets for uncompressed video Solution: Not acceptable because packet loss ≤ 10-2 is acceptable for

uncompressed video. v. End-to-end delay of 60 ms acceptable for telephone speech

41

Solution: Acceptable because around 250 ms end-to-end delay is acceptable for telephone speech.

d. Write item by item whether the following concepts are related to Next Generation

Networks or not. i. IP based services Ans: Related. ii. SS7 Ans: Not. iii. 3-D holography Ans: Related iv. IP/ATM Network Ans: Related v. Intelligent Networks Ans: Not.

e. When do you need a VPN connection and what are the 3 points that must be

fulfilled in order to have VPN.

VPN is needed when a remote fixed or mobile user wants to be connected to a LAN. 3 points that must be fulfilled in order to have VPN are encryption, autentication and tunneling.

42


1. a. What comes to your mind first when you read i. All optical networks. Ans: Very high rate communication. ii. Protocols in packet switched networks. Ans: Set of rules to achieve certain functionaly. iii. Voice over IP. Ans: Digitized voice carried in IP. iv. Encapsulation. Ans: Putting one packet frame into another packet frame. v. Quality of service. Ans: Technical specifications for an application of

telecommunications b. Which one gives the highest rate and the lowest rate transmissions among E5,

STM-4, G3 mobile, GSM, GPRS. Ans: STM-4 gives the highest rate transmission, GSM gives the lowest rate transmission.

c. Write 5 concepts or systems you think would be the most important in future telecommunication networks.

Ans: All optical networks, very high rate communications, all digital networks,

common protocols, next generation network applications such as virtual reality, 3-D holographic TV.

d. In each of the following items (i, ii, iii, iv, v) 5 concepts are named. For each item,

write the name of the concept which is unrelated to the other 4 concepts. i. Multiple Access Systems, FDMA, MEMS, CDMA, TDMA. Ans: MEMS ii. Authentication, Encryption, EDFA, VPN, Tunneling. Ans: EDFA iii. PSTN, PPTP, VPN, L2TP, Internet Infrastructure. Ans: PSTN iv. Encapsulation, IP/ATM, TCP in IP, HTTP in TCP, E1. Ans: E1 v. Wavelength Switching, QoS, All Optical Networks, EDFA, DWDM. Ans: QoS.

e. In all optical networks, write 5 functions that need to be performed optically. Ans: Amplification, multiplexing, packet switching, transmission, routing. 2. Below structure shows an IP packet frame. Options part in the header is 3 bytes. This

IP packet is sent in 1 msec.

a. What is the total number of bits in this IP packet?

43

Ans: Total length part of the packet is 0000010011100010 = 1x210 + 0x29 + 0x28 + 1x27 + 1x26 + 1x25 + 0x24 + 0x23 + 0x22 + 1x21 + 0x20 = 1024 + 128 + 64 + 32 + 2 = 1250 bytes = 1250 bytes x 8 bits / byte = 10000 bits

b. Find the rate of this IP packet. Ans: Packet has 10000 bits which is sent in 1 msec, i.e., 10000 bits/ msec x 1000 sec / msec = 10000000 bits / sec = 10 Mbps c. How many ATM-AAL1 cells are required for carrying this IP packet?

Ans: There are 1250 total bytes (header + data) in this IP packet frame. In ATM-AAL1, there are 47 bytes of payload in one cell. Thus, the number of ATM-AAL1 cells required for carrying this IP packet is 1250 bytes / 47 bytes, i.e. > 26, i.e., 27 cells.

d. Write the appropriate digits to replace the 5 digits denoted as “?” in this IP packet.

Ans: Since this part is the zero padding, all the digits indicated by “?” are “0”. e. Write the possible protocol which starts in the 7 th row of this IP packet.

Ans: From IHL part which is 0110 = 0x23 + 1x22 + 1x21 + 0x20 = 4 + 2 = 6 bytes i.e., there are 6 rows of IP header which means that IP payload starts in the7 th

row. Thus the 7 th row of this IP packet possibly will carry TCP. f. What is the destination IP address in dotted decimal notation? Ans: 125.130.18.50

3. Chip codes in a 3 user CDMA system are given below. (1)

c , (2)

c and (3)

c are the chip sequences for users 1, 2 and 3, respectively. Information rate is 100 kbit/sec. One information bit is multiplied by 8 chips.

44

a. How many times does the bandwidth of the information signal is spread in this CDMA system? Ans: Spreading factor = Tb / Tc Tb = 1 / (100 kbit/sec) = 10 -5 sec Tc = 2 microsec = 2 x 10 -6 sec Spreading factor = 10 -5 sec / 2 x 10 -6 sec = 5 times

b. Knowing that when the condition

7( ) ( )

0

c c 0i j

n n

n=

=∑ is fulfilled, ( )

ci

n and ( )

cj

n can be

used as spreading codes. Utilize this condition to find which of the chip

sequences, (1)

c , (2)

c , (3)

c can be used in CDMA.

Ans:

7(1) (2)

0

c c 0n n

n=

=∑ ,

7(1) (3)

0

c c 0n n

n=

=∑ ,

9(2) (3)

0

c c 0n n

n=

=∑ so all the chip

sequences, (1)

c , (2)

c , (3)

c can be used in CDMA. c. If the amplitude of the information bit symbol is 1 (representing 1), plot

the spread signal from t=0 to t=16 µsec for all the 3 users. Solution: They will be the same as the chip sequences, i.e.,

45

d. If the amplitude of the information bit symbol is - 1 (representing 0), plot

the spread signal from t=0 to t=16 µsec for all the 3 users. Solution: They will be the same as the chip sequences except all the bits are

multiplied by -1, i.e.,

e. Why do you need different chip sequences for different users in CDMA? Ans: We need different chip sequences for different users in CDMA to separate each users information signal in a multi user mobile system. 4. a. Are the following comments correct or wrong?

i. HTTP is in the payload of IP. Ans: Correct

46

ii. TCP payload is in the payload of IP. Ans: Correct iii. HTTP is in the payload of PPP. Ans: Correct iv. IP header is in the payload of TCP. Ans: Wrong v. IP header is in the payload of PPP.

Ans: Correct b. What are the two points that distinguish a CDMA system from other

telecommunication systems? Ans: - Information signal is intentionally spread in spectrum. - Clean information signal is intentionally multiplied by a noise like PN sequence

c. Write 5 systems that you can make LAN to LAN connection.

Ans: VPN, Frame relay, ATM, X.25, SDH

d. Write item by item whether the following concepts are related to Quality of Service

(QoS) or not. i. Technical specifications of an application. Ans: Related. ii. 64 kbps PSTN telephone voice channel. Ans: Related. iii. MEMS. Ans: Not related iv. SS7. Ans: Not related v. 200 msec delay in video transmission. Ans: Related.

47


1. An IP packet, having a total of 1000 bytes, is transmitted in 100 µsec. There are 24 bytes of header in one IP packet.

a. What is the rate of this IP transmission? Ans: IP Packet has 1000 bytes x 8 bits / byte = 8000 bits which is sent in 100 µsec., i.e., 8000 x 104 = 80 Mbps b. Find the number of ATM-AAL1 cells needed to carry one IP packet.

Ans: There are 1000 total bytes (header + data) in one IP packet frame. In ATM-AAL1, there are 47 bytes of payload in one cell. Thus, the number of ATM-AAL1 cells required for carrying one IP packet is 1000 bytes / 47 bytes, i.e. > 21, i.e., 22 cells.

c. Find the total number of ATM-AAL1 bits needed to carry one IP packet.

Ans: 53 bytes / cell x 22 cells = 1166 bytes = 1166 bytes x 8 bits / byte = 9328 bits.

d. How many total header bits are used in this transmission?

Ans: For the first ATM-AAL1 cell, there are: 1 cell x 5 ATM header bytes / cell = 5 ATM header bytes 1 cell x 1 AAL1 header byte / cell = 1 AAL1 header byte 1 cell x 24 IP header bytes / cell = 24 IP header bytes For the remaining 21 ATM-AAL1 cells, there are: 21 cells x 5 ATM header bytes / cell = 105 ATM header bytes 21 cells x 1 AAL1 header bytes / cell = 21 AAL1 header bytes

No IP header bytes So, the total header bits used in this transmission, i.e., in 22 ATM-AAL1 cells is

(5 + 1 + 24 + 105 + 21) x 8 bits / byte = 156 header bytes x 8 bits / byte = 1248 header bits.

e. Repeat part d if the IP packet has a total of 1000 bytes but with 32 bytes of header in one IP packet?

Ans: The only change will be in the first ATM-AAL1 cell in which there are: 1 cell x 5 ATM header bytes / cell = 5 ATM header bytes 1 cell x 1 AAL1 header byte / cell = 1 AAL1 header byte 1 cell x 32 IP header bytes / cell = 32 IP header bytes For the remaining 21 ATM-AAL1 cells, there are the same number of header bytes as in part d,

So, the total header bits used in this transmission, i.e., in 22 ATM-AAL1 cells is (5 + 1 + 32 + 105 + 21) x 8 bits / byte = 164 header bytes x 8 bits / byte = 1312 header bits.

2. An IP packet has a header of 192 bits and a payload of 4416 bits. Options part in the

header is one byte and has the sequence 01101001. It is transmitted from a PC with IP address 212.20.202.16 to another PC with IP address 136.33.125.46. It is the first

48

fragment, will fragment again and the packet reached a stage in the network that it must be destroyed. In your answers below, each wrong bit will be graded as -1.

a. Fill in the bits in the first 2 rows of the IP packet shown below:

Ans:

b. Fill in the bits in rows 3, 4 and 5 of the IP packet shown below:

Ans:

c. Fill in the bits in rows 6 and 7 of the IP packet shown below:

49

Ans:

d. Write the first 4 bytes of the header of the protocol which is encapsulated by this IP

packet. Ans:

e. If a 24 bytes header TCP carrying HTTP is transmitted with this IP, find the number

of HTTP bits carried by this IP packet. Note that the total HTTP size is larger than 600 bytes.

Ans: IP payload size is 576-24 = 552 bytes 24 bytes is TCP header so the number of HTTP bits carried by this IP packet = (552 – 24) x 8 = 4224 bits

3. Chip codes in a 3 user CDMA system are given below. (1)

c , (2)

c and (3)

c are the chip sequences for users 1, 2 and 3, respectively. All the chip sequences are periodic with period of 16 µsec. Information rate is 50 kbit/sec.

50

a. What is the rate of the modulated signal in this CDMA system? Ans: Spreading factor = Tb / Tc Tb = 1 / (50 kbit/sec) = 0.2 x 10 -4 sec Tc = 1 microsec = 1 x 10 -6 sec Spreading factor = 0.2 x 10 -4 sec / 1 x 10 -6 sec = 20 Thus, the rate of the modulated signal = 20 x 50 kbit/sec = 1 Mbits/sec b. If the carrier in part a is 1 GHz, how many cycles of carrier will be present in one

bit of the modulated signal? Solution: The duration of one bit of the modulated signal is 1 µsec. There are 109

cycles of carrier in one second. Thus, the number of cycles of carrier that will be present in one bit of the modulated signal is 109 x 10-6 = 103 = 1000

c. If the amplitude of the information bit symbol is 1 (representing 1), plot the spread

signals ( ) ( ) ( )( )1 2 3

. ., , ,i e m m m from t=0 to t=20 µsec for all the 3 users.

Solution:

51

d. If the amplitude of the information bit symbol is – 1 (representing 0), plot the

spread signals ( ) ( ) ( )( )1 2 3

. ., , ,i e m m m from t=0 to t=20 µsec for all the 3 users.

Solution:

52

e. Why can’t you use the same chip sequences for different users in CDMA? Your answer should be one sentence. Ans: We can’t you use the same chip sequences for different users in CDMA because otherwise we can not differentiate the different users information signal

53

in a multi user mobile system. 4. a. Are the following comments correct or wrong?

i. IP payload can be encapsulated in ATM payload. Ans: Correct ii. IP header can be encapsulated in ATM payload. Ans: Correct iii. IP header can be encapsulated in ATM header. Ans: Wrong iv. IP payload can be encapsulated in ATM header. Ans: Wrong v. HTTP header can be encapsulated in ATM payload.

Ans: Correct b. In each of the following items (i, ii, iii, iv, v) 5 concepts are named. For each item, write the name of the concept which is unrelated to the other 4 concepts.

i. CBR, VBR, AAL, EIR, UBR. Ans: EIR ii. LAN, GRE, VPN, PPTP, E-hierarchy. Ans: E-hierarchy iii. DWDM, X.25, Optical Fiber, FSO, Holographic Switch. Ans: X.25.

iv. Delay, Tunneling, Jitter, Packet Rate, Packet Loss. Ans: Tunneling v. Frame Relay, ATM, PSTN, X.25, IP. Ans: PSTN

c. Write one Quality of Service (QoS) item for the following applications:

i. One telephone channel in PSTN. Ans: 64 kbps ii. Video transmission in HDTV format. Ans: 60 fps iii. Compressed video transmission. Ans: Packet loss ≤ 10-11.

iv. Audio transmission. Ans: Sample size 8-bit or sample rate 8 kHz, intermediate delay 125 µs, playback point ~100 to 150 ms.

v. Video transmission in PAL format. Ans: 25 fps d. Write the name of the application, system, concept or network whose descriptions are given below:

i. Digitized voice carried in IP. Ans: Voice over IP. ii. Inserting one packet into the payload of another packet. Ans: Encapsulation. iii. Technical specifications for an application of telecommunications. Ans: Quality of service. iv. Establishing LAN in remote sites. Ans: VPN. v. Multiple access mobile communication in which different time slots are

assigned for each user. Ans: TDMA. e. Write the most important three points which makes VPN possible. What is achieved by VPN? Ans: Autentication, encryption, tunneling. Establishing LAN in remote sites.

54


1. a. In the PSTN network in which you carry classical telephone voice signal, i. What is the maximum frequency taken for one channel voice signal? Answer: 4

kHz. ii. What is the sampling rate applied? Answer: 8000 samples / sec. iii. How many quantization levels are used? Answer: 256 level. iv. How many bits are used to to represent one sample value? 8 bits/sample. v. What is the rate of one voice channel? Answer: 8000 samples / sec x 8

bits/sample = 64 kbps. b. The statement below describes which one among VPN, PSTN, FSO, Q0S, PAL, Next Generation Network, FIFO, the best? i. Extends LAN coverage by using WAN infrastructure. Answer: VPN ii. High rate access system. Answer: FSO iii. Voice infrastructure. Answer: PSTN iv. Technical specification in video transmission. Answer: PAL v. A method used to fulfil the technical specifications in an application. Answer: FIFO c. i. Write 5 different packet types. Ans: IP, ATM, Frame Relay, X.25, Ethernet

ii. Write 3 systems that are used for LAN transmission. Ans: Frame Relay, ATM, VPN iii. Write 5 of the all optical network components. Ans: Fiber, FSO, EDFA, DWDM, Optical Switch iv. Write 5 different rates used in E hierarchy. Ans: 2 Mbps, 8 Mbps, 34 Mbps, 140 Mbps, 565 Mbps v. Write 5 different protocols: Ans: IP, TCP, PPP, PPTP, L2TP

d. In each of the following items (i, ii, iii, iv, v) 5 concepts are named. For each item, write the name of the concept which is unrelated to the other 4 concepts.

i. L2TP, GRE, FTP, VPN, PPTP. Ans: FTP ii. VBR, UNI, ATM-AAL 2, CIR, CBR. Ans: CIR iii. Dispersion, Attenuation, PSK, BER, Bit Rate. Ans: PSK iv. 2G, GSM, FDMA, CDMA, TDMA. Ans: CDMA v. Next Generation Network, IP/ATM Networks, VPN, 3-D Holography, VOIP. Ans:

VPN e. i. Write the number of maximum total bits in one IP packet. Ans: 8 x 216 =524288

ii. Write 5 different types of optical exchanges. Ans: MEMS, Liquid Crystal, Bubble, Thermo-optical, Wavelength

iii. In compressed video transmission, is it acceptable if one packet out of 1 billion packets is lost? Ans: No, packet loss ≤ 10-11.

iv. What video transmission format does Q0S of 60 fps represent? Ans: HDTV v. What is digitized voice carried in IP called? Ans: Voice over IP

2. Assume that in a CDMA system there are 3 users. (1)

c , (2)

c and (3)

c are the chip sequences for users 1, 2 and 3, respectively. All the chip sequences are periodic with period of 20 µsec. Each information bit is multiplied by the 20 chip bit sequence. Information rate is 50 kbit/sec.

55

a. What is the rate of the CDMA signal? Answer: Tb = 1 / (50 kbit/sec) = 0.2 x 10 -4 sec Tc = 1 microsec = 1 x 10 -6 sec Tb / Tc = 0.2 x 10 -4 sec / 1 x 10 -6 sec = 20 Thus, the rate of the modulated signal = 20 x 50 kbit/sec = 1 Mbits/sec

b. Plot the CDMA signals ( ) ( ) ( )( )1 2 3

. ., , ,i e m m m from t=0 to t=20 µsec for all the 3

users, if the amplitude of the information bit symbol is 2 (representing 1).

Answer:

56

c. Plot the CDMA signals ( ) ( ) ( )( )1 2 3

. ., , ,i e m m m from t=0 to t=20 µsec for all the 3

users, if the amplitude of the information bit symbol is -2 (representing 0).

Answer:

57

d. Find the number of cycles of carrier that is present in one bit of the modulated

signal, if the carrier in part a is 2 GHz. Answer: The duration of one bit of the modulated signal is 1 µsec. There are 2 x

109 cycles of carrier in one second. Thus, the number of cycles of carrier that will be present in one bit of the modulated signal is 2 x 109 x 10-6 = 2 x 103 = 2000

e. Are the following statements for CDMA correct or wrong? i. CDMA is used in 2G and 3G. Answer: Wrong ii. Rate of the information signal is higher than the modulated signal. Answer:

Wrong iii. Rate of the chip signal is smaller than the modulated signal. Answer: Wrong iv. Chip signal is directly modulated. Answer: Wrong v. Pseudonoise is only one representation of chip codes, but in general chip

codes can be formed by arbitrary digital sequences. Answer: Wrong 3. The total bytes of an IP packet is 576. There are 32 bytes of header in one IP packet.

This IP packet is carried by ATM-AAL1 cells. Bit duration of this ATM cell is 4.24 x 10-8 sec.

a. How many ATM-AAL1 cells can carry one of such IP packet. Ans: There are 576 total bytes (header + data) in one IP packet frame. In ATM-AAL1, there are 47 bytes of payload in one cell. Thus, the number of ATM-AAL1 cells required for carrying one IP packet is 576 bytes / 47 bytes, i.e. > 12, i.e., 13 cells.

b. What is the rate of this ATM transmission? Ans: 1 bit is transmitted in 4.24 x 10-8 sec. Thus the rate of ATM is 1 / 4.24 x 10-8

sec = 23.58 Mbps

58

c. Is the rate of this IP packet smaller, the same or larger than the rate of this ATM cell? Explain. Ans: Smaller, because within the duration of one ATM cell, there are smaller number of IP bits than the ATM bits.

d. In the transmission of this one IP packet by the ATM-AAL 1 cells, find the number

of total (IP + ATM-AAL 1) header bytes. Ans: For the first ATM-AAL1 cell, there are: 1 cell x 5 ATM header bytes / cell = 5 ATM header bytes 1 cell x 1 AAL1 header byte / cell = 1 AAL1 header byte 1 cell x 32 IP header bytes / cell = 32 IP header bytes For the remaining 12 ATM-AAL1 cells, there are: 12 cells x 5 ATM header bytes / cell = 60 ATM header bytes 12 cells x 1 AAL1 header bytes / cell = 12 AAL1 header bytes

No IP header bytes So, the total header bytes used in this transmission, i.e., in 13 ATM-AAL1 cells is 5 + 1 + 32 + 60 + 12 = 110 header bytes

e. If this IP packet carries a TCP packet which has 24 bytes header, find the number of the total header bytes (IP + ATM-AAL 1 + TCP) existing in the first ATM-AAL 1 cell of this transmission.

Ans: For the first ATM-AAL1 cell, there are: 1 cell x 5 ATM header bytes / cell = 5 ATM header bytes 1 cell x 1 AAL1 header byte / cell = 1 AAL1 header byte 1 cell x 32 IP header bytes / cell = 32 IP header bytes 1 cell x 24 TCP header bytes / cell = 24 TCP header bytes

So, the total header bytes existing in the first ATM-AAL 1 cell of this Transmission is 5 + 1 + 32 + 24= 62 header bytes. However, the first ATM -AAL 1 cell has only 53 bytes, thus, the total header bytes existing in the first ATM-AAL 1 cell of this transmission is 53 bytes. 4. Header of an IP packet is 224 bits. Thus the payload is 4480 bits. Options part in the

header is 6 bytes. Padding is 2 bytes. The last 5 digits of the options part is all 1. IP is transmitted from a PC with IP address 118.19.152.22 to another PC with IP address 226.16.215.41. It is the first fragment, will fragment again and the packet reached a stage in the network that it must be destroyed. This IP carries a TCP packet which has a header of 28 bytes. In the payload of the TCP packet, HTTP is carried. The total HTTP size is larger than 10000 bits. The first 10 bits of the TCP header carried by this IP packet are 1100110011. Note that, in your answers below, each wrong bit will be graded as -1.

a. Fill in the bits in the first 2 rows of the IP packet shown below:

59

Ans:

b. Fill in the bits in rows 3, 4 and 5 of the IP packet shown below:

Ans:

c. Fill in the bits in rows 6 and 7 of the IP packet shown below:

Ans:

60

d. Fill in the bits in row 8 of the IP packet shown below:

Ans:

e. Find the number of HTTP bits carried by this IP packet. Ans: IP payload size is 588-28 = 560 bytes 28 bytes is TCP header so the number of HTTP bits carried by this IP packet = (560 – 28) x 8 = 4256 bits

61


1. In an IP packet there are total of 1088 bytes, out of which 36 bytes are IP header. ATM-AAL1 cells that carry these IP packets have a rate of 100 Mbps. In the IP packet, TCP packet with a header of 24 bytes is carried.

a. Find the duration of one bit of ATM-AAL1 cell. Ans: ATM-AAL1 rate is 100 Mbps. Thus, the duration of one bit of ATM-AAL1 cell

is 1 bit / 100 Mbps = 1 bit / 108 bps =10-8 s = 10 nsec b. Find the number of ATM-AAL1 cells needed to carry one of this IP packet.

Ans: There are 1088 total bytes (header + data) in one IP packet frame. In ATM-AAL1, there are 47 bytes of payload in one cell. Thus, the number of ATM-AAL1 cells needed for carrying one IP packet is 1088 bytes / 47 bytes, i.e. > 23, i.e., 24 cells.

c. How many total header bytes (ATM-AAL 1 + IP + TCP) are there in the first ATM- AAL 1 cell?

Ans: In the first ATM-AAL1 cell, there are: 5 ATM header bytes 1 AAL1 header byte 36 IP header bytes 24 TCP header bytes, i.e., total of 5+1+36+24 = 66 header bytes

However, the first ATM-AAL 1 cell has only 53 bytes. Thus, the total header bytes existing in the first ATM-AAL 1 cell is 53 bytes. d. If the TCP carries http, find the number of http bits in the first ATM- AAL 1 cell.

Ans: As found in part c, first ATM- AAL 1 cell is fully composed of header bits. Thus, the number of http bits in in the first ATM- AAL 1 cell is zero.

e. If the TCP carries http, find the number of http bits in the second ATM- AAL 1 cell.

Ans: In part c, 66 header bytes are found for the first ATM- AAL 1 cell. However, only 53 of these header bytes are used in the first ATM- AAL 1 cell. Thus for the second ATM AAL 1, we have 66-53=13 header bytes. Also from the second ATM- AAL 1 cell, we have 5 ATM header bytes + 1 AAL1 header byte = 6 header bytes more so in the second ATM AAL 1, we have a total of 13 + 6 = 19 header bytes. Thus we are left with 53 – 19 = 34 payload bytes carrying http = 34 x 8 = 272 http bits.

2. An IP header is given below. Padding is 1 byte.

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a. Find the number of “Payload” bytes.

Ans: From the first row, bits 16-31 represent the total length which is

, i.e.,

which is 588 bytes. Header is 7 x 4 = 28 bytes, so the number of payload bytes = 588 – 28 = 560 bytes. b. Find the number of “Options” bytes.

Ans: “Options” field starts from the sixth row. Since padding is 1 byte, number of

“Options” bytes is 2 x 4 - 1 = 7 bytes. c. Write the “Source IP Address” in dotted decimal notation.

Ans: The fourth row represents the Source IP Address, which is

Dotted decimal notation form of this row is 118.19.152.22. d. Write the “Destination IP Address” in dotted decimal notation.

Ans: The fifth row represents the Destination IP Address, which is

Dotted decimal notation form of this row is 226.16.215.41. e. Can this IP packet proceed to the next node? Why?

Ans: This IP packet can not proceed to the next node because the “Time To Live”

field which is the bits 0-7 are all zero which means Time To Live = 0. If this field contains the value zero, then the packet must be destroyed.

3. a. In each of the following items (i, ii, iii, iv, v) 5 concepts are named. For each item, write the name of the concept which is unrelated to the other 4 concepts.

i. Dial up, Leased Line, PVC, SS7, SVC. Ans: SS7 ii. IP, PPTP, PPP, L2TP, AIN. Ans: AIN iii. CIR, GRE, L2TP, PPTP, VPN. Ans: CIR iv. Spread Spectrum, TDMA, ATM, FDMA, CDMA. Ans: ATM v. ATM-AAL 1, VBR, PPP, NNI, CBR. Ans: PPP

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b. Write 5 different types of i. QoS parameters. Ans: Sampling rate, frames per second, quantization level, rate, delay

ii. Protocols. Ans: IP, TCP, PPP, PPTP, L2TP iii. Packets. Ans: IP, ATM, Frame Relay, X.25, Ethernet iv. Rates used in E hierarchy. Ans: 2 Mbps, 8 Mbps, 34 Mbps, 140 Mbps, 565 Mbps v. All optical network components. Ans: Fiber, FSO, EDFA, DWDM, Optical

Switch c. Write i. The highest rate core medium. Answer: Fiber optics. ii. The name of the classical public telephone infrastructure. Answer: PSTN iii. The system which allows a remote user to become a user of a local LAN. Answer: VPN iv. One QoS for video application. Answer: PAL v. An access network system. Answer: FSO d. Are the following comments correct or wrong?

i. IP is in the payload of http. Ans: Wrong ii. TCP header is in the payload of IP. Ans: Correct iii. TCP is in the payload of http. Ans: Wrong iv. IP header is in the payload of PPP. Ans: Correct v. IP header is in the payload of TCP. Ans: Wrong

e. In general, which one indicates better QoS?

i. 90 Mbps video or 140 Mbps video. Ans: 140 Mbps video ii. 10 msec delay or 20 msec delay. Ans: 10 msec delay iii. 256 levels of quantization or 512 levels of quantization. Ans: 512 levels of

quantization iv. Transmission of the signal in compressed or uncompressed form. Ans: Uncompressed form v. CBR or VBR. Ans: CBR

4. In a CDMA system, assume that the below given (1)

c , (2)

c and (3)

c are the chip sequences belonging to users 1, 2 and 3, respectively. Each chip bit is 1 µsec long and the period of all the chip sequences are 40 µsec. One information bit is multiplied by the 40 chip bit sequence. Amplitude of the information bit is -5 representing level 0 and +5 representing level 1. Information rate is 25 kbit/sec. The carrier frequency is 1 GHz.

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a. Find the Processing Gain (or the Spreading Factor). Answer: Since one information bit is multiplied by the 40 chip bits, the Processing Gain (or the Spreading Factor) is 40. b. Find the rate of the CDMA signal. Answer: Information rate is 25 kbit/sec. Processing Gain (or the Spreading Factor) is 40. Thus, the rate of the CDMA signal is 25 kbit/sec x 40 = 1 Mbit/sec. OR: Chip bit is 1 µsec long, so the rate of the CDMA signal is 1 / 1 µsec = 1 Mbit/sec

c. For all the 3 users, draw ( ) ( ) ( )1 2 3

, ,m m m (i.e., CDMA signals) from t=0 to t=40 µsec,

if level 0 is transmitted.

Answer:

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d. For all the 3 users, draw ( ) ( ) ( )1 2 3

, ,m m m (i.e., CDMA signals) from t=0 to t=40 µsec,

if level 1 is transmitted..

Answer:

e. How many cycles of carrier exist in one bit of the modulated signal? Answer: The duration of one bit of the modulated signal is 1 µsec. Since the

carrier frequency is 1 GHz, there are 1 x 109 cycles of carrier in one second. Thus, the number of cycles of carrier that will be present in one bit of the modulated signal is 1 x 109 x 1 x 10-6 = 1 x 103 = 1000.

ANSWERS FOR MISCELLANEOUS HOMEWORK-1 374...1 ANSWERS FOR MISCELLANEOUS HOMEWORK-1 1. There are two...

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