Class: TE Computer Academic Year: 2018-19 SEM:- I Subject

1

Q1) Match the following to one or more layers of the OSI model.[3M]

Ans:

i. Transmission of bit stream across physical medium- Physical Layer

ii. Defines frames- Data Link Layer

iii. Error Correction & retransmission- Data Link Layer & Transport Layer

iv. Reliable process-to-process delivery- Transport layer

v. Route selection – Network Layer

vi. Provides user services such as e-mail and file transfer- Application Layer

Q2) Define FHSS and Explain how it achieves bandwidth spreading [5M]

Ans:

Frequency Hopping Spread Spectrum (FHSS):

Signal is broadcast over seemingly random series of radio frequencies

Signal hops from frequency to frequency at fixed intervals

Receiver, hopping between frequencies in synchronization with transmitter, picks up message

Fig: FHSS

Working:

Frequency hopping spread spectrum is a Spread Spectrum modulation technique in which the data

signal is transmitted over different frequencies. The data signal is hopped from one frequency to

another frequency for a specific interval of time.

Class: TE Computer Academic Year: 2018-19 SEM:- I

Subject : Computer Networks Exam: SPPU Aug INSEM Marks:30Date: 11/8/2018 Paper Solution

2

Pseudo random code generator generates a random number every time which is used to pick a

frequency from frequency table.

Then this frequency is given to Frequency synthesizer.

Modulator modulates original signal using the selected frequency which results to spread signal.

Advantages

Efficient utilization of available bandwidth

Eavesdropper hear only unintelligible blips

Attempts to jam signal on one frequency succeed only at knocking out a few bits

Q1 c) What is three major classes of guided media [2M]

Ans

Three major classes of guided media are as below.

Q 2a) what is difference between a port address, a logical address and a physical address [2M]

Ans:

Logical Address: An IP address of the system is called logical address. This address is the combination of Net ID and Host ID. This address is used by network layer to identify a particular network (source to destination) among the networks. This address can be changed by changing the host position on the network. So it is called logical address.

Guided Media

Twisted Pair Cable

STP

UTP

Coaxial Cable

Fiber Optic Cable

Class: TE Computers Academic Year: 2018-19 SEM:- I



3

Physical address: Each system having a NIC(Network Interface Card) through which two systems physically connected with each other with cables. The address of the NIC is called Physical address or mac address. This is specified by the manufacturer company of the card. This address is used by data link layer.

Port Address: There are many application running on the computer. Each application run with a port no.(logically) on the computer. This port no. for application is decided by the Kernal of the OS. This port no. is called port address.

Q2b)For the bit sequence 10000101111 draw the waveform for [4M]

i. Manchester Encoding

ii. Differential Manchester encoding

Ans: Manchester Encoding

1 0 0 0 0 1 0 1 1 1 1

i. Differential Manchester encoding

1 0 0 0 0 1 0 1 1 1 1

Subject : Computer Networks Exam: SPPU Aug INSEM Marks:30ate: 11/8/2018 Paper Solution

4

Q2 c) Explain in short, various networking devices Bridge,Switch,Router and Access Point[4M]

Ans:

Network Devices

Working

Routers Works at Network Layer They process logical addressing information in the Network header of a packet such as IP Addresses.

Functionality: To connect different network segments. To connect different network protocols such as IP and IPX. To Provide shortest path between sender and receiver.

Bridges Works at Data Link Layer

Functionality: A bridge is a computer networking device that builds the connection with

the other bridge networks which use the same protocol. It works at the Data Link layer of the OSI Model and connects the different

networks together and develops communication between them. It connects two local-area networks; two physical LANs into larger logical

LAN or two segments of the same LAN that use the same protocol. Types of Bridges:

1. Transparent Bridge 2. Source Route Bridge 3. Translational Bridge

Switches Works at Data Link Layer Functionality:

Switches are connected to them through twisted pair cabling. Switch transfers data only to that port which is connected to the

destination device.

Types of Switches: Cut-through transmission: It allows the packets to be forwarded as soon

as they are received. Store and forward: In this switching environment the entire packet are

received and ‘checked’ before being forwarded ahead. Fragment Free: In a fragment free switching environment, a greater part

of the packet is examined so that the switch can determine whether the packet has been caught up in a collision.

Access Point Functionality: Wireless access point (WAP), or more generally just access point (AP), is a

networking hardware device that allows a Wi-Fi device to connect to a wired network.

The AP usually connects to a router (via a wired network) as a standalone device, but it can also be an integral component of the router itself.

An AP is differentiated from a hotspot, which is the physical location where Wi-Fi access to a WLAN is available.




5

Q3 a) Compare & contrast the Go-back-N ARQ protocol with Selective Repeat ARQ. [4M]

Ans:

Basis for Comparison

Go-Back-N Selective Repeat

Basic Retransmits all the frames that sent after the frame which suspects to be damaged or lost.

Retransmits only those frames that are suspected to lost or damaged.

Bandwidth Utilization

If error rate is high, it wastes a lot of bandwidth.

Comparatively less bandwidth is wasted in retransmitting.

Complexity Less complicated. More complex Window size N-1 <= (N+1)/2

Sorting Sorting is not required Required Storing Receiver do not store the frames received

after the damaged frame until the damaged frame is retransmitted.

Receiver stores the frames received after the damaged frame in the buffer until the damaged frame is replaced.

ACK Numbers NAK number refer to the next expected frame number.

NAK number refer to the frame lost.

Q 3b) Explain PPP in detail. [4M]

Ans: PPP provides several services:

PPP defines the format of the frame to be exchanged between devices.

PPP defines how two devices can negotiate the establishment of the link and the exchange of data.

PPP defines how network layer data are encapsulated in the data link frame.

PPP defines how two devices can authenticate each other.

PPP provides multiple network layer services supporting a variety of network layer protocols.

PPP provides connections over multiple links.

PPP provides network address configuration. This is particularly useful when a home user needs a

temporary network address to connect to the Internet.

Framing

PPP is a byte-oriented protocol. Framing is done according to the discussion of byte-oriented protocols at

the beginning of this semester.

Frame Format

Fig: PPP Frame Format



6

Flag: A PPP frame starts and ends with a 1-byte flag with the bit pattern 01111110. Although this

pattern is the same as that used in HDLC, there is a big difference. PPP is a byte-oriented protocol;

HDLC is a bit-oriented protocol. The flag is treated as a byte, as we will explain later.

Address: The address field in this protocol is a constant value and set to 11111111 (broadcast

address). During negotiation, the two parties may agree to omit this byte.

Control: This field is set to the constant value 11000000 (imitating unnumbered frames in HDLC).

As we will discuss later, PPP does not provide any flow control.

Protocol: The protocol field defines what is being carried in the data field: either user data or other

information. This field is by default 2 bytes long, but the two parties can agree to use only 1 byte.

Payload field: This field carries either the user data or other information.

FCS: The frame check sequence (FCS) is simply a 2-byte or 4-byte.

Q3c) what are various design issues of data link layer. [2]

Ans:

Various design issues of data link layer are:

1. Services Provided to the Network Layer : A well-defined serve interface in the network layer. The principle service is transferring data from the network layer on source machine to the network layer on destination machine.

2. Frameing : The source machine send data in blocks called frames to be the destination machine. The starting and ending of each frame should be recognized by the destination machine.

3. Flow Control : The source machine must not be send data frames at a rate faster than the destination machines must be can accepted them.

4. Error Control : The errors mode in bits during transmission from source to destination machines must be detected and corrected.

5. Addressing: On a multipoint line, such as many machine connected together (LAN), the identity of the individual machine must be specified while transmitting the data frames.

Q4a) Explain the working of CRC using the following. Example (Show the complete steps of

division)[4M]

Data bit: 1101110110

Generator Polynomial: x3+x+1

Write the redundant bits that will be sent along with the data bits.Suppose the 2nd bit from the left is

inverted during transmission. Show that error is detected at the receiver’s end.

Ans:

Data word-1101110110 Divisor: x3+x+1-> 1011



7

Sender Side

Augmented bit: no of divisor bits-1= 4-1=3. Append 3 zeros at the end of data word.

______________________________________________________

1011) 1101110110000

1011 ______________ 1101 1011 _____________ 1101

1011

_______________

1100

1011

_______________ 1111

1011

______________ 1001

1011

_____________ 0100

0000

_____________ 1000

1011

___________ 0110

0000

_______________ 1100

1011

_____________ 111 Reminder

Code word= Data word + Remainder(Redundant bits)

Code word= 1101110110111



8

Receiver Side

Codeword received with 2nd bit from left is inverted= 1001110110111 If remainder is non-zero it indicated there is error ______________________________________________________

1011) 1001110110111

1011 ______________ 0101 0000 _____________ 1011

1011

_______________

0000

0000

_______________ 0001

0000

______________ 0011

0000

_____________ 0110

0000

_____________ 1101

1011

___________ 1101

1011

_______________ 1101

1011

_____________ 110 Reminder

As reminder is non-zero it indicates there are errors in received data Q 4b) Explain control field of HDLC w.r.t. I-frame, U-frame and S-frame. [6M] Ans:

There are three frame formats in HDLC: I-frame,S-frame and U-frame. Their control fields differs with each

other.



9

Control field format for the different frame types

Control Field for I-Frames:

I-frames are designed to carry user data from the network layer. In addition, they can include flow and error

control information (piggybacking). The subfields in the control field are used to define these functions.

1. The first bit defines the type. If the first bit of the control field is 0, this means the frame is an I-frame.

2. The next 3 bits, called N(S), define the sequence number of the frame. Note that with 3 bits, we can

define a sequence number between 0 and 7

3. The last 3 bits, called N(R), correspond to the acknowledgment number when piggybacking is used.

4. The single bit between N(S) and N(R) is called the P/F bit. The P/F field is a single bit with a dual

purpose. It has meaning only when it is set (bit = 1) and can mean poll or final. It means poll when the

frame is sent by a primary station to a secondary (when the address field contains the address of the

receiver). It means final when the frame is sent by a secondary to a primary (when the address field

contains the address of the sender).

Control Field for S-Frames:

Supervisory frames are used for flow and error control .

1. If the first 2 bits of the control field is 10, this means the frame is an S-frame.

2. The last 3 bits, called N(R), corresponds to the acknowledgment number (ACK) or negative

acknowledgment number (NAK) depending on the type of S-frame.

3. The 2 bits called code is used to define the type of S-frame itself. With 2 bits, we can have four

types of S-frames, as described below:

Receive ready (RR),. Receive not ready (RNR),. Reject (REJ), Selective reject (SREJ)

Control Field for U-Frames:

Unnumbered frames are used to exchange session management and control information between

connected devices. Unlike S-frames, U-frames contain an information field, but one used for system

management information, not user data. U-frame codes are divided into two sections: a 2-bit prefix before



10

the P/F bit and a 3-bit suffix after the P/F bit. Together, these two segments (5 bits) can be used to create up

to 32 different types of U-frames.

Q5 a) Measurement of a slotted ALOHA channel with an infinite number of users, show that 10 percent of the slots are idle: [5M]

i. What is Channel load? ii. What is throughput iii. Is the channel underload or overload?

Ans:

Measurement of a slotted ALOHA channel with an infinite number of users shown that 10 percent of

the slots are idle.

(a) What is the channel load, G?

Ans: When a slot is idle, there is 0 frame generated in that frame time. Therefore Pr[0]=0.1.

Pr[0]=e-G=0.1; -G=ln(0.1); G=2.303.

(b) What is the throughput?

Ans: S=G*e-G=2.303*0.1=0.2303.

(c) Is the channel underloaded or overloaded?

Ans: When G=1, the slotted Aloha obtains the optimal throughput. G>1, we have too many generated

frame in a slot. It is an overloaded situation. Here we have G=2.303; S=0.2303<Smax=0.368. G>S.

Therefore the channel is overloaded.

Q5 b) Explain 802.3 MAC frame format. [3M]

Ans:

1. Preamble – 8 bytes

Start of Frame Delimiter (802.3).



11

2. Two addresses each 6 bytes – destination + source

First bit of the destination address is 0 for ordinary addresses and 1 for group addresses.

3. Type or Length field.

Depending whether the frame is Ethernet or IEEE 802.3

Ethernet uses a Type field to tell the receiver what to do with the frame.

4. Data Field

Up to 1500 bytes.

Q5 c) Draw flowchart for CSMA/CA [2M]

Ans:



12

Q6a) Explain 802.11 wireless LAN frame format [4M]

Ans:

Protocol Version: zero for 802.11 standard

Type= frame type: data, management, control . Subtype = frame sub-type:

ToDS: When bit is set indicate that destination frame is for DS

FromDS: When bit is set indicate frame coming from DS

Retry: Set in case of retransmission frame

More fragments: Set when frame is followed by other fragment

Power Management bit set when station go Power Save mode (PS)

More Data: When set means that AP have more buffered data for a station in Power Save mode

WEP:When set indicate that in the Frame Body field there are datas need to processed by WEP

algorithm.

Order:When set indicate restrictions for transmission

Q6 b) A Slotted ALOHA network transmits 200-bit frames using a shared channel with a 200-kbps

bandwidth. Find the throughput if the system(all stations together) produces[4M]

i. 1000 frames per second

ii. 500 frames per second

Ans:

The throughput for pure ALOHA is S = G x e-2G

Transmission time of frame = 200 bits / 200 kbps = 1ms



13

i. Now, system as taken altogether produces 1000 frames per sec i.e , 1 frame per ms which is

double the load factor of transmission. Now, as load factor G = 1. In this case S = G x e-G or S

=0.368(36.8percent).This means that the throughput is 1000 × 0.368 = 368 frames.

ii. Now, system as taken altogether produces 500 frames per sec i.e , 1/2 frame per ms which

is load factor of transmission. Now, as load factor G = 1/2 , so S= G x e-G =0.303(30.3

%) .This means that the throughput is 500 × 0.303 = 151 frames.

Q6c) What are various common Fast Ethernet implementations?[2M]

Ans:


Documents

Class: TE Computer Academic Year: 2018-19 SEM:- I Subject