Solution

i. True - Bandwidth is high then stop and wait is useless. Because it can send 1 data out of 1+2a packet.

ii. True - Packet error is high then Go Back N is useless. Because on every side, there will be a lot of data retransmission again.

iii. False - In stop and wait, there is no negative acknowledgement.

iv. False - Go Back N, there is no negative acknowledgement. SR has only Negative acknowledgement.

Solution

Selective Reject (or Selective Repeat) protocol is one of the automatic repeat request (ARQ) techniques used for communications.

In SR, if X bits is maximum available sequence number then, the window size of both sender and receiver = $2^{X-1}$

In GBN, if X bits is the maximum available sequence number then sender window size = $2^X$ - 1 and receiver window size = 1.

Solution

From the questions, Transmission delay = Propagation delay

According to formula, Length(L) / B.W(B) = Distance (M)/ Velocity(V)

M = (L / B) x V = ( 1024 x 8 / 512 x 10^9 ) x 2 x 10^9 = 32 meter

Solution

Efficiency formula is 1/1+2a, a=Tp/Tt =>15 / 10 = 1.5

1/ (1 + 2*1.5) = 1/4 = 0.25

Solution

Stop-and-Wait ARQ is a special case of Go-Back-N ARQ in which the size of the sender window is 1.

Solution

Total number of transmission: 1 2 3 4 4 5 6 7 7 8 9 10 10 11 12= 15

Total number of Re-transmission: 3

Solution

Given, Tp = 1.25 sec,

Tt = L / B = 2 KB / 1 Mbps => 2 x 1024 x 8 / 106 = 0.016384 sec

No. of bits required in sequence number always equals to the sender window size(SWS)

SWS = 1+2a (a=Tp/Tt), a = 1.25/0.016384 =76.293

SWS = 1+2*76.293 = 153.58

By default sliding window means Go_Back N ARQ so here N = ⌈log₂(153.58 )⌉ = 8 bits

Solution

Efficiency of ethernet = 1/ 1+6.44a

And a= propagation delay (Tp) / Transmission delay (Tt)

Since a is in the denominator, its value should be lower in order to get high efficiency.

Solution

• Framing: Data-link layer takes packets from the Network Layer and encapsulates them into Frames. Then, it sends each frame bit-by-bit on the hardware. At the receiver end, the data link layer picks up signals from hardware and assembles them into frames.
• Addressing: Data-link layer provides layer-2 hardware addressing mechanism. The hardware address is assumed to be unique on the link. It is encoded into hardware at the time of manufacturing.
• Synchronization: When data frames are sent on the link, both machines must be synchronised in order to transfer to take place.
• Error Control: Sometimes signals may have encountered the problem in transition, and the bits are flipped. These errors are detected and attempted to recover actual data bits. It also provides error reporting mechanisms to the sender.
• Flow Control: Stations on the same link may have different speed or capacity. Data-link layer ensures flow control that enables both machines to exchange data at the same speed.
• Multi-Access: When a host on the shared link tries to transfer the data, it has a high probability of collision. Data-link layer provides a mechanism such as CSMA/CD to equip capability of accessing a shared media among multiple Systems.

Channel coding: It is the function of the physical layer. Data Link Layer is the layer where the packets are encapsulated into frames.

Solution

Given message M = 1010001011

Divisor x^5 + x^3 + x^2 +1 = 101101

Perform exclusive OR or MOD 2 division

Diagram

Solution

Delimiter flag: 01111

Data sequence: 0111110101

Given a flag of 5 bits so that we will stuff bit 0 in the data sequence with the pattern of 4 bits, i.e. 0111 matches. Therefore output data sequence: 01110110101

Solution

(A) One of the main functions of a Network Layer is Routing. So it is CORRECT.

(B) Bit Synchronization is always handled by the Physical Layer of the OSI model but not Data Link Layer. So it is INCORRECT.

(C) End–to–end Process Communication is handled by the Transport Layer. So it is CORRECT.

(D) MAC sublayer has three types of protocols (Random, Controlled and Channelized Access).

Solution

Signal propagation speed is 55% of the speed of light = (55/100) x 3 x10^8 m/s

= 1.65 × 10^8 m/sec.

Signal propagation time for 50 km is d / v = (50 × 10^3) m / (1.65 × 10^8 ) m/sec

= 30.303 × 10^-5sec = 303.03 μ sec

Contention slots are those slots which are not able to transmit their journey due to the collision. Suppose A station transmitted data but collided and worst-case time wasted is 2 x Tp.

So, The length of contention slot is 2 xTp = 2 x 303.03 μ sec = 606.06 μsec

Solution

Sender window size = 31

For Go-Back-n 2^n -1 = 31 => Therefore, n = 5 bits

For Selective repeat 2^(n-1) = 31 => Therefore, n = 6 bits

Solution

Tt = L / B.W. = 5000 / 4 x 10^3 = 5 / 4 sec

Tp = D / V = 20 x 10^3 / 200 = 100 sec

N = 127

Efficiency (η) = N / 1+2a { a = Tp / Tt = (100/5) x 4 = 80}

= 127 / (1+2 x 80) = 127 / 161= 0.7888 => 78.88%