Flow Control MCQ Quiz in తెలుగు - Objective Question with Answer for Flow Control - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

For any sliding window protocol,

Available Sequence Numbers ≥  Sender window size + Receiver window size

In Selective repeat Protocol, sender  window size = receiver window size = W

Calculation:

With 16 bits, the total number of sequence numbers possible = 2¹⁶.

Available Sequence Numbers ≥  Sender window size + Receiver window size

2¹⁶ ≥  W + W

∴ W ≤ \(\frac{2^{16}}{2}\) 

W_max = 2¹⁵ = 32768

The maximum window size for the data transmission is 32768.

Bandwidth (B) = 1 kilobytes/sec = \(8 \times10^3\)bits per sec

No. of bytes in the information frame (L) = 5000 bytes = 5000 \(\times\)8 bits

Propagation Delay (\(Tp\)) = 0.08 sec

Processing time (\(Tproc\)) = 0.04 sec

Efficiency = ɳ = 80% = 0.8

Transmission time (\(Tx\)) =\(\frac{L}{B}\)= \(\frac{40000}{8000}\)= 5 sec

Transmission time for acknowledge (\(Tack\)) = \(\frac{Lack}{B}\) = \(\frac{Lack}{8\times10^3}\)

Efficiency(ɳ) = \(\frac{Tx}{Tx+Tack+2Tp+Tproc}\)

0.8 = \(\frac{5}{5+\frac{Lack}{8\times10^3}+2\times0.08+0.04}\)

\(Lack\) = 1.05 \(\times\)8\(\times\)\(10^3\) = 1050 ​\(\times\)8 bits = 1050 bytes

Concept:

The ‘Ping’ command is a Command Prompt command used to test the ability of the source computer to reach a specified destination computer. Therefore, the uses of 'Ping' command is to test a device on the network is reachable.

Important Point

Ping - t means will ping the target until you force it to stop

ping - t 127.0.0.1

A correct answer is an option (3)

Bandwidth

Concept:-

The amount of data that may be transported from one point to another inside a network in a given amount of time is referred to as bandwidth.
Key Points

• The term bandwidth refers to a connection's transmission capacity and is an important aspect in evaluating a network's or internet connection's quality and speed.
• Typically, bandwidth is measured in bits per second and expressed as a bitrate (bps).
• The movement of data from one digital device to another is known as data transmission. This is accomplished through the use of point-to-point data streams or channels.
• The amount of disc space provided by one or more storage devices is referred to as storage capacity. It is a measurement of how much data a computer system can hold.

The correct answer is option 3.

Concept:

To determine how many characters per second can be transmitted over a 2600 b/s (bits per second) line in synchronous mode with 7 bits per character plus parity, and considering one start and one stop bit, we need to calculate the total bits per character and then find how many such characters fit into the 2600 b/s bandwidth.

In synchronous transmission, the concept of start and stop bits is typically not used in the same way as it is in asynchronous transmission. Instead, synchronous transmission relies on a continuous stream of data, where timing signals are used to indicate the beginning and end of each character or frame.

Steps for Calculation:

1. Bits per Character:

•     7 bits for the character
•     1 parity bit
•     1 start bit
•     1 stop bit

   Therefore, total bits per character = 7 + 1 = 8 bits (synchronous transmission, start and stop bits are not added )

2. Characters per Second:
    Given the line speed is 2600 b/s, we divide the total bit rate by the bits per character:
\( \text{Characters per second} = \frac{\text{Total bits per second}}{\text{Bits per character}}\)   
 
\( = \frac{2600 \text{ b/s}}{8 \text{ bits/character}} = 325 \text{ characters/second}\)
   

Therefore, the correct answer is: 2) 325

Data:

Propagation speed of the signal = 2 x 10⁸ m/s

Token holding time = 2 μ sec

Token ring network length = 2 Km

Concept:

Minimum waiting time = Prorogation delay or Ring latency + Token holding time for every station

Minimum waiting time = T_p + n xTHT

Calculation:

\({{\rm{T}}_{\rm{P}}} = \frac{{{\rm{Length}}}}{{{\rm{Speed}}}} = \frac{{2 \times {{10}^3}}}{{2 \times {{10}^8}}} = {10^{ - 5}}{\rm{\;}}\frac{{10}}{{10}} = 10{\rm{\;\mu \;sec\;}}\) (1 μ sec = 10^-6 sec )

Minimum waiting time = 10 + 10 x 2 = 30 μ sec              

Note: 

28 μ sec is also possible answer because suppose station 1 is holding token then after releasing it only 9 station will hold the token for 2 μ sec to get back to station 1 or token is to lost so for waiting time for a station 1 is only 18 μ sec + ring latency = 28 μ sec   

Data:

Length of packet = L = 2 × 106 bytes = 16 × 106 bits

Bandwidth (BW) = 4 × 107 b/s

Efficiency = η = ½

Formula:

\({\rm{\eta }} = \frac{1}{{1 + \frac{{2{T_p}}}{{{T_t}}}}}\)

\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)

Calculation:

\({{\rm{T}}_{\rm{t}}} = \frac{{16 \times {{10}^6}}}{{4 \times {{10}^7}}} = 0.4\;s = 400\;ms\)

Since efficiency is maximum

\(\frac{1}{2} = \frac{1}{{1 + \frac{{2{T_p}}}{{400}}}}\)

\(2 = 1 + \frac{{2{T_p}}}{{400}}\)

Data:

packet size = 1000 bytes

Transmission speed = 10⁶ bits/sec

Efficiency = 25 %

Formula:

\({\rm{Efficiency}} = \frac{{{\rm{transmission\;time}}}}{{{\rm{RTT}} + {\rm{transmissition\;time}} + {\rm{acknowledgement\;time}}}}\)

(if acknowledgment time is negligible), then

\({\rm{Efficiency}} = \frac{{{\rm{transmission\;time}}}}{{{\rm{RTT}} + {\rm{transmissition\;time}}}}\)

\({\rm{transmission\;time}} = \frac{{{\rm{packet\;size}}}}{{{\rm{transmissition\;speed}}}}\)

Calculation:

\({\rm{transmission\;time}} = {\rm{\;}}\frac{{1000 \times 8}}{{{{10}^6}}} = 8\;ms\)

0.25 = \(\frac{8}{{8 + 2 \times {\rm{propagation\;delay}}}}\)

2 + 0.5 (propagation delay) = 8

0.5 (propagation delay) = 6

Propagation delay = 12 milliseconds

The value of the one-way propagation delay is 12 ms.

Note:

RTT → round trip time or two-way propagation delay

BW = 8 kbps = 10³ B

T_p = 30 ms = 30 × 10^-3 s

Efficiency of stop and wait protocol:

\(\eta = \frac{{{T_t}}}{{{T_t}\; + \;2 \times {T_p}}}\)

\(\eta = \frac{1}{{1\; + 2\;a}}\)

\(\eta \ge \frac{{60}}{{100}}\)

\( \frac{1}{{1 + 2 \times a}} \ge0.6\;\)

\(a = \frac{{{T_p}}}{{{T_t}}} = 30\times 10^{-3} \times \frac{{BW}}{L}\)

\(1 + 2a \le \frac{5}{3}\)

\(a \le \frac{1}{3}\)

\(30 \times {10^{ - 3}} \times \frac{{BW}}{L} \le \frac{1}{3}\)

\(30 \times {10^{ - 3}} \times \frac{{{{10}^3}}}{L} \le \frac{1}{3}\)

\(\frac{{30}}{L} \le \frac{1}{3}\)

Concept:

A sliding window protocol is a feature of packet-based data transmission protocols.

Sliding window protocols are used where reliable in-order delivery of packets is required.

Go-Back-N is the sliding window protocol. In this maximum window size is possible

Data

In GBN,

Number of bits = n = 4

Formula:

window size of sender + window size of receiver = Total size

∴ Ws +Wr   =  2n 

Calculation

Wr = 1

Ws= 2n – 1

∴ Ws= 24 – 1 = 15

Maximum window size possible is 15