Access-Control-Methods

Question 1

Let us consider a statistical time division multiplexing of packets. The number of sources is 10. In a time unit, a source transmits a packet of 1000 bits. The number of sources sending data for the first 20 time units is 6, 9, 3, 7, 2, 2, 2, 3, 4, 6, 1, 10, 7, 5, 8, 3, 6, 2, 9, 5 respectively. The output capacity of multiplexer is 5000 bits per time unit. Then the average number of backlogged of packets per time unit during the given period is

A
5
B
4.45
C
3.45
D
0
Computer-NetworksAccess-Control-MethodsGATE 2007-IT
Question 1 Explanation: 
The capacity of multiplexer is 5000 bits per time unit. This means there are 5 packets per unit time since each source transmits a packet of 1000 bits in a unit time.
If the no. of packets transmitted is larger than 5 then the extra packets are backlogged. This means gets added to the next number and further backlog is calculated.

Average no. of backlogged packets = 89/20 = 4.45
Question 2

A broadcast channel has 10 nodes and total capacity of 10 Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 80 μs to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 bytes. The maximum throughput of the broadcast channel is

A
1 Mbps
B
100/11 Mbps
C
10 Mbps
D
100 Mbps
Computer-NetworksAccess-Control-MethodsGATE 2007-IT
Question 2 Explanation: 
Efficiency = Transmission time(Tt)/Transmission time+Polling delay
(Tt) = 1000bytes/10×106bits/sec = 800μs
Polling delay = 80 μs
Efficiency = 800/800+80 = 800/880 = 10/11
Maximum throughput is
= 10/11 × 10 Mbps
= 100/11 Mbps
Question 3
If link transmits 4000 frames per second and each slot has 8 bits, the transmission rate of circuit of this TDM is ______.
A
64 Kbps
B
32 Mbps
C
32 Kbps
D
64 Mbps
Computer-NetworksAccess-Control-MethodsUGC NET CS 2016 July- paper-3
Question 3 Explanation: 
Given data,

-- Link transmits frame per second= 4000 Frames per second

-- Each slot bits= 8 bits

-- Transmission rate of circuit of this TDM=?

Step-1: Transmission rate of circuit of this TDM= Link transmits frame per second * Each slot bits

= 4000 * 8

= 32000 bits (or 32Kbps)

Question 4
Given the following statements:

(A) Frequency Division Multiplexing(FDM) is a technique that can be applied when the bandwidth of a link is greater than combined bandwidth of signals to be transmitted.

(B) Wavelength Division Multiplexing (WDM) is an analog multiplexing Technique to combine optical signals.

(C) WDM is a Digital Multiplexing Technique.

(D) TDM is a Digital Multiplexing Technique.

Which of the following is correct?

A
(A), (B), (C) and (D) are true.
B
(A), (B), (C) and (D) are false.
C
(A), (B) and (D) are false; (C) is true.
D
(A), (B) and (D) are True; (C) is false.
Computer-NetworksAccess-Control-MethodsUGC NET CS 2016 July- paper-3
Question 4 Explanation: 
In this question, if Statement-B is true then definitely Statement-C should be FALSE.

If Statement-B is FALSE then definitely Statement-C should be TRUE.

Option-A: We can ruled out based on the above two constraints.

Option-B: We can ruled out based on the above two constraints.

Option-C: Statement-B is FALSE because Wavelength Division Multiplexing (WDM) is an analog multiplexing Technique to combine optical signals.



There are 4 questions to complete.

Access quiz wise question and answers by becoming as a solutions adda PRO SUBSCRIBER with Ad-Free content

If you have registered and made your payment please contact solutionsadda.in@gmail.com to get access