Data Communications and Networking : Multiple Access Questions and Answers

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List three categories of multiple access protocols discussed in this chapter. 
random access protocols, controlled access protocols, and channelization protocols.

Define random access and list three protocols in this category. 
In random access or contention methods, no station is superior to another station and none is assigned the control over another. No station permits, or does not permit, another station to send. In a random access method, each station has the right to the medium without being controlled by any other station. There is no scheduled time for a station to transmit. Transmission is random among the stations. That is why these methods are called random access. Also, no rules specify which station should send next. Stations compete with one another to access the medium. That is why these methods are also called contention methods. Protocols in this category : ALOHA, CSMA, CSMA/CD, CSMA/CA.

Define controlled access and list three protocols in this category. 
In controlled access, the stations consult one another to find which station has the right to send. A station cannot send unless it has been authorized by other stations. Protocols in this category: reservation, polling, token passing.

Define channelization and list three protocols in this category. 
Channelization is a multiple-access method in which the available bandwidth of a link is shared in time, frequency, or through code, between different stations. Protocols in this category: FDMA, TDMA, and CDMA.

Explain why collision is an issue in a random access protocol but not in controlled access or channelizing protocols. 
In random access methods, there is no access control (as there is in controlled access methods) and there is no predefined channel (as in channelization). Each station can transmit when it desires. This liberty may create collision.

Compare and contrast a random access protocol with a controlled access protocol. 
 In random access method, there is no control over which station holds the link, any station can randomly use the link, which may even cause collision. In controlled access protocol, there is a primary station that controls the link, and decides which station gets the link at a given time, thus reducing chances of collision.

Compare and contrast a random access protocol with a channelizing protocol. 
In a random access method, the whole available bandwidth belongs to the station that wins the contention; the other stations need to wait. In a channelization method, the available bandwidth is divided between the stations. If a station does not have data to send, the allocated channel remains idle.

Compare and contrast a controlled access protocol with a channelizing protocol. 
In controlled access protocol, the whole available bandwidth belongs to the primary station (the sation with control to the link). In channelization protocol, available bandwidth is divided between stations. If a station does not have data to send the allocated channel remains idle.

Do we need a multiple access protocol when we use the local loop of the telephone company to access the Internet? Why? 
 We do not need a multiple access method in this case. The local loop provides a dedicated point-to-point connection to the telephone office. 

Do we need a multiple access protocol when we use one CATV channel to access the Internet? Why? 


We have a pure ALOHA network with 100 stations. If Tfr = 1 μs, what is the number of frames/s each station can send to achieve the maximum efficiency. 
For maximum efficiency in pure ALOHA, G = 1/2 (max throughput at G=1/2 is 0.184). G = N × fps × Tfr => 100 × fps × 1 μs = 1/2 → nfs = 5000 frames/s

Repeat Exercise 11 for slotted ALOHA. 
For maximum efficiency in slotted ALOHA, G = 1 (max throughput at G=1 is 0.368). If we let ns to be the number of stations and nfs to be the number of frames a station can send per second. G = N × fps × Tfr => 100 × fps × 1 μs = 1 → nfs = 10000 frames/s. It is double of that in pure ALOHA.

One hundred stations on a pure ALOHA network share a l-Mbps channel. If frames are 1000 bits long, find the throughput if each station is sending 10 frames per second. 
G = N × fps × Tfr . Here N = number of stations = 100, fps = frames per second = 10, Tfr = transmission time = 1000bits/1Mbps = 1ms. Therefore, G = 100 X 10 X 10-3   = 1. Throughput = S = G X e-2G ≈ 13.53 % . Therefore, only 1.353 frames per second can be successfully transmitted.  

Repeat Exercise 13 for slotted ALOHA. 

G = N × fps × Tfr . Here N = number of stations = 100, fps = frames per second = 10, Tfr = transmission time = 1000bits/1Mbps = 1ms. Therefore, G = 100 X 10 X 10-3   = 1. Throughput = S = G X e-G ≈ 36.8 % . Therefore, only 3.68 frames per second can be successfully transmitted.  

In a CDMA/CD network with a data rate of 10 Mbps, the minimum frame size is found to be 512 bits for the correct operation of the collision detection process. What should be the minimum frame size if we increase the data rate to 100 Mbps? To 1 Gbps? To 10 Gbps? 

Tfr = 2 X Tp = min frame size/data rate = 512bits/10Mbps = 51.2 μs.
min frame size = Tfr X Data Rate = 51.2 X 100 = 5120 bits. (μ and M cancel out), 51.2 X 1000 = 51200 bits, 51.2 X 10000 = 512000 bits.

In a CDMA/CD network with a data rate of 10 Mbps, the maximum distance between any station pair is found to be 2500 m for the correct operation of the collision detection process. What should be the maximum distance if we increase the data rate to 100 Mbps? To 1 Gbps? To 10 Gbps? 

Tfr = (frame size) / (data rate) = 2 × Tp = 2 × distance / (propagation speed) or distance = [(frame size) (propagation speed)] / [2 × (data rate)] or distance = K / (data rate) This means that distance is inversely proportional to the data rate (K is a constant). When the data rate is increased, the distance or maximum length of network or collision domain is decreased proportionally. Data rate = 10 Mbps → maximum distance = 2500 m Data rate = 100 Mbps → maximum distance = 250 m Data rate = 1 Gbps → maximum distance = 25 m Data rate = 10 Gbps → maximum distance = 2.5 m 

(To be updated soon with more questions)

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