Multiple Access-Data Communication Systems-Assignment Solution, Exercises of Data Communication Systems and Computer Networks

This file contains solution to problems related Data Communication Systems. Mr. Prajin Ahuja assigned task at Birla Institute of Technology and Science. Its main points are: Multiple, Access, Dom, Access, Controlled, Channelization, Reservation, Polling, Token, Passing

Typology: Exercises

2011/2012

Uploaded on 07/26/2012

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CHAPTER 12
Multiple Access
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. The three categories of multiple access protocols discussed in this chapter are ran-
dom access, controlled access, and channelization.
3. In controlled access methods, the stations consult one another to find which sta-
tion has the right to send. A station cannot send unless it has been authorized by
other stations. We discuss three popular controlled-access methods: reservation,
polling, and token passing.
5. In random access methods, there is no access control (as there is in controlled
access methods) and there is no predefined channels (as in channelization). Each
station can transmit when it desires. This liberty may create collision.
7. In a random access method, the whole available bandwidth belongs to the station
that wins the contention; the other stations needs 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.
9. 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.
Exercises
11. To achieve the maximum efficiency in pure ALOHA, G = 1/2. If we let ns to be the
number of stations and nfs to be the number of frames a station can send per sec-
ond.
G = ns × nfs × Tfr = 100 × nfs × 1 μs = 1/2 nfs = 5000 frames/s
The reader may have noticed that the Tfr is very small in this problem. This means
that either the data rate must be very high or the frames must be very small.
13. We can first calculate Tfr and G, and then the throughput.
Tfr = (1000 bits) / 1 Mbps = 1 ms
G = ns × nfs × Tfr = 100 × 10 × 1 ms = 1
For pure ALOHA S = G × e2G 13.53 percent
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CHAPTER 12

Multiple Access

Solutions to Odd-Numbered Review Questions and Exercises

Review Questions

  1. The three categories of multiple access protocols discussed in this chapter are ran- dom access , controlled access , and channelization.
  2. In controlled access methods , the stations consult one another to find which sta- tion has the right to send. A station cannot send unless it has been authorized by other stations. We discuss three popular controlled-access methods: reservation , polling , and token passing.
  3. In random access methods, there is no access control (as there is in controlled access methods) and there is no predefined channels (as in channelization). Each station can transmit when it desires. This liberty may create collision.
  4. In a random access method, the whole available bandwidth belongs to the station that wins the contention; the other stations needs 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.
  5. 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.

Exercises

  1. To achieve the maximum efficiency in pure ALOHA, G = 1/2. If we let ns to be the number of stations and nfs to be the number of frames a station can send per sec- ond. G = ns × nfs × T (^) fr = 100 × nfs × 1 μs = 1/2 → nfs = 5000 frames/s The reader may have noticed that the T (^) fr is very small in this problem. This means that either the data rate must be very high or the frames must be very small.
  2. We can first calculate T (^) fr and G, and then the throughput. Tfr = (1000 bits) / 1 Mbps = 1 ms G = ns × nfs × Tfr = 100 × 10 × 1 ms = 1 For pure ALOHA → S = G × e−2G^ ≈ 13.53 percent

This means that each station can successfully send only 1.35 frames per second.

  1. Let us find the relationship between the minimum frame size and the data rate. We know that T (^) fr = (frame size) / (data rate) = 2 × Tp = 2 × distance / (propagation speed) or (frame size) = [2 × (distance) / (propagation speed)] × (data rate)] or (frame size) = K × (data rate) This means that minimum frame size is proportional to the data rate (K is a con- stant). When the data rate is increased, the frame size must be increased in a net- work with a fixed length to continue the proper operation of the CSMA/CD. In Example 12.5, we mentioned that the minimum frame size for a data rate of 10 Mbps is 512 bits. We calculate the minimum frame size based on the above pro- portionality relationship
  2. We have t 1 = 0 and t 2 = 3 μs a. t 3 − t 1 = (2000 m) / (2 × 10 8 m/s) =10 μs → t 3 = 10 μs + t 1 = 10 μ s b. t 4 − t 2 = (2000 m) / (2 × 10 8 m/s) =10 μs → t 4 = 10 μs + t 2 = 13 μ s c. T (^) fr(A) = t 4 − t 1 = 13 − 0 = 13 μs → Bits (^) A = 10 Mbps × 13 μs = 130 bits d. T (^) fr(C) = t 3 − t 2 = 10 − 3 = 07μs → Bits (^) C = 10 Mbps × 07 μs = 70 bits
  3. See Figure 12.1.

Data rate = 10 Mbps → minimum frame size = 512 bits Data rate = 100 Mbps → minimum frame size = 5120 bits Data rate = 1 Gbps → minimum frame size = 51,200 bits Data rate = 10 Gbps → minimum frame size = 512,000 bits

Figure 12.1 Solution to Exercise 19

+1 + +1 −

+1 −

+1 +1 +1 + +1 − −1 − −1 +

+1 + +1 −

+1 −

+1 +1 +1 + +1 − −1 − −1 +

+1 + +1 −

+1 −

+1 +1 +1 + +1 − −1 − −1 +

−1 − −1 +

−1 +

−1 −1 −1 − −1 + +1 + +1 −

W 8 =

Frame 4 for all four stations: 4 × [poll + frame + ACK)] Frame 5 for all four stations: 4 × [poll + frame + ACK)] Polling and Sending NAKs Station 1: [poll + NAK] Station 2: [poll + NAK] Station 3: [poll + NAK] Station 4: [poll + NAK] Total Activity: 24 polls + 20 frames + 20 ACKs + 4 NAKs = 21536 bytes We have 1536 bytes of overhead which is 512 bytes more than the case in Exercise

  1. The reason is that we need to send 16 extra polls.