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(ELEC315)mid04_sol.pdf
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Reference Solution to Midterm Examination 2004
(1) Solution:


Yes, the pattern can be detected. We can represent the error as


Since the cannot divide

, we only concentrate on the 2nd part of , by doing the long division of

, we find it cannot be divided exactly by . As a result, the error pattern would be able to be detected.


(2) Solution:


Since the receiver window is four, Selective Repeat scheme is used we have the following diagrams. (Two packets sent, ACK0 lost and ACK1 received)








sender











receiver



(a) (b) (c) (d)

(a) Initial state, both Ws and Wr are at 0,1,2 & 3.
(b) Packet 0 is correctly transmitted and received by the receiver, Wr advance by 1 (1,2,3 & 4) and ACK 0 sent to the transmitter.
(c) Since ACK 0 is lost, Ws remain the same. Packet 1 is correctly transmitted and received by the receiver, Wr advance by 1 (2,3,4 & 5) and ACK 1 sent to the transmitter.
(d) ACK 1 is received by the transmitter, Ws advances by 2 since cumulative acknowledgment is used.

( 3 ) Solution:
0

1

2

3

4

5

6

7

0

1

2

3

4

5

6

7

The cycle is shows as follows


Efficiency = Packet Transmission Time / Cycle Time
For Go-Back-N, ack 0 must received before Ws advance by one, so
Efficiency = Packet Transmission Time x Window Size / Cycle Time

If Efficiency = 90%, Let Ws = Sending Window Size







Therefore, the sending window size (Ws) should be 99 in order to achieve the efficiency of 90%.


(4) Solution: For Pure Aloha the ideal utilization is 0.184, the maximum available bandwidth is B=10kbps*0.184=1.84kbps, and for every user the bandwidth required is


W=120/(10*60*1000)=0.0002kbps
So the maximum available number of users will be
N=B/W=9200


(5) Solution: A and B will choose from {0 1 2 3} and C will choose from {0 1}

a) in the following cases, the next transmission will be from Station A:



A: 0 1
B: x 1/2/3
C: x 1
Prob. Of the case: 1/4 1/4*3/4*1/2
Here x represents any possible slot

The total probability of these cases will be
1/4 +1/4*3/4*1/2 = 11/32 = 0.34375
b) In the following cases, the next transmission will be a success (from station A or B or C):


A: 0 1/2/3 1/2/3 2/3
B: 1/2/3 0 1/2/3 2/3
C: 1 1 0 1
Prob. Of the case: 1/4*3/4*1/2 3/4*1/4*1/2 3/4*3/4*1/2 1/2*1/2*1/2

The total probability of these cases will be:
(1/4*3/4*1/2) + (3/4*1/4*1/2) + (3/4*3/4*1/2) + (1/2*1/2*1/2)
= 3/32+3/32+9/32+1/8
= 19/32 = 0.59375

(6) Solution:


Address sent to port 3: 11101010.00111000. 00011xxx.xxxxxxxx
Address sent to port 2: 11101010.00111000.0001xxxx.xxxxxxxx
All the other addresses should be sent to port 1.
(a) 234.56.30.22= 11101010.00111000.00011110.00010110-> port 3

(b) 245.255.254.127=11110101.11111111.11111110.01111111-> port 1