Telecommunications architectures

2001: Home assignment 2 - Model answer

Notice! The role of this model answer is instructional only. The answers might not cover everything in detail that the questions asks. The answers give you the content of the answer and idea what it should be. The things that are important in the grading of the homework have been emphasized. If there is a mistake in the model answer in your opinion, please let the course staff know!

1. CRC-Checksum (3 Points)

1. (1 Point)
   frame (4 zeros added) -> 110111010000
   generator polynomial 11001
   

              _____________
         11001|110111010000
               11001
               -----
               00010101
                  11001
                  -----
                  011000
                   11001
                   -----
                   00001000
   

   -> remainder 1000
   -> frame to send: 110111011000
   

2. (1 Point)
   frame: 111001011011
   generator: 10001
   

               _____________
          10001|111001011011
                10001
                -----
                011011
                 10001
                 -----
                 010100
                  10001
                  -----
                  0010111
                    10001
                    -----
                    0011001
                      10001
                      -----
                      010001
                       10001
                       -----
                       00000
   

   no remainder -> no errors :) Should be accepted

3. (1 Point)
   There's no way - the frame should be longer than the generator polynomial

2. Flow control (6 Point)

Values given:
frame size 2048
connection speed A 56kbps = 56000 b/s
connection speed B 2 Mbps = 2x10**6 b/s
propagation speed 1 2x10**8m/s
propagation speed 2,3 3x10**8m/s (2x10**8m/s also accepted)
distance 1 3000m
distance 2 20000m
distance 3 50x10**6m
BER 10**-5

One frame transmission time Tx is the frame size divided by the connection speed.
TxA 2048/56000 = 0.0366 s
TxB 2048/2x10**6 = 0.001024 s
Transmission delay Tp is the distance in meters divided by the propagation speed.

Tp1 3000/2x10**8=1.5e-5s
Tp2 20000/3x10**8=6.67e-5s
Tp3 50x10**6/3x10**8=0.167s
The link utilization U with errors is
U=(1-Pf)/(1+2*a)
a = Tp/Tx
Pf ~ N*P = 2048x10**-5=0.02048

   1. A.   a=Tp1/TxA=4.0983e-4
           U=(1-Pf)/(1+2*a)=0.98
      B.   a=Tp1/TxB=0.01465
           U=(1-Pf)/(1+2*a)=0.95
   2. A.   0.98
      B.   0.86
   3. A.   0.1
      B.   0.003

3. IP Addressing (3 points)

1. (1 Point)
   1. 130.233.220.31 = B
   2. D8EF2764 = C
   3. 11100110.10001110.00011110.00000100 = D

2. (1 Point)
   A subnet mask is 255.255.255.240:
   240(D)=11110000(B)->4 bits for hosts
   2^4 = 16, 16-2=14 (two received)

3. (1 Point)
   

     Case one:
             173.223.95.105(D)=1010 1101.1101 1111.0101 1111.0110 1001(B)
             173.223.90.125(D)=1010 1101.1101 1111.0101 1010.0111 1101(B)
                           XOR -----------------------------------------
                               0000 0000.0000 0000.0000 0101.0001 0100
                   subnet mask 1111 1111.1111 1111.1111 1000.0000 0000
                           AND ---------------------------------------
                               0000 0000.0000 0000.0000 0000.0000 0000 
             
             So, they are in the same subnet                              
     Case Two:
             173.223.95.105(D)=1010 1101.1101 1111.0101 1111.0110 1001(B)
                   subnet mask=1111 1111.1111 1111.1111 1000.0000 0000
                           AND ---------------------------------------
                               1010 1101 1101 1111 0101 1000 0000 0000 
   
   
             173.223.90.125(D)=1010 1101.1101 1111.0101 1010.0111 1101(B)
                   subnet mask 1111 1111.1111 1111.1111 1000.0000 0000
                           AND ---------------------------------------
                               1010 1101 1101 1111 0101 1000 0000 0000 
   
            They share the same subnet number.
            So they are in the same subnet.
   
            These two methods are both accepted for this question.
   

4. LANs (6 points)

1. (2 Points)
   distance d = 1000 m
   data rate R = 1 Gbps = 10**9 b/s
   propagation speed V = 2x10**8 m/s
   propagation time = 2d/V = 2000m/2x10**8m/s=10**-5 s
   So min. frame size F = R2d/V = 10**9b/s*10**-5s=10**4b=10 000b
   

2. (2 Points)
   from previous: F=R2d/V <=> d = FV/2R
   d = (512b*8*2x10**8m/s)/(2*10**9b/s)=410m

   (Notice: If you used bits, the lenght should be 51m.)

3. (2 Points)
   data rate R = 10 Mbps
   propagation speed V = 2x10**8 m/s
   bits on ring b = 15
   "bit length" l = V/R
   min. distance d = lb=Vb/R=(2x10**8m/s*15b)/(10**7b/s)=300m