2- Solutions of Second Semester Examination 2014-2015

Al-Mustaqbal University College
Computer Engineering Techniques Department
2- Solutions of Second Semester Examination 2014-2015



4th class – Computer Networks – Almustakbel College
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Q1: If the hamming code word received by receiver is ? 010010001011 (Data bits =8), assuming the even parity, state whether the received code word is correct or wrong? If wrong locate the bit having error?
The received data is 010010001011
2k-1 >= m+k , 24 – 1 >= 12 , 15 >=12
C1= 010010001011 ?Not Even
C2 = 010010001011 ? Even
C4 = 010010001011 ? Even
C8 = 010010001011 ? Not Even
Bit error = 1+8 = 9
The correct data 010010000011
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Q2: Determine whether the received data is accepted or not using Check Sum method, 01010101, 01011010, 10101011, 10110011 and the check sum is 11000101.?
01010101 + 01011010= 10101111 + 10101011= 01011010 +1 = 01011011 =10110011= 00001110 +1 =00001111 + 11000101 = 11010100 ( Data not accepted)

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Q3: If a Mobile is connected via point to point with Youtube Server where the Bandwidth is low and most of packets are reaching out of order with high error, which error control protocol is suitable to be used in this case? Selective and Repeat SR
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Q4: In GB3 if every 4th packet that is being transmitted is list, and if we have to send 12 packets, then how many transmissions are required ? Compare it with Selective and Repeat and Stop and Wait methods?
1 2 3 4 5 6 7 8 9 10 11 12
4 5 6 7
5 6 7 8
6 7 8 9
7 8 9 10
8 9 10 11
9 10 11 12
10 11 12
11 12
The number of transmission in GBN = 35
In S & W = 15 , In SR = 15
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Q5: Us CRC method to check whether the received data 1000101110 is correct or not where G=1010?
1000101110
1010
0010101110
1010
0 0 0 0 1110
1010
0100 ? Data not accepted
Q6: What is the maximum window size of receiver where the transmission time is 1msec and propagation delay time is 49.5msec . In case of GBN is GB10 and Bandwidth is 30 mbps , find the efficiency and the throughput ?
Max Window Size =1+2*a = 1+2* (Tp*Tt)= 1+ 2 * (49.5* 1) = 100
Efficiency = Ws / Max Window Size = 10 / 100 = 10%
Throughput = Bandwidth BW * Efficiency= 30 * (10 / 100) = 3 mbps
Q7: Draw Fiber Optic cable .

Q8: What are Disadvantage of Optical Fiber ?
Disadvantages of Optical Fiber
The main disadvantages of fiber optics are cost, installation/maintenance, and fragility.
-Cost. Fiber-optic cable is expensive. Also, a laser light source can cost thousands of dollars, compared to hundreds of dollars for electrical signal generators.
-Installation/maintenance Fragility. Glass fiber is more easily broken than wire, making it less useful for applications where hardware portability is required.
Q9: What is Interleaving?
Interleaving
The process of taking a group of bits from each input line for multiplexing is called interleaving. We interleave bits (1 - n) from each input onto one output.


Q10 : What are the advantage of TDM?
Advantage of TDM are: Digital signals , Relatively Simple , used with T-1, ISDN.

Q11: What are the disadvantage of FDM?
Disadvantage are: Noise problems due to analog signals, Wastes Bandwidth , Limited by frequency ranges



Dr. Rafah M. Almuttairi