3-Solutions of Second Semester Examination 2014-2015

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



Q1: If the 7 bit hamming code word received by receiver is 1011011 , 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 1101101
2k-1 >= m+k , 23 – 1 >= 4+3 , 7=7
C1=1011 ? Odd
C2 = 1001 ? Even
C4 =1101 ? Odd
Bit error = 1+4 = 5
The correct data 1001011
Q2: Determine whether the received data is accepted or not using Check Sum method, 01010101, 01011010, 10101011, 10110011 and the check sum is 01000101.
01010101 +
01011010
-------------------
10101111 +
10101011
-------------------
01011010 +
1
-------------------
01011011
10110011
-------------------
00001110
1
-------------------
00001111
01000101
-------------------
01010100 Data is not accepted
Q3: If the link connected two routers is wired where the Bandwidth is high , which error control protocol is suitable to be used in Data Link Layer? Go Back N
Q4: In GB3 if every 4th packet that is being transmitted is list, and if we have to send 10 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
4 5 6 7
5 6 7 8
6 7 8 9 10
7 8 9 10
8 9 10
9 10
GBN = 27 , SR = 13 , S & W = 13
Q5: Us CRC method to check whether the received data 1100101 is correct or not where G=1010?
1100101
1010
------------
0110101
1010
-------------
0011101
1010
-------------
0001001
1010
-------------
0000011 ? Data is not Correct


Q6: What is the maximum window size of receiver where the transmission time is 2msec and propagation delay time is 24.75msec . In case of GBN is GB20 and Bandwidth is 30 mbps , find the efficiency and the throughput ?
Max Window Size =1+2*a = 1+2*(Tp *Tt )= 1+ 2 * (24.75 * 2) = 100
Efficiency = Ws / Max Window Size = 20 / 100 = 20%
Throughput = Efficiency * BW =( 20 / 100) * 30 = 6 mbps
Q7: Compare between infrared and microwave transmission?
One important difference between infrared and microwave transmission is that the former does not penetrate walls. Thus the security and interference problems encountered in microwave systems are not present. Furthermore, there is no frequency allocation issue with infrared, because no licensing is required.

Q 8: What are Advantages of Optical Fiber ?
The adv. of O. F. are :
1- Noise Resistance: Because fiber-optic transmission uses light rather than electricity, noise is not a factor. External light, the only possible interference, is blocked from the channel by the outer jacket.
2- Less signal attenuation : Fiber-optic transmission distance is significantly greater than that of other guided media. A signal can run for miles without requiring regeneration.
3- Higher bandwidth: Currently, data rates and bandwidth utilization over fiber-optic cable are limited not by the medium but by the signal generation and reception technology available.

Q8: Fill the following
Classes of Transmission media are Wired (Guided ), Wireless ( Unguided )
1- Coaxial cable consist of Insulator, Plastic Cover, Outer Conductor, Inner Conductor
2- Metal cables transmit signals in form of electric current, but, the Optical Fiber transmits signals in form of light
3- Optical Fiber is made of glass or plastic
4- Fiber Optic cable consist of Core, Cladding, Coating, Strengthening Fiber, Cable Jacket
5- Disadvantage of Optical Fiber are Cost, Installation
6- Propagation methods are Ground, Sky, Line of Sight
7- Transmissions of wireless are Radio, Microwave, Infrared
8- Multiplexing is : Categories of multiplexing are : FDM, WDM, TDM
9- There are three strategies that can be used to overcome the data rate : Multi level , Multi Slot, Pulse Stuffing
10- Framing: data are sent in blocks called frames, the beginning and end of each frame must be recognized by the receiver.
11- Error control: bit errors introduced by the transmission system should be detected and/or corrected.
12- Flow control: the sending station must not send frames at a rate faster than the receiving station can absorb them.
13- Data link control protocol provides functions such as flow control, error detection, and error control.
14- Radio waves are used for multicast communications, such as radio and television, and paging systems. They can penetrate through walls. Highly regulated. Use omni directional antennas
15- Microwaves are used for unicast communication such as cellular telephones, satellite networks, and wireless LANs. Higher frequency ranges cannot penetrate walls. Use directional antennas - point to point line of sight communications
16- Infrared signals can be used for short-range communication in a closed area using line-of-sight propagation.
17- The major advantages offered by fiber-optic cable over twisted-pair and coaxial cable are noise resistance, less signal attenuation, and higher bandwidth.
18- It is possible that the flag byte’s bit pattern occur in the data, two popular solutions: Byte stuffing and Bit stuffing
19- The effect of Automatic Repeat Request (ARQ) is to turn an unreliable data link into a reliable one, three versions of ARQ which are: Stop-and-wait, Go-back-N, Selective-reject (or, selective repeat)
20- Inefficient use of Bandwidth means :Sometimes an input link may have no data to transmit. one or more slots on the output link will go unused. That is wasteful of bandwidth.
21- Advantage of TDM are: Digital signals , Relatively Simple , used with T-1, ISDN. Disadvantage is : Wastes Bandwidth
22- Advantage of FDM are : Simple , used with Radio, TV, cable TV . All the receiever, such as cellular telephones, do not need to be at the same location. Disadvantage are: Noise problems due to analog signals, Wastes Bandwidth , Limited by frequency ranges
23- Advantage of WDM are: Very high capacities over fiber , Signals can have varying speeds , Scalable . Disadvantage are : Cost and Complexity