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Sheet (2)
1) Bit Stuffing. Suppose the following bit strings are received by the data
link layer from the network layer: a) 01110111101111101111110. b)
0110111111111100 c) 00011101111111111110.
What is the resulting string after bit stuffing and framing? Bold each bit that has been
added where flag is 01111110.
2) A slotted ALOHA network transmits 100-bit frames using a shared
channel with a 100-kbps bandwidth. Find the throughput if the system
(all stations together) produces
a. 1000 frames per second
c. 500 frames per second
b. 750 frames per second
d. 250 frames per second
3) The following HDLC frames are sent from primary to secondary.
a) 01111110 00001111 10001011 FCS 01111110
b) 01111110 00001111 00001011 00110011 FCS 01111110
c) 01111110 00001111 11001011 00100111 FCS 01111110
Answer the following questions for each:
- What is the type of frame (I-frame, S-frame, U-frame)?
- What is the address of secondary?
- Does the frame carry user data? If yes what is the value?
Does the frame carry management data? If yes what is the value?
4) The timer of a system using the Stop-and-Wait ARQ Protocol has a time-out of 4
ms, Draw the flow diagram for four frames if the round trip delay is 6 ms. Assume
no data frame or control frame is lost or damaged.
5) Sketch in the diagram all collision domains.
6) (Addressing in the Data Link Layer)
a) Data Link Layer protocols specify the format of. . .
1) physical network addresses
2) logical network addresses
b) Give the name (technical term) of physical network addresses (Data Link Layer
addresses).
c) Name the protocol that is used by Ethernet for the address resolution.
d) Which
devices receive a frame with the destination address
FF-FF-FF-FF-FF-FF.
e) Explain what MAC spoofing is.
7) (Byte Stuffing)
The Data Link Layer splits the bit stream from the Physical Layer into frames. The characteroriented protocol BISYNC uses control characters to mark the structure of the frames. The start
of a frame highlights the character SYN. The start of the header highlights the character SOH
(Start of header). The payload is located between STX (Start of text) and ETX (End of text).
The figure shows the structure of BISYNC frames:
Control character
Hexadecimal notation
SOH STX ETX DLE SYN
01
02
03
10
16
If the payload (body) contains the control characters ETX and DLE (Data Link Esca- pe), they
are protected (escaped) by the Data Link Layer protocol with a stuffed DLE caracter. A
single ETX in the payload area is represented by the sequence DLE ETX. The DLE
character itsef is represented by the sequence DLE DLE.
Mark the payload inside the following BISYNC frames.
1. 16 16 01 99 98 97 96 95 02 A1 A2 A3 A4 A5 03 A0 B7
2. 16 16 01 99 98 97 96 95 02 05 04 10 03 02 01 03 76 35
3. 16 16 01 99 98 97 96 95 02 10 03 10 10 10 03 03 92 55
4. 16 16 01 99 98 97 96 95 02 10 10 10 10 10 03 01 02 A1 03 99 B2
8) (Bit Stuffing)
The Data Link Layer protocol HDLC (High-Level Data Link Control) uses Bit Stuf- fing. If
the sender discovers 5 consecutive 1 bits in the bitstream from the Network Layer, it stuffs
a single 0 bit into the outgoing bit stream. If the receiver discovers 5 consecutive 1 bits,
followed by a single 0 bit in the bit stream from the Physical Layer, it removes (destuffs)
the 0 bit.
Give the encoding for each one of the following bit sequences, when the sender stuffs
after 5 consecutive 1 bits a single 0 bit into the bit stream from the Network Layer.
1. 01111110 10100111 11111000 11110010 10011111 10111111 11100101
2. 00111111 01110001 11110011 11111100 10101010 11001111 11100001
3. 11111111 11111111 11111111 11111111 11111111 11111111 11111111
9) (Error Detection – CRC)
1. Calculate the frame to be transferred.
Generator polynomial: 100101
Payload: 11010011
2. Check, if the received frame was transmitted correctly.
Transferred frame: 1101001110100
Generator polynomial: 100101
3. Check, if the received frame was transmitted correctly.
Transferred frame: 1101001111100
Generator polynomial: 100101
4. Calculate the frame to be transferred.
Generator polynomial: 100101
Payload: 10110101
5. Check, if the received frame was transmitted correctly.
Transferred frame: 1011010110110
Generator polynomial: 100101
6. Check, if the received frame was transmitted correctly.
Transferred frame: 1011010110100
Generator polynomial: 100101
7. Check, if the received frame was transmitted correctly.
Transferred frame: 1010010110100
Generator polynomial: 100101
8. Calculate the frame to be transferred.
Generator polynomial: 100000111
Payload: 1101010101110101
9. Check, if the received frame was transmitted correctly.
Transferred frame: 110101010111110110110111
Generator polynomial: 100000111
10) (Media Access Control)
1. Explain why computer networks use protocols for media access control.
2. Explain why do Ethernet and WLAN use different media access control
methods.
3. Explain how Ethernet devices react, when they detect a collision.
11) Sketch in the diagram all collision domains and all broadcast domains.
12) Addressing in the Network Layer
1. Explain the meaning of Unicast in the Network Layer of computer networks.
2. Explain the meaning of Broadcast in the Network Layer of computer networks.
3. Explain the meaning of Anycast in the Network Layer of computer networks.
4. Explain the meaning of Multicast in the Network Layer of computer networks.
5. Explain why the IPv4 address space does contain only 4,294,967,296
addresses.
Imagine, NASA sent a spacecraft to planet Mars, which landed there. A 128 kbps (kilobit
per second) point-to-point link is set up between planet Earth and the spacecraft.
The distance between Earth and Mars fluctuates between approx. 55,000,000km and
approx.
400,000,000 km. For the further calculations, we use the 55,000,000 km, which is the
distance from Earth to Mars, when they are closest together.
The signal propagation speed is 299, 792, 458 m/s, which is the speed of light.
a) Calculate the Round Trip Time (RTT) for the link.
b) Calculate the bandwidth-delay product for the link to find out what is the maximum
number of bits, that can reside inside the line between the sender and receiver?
13) The stations on a wireless ALOHA network is a maximum of 600 km apart. If we
assumethat signals propagate at 3 x 108 mis, we find Tp = (600 x 105) / (3 x 108) = 2 ms.
Now we can find the value of TB for different values of K=1,2,3.
14) A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps.
Whatis the requirement to make this frame collision-free?
15) A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps.
Whatis the throughput if the system (all stations together) produces?
a. 1000 frames per second
b. 500 frames per second
c. 250 frames per second
16) A slotted ALOHA network transmits 200-bit frames using a shared channel with a
200-kbps bandwidth. Find the throughput if the system (all stations together) produces
a. 1000 frames per second
b. 500 frames per second
c. 250 frames per second
17) Define the type of the following destination addresses:
a. 4A:30:10:21:1O:1A
b. 47:20:1B:2E:08:EE
c. FF:FF:FF:FF:FF:FF
18) Show how the address 47:20:1B:2E:08:EE is sent out on line.
19) Check, if the received frame was transmitted correctly.
Transferred frame: 110101010111010110110111
Generator polynomial: 100000111
20) Transmission errors can be detected via CRC checksums. If it is important to not only
recognize errors, but also to be correct them, then the data to be transmitted must be encoded
in a way, that error-correction is possible. Error correction can be realized e.g. via the
Simplified Hamming Code we discussed in the computer networks course.
1. A message of 8 bits payload (10011010) needs to be transferred. Calculate the
message, that will be transmitted (payload inclusive parity bits).
2. The following messages have been received. Verify, if they were transmitted
correctly.
a) 00111101
b) 101110100010
c) 001101100100
d) 0001101100101101
21) (Media Access Control)
1. Explain why computer networks use protocols for media access control.
2. Explain why do Ethernet and WLAN use different media access control
methods.
3. Explain how Ethernet devices react, when they detect a collision.