2011-03-16

Practice Midterm Problems (Problem 5).

Originally Posted By: schroeder
5. Briefly describe Manchester Encoding, 4B/5B encoding, what a PPP frame looks like.
Manchester encoding:
Sends the XOR of the clock cycle & NRZ-encoded data.
0's become low-to-high
1's become high-to-low
Effectively doubles the baud rate of a single bit.
See the diagram here:
http://www.cs.sjsu.edu/faculty/pollett/ ... l#%2817%29

4B/5B encoding:
Every 4 bits of actual data are encoded into a 5 bit code.
The 5-bit codes are selected in a way so that they contain:
* No more than ONE leading 0
* No more than TWO trailing 0's
The results contain no more than 3 consecutive 0's (solving the problem of consecutive 0's)
Transmitted using NRZI encoding (solving the problem of consecutive 1's)

PPP contains:
an 8 bit flag (sentinel, 01111110),
followed by 8 bits for an address,
4 8 bit control,
16 bits indicating protocol (such as IP or IPX),
a payload of up to 1500 bytes (including char stuffing),
a 16 bit checksum,
and lastly an 8 bit flag (01111110)
'''Originally Posted By: schroeder''' 5. Briefly describe Manchester Encoding, 4B/5B encoding, what a PPP frame looks like.<br>Manchester encoding:<br>Sends the XOR of the clock cycle & NRZ-encoded data.<br>0's become low-to-high<br>1's become high-to-low<br>Effectively doubles the baud rate of a single bit. <br>See the diagram here:<br>http://www.cs.sjsu.edu/faculty/pollett/ ... l#%2817%29<br><br>4B/5B encoding:<br>Every 4 bits of actual data are encoded into a 5 bit code.<br>The 5-bit codes are selected in a way so that they contain:<br>* No more than ONE leading 0<br>* No more than TWO trailing 0's<br>The results contain no more than 3 consecutive 0's (solving the problem of consecutive 0's)<br>Transmitted using NRZI encoding (solving the problem of consecutive 1's)<br><br>PPP contains:<br>an 8 bit flag (sentinel, 01111110),<br>followed by 8 bits for an address,<br>4 8 bit control,<br>16 bits indicating protocol (such as IP or IPX),<br>a payload of up to 1500 bytes (including char stuffing),<br>a 16 bit checksum,<br>and lastly an 8 bit flag (01111110)

-- Practice Midterm Problems (Problem 5)
Originally Posted By: RobertToney
Problem 6:

Suppose we are computing a CRC-15 of 1KB of data using the generating polynomial x^15+ x^14+1. Based on what we learned in class what kinds of errors would we be able to detect?

ANSWER
CRC-15 of 1KB using x^15+ x^14+1 will be able to detect 2 types of errors:
-Double bit errors, since x^15+x^14+1 does not divide x^k + 1 for k RobertToney — Wed Mar 16, 2011 6:29 pm <hr>
'''Originally Posted By: RobertToney''' Problem 6:<br><br>Suppose we are computing a CRC-15 of 1KB of data using the generating polynomial x^15+ x^14+1. Based on what we learned in class what kinds of errors would we be able to detect?<br><br>ANSWER<br>CRC-15 of 1KB using x^15+ x^14+1 will be able to detect 2 types of errors:<br>-Double bit errors, since x^15+x^14+1 does not divide x^k + 1 for k RobertToney &mdash; Wed Mar 16, 2011 6:29 pm <hr>

-- Practice Midterm Problems (Problem 5)
Originally Posted By: Jay_Reynolds_Freeman
Problem 3:

Without using try/catch, our solution was as follows:

Socket sock = new Socket( "somewhere.com", 1234 );
InputStream is = sock.getInputStream();
InputStreamReader ir = new InputStreamReader( is );
BufferedReader rd = new BufferedReader( ir );
String str = rd.readLine();
'''Originally Posted By: Jay_Reynolds_Freeman''' Problem 3:<br><br>Without using try/catch, our solution was as follows:<br><br>Socket sock = new Socket( &quot;somewhere.com&quot;, 1234 );<br>InputStream is = sock.getInputStream();<br>InputStreamReader ir = new InputStreamReader( is );<br>BufferedReader rd = new BufferedReader( ir );<br>String str = rd.readLine();
2011-03-17

-- Practice Midterm Problems (Problem 9)
Originally Posted By: eaon168
So for problem 9, each message has 3 columns: sender ID, root ID, distance to the root

Round 1:
node 1: 1, 1, 0
node 2: 2, 2, 0
node 3: 3, 3, 0
node 4: 4, 4, 0
node 5: 5, 5, 0
node 6: 6, 6, 0

Round 2:
node 1: 1, 1, 0
node 2: 2, 1, 1
node 3: 3, 2, 1
node 4: 4, 3, 1
node 5: 5, 4, 1
node 6: 6, 1, 1

Round 3:
node 1: 1, 1, 0
node 2: 2, 1, 1
node 3: 3, 1, 2
node 4: 4, 2, 2
node 5: 5, 1, 2
node 6: 6, 1, 1

Round 4:
node 1: 1, 1, 0
node 2: 2, 1, 1
node 3: 3, 1, 2
node 4: 4, 1, 3
node 5: 5, 1, 2
node 6: 6, 1, 1


The resulting tree will look as follow(should look like a ring with the edge between 4 and 5 removed):

1--2--3--4
|
6
|
5
'''Originally Posted By: eaon168''' So for problem 9, each message has 3 columns: sender ID, root ID, distance to the root<br><br>Round 1:<br>node 1: 1, 1, 0<br>node 2: 2, 2, 0<br>node 3: 3, 3, 0<br>node 4: 4, 4, 0<br>node 5: 5, 5, 0<br>node 6: 6, 6, 0<br><br>Round 2:<br>node 1: 1, 1, 0<br>node 2: 2, 1, 1<br>node 3: 3, 2, 1<br>node 4: 4, 3, 1<br>node 5: 5, 4, 1<br>node 6: 6, 1, 1<br><br>Round 3:<br>node 1: 1, 1, 0<br>node 2: 2, 1, 1<br>node 3: 3, 1, 2<br>node 4: 4, 2, 2<br>node 5: 5, 1, 2<br>node 6: 6, 1, 1<br><br>Round 4:<br>node 1: 1, 1, 0<br>node 2: 2, 1, 1<br>node 3: 3, 1, 2<br>node 4: 4, 1, 3<br>node 5: 5, 1, 2<br>node 6: 6, 1, 1<br><br><br>The resulting tree will look as follow(should look like a ring with the edge between 4 and 5 removed):<br><br>1--2--3--4<br>|<br>6<br>|<br>5

-- Practice Midterm Problems (Problem 5)
Originally Posted By: ahii
Problem 1:

Packet-switched networks use a strategy called store-and-forward. Each node in a store-and-forward network first receives a complete packet over some link, stores the packet in its internal memory, and then forwards the complete packet to the next node.
A circuit-switched network first establishes a dedicated circuit across a sequence of links and then allows the source node to send a stream of bits across this circuit to a destination node.
Synchronous time-division multiplexing (STDM) is to divide time into equal-sized quanta and, in a round-robin fashion, give each flow a chance to send its data over the physical link.
Frequency-division multiplexing (FDM) is to transmit the data of different users over the channel at different frequencies.
'''Originally Posted By: ahii''' Problem 1:<br><br>Packet-switched networks use a strategy called store-and-forward. Each node in a store-and-forward network first receives a complete packet over some link, stores the packet in its internal memory, and then forwards the complete packet to the next node. <br>A circuit-switched network first establishes a dedicated circuit across a sequence of links and then allows the source node to send a stream of bits across this circuit to a destination node.<br>Synchronous time-division multiplexing (STDM) is to divide time into equal-sized quanta and, in a round-robin fashion, give each flow a chance to send its data over the physical link.<br>Frequency-division multiplexing (FDM) is to transmit the data of different users over the channel at different frequencies.
2011-03-19

-- Practice Midterm Problems (Problem 5)
Originally Posted By: sandybb
Can anyone post the answer for the following Q -

Describe how a sender knows a packet has arrived in a token ring.
'''Originally Posted By: sandybb''' Can anyone post the answer for the following Q -<br><br>Describe how a sender knows a packet has arrived in a token ring.

-- Practice Midterm Problems (Problem 7)
Originally Posted By: razgriz
(7) Describe how the Sliding Window ARQ algorithm works.

An ARQ is an automatic repeat request, an algorithm that uses acknowledgements and timeouts to determine which frames are requested again from the sender.

The sliding window algorithm works by limiting the number of unacknowledged frames that can be sent at a time due to buffer constraints based on the sequence number assigned to each frame. This maximum limit is the window size available both to the sender and receiver.

The sender's window size / send limit is chosen such that it will keep the transmission pipe as full as possible. The sender can continue sending frames so long as it maintains that the sequence number for the last frame sent minus the sequence number for the last acknowledgement received from the receiver is less than the sender's window size. Thus the sender will not send a packet that has a sequence number outside of its window limit. Any acknowledgement the sender receives is also taken as a promise from the receiver that any frame with a lower sequence number than the acknowledgement is also received and in order.

Similarly, the receiver maintains a window size that represents the maximum number of out-of-order frames that the receiver is willing to take. Any frame received that is within the receiver window is accepted. Any frame outside the window is discarded. The receiver maintains that the largest acceptable frame, the highest sequence number acceptable, minus the last frame received is less than its window size.



Ref: 16 February 2011 lecture "More Error Detection, Reliability, Ethernet"
Sorry it took so long to get posted.
'''Originally Posted By: razgriz''' (7) Describe how the Sliding Window ARQ algorithm works.<br><br>An ARQ is an automatic repeat request, an algorithm that uses acknowledgements and timeouts to determine which frames are requested again from the sender.<br><br>The sliding window algorithm works by limiting the number of unacknowledged frames that can be sent at a time due to buffer constraints based on the sequence number assigned to each frame. This maximum limit is the window size available both to the sender and receiver.<br><br>The sender's window size / send limit is chosen such that it will keep the transmission pipe as full as possible. The sender can continue sending frames so long as it maintains that the sequence number for the last frame sent minus the sequence number for the last acknowledgement received from the receiver is less than the sender's window size. Thus the sender will not send a packet that has a sequence number outside of its window limit. Any acknowledgement the sender receives is also taken as a promise from the receiver that any frame with a lower sequence number than the acknowledgement is also received and in order.<br><br>Similarly, the receiver maintains a window size that represents the maximum number of out-of-order frames that the receiver is willing to take. Any frame received that is within the receiver window is accepted. Any frame outside the window is discarded. The receiver maintains that the largest acceptable frame, the highest sequence number acceptable, minus the last frame received is less than its window size.<br><br><br><br>Ref: 16 February 2011 lecture &quot;More Error Detection, Reliability, Ethernet&quot;<br>Sorry it took so long to get posted.

-- Practice Midterm Problems (Problem 5)
Originally Posted By: eaon168
Anyone has the answer for problem 4,10 and want to share? Thanks
'''Originally Posted By: eaon168''' Anyone has the answer for problem 4,10 and want to share? Thanks

-- Practice Midterm Problems (Problem 8)
Originally Posted By: eaon168
8. How can one do a broadcast in Ethernet? What is 1-persistence? Describe how a sender knows a packet has arrived in a token ring.

Broadcast in Ethernet
FF:FF:FF:FF:FF:FF

1 persistence
When a send has something to send and the line is busy, the sender will send
immediately after the line becomes idle. The probability that the sender will send
data after the line is idle is 1. It is called 1 persistence.

A sender knows a packet has arrived in a token ring by inspecting the A and C bit
If A is set to 1, the receiver sees the packet. When C is set to 1, the receiver has
copied the data into his buffer.
'''Originally Posted By: eaon168''' 8. How can one do a broadcast in Ethernet? What is 1-persistence? Describe how a sender knows a packet has arrived in a token ring.<br><br>Broadcast in Ethernet<br>FF:FF:FF:FF:FF:FF<br><br>1 persistence<br>When a send has something to send and the line is busy, the sender will send<br>immediately after the line becomes idle. The probability that the sender will send<br>data after the line is idle is 1. It is called 1 persistence.<br><br>A sender knows a packet has arrived in a token ring by inspecting the A and C bit<br>If A is set to 1, the receiver sees the packet. When C is set to 1, the receiver has<br>copied the data into his buffer.
2011-05-22

-- Practice Midterm Problems (Problem 5)
Originally Posted By: eturkov
eaon168 wrote:
Anyone has the answer for problem 4,10 and want to share? Thanks


Here is problem 10 (at least part of it):

Q: Give two different ways frame synchronization can be done in ATM. What is a self-routing fabric?

A: Self routing fabrics rely on some information in the packet header to direct each packet to its correct output. Usually a special "self-routing header” is appended to the packet by the input port after it has determined which output port the packet needs to go to. This extra header is removed before the packet leaves the switch.
'''Originally Posted By: eturkov''' eaon168 wrote:<br>Anyone has the answer for problem 4,10 and want to share? Thanks<br><br><br>Here is problem 10 (at least part of it):<br><br>Q: Give two different ways frame synchronization can be done in ATM. What is a self-routing fabric?<br><br>A: Self routing fabrics rely on some information in the packet header to direct each packet to its correct output. Usually a special &quot;self-routing header&rdquo; is appended to the packet by the input port after it has determined which output port the packet needs to go to. This extra header is removed before the packet leaves the switch.
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