a.Consider a situation in which a cyberterrorist makes all the DNS servers in the world crash
simultaneously. How does this change one’s ability to use the Internet? (Computer Networks A top
Down Approach)
The DNS servers provide a mapping between domain names and IP ad- dresses, such that when a
request for a Web page is received, the browser can look up in the DNS server the IP address
corresponding to the domain name of the requested page, and then download the requested page from
that IP ad- dress. If all the DNS servers in the world were to crash at the same time, one would not be
able to map between domain names and IP addresses. Therefore, the only way to access Web pages
would be by using the IP address of the host server instead of the domain name. Since most of us do not
know the IP ad- dresses of the servers we access, this type of situation would make use of the Internet
extremely inefficient, if not virtually impossible for most users.
b. Suppose that the TCP congestion window is set to 18 KB and a timeout occurs. How big will the
window be if the next four transmission bursts are all successful? Assume that the maximum segment
size is 1 KB.
When a timeout occurs, three things happend. First, slow start will be initiated. Second, the congestion
window would start at 1. Third, the threshold will be reset to 18KB/2=9KB. If the next four transmission
are all successful, then

1st transmission: 1 segment, 1KB

2nd transmission: 2 segments, 2KB

3rd transmission: 4 segments, 4KB

4th transmission: 8 segments, 8KB
After these four successful transmissions, the window size is supposed to be 16. However, since the
threshold is 9KB, the window size can only be 9KB.
c. What is Silly Window Syndrome? Explain different algorithms that can be used to avoid the problem
of silly window syndrome.



“Silly Window” occurs when small-sized segments are transmitted, resulting in inefficient
use of the network pipe
For e.g., suppose that TCP sender generates data slowly, 1-byte at a time
Solution: wait until sender has enough data to transmit – “Nagle’s Algorithm”
Nagle’s Algorithm
1. TCP sender sends the first piece of data obtained from the application (even if data is only a few
bytes).
2. Wait until enough bytes have accumulated in the TCP send buffer or until an ACK is received.
3. Repeat step 2 for the remainder of the transmission.
•
Suppose that the receiver consumes data slowly
–
•
Receive Window opens slowly, and thus sender is forced to send small-sized segments
Solutions
–
Delayed ACK
–
Advertise Receive Window = 0, until reasonable amount of space available in receiver’s
buffer
d. What is the polynomial representation of 101110?
101110 → x5 + x3 + x2 + x
2. Consider sending a 8000-byte datagram into a link that has an MTU of 500 bytes. Suppose that the
original datagram has the identification number 510.
a) How many fragments are generated?
b). What are the values of Identification,DM, MF and fragment offset?
Solution: Maximum Size of data in fragment = 500 – 20 = 480.
a)
Required number of fragments (i.e # of fragments generated) = ceil(8000-2/480) = 17
Offset = 480/8 = 60
b)
Length
480
480
480
480
480
480
480
480
480
480
480
480
480
480
480
480
300
ID
510
510
510
510
510
510
510
510
510
510
510
510
510
510
510
510
510
Fragment Flag
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Offset
0
60
120
180
240
300
360
420
480
540
600
660
720
780
840
900
960
3.
a. Leaky Bucket.
//see slides to give reason
b. Token Bucket.
//same as above
4.
14 bit sequence = 11000011110101
D19
D18
D17
P16
D15
D14
D13
D12
1
1
D11
D10
D9
P8
D7
D6
1
1
P8
0
1
0
D5
P4
D3
P2
P1
Insert Data
1
1
0
P16
0
0
0
Finding Parity Bits (Following Even Parity)
Even Parity: If count of 1’s is odd then parity bit is 1 else 0.
For P1
Count one location and skip one location starting from P1
P1 = 0
For P2
Count two locations and skip two location starting from P2
P2 = 0
For P4
Count four locations and skip four locations starting from P4
P4 = 0
For P8
Count eight locations and skip eight locations starting from P8
For P16
Count Sixteen locations from P16 and Skip sixteen locations
P4
1
P2
P1
P16 = 0
New Sequence
1
1
0
0
0
0
0
1
1
1
1
0
0
1
0
0
1
0
0
For decoding, assume a single bit error in above sequence. Then for that sequence, find parity values.
After finding the parity values write them in this order : P16 P8 P4 P2 P1
This sequence will give the location of that bit which contains error.
5.
Solution.
6.
Solution.
The total number of addresses in this block is 232-16 = 65536. The ISP can divide
this large block in several ways depending on the predicted needs of its customers
in the future. We assume that the future needs follow the present pattern. In other
words, we assume that the ISP will have customers that belong to one of the
present groups. We design four ranges: group 1, group 2, group 3, and one reserved
range of addresses as shown in Figure 19.1.