1. Give three examples of protocol parameters that might be negotiated when a connection is
set up
The negotiation can be on setting the window size, maximum packet size (MTU
or MSS), and timer values.
3.A computer on a 6-Mbps network is regulated by a token bucket. The token bucket is
filled at a rate of 1 Mbps. It is initially filled to capacity with 8 megabits. How long can
the computer, transmit at the full 6 Mbps?
let T be the taken to transmit full 6Mbps
then maximum number of capacity of bucket in Tsec is c+rT.
then in 1sec=c+rT/T=maximum capacity(M)
therefore,T=c/M-r
=8Mbps/6Mbps-1Mbps
=1.6sec,Ans
4. Is TCP checksum necessary or could TCP allow IP to checksum the data
TCP checksums are identical to UDP checksums, with the exception
that checksums are mandatory with TCP (instead of being optional, as
they are with UDP). Furthermore, their usage is mandatory for both the
sending and receiving systems.
6. Datagram subnets route each packet as a separate unit, independent of all others.
Virtualcircuit subnets do not have to do this, since each data packet follows a
predetermined route. Does this observation mean that virtual-circuit subnets do not need
the capability to route isolated packets from an arbitrary source to an arbitrary
destination? Explain your answer
Answer:
No, it does not. In virtual-circuit subnets, routers have to route packets based on the following
pieces of information:
- Virtual-Circuit number
- Incoming line
Each packet contains a Virtual-Circuit number and the incoming line identification is, of course,
known by the router.
Finally, to re-estate the observation, virtual-circuit subnets do need the capability to route isolated
packets from an arbitrary incoming line (which is well know by the router) and with an arbitrary
circuit number (information carried in the packet). That means, knowing the original source and the
final destination is not required.
7. Suppose two TCP connections are present over some bottleneck link of rate R bps.
Both connections have a huge file to send (in the same direction over the bottleneck link).
The transmissions of the files start at the same time. What is the transmission rate that the
TCP would like to give to each of the connections?
Consider data,
Two TCP connections are present over some bottleneck link of rate R bps.
Both connections have a huge file to send. transmissions of the files start
at the same time.
Transmission rate would TCP like to give to each of the connections= R/2.
8. Consider the following design problem concerning implementation of virtual-circuit
service. If virtual circuits are used internally to the subnet, each data packet must have a
3-byte header and each router must tie up 8 bytes of storage for circuit identification. If
datagrams are used internally, 15-byte headers are needed, but no router table space is
required. Transmission capacity costs 1 cent per 10 6 bytes, per hop. Very fast router
memory can be purchased for 1 cent per byte and is depreciated over two years, assuming
a 40-hour business week. The statistically average session runs for 1000 sec, in which
time 200 packets are transmitted. The mean packet requires four hops. Which
implementation is cheaper, and by how much?
mean packet requires 4 hops means 3 router in between the source and destination.
header size =3 bytes for each packet so overhead for 200 packets=200*3=600 bytes
routers have 8 bytes for circuit identification means total=8*3=24 bytes
so total overhead=600+24=624 bytes
let the data size=D,so packet size=(D+3) bytes
transmission cost =1 cent per 10^6 bytes per hop=>transmission cost for 1 Byte,per
hop=10^-6 cent
transmission cost for 1 packet=(D+3)*number of hops*transmission cost of 1 hop
=>(D+3)*4*10^-6 cents
so total costs=200*(D+3)*4*10^-6 cents=>8*(D+3)*10^-4 cents
in the second case header size is 15 bytes for each packet so total header
overhead=200*15=3000 bytes
no router table space is required means space=0 byte
so total overhead=3000 bytes
cost for 1 packet=(D+15)*4*10^-6 cents
=>for 200 packets=>200*(D+15)*4*10^-6 cents=8*(D+15)*10^-4 cents
so from above discussion we can say first one implementation is cheaper
12. Suppose nodes A, B, and C each attach to the same broadcast LAN (through their
adapters). If A sends thousands of frames to B with each frame addressed to the LAN
address of B, will C's adapter process these frames? If so, will C's adapter pass the IP
datagram’s in these frames to C (i.e., the adapter's parent node)? How will your answers
change if A sends frames with the LAN broadcast address?
C’s adapter will process the frames, but the adapter will not
pass the datagrams up the protocol stack. If the LAN broadcast
address is used, then C’s adapter will both process the frames
and pass the datagrams up the protocol stack..
14. Suppose there are three routers between a source host and destination host. Ignore
fragmentation. An IP datagram sent from the source host to the destination host will
travel over how many interfaces? How many forwarding table will be indexed to move
the datagram from the source to destination?


An IP datagram sent from the source host to the destination host
will travel over 8 interfaces.
3 forwarding tables will be indexed to move the datagram from the
source to the destination
15. A router has just received the following new IP addresses: 57.6.96.0/21,
57.6.104.0/21, 57.6.112.0/21, and 57.6.120.0/21. If all of them use the same outgoing
line, can they be aggregated? If so, to what? If not, why not?
They can be aggregated into 57.6.96.0/19.
17. Consider sending a 2400-byte datagram into a link that has an MTU of 700 bytes.
Suppose the original datagram is stamped with the identification number 422. How many
fragments are generated? What are the values in the various fields in the IP datagram(s)
generated related to fragmentation?
datagram size is 2400 bytes including ip header size.so only payload size is 2380bytes.
MTU 700B also includes ip header.allowable size 700-20=680B.
1)first fragment is of size 680B.offset is (0-84)
2)second fragment is also size 680B.it again fragmented because it is greater than
MTU size.MTU ip header,allowable datasize is 380B.
i)first fragment size in second one is 376B.offset is (85-131)
ii)second fragment size in second one is 304B.offset is(132-169)
3)third fragment is also size 680B. offset is (170-254)
4)fourth fragment is of size 344B.offset is (255-297).it includes 4 padding bits.
so total 5 fragments .
18. Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment
tie up reassembly buffers forever. Suppose that a datagram is fragmented into four
fragments. The first three fragments arrive, but the last one is delayed. Eventually, the
timer goes off and the three fragments in the receiver's memory are discarded. A little
later, the last fragment stumbles in. What should be done with it?
As far as the receiver is concerned, this is a part of new datagram, since no
other parts of it are known. It will therefore be queued until the rest show
up. If they do not, this one will time out too.
19. What is the fastest line speed at which a host can blast out 1500-byte TCP payloads
with a 120-sec maximum packet lifetime without having the sequence numbers wrap
around? Take TCP, IP, and Ethernet overhead into consideration. Assume that Ethernet
frames may be sent continuously.
Case 1: Assume that there is no segmentation needed.
IP header = 20 bytes
TCP header = 20 bytes
TCP payload can be 1500 bytes.
Ethernet header = 8 + 6 + 6 + 2 bytes
Ethernet checksum = 4 bytes
Therefore, the size of first frame is 1566 bytes.
The time to live = 120 sec.
The length of sequence number = 32 bits.
Therefore, the maximum number of bytes that a host can send out without
wrapping around the sequence number is 2 ^32/ 1500.
Since Ethernet frames may be sent continuously, the maximum line speed, V can
be:
V <= 1566 * 8 * (2^32 / 1500) (bits)/120 (sec)
= 298929723.8016 bits/sec
= 298.9 Mbps
The maximum line speed is 298.9 Mbps.
Case 2: Since the maximum size of Ethernet payload is 1500 bytes, the TCP
payload needs to be segmented as followings.
Frame 1:
IP header = 20 bytes
TCP header = 20 bytes
Since the maximum Ethernet Payload is 1500 bytes including IP header and TCP
header, TCP payload can be 1460 bytes.
Ethernet header = 8 + 6 + 6 + 2 bytes
Ethernet checksum = 4 bytes
Therefore, the size of first frame is 1526 bytes
Frame 2:
IP header = 20 bytes
TCP header = 20 bytes
TCP payload remained after segmentation = 40 bytes (= 1500 – 1460 bytes)
Ethernet header = 8 + 6 + 6 + 2 bytes
Ethernet checksum = 4 bytes
Therefore, the size of second frame is 106 bytes
The time to live = 120 sec.
The length of sequence number = 32 bits.
Therefore, the maximum number of bytes that a host can send out without
wrapping around the sequence number is 2 ^32/ 1500.
Since Ethernet frames may be sent continuously, the maximum line speed, V can
be:
V <= (1526 + 106) * 8 * (2^32 / 1500) (bits)/120 (sec)
= 1632 * 8 * (2^32 /1500) (bits)/ 120 (sec)
= 311528294.5365 bits/sec
= 311.5 Mbps
The maximum line speed is 311.5 Mbps.
20. When the IPv6 protocol is introduced, does the ARP protocol have to be changed? If
so, are the changes conceptual or technical?
There are no changes conceptually. Technically, the IP addresses requested are
now bigger, so bigger fields are needed.
21. Suppose Host A sends two TCP segments back to back to Host B over a TCP
connection. The first segment has sequence 200; the second segment has sequence
number 2024. a. How much data is there in the first segment? b. Suppose that the first
segment is lost but the second segment arrives at B. In the acknowledgement that Host B
send to Host A, what will be the acknowledgement number?
a)
Consider sequence numbers,First segment=90
Second segment=110
Data in the first segment=110-90
=20
b) Consider the first segment is lost but the second segment arrives at B.
In the acknowledgment that Host B sends to Host A, then the
acknowledgment number will be first segment of sequence number, that
is 90.
22. Consider the effect of using slow start on a line with a 10-msec round-trip time and no
congestion. The receive window is 24 KB and the maximum segment size is 2 KB. How
long does it take before the first full window can be sent?
SolutionGiven


Receiver window size = 24 KB
Maximum Segment Size = 2 KB
RTT = 10 msec
Receiver Window SizeReceiver window size in terms of MSS
= Receiver window size / Size of 1 MSS
= 24 KB / 2 KB
= 12 MSS
Slow Start ThresholdSlow start Threshold
= Receiver window size / 2
= 12 MSS / 2
= 6 MSS
Slow Start Phase-




Window size at the start of 1st transmission = 1 MSS
Window size at the start of 2nd transmission = 2 MSS
Window size at the start of 3rd transmission = 4 MSS
Window size at the start of 4th transmission = 6 MSS
Since the threshold is reached, so it marks the end of slow start phase.
Now, congestion avoidance phase begins.
Congestion Avoidance Phase





Window size at the start of 5th transmission = 7 MSS
Window size at the start of 6th transmission = 8 MSS
Window size at the start of 7th transmission = 9 MSS
Window size at the start of 8th transmission = 10 MSS
Window size at the start of 9th transmission = 11 MSS
Window size at the start of 10th transmission = 12 MSS
From here,


Window size at the end of 9th transmission or at the start of 10th transmission is 12
MSS.
Thus, 9 RTT’s will be taken before the first full window can be sent.
So,
Time taken before the first full window is sent
= 9 RTT’s
= 9 x 10 msec
= 90 msec
24. Consider a TCP connection between Host A and Host B. Suppose that the TCP
segments traveling from Host A to Host B have source port number x and destination port
number y. What are the source and destination port numbers for the segments traveling
from Host B to Host A?
The source and destination port numbers for the segments travelling
from Host B to Host A: Source port number y and destination port
number x.
25. Describe why an application developer might choose to run an application over UDP
rather than TCP.
Transmission Control Protocol (TCP): Transmission Control Protocol is a
connection oriented protocol that establishes the connections between the
computers before sending the data.
User Datagram Protocol (UDP): User Datagram Protocol is a connectionless
protocol in which data is sent to destination computer without checking whether the
system is ready to receives the data or not.
The reasons why an application developer chooses to run an application over
UDP rather than TCP are as follows:
• At the time of congestion, the TCP’s congestion control suffocates the application’s
sending rate. Thus, many application developers don’t want their applications to use
TCP’s congestion control.
• When application runs over UDP, many more active clients can be supported by
the server which is devoted to a particular application.
• Even though data transfer by TCP is reliable, some applications do not need
reliable TCP data transfer. Therefore, application developers prefer UDP.
• Generally, designers of IP (internet protocol) video conference applications and IP
telephony run their applications over UDP to avoid TCP congestion control.
26. Ten thousand airline reservation stations are competing for the use of a single slotted
ALOHA channel. The average station makes 18 requests/hour. A slot is 125 µsec. What
is the approximate total channel load
Total No. of stations=10,000
Average requset/station=18 req/hr (1req/200sec)
Total No of requests=18*1000=18000 req/hr
Or 50 requests/sec
Each terminal makes one request every 200 sec, for a total load of 50 requests/sec.
Hence G = 50/8000 = 1/160.