DS420
Homework #2
1. #3.1 page 122 . A client sends a 200 byte request message to a service, which produces a
response containing 5000 bytes. Estimate the total time to complete the request in each
of the following cases, with the performance assumptions listed below:
i)
Using connectionless (datagram) communication (e.g. UDP)
ii)
Using connection-oriented communication (e.g. TCP);
iii)
The server process is in the same machine as the client
[ Latency per packet (local or remote, incurred on both send
and received) : 5ms
Connection setup time (TCP only) : 5 ms
Data transfer rate : 10 Mbps
MTU : 1000 bytes
Server request processing time : 2 ms
Assume that the network is lightly loaded. ]
UDP:
5ms + [(200byte * 8bit/byte)/(10*10^6bit/sec)*10^3]ms + 2ms + 5frames * [5ms
+ (1000byte * 8bit/byte)/(10*10^6bit/sec)*10^3]ms
=5ms+1600bit/10*10^3bit/ms+2ms+5frame*(5ms+8000bit/byte/10*10^3bit/ms)
=5ms+0.16ms+2ms+5frame*(5ms+0.8ms)
=7.16ms+5frame*5.8ms
=7.16ms+29ms
=36.16 ms
TCP:
connection setup time + UDP
= 5ms + 36.16ms
= 41.16ms
Same Machine : the messages can be sent by a single in memory copy; estimate interprocess
data transfer rate is m Mbps (should be faster than 10 mbps data transfer rate).
The total time = 5.0 + 200*8/(m * 10000) + 2.0 + 5.0 + 5000*8 / (m * 10000) ms
(there is no connection setup time)
2. #3.7 page 123. Compare connectionless (UDP) and connection-oriented (TCP)
communication for the implementation of each of the following application-level or
presentation-level protocols:
i)
virtual terminal access (e.g. Telnet )
ii)
file transfer (e.g. FTP )
iii)
user location (e.g. rwho , finger);
iv)
information browsing (e.g.HTTP)
v)
remote procedure call
i)
ii)
iii)
iv)
v)
The long duration of sessions, the need for reliability and the unstructured sequences of
characters transmitted make connection-oriented communication most suitable for this
application. Performance is not critical in this application, so the overheads are of little
consequence.
File calls for the transmission of large volumes of data. Connectionless would be o.k. if
error rates are low and the messages can be large, but in the Internet, these requirements
aren't met, so TCP is used.
Connectionless is preferable, since messages are short, and a single message is sufficient
for each transaction.
Either mode could be used. The volume of data transferred on each transaction can be
quite large, so TCP is used in practice.
RPC achieves reliability by means of timeout and re-trys, so connectionless (UDP)
communication is often preferred.
3. #3.9 page 123. A specific problem that must be solved in remote terminal access
protocols such as Telnet is the need to transmit exceptional events such as 'kill signals'
from the 'terminal' to the host in advance of previously-transmitted data. Kill signals
should reach their destination ahead of any other ongoing transmissions. Discuss the
solution of this problem with connection-oriented and connectionless protocols.
This problem is that a kill signal should reach the receiving process quickly even when there is
buffer (e.g. caused by an infinite loop in the sender) or other exceptional conditions at the
receiving host.
With a connection-oriented , reliable protocol such as TCP, all packets must be received and
acknowledged in the order in which they are transmitted. Thus a kill signal cannot overtake
other data already in the stream. To overcome this, an out-of-band signalling mechanism must be
provided. In TCP, this is called the URGENT mechanism. Packets containing data that is flagged
as URGENT bypass the flow-control mechanisms at the receiver and are processed immediately.
With connectionless protocols, the process at the sender simply recognized the event and
sends a message containing a kill signal in the next outgoing packet. The message must be resent
until the receiving process acknowledges it.
3. #4.7 page 162 Sun XDR marshals data by converting it into a standard big-endian form
before transmission. Discuss the advantages and disadvantages of this method when
compared with CORBA's CDR.
The XDR method which uses a standard form is inefficient when communication takes
place between pairs of similar computers whose byte orderings differ from the standard. It is
efficient in networks in which the byte-ordering used by the majority of the computers is the
same as the standard form. The conversion by senders and recipients that use the standard form
is in effect a null operation.
In CORBA CDR senders include an identifier in each message and recipients convert the
bytes to their own ordering if necessary. This method eliminates all unnecessary data
conversions, but adds complexity in that all computers need to deal with both variants.
4. #4.8 page 163 Sun XDR aligns each primitive value on a four byte boundary, whereas
CORBA CDR aligns a primitive value of size n on an n-byte boundary. Discuss the
trade-offs in choosing the sizes occupied by primitive values.
Marshalling is simpler when the data matches the alignment boundaries of the computers
involved. Four bytes is large enough to support most architectures efficiently, but some space
is wasted by smaller primitive values. The hybrid method of CDR is more complex to
implement, but saves some space in the marshalled form.
6. a) Describe and illustrate the process of decrypting base64 encoded text, using as an
example the encoded text
QSBzaG9ydCBzYW1wbGU=
The RFC 2045 "MIME message formats" will also be helpful.
Step 1:
Convert the letters into base64 codes
16,18,1,51,26,6,61,50,29,2,1,51,24,22,53,48,27,6,20
Step 2:
Convert base64 codes into binary numbers
01000001,00100000,01110011,01101000,01101111,01110010,
01110100,00100000,01110011,01100001,01101101,01110000,
01101100,01100101,00
Step 3:
Convert binary numbers into decimal numbers
65,32,115,104,111,114,116,32,115,97,109,112,108,101
Step 4:
Convert decimal numbers to ASCII characters
Ans: A short sample
b) A given file is 3072 bytes long. How many bytes long will it be if it is encoded into base64
encoding, where a CR-LF pair is inserted after every 80 output bytes and at the end?
Step 1:
3072 bytes long file becomes 3072 *(4/3) = 4096 bytes in bsae64 format.
No. of 80-character line =ceiling 4096/80 = 52 lines
 52 pairs of CR-LF are needed.
Step 2:
Total size of the file = 4096 + 52* 2 = 4200
Ans: 4200 bytes long will it be if it is encoded into base64 encoding.