# Computer Networks - Online Test

Q1. The transport layer protocols used for real time multimedia, file transfer, DNS and email, respectively are
Explaination / Solution:

Real time multimedia needs connectionless service, so under lying transport layer protocol used is UDP File transfer rums over TCP protocol with port no-21 DNS runs over UDP protocol within port no-53 Email needs SMTP protocol which runs over TCP protocol within port no – 25

Q2.  Match the problem domains in Group I with the solution technologies in Group II
Group I                                                                     Group II
(p) Services oriented computing                                      (1) Interoperability
(q) Heterogeneous communicating systems                    (2) BPMN
(R) Information representation                                         (3) Publish-find bind
(S) Process description                                                    (4) XML
Explaination / Solution:
No Explaination.

Q3. In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are
Explaination / Solution:

M= 0 – Means there is no fragment after this, i.e. Last fragment
HLEN=10 - So header length is 4×10=40, as 4 is constant scale factor
Fragment Offset = 300, that means 300×8 Byte = 2400 bytes are before this last fragment
So the position of datagram is last fragment
Sequence number of First Byte of Payload = 2400 (as 0 to 2399 Sequence no are used)
Sequence number of Last Byte of Payload = 2400+360-1=2759

Q4. Determine the maximum length of cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s
Explaination / Solution:

Q5.  Consider the following sequence of micro–operations
MBR ← PC
MAR ← X
PC ← Y
Memory ← MBR
Which one of the following is a possible operation performed by this sequence?
Explaination / Solution:

PC content is stored in memory via MBR and PC gets new address from Y. It represents a function call (routine), which is matching with interrupt service initiation

Q6. In the following pairs of OSI protocol layer/sub-layer and its functionality, the INCORRECT pair is
Explaination / Solution:

(a) One of the main functionality of Network Layer is Routing. So Option (a) is CORRECT. (b) Bit Synchronization is always handled by Physical Layer of OSI model but not Data Link Layer. So Option (b) is INCORRECT. (c) End – to – End Process Communication is handled by Transport Layer. So Option (c) is CORRECT. (d) MAC sub layer have 3 types of protocols (Random, Controlled and Channelized Access).

Q7. A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is
Explaination / Solution:

Given 8 – bit delimiter pattern of 01111110.
Output Bit string after stuffing is 01111100101
⇓
StuffedBit
Now, Input String is 0111110101

Q8. Host A (on TCP/IP v4 network A) sends an IP datagram D to host B (also on TCP/IP V4 network B). Assume that no error occurred during the transmission of D. When D reaches B, which of the following IP header field(s) may be different from that of the original datagram D? (i) TTL (ii) Checksum (iii) Fragment Offset
Explaination / Solution:

While an IP Datagram is transferring from one host to another host, TTL, Checksum and Fragmentation Offset will be changed.

Q9. An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are
Explaination / Solution:

Q10. Consider a simple checkpointing protocol and the following set of operations in the log.
(Start, T4); (write, T4,y,2,3); (Start, T1); (commit,T4); (write, T1,z,5,7);
(checkpoint);
(Start,T2); (write, T2,x,1,9); (commit,T2); (start,T3), (write,T3,z,7,2);
If a crash happens now and the system tries to recover using both undo and redo operations, what are the contents of the undo lists and the redo list?
Explaination / Solution:

As T1 & T3are not yet committed they must be undone. The transactions which are after the latest checkpoint must be redone. So T2 must be redone. No need to redo the records which are before last checkpoint, so T4 need not be redone.