ISRO CSE 2014 | Question: 58

Question

Assume the following information.

• Original timestamp value $= 46$
• Receive  timestamp value $= 59$
• Transmit timestamp value $= 60$
• Timestamp at arrival of packet $= 69$

​Which of the following statements is correct?

• A. Receive clock should go back by $3$ milliseconds
• B. Transmit and Receive clocks are synchronized
• C. Transmit clock should go back by $3$ milliseconds
• D. Receive clock should go ahead by $1$ milliseconds

Comments

ARRIVAL OF PACKET SHUD BE 67....

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This is from Network Time Protocol (NTP), specifically clock synchronization which is NOT in the current GATE CS syllabus

Answer 1

Ans a.

This topic is from ICMP and I dont think its in GATE syllabus anymore. Still it goes like this :

Note : TS -> Timestamp

assume there is a sender and a reciever.

Original TS : Time at which sender sent packet acc to its own clock.

Recieving TS : Time at which reciever recieves that packet acc to its own clock.

Transmit TS : Time at which reciever sent a reply of that packet acc to its own clock.

Returned TS : Time at which sender of recieves the reply acc to its own clock.

=> Sending time = Recieving time - Original time

=> Recieving time = Return time - Transmit time

=> RTT = Sending Time + Recieving time (Think why ?)

=> One way time should be $\frac{RTT}{2}$ => 10ms

=> Sender sent the packet at 46 and reciever should have recieved it at 56 , but its recieving it at 59, => its 3 ms ahead. So it should set its clock 3 ms behind to sync with sender.

Comments

but RTT =22 not 20.

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Exactly.

Answer 2

Given the following information:

Value of original timestamp: 46

Value of receive timestamp: 59

Value of transmit timestamp: 60

Time the packet arrived: 67

We can calculate:

Sending time = 59 − 46 = 13 milliseconds

Receiving time = 67 − 60 = 7 milliseconds

Round-trip tim e = 13 + 7 = 20 milliseconds

Time difference = receive timestamp − ( original timestamp field + one-way time duration)

Time difference = 59 − (46 + 10) = 3

So,answer is (A)

Comments

@ Divya Bharti

 u have taken the Time the packet arrived as 67 but it is given as 69 ,

then the round trip time will be 22ms and for one way it will be 11.

 then,

 time difference = 59 − (46 + 11) = 2