1
GATE CSE 2015 Set 2
MCQ (Single Correct Answer)
+2
-0.6
Consider the intermediate code given below.
(1)  i = 1
(2)  j = 1
(3)  t1 = 5 ∗ i
(4)  t2 = t1 + j
(5)  t3 = 4 ∗ t2
(6)  t4 = t3
(7)  a[t4] = -1
(8)  j = j + 1
(9)  if j<=5 goto (3)
(10) i=i+1
(11) if i<5 goto (2)

The number of nodes and edges in the control-flow-graph constructed for the above code, respectively, are

A
5 and 7
B
6 and 7
C
5 and 5
D
7 and 8
2
GATE CSE 2015 Set 2
Numerical
+1
-0
A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgement has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at the nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is ___________.
Your input ____
3
GATE CSE 2015 Set 2
Numerical
+2
-0
Two hosts are connected via a packet switch with 107 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 1000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last of the data in microsecond is _________.
Your input ____
4
GATE CSE 2015 Set 2
MCQ (Single Correct Answer)
+2
-0.6
Host A sends a UDP datagram containing 8880 bytes of user data to host B over an Ethernet LAN. Ethernet frames may carry data up to 1500 bytes (i.e. MTU = 1500 bytes). Size of UDP header is 8 bytes and size of IP heard is 20 bytes.There is no option field in IP header How many total number of IP fragments will be transmitted and what will be the contents of offset field in the last fragment?
A
6 and 925
B
6 and 7400
C
7 and 1110
D
7 and 8880
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