PREVIOUS YEAR SOLVED PAPERS MCQ - GATE 2020

The value returned by pp(3, 4) is ______.

Note – Numerical Type question

PP(3, 4) = _____
PP function:
Gate2020 cs

Len = tob(b, arr)
tob functions:
Gate2020 cs

i = 0 4 > 0, (with b = 4)
if (4 % 2) False, so else gets executed
So, arr[0] = 0,
if t b = b/2
i = 1 2 > 0 (with b = 2)
if (2 % 2) false
So, else gets executed ⇒ arr [1] = 0, i = i + t b = b/2
i = 2 1 > 0 (with b = 1)
if (1 % 2) True arr [2] = 1 i + 1, b = b/2
i = 30 > 0 (with b = 0) false
return I, returns ‘3’ to len, in PP function.
In PP function: len = 3
For loop.

i = 0	i = 1	i = 2
ex = 3 x 3	ex = 9 x 3	tot = t * t *ex
= 9	= 81	= 1 * 81 = 81

62 Consider the following C functions.

The return value of fun2(5) is _______.

Note – Numerical Type question

63 Consider the following set of processes, assumed to have arrived at time Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order P₁, P₂, P₃, P₄.

Processes	P₁	P₂	P₃	P₄
Burst time (in ms)	8	7	2	4

If the time quantum for RR is 4 ms, then the absolute value of the difference between the average turnaround times (in ms) of SJF and RR (round off t0 2 decimal places) is ……………...

Note – Numerical Type question

Average turnaround time
P₁ = 21, P₂ = 13, P₃ = 2, P₄ = 6
Avg = (21+13+2+6)/4 = 42/4 = 10.5
Gate2020 cs

Average turnaround time
P₁ = 18, P₂ = 21, P₃ = 10, P₄ = 14
Avg = (18+21+10+14)/4 = 63/4 = 15.75
Difference = 15.75 – 10.5 = 5.25

64 Consider a database implemented using B+ tree for file indexing and installed a disk drive with block size of 4 KB. The size of the search key is 12 bytes and size of tree/disk pointer is 8 bytes. Assume that the database has one million records. Also assume that no node of the B+ tree and no records are present initially in main memory. Consider that each record fits into one disk block. The minimum number of disk accesses required to retrieve any record in the database is ______.

Note – Numerical Type question

No. of records = 10⁶
Block size = 8 KB
Search key = 13 bytes
Block pointer size = 8 bytes
Balancing factor of index file = [Block size /(search key + Block pointer)]
= [4KB/(12+8)] = [2¹²/20]
Balancing Factor of index file =204
Gate2020 cs

65 Consider a TCP connection between a client and a server with the following specifications: the round trip time is 6 ms, the size of the receiver advert window is 50 KB, slow-start threshold at the client is 32 KB, and the maximum segment size is 2 KB. The connection is established at time t = Assume that there are no timeouts and errors during transmission. Then the size of the congestion window (in KB) at time t + 60 ms after all acknowledgments processed is _______.

Note – Numerical Type question

TCP connections at t = 0
1MSS = 2KB Threshold = 32KB (slow start)
RTT = 6ms. Then how big packet it (packet size) may deliver after t + 60 ms
Gate2020 cs

1 RTT = 6 ms
⇒ 10 RTT = 60 ms
t + 60 = 0 + 60 = 60 ms
At 60 ms, CWND = 22 MSS
1 MSS = 2KB
⇒ 22 MSS = 22 × 22 = 44 KB