PREVIOUS YEAR SOLVED PAPERS - GATE 2020
- Option : A
- Explanation : PP(3, 4) = _____
PP function:
Len = tob(b, arr)
tob functions:
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
| Processes | P1 | P2 | P3 | P4 |
| Burst time (in ms) | 8 | 7 | 2 | 4 |
Note – Numerical Type question
- Option : A
- Explanation : Average turnaround time
P1 = 21, P2 = 13, P3 = 2, P4 = 6
Avg = (21+13+2+6)/4 = 42/4 = 10.5
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
- Option : A
- Explanation : 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
- Option : B
- Explanation : 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
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
