PREVIOUS YEAR SOLVED PAPERS - GATE 2020
Therefore 4 rows returned by the above query.- Option : B
- Explanation : Statement 1: False, Because, Intermediate router may modify fields like TTL, offset checksum. Statement 2: True, router may not implement routing protocol, for static routing. Statement 3: False, It is not mandatory to reassemble.
- A
The hole created by the best fit is never larger than the hole created by first fit
- B
The hole created by the best fit is always larger than the hole created by next fit
- C
The hole created by the next fit is never larger than the hole created by best fit
- D
The hole created by worst fit is always larger than the hole created by first fit
- Option : A
- Explanation : Let us consider a new process required 120 kb memory and existing holes in the memory are 200, 300, 150 kb as shown in the diagram in the same order.
Now, when we allocate, this new process a memory using different algorithms, it would be like given below
Algorithm Allocated partition size of new tube
First Fit 200 KB 80 KB
Best Fit 150 KB 30 KB
Worst Fit 300 KB 180 KB
Next Fit 300 KB 180 KB
- Option : B
- Explanation : Let L1 = an bn⇒ CFL L2 = Σ* ⇒ Regular L1 ∪ L2 = (an bn) ∪ Σ* = Σ* ⇒ Regular Since L1 ∪ L2 is regular but L1 is not regular, Hence statement I is false. Regular language is not closed under infinite union. Hence, statement-II is also false. ∴ Neither I nor II is true.
