Operating Systems | Set 14

2.5

Following questions have been asked in GATE CS 2006 exam.

1) Consider three CPU-intensive processes, which require 10, 20 and 30 time units and arrive at times 0, 2 and 6, respectively. How many context switches are needed if the operating system implements a shortest remaining time first scheduling algorithm? Do not count the context switches at time zero and at the end.
(A) 1
(B) 2
(C) 3
(D) 4

Answer (B)
Let three process be P0, P1 and P2 with arrival times 0, 2 and 6 respectively and CPU burst times 10, 20 and 30 respectively. At time 0, P0 is the only available process so it runs. At time 2, P1 arrives, but P0 has the shortest remaining time, so it continues. At time 6, P2 arrives, but P0 has the shortest remaining time, so it continues. At time 10, P1 is scheduled as it is the shortest remaining time process. At time 30, P2 is scheduled. Only two context switches are needed. P0 to P1 and P1 to P2.



2) A computer system supports 32-bit virtual addresses as well as 32-bit physical addresses. Since the virtual address space is of the same size as the physical address space, the operating system designers decide to get rid of the virtual memory entirely. Which one of the following is true?
(A) Efficient implementation of multi-user support is no longer possible
(B) The processor cache organization can be made more efficient now
(C) Hardware support for memory management is no longer needed
(D) CPU scheduling can be made more efficient now

Answer (C)
For supporting virtual memory, special hardware support is needed from Memory Management Unit. Since operating system designers decide to get rid of the virtual memory entirely, hardware support for memory management is no longer needed



3) A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation look-aside buffer (TLB) which can hold a total of 128 page table entries and is 4-way set associative. The minimum size of the TLB tag is:
(A) 11 bits
(B) 13 bits
(C) 15 bits
(D) 20 bits

Answer (C)
Size of a page = 4KB = 2^12
Total number of bits needed to address a page frame = 32 – 12 = 20
If there are ‘n’ cache lines in a set, the cache placement is called n-way set associative. Since TLB is 4 way set associative and can hold total 128 (2^7) page table entries, number of sets in cache = 2^7/4 = 2^5. So 5 bits are needed to address a set, and 15 (20 – 5) bits are needed for tag.


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