1
GATE CSE 1996
Subjective
+5
-0
A demand paged virtual memory system uses $$16$$ bit virtual address, page size of $$256$$ bytes, and has $$1$$ Kbyte of main memory. $$LRU$$ page replacement is implemented using a list, whose current status (page numbers in decimal ) is GATE CSE 1996 Operating Systems - Memory Management Question 22 English
$$\eqalign{ & \,\, \uparrow \cr & LRU\,Page \cr} $$
For each hexa decimal address in the address sequence given below,
$$00FF,$$ $$010D,$$ $$10FF,$$ $$11B0$$
Indicate,
i) The new status of the list
ii) Page faults, if any, and
iii) Page replacements, if any
2
GATE CSE 1996
Subjective
+5
-0
A file system with a one-level directory structure is implemented on a disk with disk block size of $$4$$ K bytes. The disk is used as follows:

Disk-block $$0:$$ File Allocation Table, consisting of one $$8$$-bit entry per date block, representing the data block address of the next date block in the file:

Disk block $$1:$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ Directory, with one $$32$$ bit entry per file:
Disk block $$2:$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ Data block $$1;$$
Disk block $$3:$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ Data block $$2;$$ etc.

(a) What is the maximum possible number of files?
(b) What is the maximum possible file size in blocks?

3
GATE CSE 1996
Subjective
+5
-0
A computer system uses the Banker’s Algorithm to deal with deadlocks. Its current state is shown in the tables below, where P0, P1, P2 are processes and R0, R1, R2 are resource types. GATE CSE 1996 Operating Systems - Deadlocks Question 8 English
(a) Show that the system can be in this state.
(b) What will the system do on a request by process P0 for one unit of resource type R1?
4
GATE CSE 1996
MCQ (Single Correct Answer)
+2
-0.6
The correct matching for the following pairs is

List - I
(A) Activation record
(B) Location counter
(C) Reference counts
(D) Address relocation

List - II
(1) Linking loader
(2) Garbage collection
(3) Subroutine call
(4) Assembler

A
A – 3 B – 4 C – 1 D - 2
B
A – 4 B – 3 C – 1 D - 2
C
A – 4 B – 3 C – 2 D - 1
D
A – 3 B – 4 C – 2 D - 1
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