UGC NET CS 2014 Dec – III

Question 1

A hierarchical memory system that uses cache memory has cache access time of 50 nano seconds, main memory access time of 300 nano seconds, 75% of memory requests are for read, hit ratio of 0.8 for read access and the write-through scheme is used. What will be the average access time of the system both for read and write requests ?

• A. 157.5 n.sec.
• B. 110 n.sec.
• C. 75 n.sec.
• D. 82.5 n.sec.

Answer: (A)

UGC NET CS 2014 Dec - III Computer Organization and Architecture
Discuss it

cache access time = 50 ns; main memory access time = 300 ns. Since the system is hierarchical: t_read = hit ratio * cache access time + (1 - hit ratio) *(main memory access time + cache access time) = 0.8 * 50 + 0.2 * (300 + 50) = 110 ns t_write = 1 * max(main memory access time, cache access time) = 1 * max(300,50) = 1* 300 = 300 ns. read frequency = 75%:; write frequency = 25%. Now, average access time for both read and write requests = read frequency * t_read + write frequency * t_write 0.75 * 110 + 0.25 * 300 = 157.5 ns So, option (A) is correct.

Question 2

For switching from a CPU user mode to the supervisor mode following type of interrupt is most appropriate

• A. Internal interrupts
• B. External interrupts
• C. Software interrupts
• D. None of the above

Answer: (C)

UGC NET CS 2014 Dec - III OS Process Management
Discuss it

For switching from a CPU user mode to the supervisor mode Software interrupts occurs. Software interrupts is internal interrupt triggered by some some software instruction. And external interrupt is caused by some hardware module. For more information on Interrupt Refer:Interrupts in 8085 microprocessor Option (C) is correct.

Question 3

In a dot matrix printer the time to print a character is 6 m.sec., time to space in between characters is 2 m.sec., and the number of characters in a line are 200. The printing speed of the dot matrix printer in characters per second and the time to print a character line are given by which of the following options?

• A. 125 chars/second and 0.8 seconds
• B. 250 chars/second and 0.6 seconds
• C. 166 chars/second and 0.8 seconds
• D. 250 chars/second and 0.4 seconds
• E. 125 chars/second and 1.6 seconds

Answer: (E)

UGC NET CS 2014 Dec - III OS Process Management
Discuss it

Total no of characters = 200 (each character having one space) Time taken to print one character = 6 ms; Time taken to print one space = 2 ms. character printing = 200 * 6 = 1200 ms space printing = 200 * 2 = 400 ms total printing time = 1200 + 400 = 1600 ms = 1.6 s. The printing speed of the dot matrix printer in characters per second = 200 / 1.6 = 125 / s. So, option (E) is correct.

Question 4

Match the following:

List - I	List - II
(a)XCHG	(i)only carry flag is affected.
(b)SUB	(ii) no flags are affected.
(c)STC	(iii) all flags other than carry flag are affected.
(d)DCR	(iv)all flags are affected.

codes:

	(a)	(b)	(c)	(d)
(1)	(iv)	(i)	(iii)	(ii)
(B)	(iii)	(ii)	(i)	(iv)
(C)	(ii)	(iii)	(i)	(iv)
(D)	(ii)	(iv)	(i)	(iii)

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Answer: (D)

UGC NET CS 2014 Dec - III Microprocessor
Discuss it

In XCHG instruction exchange of data takes place so, no flags are affected.

In SUB instruction subtraction occurs so, all flags are affected.

STC is set carry flag so, only carry flag will be affected

DCR is decrement flag all flag other then carry will be affected.

So, option (D) is correct.

Question 5

How many times will the following loop be executed?  LXI B, 0007 H LOP : DCX B   MOV A, B   ORA C   JNZ LOP

• A. 05
• B. 07
• C. 09
• D. 00

Answer: (B)

UGC NET CS 2014 Dec - III Microprocessor
Discuss it

Question 6

Specify the contents of the accumulator and the status of the S, Z and CY flags when 8085 microprocessor performs addition of 87 H and 79 H.

• A. 11, 1, 1, 1
• B. 10, 0, 1, 0
• C. 1, 1, 0, 0
• D. 00, 0, 1, 1

Answer: (D)

UGC NET CS 2014 Dec - III Microprocessor
Discuss it

Question 7

Location transparency allows: I.Users to treat the data as if it is done at one location. II. Programmers to treat the data as if it is at one location. III. Managers to treat the data as if it is at one location. Which one of the following is correct?

• A. I, II and III
• B. I and II only
• C. II and III only
• D. II only

Answer: (A)

UGC NET CS 2014 Dec - III Distributed system
Discuss it

In distributed databases, Location transparency allows: I.Users to treat the data as if it is done at one location. II. Programmers to treat the data as if it is at one location. III. Managers to treat the data as if it is at one location. All statement are correct. So, option (A) is correct.

Question 8

Which of the following is correct ? I.Two phase locking is an optimistic protocol. II. Two phase locking is pessimistic protocol III. Time stamping is an optimistic protocol. IV. Time stamping is pessimistic protocol.

• A. I and III
• B. II and IV
• C. I and IV
• D. II and III

Answer: (D)

UGC NET CS 2014 Dec - III Transactions and concurrency control
Discuss it

Two phase locking is pessimistic protocol and Time stamping is an optimistic protocol. For more information On two phase locking protocol and time stamp protocol Refer:DBMS | Concurrency Control Protocol | Two Phase Locking (2-PL)-III and DBMS | Concurrency Control Protocols | Timestamp Ordering Protocols Option (D) is correct.

Question 9

__________ rules used to limit the volume of log information that has to be handled and processed in the event of system failure involving the loss of volatile information.

• A. Write-ahead log
• B. Check-pointing
• C. Log buffer
• D. Thomas

Answer: (B)

UGC NET CS 2014 Dec - III Transactions and concurrency control 50 DBMS MCQs with Answers
Discuss it

• Check-pointing rules used to limit the volume of log information that has to be handled and processed in the event of system failure involving the loss of volatile information.
• Write-ahead log provide atomicity and durability in transaction.
• Log buffer is a small contiguous memory which holds memory transaction logs per database.
• Thomas write rule is a timestamp based concurrency control protocol.

For more information on Thomas write rule Refer: DBMS | Concurrency Control Protocol | Thomas Write Rule

Option (B) is correct.

Question 10

Let R = ABCDE is a relational scheme with functional dependency set F = {A → B, B → C, AC → D}. The attribute closures of A and E are

• A. ABCD, φ
• B. ABCD, E
• C. Φ, φ
• D. ABC, E

Answer: (B)

UGC NET CS 2014 Dec - III Database Design(Normal Forms) 50 DBMS MCQs with Answers
Discuss it

The attribute closures of A: A⁺ = ABCD The attribute closures of E: E⁺ = E. So, option (B) is correct.