In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. If an empty frame is available or if the replaced page is not modified, and it takes 20 m.sec., if the replaced page is modified. What is the average access time to service a page fault assuming that the page to be replaced is modified 70% of the time?
| A. | 11.6 m.sec. |
| B. | 16.4 m.sec. |
| C. | 28 m.sec. |
| D. | 14 m.sec. |
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Option: A Explanation : Click on Discuss to view users comments. Muthu Mariyappan said: (4:55pm on Monday 29th December 2014)
c
Deepak said: (7:22pm on Thursday 23rd November 2017)
Effective Memory Access Time = P*(Page fault occurs) (1-p)* Memory Access TimeP = 70 %therefore , Effective M.A.T = 70/100*8 30/100*20 = 11.6 msec
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Which one of the following is used to compute cyclomatic complexity?
| A. | The number of regions - 1 |
| B. | E - N + 1, where E is the number of flow graph edges and N is the number of flow graph nodes. |
| C. | P – 1, where P is the number of predicate nodes in the flow graph G. |
| D. | P + 1, where P is the number of predicate nodes in the flow graph G. |
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Option: D Explanation :
cyclomatic complexity formulas
C(n)=E+N-2P
C(n)=P+1
Click on Discuss to view users comments. Muthu Mariyappan said: (4:56pm on Monday 29th December 2014)
b
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The directory can be viewed as _______ that translates filenames into their directory entries.
| A. | Symbol table |
| B. | Partition |
| C. | Swap space |
| D. | Cached |
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Option: A Explanation : A directory can be viewed as a symbol table that translates file names into directory entries.
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What does the following expression means?
char *(*(*a[N]) ()) ();
| A. | a pointer to a function returning array of n pointers to function returning character pointers. |
| B. | a function return array of N pointers to functions returning pointers to characters. |
| C. | an array of n pointers to function returning pointers to characters. |
| D. | an array of n pointers to function returning pointers to functions returning pointers to characters. |
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Option: D Explanation : Click on Discuss to view users comments. |
Consider the following S1 and S2 :
S1 : A hard handover is one in which the channel in the source cell is retained and used for a while in parallel with the channel in the target cell.
S2 : A soft handover is one in which the channel in the source cell is retained and only then the channel in the target cell is engaged.
| A. | S1 is true and S2 is not true. |
| B. | S1 is not true and S2 is true. |
| C. | Both S1 and S2 are true. |
| D. | Both S1 and S2 are not true. |
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Option: D Explanation :
Answer is D
A hard handover is one in which the channel in the source cell is released and only then the channel in the target cell is engaged. Thus the connection to the source is broken before or 'as' the connection to the target is made—for this reason such handovers are also known as break-before-make
A soft handover is one in which the channel in the source cell is retained and used for a while in parallel with the channel in the target cell. In this case the connection to the target is established before the connection to the source is broken, hence this handover is called make-before-break. The interval, during which the two connections are used in parallel, may be brief or substantial.
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