Consider six files F1, F2, F3, F4, F5 and F6 of corresponding sizes 100, 200, 70, 40, 250 and 50 respectively. If the files are stored in such a manner to optimize access time, what will be the approximate average access time of a record from one of the six files on the sequential device ?
| A. | 131 |
| B. | 198 |
| C. | 286 |
| D. | 433 |
|
Option: C Explanation : Click on Discuss to view users comments. |
In MS-DOS, relocatable object files and load modules have extensions
| A. | .OBJ and .COM or .EXE respectively |
| B. | .COM and .OBJ respectively |
| C. | .EXE and .OBJ respectively |
| D. | .DAS and .EXE respectively |
|
Option: A Explanation : The Relocatable Object Module Format (OMF) is an object file format used primarily for software intended to run on Intel 80x86 microprocessors. It was originally developed by Intel under the name Object Module Format, and is perhaps best known to DOS users as an .OBJ file. COM file is a type of simple executable file. On the Digital Equipment operating systems of the 1970s, .COM was used as a filename extension for text files containing commands to be issued to the operating system (similar to a batch file). With the introduction of CP/M (a microcomputer operating system), the type of files commonly associated with COM extension changed to that of executable files. This convention was later carried over to MS-DOS. Even when complemented by the more general .exe file format for executables, Click on Discuss to view users comments. |
Primitive disk operating system (POS) manages its disk files in contiguous blocks. A file is saved to the first available space that is large enough to hold the file. Assume that a disk has only 10 contiguous blocks of available free space. Which of the following set of file operations Fa=1, Fb=3, Fc=5, Fd=6 can not be completed given the set of files and file sizes specified below
| A. | save (ƒa), save (ƒb), save (ƒd), delete (ƒd), save (ƒc) |
| B. | save (ƒc), save (ƒa), delete (ƒc), save (ƒd) |
| C. | save (ƒd), save (ƒb), delete (ƒd), save (ƒc) |
| D. | save (ƒb), save (ƒa), delete (ƒb), save (ƒd) |
|
Option: B Explanation : Given Fa=1, Fb=3, Fc=5, Fd=6 Option (a) save (ƒa), save (ƒb), save (ƒd), delete (ƒd), save (ƒc) is possible as we can save 1,3,6 in 9 contiguous block and then delete ( fd) to get 6 contiguous space. After that we can save ( fc ) = 5 in remaining 7 blocks.. So option (a) is possible. Similarly we can execute Option (c) and option (d) where as option (b) is not possible Click on Discuss to view users comments. nirav said: (6:25pm on Sunday 5th May 2013)
explain it
|
A file sometimes called a
| A. | collection of input data |
| B. | data set |
| C. | temporary place to store data |
| D. | program |
|
Option: B Explanation : Click on Discuss to view users comments. |
A program P reads and processes 1000 consecutive records from a sequential file F stored on device D without using any file system facilities. Given the following:
(i) Size of each record = 3200 bytes.
(ii) Access time of D = 10 m secs.
(iii) Data transfer rate of D = 800 x 103 bytes/sec
(iv) CPU time to process each record = 3 m secs.
What is the elapsed time of P if F contains unblocked records and P does not use buffering?
| A. | 12 sec |
| B. | 14 sec |
| C. | 17sec |
| D. | 21sec |
|
Option: C Explanation :
In case P uses one 'Read ahead' buffer the processing and transferring of records can be overlapped. Here Access time = 10ms (given) Transfer time = (800 x 103 )/3200 sec = 0.004 sec = 4 ms
Therefore Elapsed Time = (10 + 4 +3 ) * 1000 m sec = 17 sec. Click on Discuss to view users comments. Appasami said: (7:47pm on Saturday 2nd June 2018)
Transfer time = 3200 /(800 x 10^3 ) sec = 0.004 sec = 4 ms
|
