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An operating system supports a paged virtual memory, using a central processor with a cycle time of one microsecond. It costs an additional one microsecond to access a page other than the current one. Pages have $1000$ words, and the paging device is a drum that rotates at $3000$ revolutions per minute and transfers one million words per second. Further, one percent of all instructions executed accessed a page other than the current page. The instruction that accessed another page, $80$% accessed a page already in memory and when a new page was required, the replaced page was modified $50$% of the time. What is the effective access time on this system, assuming that the system is running only one process and the processor is idle during drum transfers ?

1. $30$ microseconds
2. $34$ microseconds
3. $60$ microseconds
4. $68$ microseconds
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Effective access time =0.99*(1misec) +0.8*.02misec+0.001*(10000 misec +1000 misec)  + 0.001 * (20000 misec +2000 misec)

=(0.99 +0.016+22.0+11.0)misec= 34 misec

bcoz .99 in memory & current page (no prob. dude)

further .8 i.e. 80% of those not current i.e. .01 are in memory(remember:It costs an additional one microsecond to access a page other than the current one.) so 2 misec.

Now, it's turn for those not in memory which's  .01*.2 i.e. 20% of 1%

"1 million words per minute means 1000 1000 pages per sec are transfered 1 page in 1msed or 1000 misec"

only the rotation part is not understood by me still 3000 per min means 50 per sec or 20000 mi sec for each rotn. so rotn + tx time =20000 misec +1000 misec (or maybe avg rotn time =10000)

.001 do this & further .001 also modify it(double & is last term above topmost eqn.). (50% 50% each half of .002 which access outside memory)

misec=microsecond
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Effective access time =0.99*(1sec +0.008*2) +0.002*(10000 sec +1000 sec)  + 0.001 * (10000 sec +1000 sec)

=(0.99 +0.016+22.0+11.0)= 34 sec