UGC NET CSE | August 2016 | Part 3 | Question: 51

Question

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 ?

• A. $30$ microseconds
• B. $34$ microseconds
• C. $60$ microseconds
• D. $68$ microseconds

Answer 1

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

Answer 2

Ok Let us first calculate the time required to access the drums for one-page access.

Time to access drum once = Time to access the sector + Time to transfer the data = Avg rotational latency + Time to transfer 1000 words

Avg rotational latency = $\frac{One.Rotation.Time}{2}$ = $\frac{1}{100}seconds$ = 10,000 micro seconds

Time to transfer one page(=1000 words) = $\frac{1}{10^{3}}$ seconds = 1000 micro seconds

Therefore, time to access the drum once = 11,000 micro seconds

Now we can easily solve this problem

Every instruction need one cycle to initially access the page = 1 micro second

1% need additional one cycle = $0.01*1microsecond=0.01 micro second$

Of those 1% only 20% causes page fault = $0.01*0.2*11000 micro seconds=22 micro seconds$

Of those one percent’s 20%, 50% need to swap with a dirty page in the memory = $0.01*0.2*0.5*11000 micro seconds=11 micro secons$

Now summing up all the cases = 1+0.01+22+11=34.01 micro second ≅34 micro second

Comments

Thanks bruh! Awesome explanation ✨🔥

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thanks

Answer 3

B

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

Comments

Can you explain the equation you are getting?

Answer 4

Effective access time $=099\times (1\;\mu \text{sec} + 0.008\times (2\;\mu\text{sec}) + 0.002 \times(10,0000\;\mu \text{sec} + 1,000\;\mu\text{sec}) + 0.001\times(10,000\;\mu\text{sec} + 1,000\;\mu\text{sec}) \\=(0.99 + 0.016 + 22.0 + 11.0) \;\mu\text{sec}\\=34.0\;\mu\text{sec}$

$\therefore$ B is the right option.

Comments

If possible can you please elaborate?