GATE CSE 2005 | Question: 70

Question

Consider a disk drive with the following specifications:

$16$ surfaces, $512$ tracks/surface, $512$ sectors/track, $1$ KB/sector, rotation speed $3000$ rpm. The disk is operated in cycle stealing mode whereby whenever one $4$ byte word is ready it is sent to memory; similarly, for writing, the disk interface reads a $4$ byte word from the memory in each DMA cycle. Memory cycle time is $40$ nsec. The maximum percentage of time that the CPU gets blocked during DMA operation is:

• A. $10$
• B. $25$
• C. $40$
• D. $50$

Comments

Disk / track parameters

• Sectors/track = 512, sector size = 1 KB ⇒ bytes per track = 512 × 1 KB = 512 KB.

• Rotation speed = 3000 rpm = 3000/60 = 50 rotations per second.

• So sustained transfer rate from one track = 50 × 512 KB = 25600 KB/s = 25600 × 1024 B/s.

Time between successive 4-byte words becoming available

• Time to produce 4 bytes = 4 bytes / (25600 × 1024 bytes/s)
    =425600×1024\;=\dfrac{4}{25600\times1024}=25600×10244​ s ≈ 152.6 ns.

DMA steals one memory cycle per 4-byte transfer

• Memory cycle time = 40 ns (that is the time DMA uses per transfer; CPU is blocked only during that time if it needs memory).

• Fraction of time CPU is blocked ≈ 40 ns / 152.6 ns ≈ 0.262 = 26.2%.

GATE answer is 25%

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@goku4199 by default we are assuming continues transfer ? Not isolated transfer right ? and preparation takes 4 cycles and dma takes 1 cycles -> total cycles = 5 out of which 1 cycle cpu got blocked => 1/5 * 100 = 20% ....
why u took 1/4

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@goku4199 thankyou very much. 

@JHighlight preparation take 4 cycles and dma take 1 cycle is correct, but in first cycle, preparatoin and transfer happen simultaneouly as you can see in goku's diagram.

 in 1st dma cycle: disk->diskBuffer & diskBuffer -> Main Memory (in which cpu is blocked).

in 2nd , 3rd , 4th cycle disk -> diskBuffer is still continue.

Answer 1

First lets calculate the disk transfer rate.

Only one surface is read/written at a time. 

• In one rotation, one track is read
• One track has $512$ sectors and $1\;KB$ per sector means $512\;KB$ per track
• Rotations Per Minute is $3000\implies 3000 \times 512\;KB$ per minute (or) $50 \times 512\;KB$ per second =  $25600\;KBps$
• Time to read $4$ bytes $ = \dfrac{4}{25600 \times 1024} s = 152.6  ns$
• Since memory cycle time is $40\; ns$ this will be $\left \lceil \dfrac{152.6}{40} \right\rceil =4$ cycles.

Now, coming to DMA, it does not block the CPU directly but it steals the memory cycles and if the CPU is needing memory it gets blocked. So, in worst case (see the usage “maximum percentage of time” in question) lets assume CPU is needing memory all the time (fully memory bound process execution). In this case the percentage of time CPU gets blocked will be the amount of time DMA is using the memory cycles. 

• Whenever $4$ bytes is ready from disk, memory cycles are taken – this needs $40\;$ ns
• When memory cycles are used, disk will continue to fetch further data – it is not getting blocked here or the disk wont be waiting for this $40\;ns$ if it has any pending read
• That is, if DMA is continuously transferring data, one in every $4$ memory cycles can get blocked.

So, the maximum percentage of time CPU gets blocked $ = \dfrac{1}{4} \times 100 =  25 \%$

Correct Option: B

Reference

Comments

@Ray Tomlinson You seem to be confusing tracks and cylinders. Modern hard disk drives use multiple platters/surfaces to increase the disk capacity but at any given time only a singular head is doing the reading or writing, hence the speed only depends on how much can be read/written on a singular track.

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perfect explanation

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Best intuitive simple explanation

Answer 2

$512$ KB-$1/50$  sec

$4$ Byte transfer will take total :$4/(512\times50\times2^{10})=152.58 \ ns$

DMA will transfer $4$B in $40$nsec

So, Cpu will be blocked $(40/152.58)=26 \% $ of time

Best matching answer is (B).

Comments

why did we consider by default pipeline here while we have not considered it here in this gate question

https://gateoverflow.in/3694/gate2004-it-51

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@Pratyush Priyam Kuan read comments in https://gateoverflow.in/3694/gate2004-it-51, it will clear your doubt.

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why are we not considering the avg. rotational latency and the seek time here.

Answer 3

B. 25

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Does 3000 rotations in 60 seconds

$\implies$ 1 rotation in 20 ms

In 1 rotation covers data in 1 track which is = 512 × 1 KB
20 ms $\longleftrightarrow$ 512 KB
1 sec $\longleftrightarrow$ 25600 KB
$\implies$ Transfer rate = 25600 KBps
Transfer Rate means we can perform read or write operations(one at a time) with this speed.
In one DMA cycle of 40 ns we are able to transfer 4 Bytes to disk.

CPU Idle Time = 40 ns
Data Prepration Time = $\frac{4 Bytes}{25600 KB}$ = 156.25 ns

% Time CPU Blocked = $\frac{40 ns}{40 ns + 156.25 ns} \times 100$ = 20.382

Comments

why haven't u multiplied by 16 bcoz in one rotation it can cover all 16 surfaces ??

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@sushmita I think the cpu should be idle. It may not access the system bus also since the DMA will be using it, and if does not access the system bus, then how can it carry out even the smaller RTL activities?

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it depends on disk architecture whether there are read-write heads on all surfaces or not. you can check it by matching options. here option only matches without taking 16 surfaces.

Answer 4

Rotation Speed = 3000 rpm = 50rps, 1round = $\frac{1}{50}$ seconds

Track Capacity = $512*1024 = 2^{19} =512KB$

Bytes to be transferred = 4B
Transfer Time =  $\frac{4}{50*512*1000}$ = 156.25ns  =4 memory cycles

So, One out of every 4 cycles will be stolen = that is 25% (CPU % Block)

It means CPU will only be blocked for 1 cycle, and for remaining 3 cycles it will be busy doing its own work!

Answer 5

Please see the solution given by an IIT professor for the same problem with a little change in parameters in the following

Answer 6

Time takes for 1 rotation = 60/3000
It reads 512*1024 Bytes in one rotation.
Time taken to read 4 bytes = 153 ns
153 is approximately 4 cycles (160ns)
Percentage of time CPU gets blocked = 40*100/160 = 25

Comments

Hi,

can you pls tell me what is happening in 40ns memory cycle and 153 ns time?i mean CPU will be idle for 40ns cycle time but what will it be doing during 153 ns??how is the total time 153??

can you pls clarify my doubts??

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time taken to transfer 4 bytes is 4/(512*50*2^10)=152.58 ns almost 153 ns.Please clear concept on how tracks and sectors works you will understand.

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actually i am confused that in cycle stealing mode,once the 1 byte is tranferred by DMA,it gives the control to CPU.so CPU is ideal for one cycle time but time taken to transfer  4 bytes is 153 by calculation ,so what will the CPU be doing in that time??

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what work will be done in 40 ns by DMA if it takes 153 to transfer 4 bytes?? are you understanding my doubt??