GATE CSE 2016 | Set 1 | Question: 31

Question

The size of the data count register of a $\text{DMA}$ controller is $16\;\text{bits}$. The processor needs to transfer a file of $29,154$ kilobytes from disk to main memory. The memory is byte addressable. The minimum number of times the $\text{DMA}$ controller needs to get the control of the system bus from the processor to transfer the file from the disk to main memory is _________.

Comments

@goku4199 how 2^10 came?

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1Kilo=2^10

when dealing with data size(bits or bytes) use 2^x. E.g. Kilo=2^10.Mega=2^20

 when dealing with time(mili sec,micro sec etc) use 10^x.eg mili sec=10^(-3)

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@pavansan

below image might help in conversion of clock freq to time and vice versa.

Answer 1

Data count register gives the number of words the DMA can transfer in a single cycle..

Here it is $16$ bits.. so max $2^{16}$ words can be transferred in one cycle..

Since memory is byte addressable.. $1 \text{ word}=1\;\text{byte}$
                                                so $2^{16}$ bytes in $1$ cycle..
Now for the given file..
                               File size $=29154\ \textsf{KB} = 29154\times 2^{10}\ \textsf{B}$
                                                  $1$ cylce $\rightarrow$ DMA transfers $2^{16}\ \textsf{B}$
i.e 
                                             $1\ B$ transfered by DMA $\rightarrow \dfrac{1}{2^{16}}$ cycles.

Now, for full file of size $29154\ \textsf{KB},$

Minimum number of cylces $=\dfrac{(29154\times 2^{10}\ B)}{2^{16}}= 455.53$

But number of cylces is asked so $455.53\rightarrow 456.$

Comments

wow. I was not taught this concept man. I was told only 2^16-1 B can be tranerred per time. Here it as irrelevant but couldve costed me marks

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max 2^16 words can be transfered in this 0 word is included and once the data count is 0 no transfer . so it should be 2^16 - 1 . Please correct me if I am wrong

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according to best of my knowledge..if data count register is of 3 bits then when its valeu is 7 then we can transfer ..when 6  wecan transfer ..........when 1 wecan transfer but when it becomes 0 we can not transfer so when data count register is of 3 bit hen we can transfer date 2^3-1 times so similarilywhen data count register is of 16 bit we can transfer data 2^16-1 times which is65535 time so for 29154KB we need controller for 12154*1024/65535  which is 456 times

Answer 2

Size of data count register of the DMA controller = 16 bits

Data that can be transferred in one go = 2¹⁶ bytes = 64 kilobytes File size to be transferred = 29154 kilobytes.

So, number of times the DMA controller needs to get the control of the system bus from the processor to transfer the file from the disk to main memory = ceil(29154/64) = 456  

Comments

c? There are no options:p. Correct answer btw:)

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Correct answer is 456 only.

Answer 3

minimum 456

Comments

An explanation would be helpful.

Answer 4

1. Convert file size to bytes:

   • 29,154 kilobytes * 1024 bytes/kilobyte = 29,805,568 bytes

2. Calculate maximum transfer size per DMA request:

   • 2^16 - 1 = 65,535 bytes (16-bit data count register)

3. Divide total file size by maximum transfer size:

   • 29,805,568 bytes / 65,535 bytes/request = 455.99 requests

4. Round up to the nearest whole request:

   • 455.99 requests ≈ 456 requests

Answer 5

Since nothing is mentioned about the mode of DMA working whether it is cycle stealing mode or burst mode, we consider it as burst mode by default.
As the data count register of the DMA is 16 bits long in burst mode DMA transfers 216 Bytes (= 64KB) once it gets the control.
To transfer 29,154 KB, no. of times DMA needs to take control
= (29,154 KB / 64KB)
= 29,154/64
= 455.53, means 456 times.

Comments

it is 2^16 means at a time can identify 64KB of data.

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Why burst mode is default?

We can do with cycle stealing(1) mode or transparent mode (0).

Answer 6

456 cycles

Comments

but if I use K=10^3 then I am getting wrong answer.This is not making any sense.I should get correct answer if I use either 2^10 or 10^3 I shuld get same answer right.Someone please clear my doubt.