CO question

Question

The average seek time and rotational delay in a disk system are 6 ms and 3 ms, respectively. The rate of data transfer to or from the disk is 30 MBps and all disk accesses are of 8 KB of data. Disk DMA controller, the processor, and the main memory are all attached to a single bus. The data bus width is 32 bits, and a bus transfer to or from the main memory takes 10 nanoseconds. How many disk units are there that can be simultaneously transferring data to or from the main memory?

• 11
• 12
• 13
• 14

Answer 1

As per my understanding, Its given that every component is attached to bus and and can transfer data through bus with 30 MBps(By single unit).

We are asked to find how many units can transfer simultaneously  with 30 MBps so that bus can do it in its capacity.

Bus is 32 bits wide and data tranfer to or from takes 10 ns, so 32 bits can be carried in 10 ns.

32 bits/10ns =4 byte/10 ns =400 MBPS possible speed.

given speed for disk unit is 30MBPS hence 400/30=13.33 which should be 13.

Comments

But disk can export only 30MB data per second, and according to your calculation disk can transfer 400MB (approx) data which is contradicting the question statement:

The rate of data transfer to or from the disk is 30 MBps

 Made easy also used the same logic. Is it possible?

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I found this example in Hamecher book also in one of the exercises.

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@Shubhanshu

400 MBps is capacity of the bus not of the disk.

One disk can transfer 30MBps.

we have to find maximum such no. of disk which can be attached to the bus.Which can only be possible when we transfer with maximum avalible bandwidth i.e.400 MBps.

so, 30 * x<=400

x<=400/30=40/3=13.34=13

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Yes 30 MB can be transferred by one unit

...number units that cam send 400MB are 13

This is what they meant...and as usual ambiguity in language is always the aim of test series questions !