Given the following set of events, show which routines the CPU is executing for

Question

Given the following set of events, show which routines the CPU is executing for times 0 to 100 ns. Each handler routine (with its interrupt request) takes 20 ns to complete. The priority of the interrupts ranges from IRQ6 as the highest priority interrupt to IRQ0 as the lowest priority interrupt. If an address bus needs to be able to address eight devices, how many conductors will be required? What if each of those devices also needs to be able to talk back to the I/O control device? Define the terms seek time, rotational delay, and transfer time. Explain their relationship. Suppose a disk drive has the following characteristics: 5 surfaces 1024 tracks per surface 256 sectors per track 512 bytes/sector Track-to-track seek time of 8 milliseconds Rotational speed of 7500 RPM. a. What is the capacity of the drive? b. What is the access time?

Explanation / Answer

ANS 2)
Here we need to consider two things:
-interrupts are pre-emptive
-whether this is a priority-based interrupt.

Assuming the system is pre-emptive and doesn't mask other interrupts,and by the data we know that an interrrupt
will complete 1 instruction in 20 ns then:-

Time Routine will be
0 Main program
0-10ns Handler for IRQ 1
10-30ns Handler for IRQ3
30-50ns Handler IRQ4
50-100ns Handler IRQ6

Finally in parallel to IRQ4, IRQ6 will start.

Ans 3)
Given an address bus needs to be able to address 16 devices.
So, Log(16) = no. of conductors
Hence LoG(16) base 2 = 4.

So number of conductors is 4.

If we want these devices to talk back to I/O Control device then we need to add one more conductor.   
So, log(32) base 2=5 conductors.

Ans 4)    Seek Time= The time taken by the read/write heat to move to the correct location on the disk
Rotational delay= Time taken by the rotational devices to rotate to the place where read/write head can acccess it.
Tranfer time= Total estimated time taken to read or write a file.

Relationship:-
Seek Time,Rotational delay and transfer time are the components of Disk access time.It is the time
when you initiate read/write to the time it is totally complete.

Ans 5)
The capacity of the disk will be:-
6 SURFACES * 1024TRACKS PER SURFACE * 128 SECTOR * 512 bytes/SECTOR
=6*128*512 KB
=384 MB.

Access Time=Seek time+ average rotational latency.
Given seek time=10ms
average roattaiona latency =5/2 ms=2.5 ms

Hence the average access time=10+2.5 ms= 12.5 ms Ans.

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