Name:_______Ferdinand Landry II_____________________CMIS 310 – Spring 2015

HOMEWORK #6 – Week #6

This homework is worth 10% of your course grade.

Read each problem carefully. Failure to follow the instructions for a problem will result in a zero score for that problem.

Submit the completed Homework via Assignment in LEO.

1. A 32-bit computer has two selector channels and one multiplexor channel. Each selector channel supports two magnetic disk and two magnetic tape units. The multiplexor channel has two line printers, two card readers, and ten VDTs terminals connected to it. Assume the following transfer rates.

Disk drive 800 Kbytes/sMagnetic tape drive 200 KbytesLine printer 6.6 Kbytes/sCard Reader 1.2 Kbytes/sVDT 1 Kbytes/s

Estimate the maximum aggregate I/O transfer rate in this system. To estimate the maxium aggregate I/O transfer rate in this system we need to estimate the maximum transfer rates for each channel. The maximum transfer rate for each selector channel is 800 (total of 1600KB/sec because only one device at a time can be transferred from on a selector channel. Multiplexor channel allow all devices to be transferred from at the same time for a total of 2(6.6) + 2(12) + 1(10) = 25.6 KB/sec. Therefore:

Maximum aggregate I/O transfer rate = sum of selector channels max transfer rates + multiplexer max transfer rate = 1600KB/sec + 25.6KB/sec = 1625.6 KB/sec

2. 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.

Time Action0 ns Start of main program10 ns IRQ120 ns IRQ335 ns IRQ450 ns IRQ6

Time Action

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0 ns: Start of Main Program10 to 20 ns IRQ120 to 35 ns IRQ335 to 50 ns IRQ450 to 70 ns IRQ670 to 75 ns IRQ475 to 80 ns IRQ380 to 90 ns IRQ190 to 100 ns main program

3. 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?

Each conductor carries 1 bit of information this implies that we need 3 conductor to address a 8 = 2^3 devices. We can add three more conductors to talk back to the I/O control device.

4. Define the terms seek time, rotational delay, and transfer time. Explain their relationship.Definitions:

Seek time = time takes to position disk arm over a specific trackRotational delay = time takes for required sector to position itself under read/write headTransfer time = access time + read data time

Relationship: Access time = Rotational delay + seek timeTherefore, transfer time = rotational delay + seek time + read data time

5. Suppose a disk drive has the following characteristics:

6 surfaces

1024 tracks per surface

128 sectors per track

512 bytes/sector

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Track-to-track seek time of 10 milliseconds

Rotational speed of 6000 RPM.

a. What is the capacity of the drive?

6(1024)(128)(512) = 402653184 bytes implies 384MB

b. What is the access time?

Rotational delay = (60sec/6000RPM) (0.5) = 0.005 seconds = 5milliseconds

Seek time = 10 milliseconds

Access time = rotational delay + seek time = 5 milliseconds + 10 milliseconds

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