Solution

Hardwired control unit has the highest performance compared to the microprogramming. Because control unit is implemented with the gates,flip-flops,decoders and other digital circuits. Since it has dedicated piece of hardware it’s performance is superior compare to microprogrammed control unit.

Among the horizontal microprogramming and vertical microprogramming horizontal microprogramming has better performance compared to vertical microprogramming.

Solution

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 18 bits long in burst mode DMA transfers 2^18 Bytes (= 256KB) once it gets the control.

To transfer 32,750 KB , no. of times DMA needs to take control = (32,750 KB / 256KB) = 32,750/256 = 127.93, means 128 times.

Solution

To access a word from disk takes the same time as accessing a sector because both involve seek time and rotational delay.

Solution

Rotational delay=½(rotational time)

It is given that the rotational rate is 6000 RPM.

6000 rotations----------->60 sec

1 rotation--------------->?

Rotational time = 60/6000 sec ⇒ 1 / 100 sec ⇒ 10 msec

Rotational delay = 10/2 = 5 msec

Solution

S1 is true as daisy chaining is used to assign priorities in attending interrupts.

S2 is false as vectored interrupt doesn’t involve polling but non-vectored interrupt involves polling.

Solution

A is true as polling means that the CPU periodically checks the status bits to know if any device needs attention.

B is true as during DMA only one of the CPU or DMA can be bus master at a time.

C is false because in cycle stealing mode DMA releases the bus control after one clock cycle.

D is true because in burst mode DMA holds the bus control for many cycles till all the required data is transferred.

Solution

It is given in the question that the DMA maintains the control of the bus until the whole block is transferred. This means that the DMA is operating in burst mode. In the burst mode the DMA is controlled by the device speed.

Time to start + end DMA = 250 + 250 = 500 nsec. Since device is slow, the DMA speed is governed by device speed.

Time taken by device to transfer 128 bytes = (128/100 * 1024) sec = 1.31msec

Total time taken = 500 nsec + 1.31 msec. ≈ 1.31 msec.

Solution

Given that the disk pack has 16 surfaces, 64 tracks per surface, 256 sectors per track and each sector size is 512 bytes.

So the disk pack capacity = 16 * 64 * 256 * 512 bytes = 128 MB

To specify a sector we need the information about surface number, track number and sector number within a track. Surface number needs 4 bits as there are 16 surfaces(2^4), track number needs 6 bits as there are 64 tracks(2^6) within a surface, within a track the sector number needs 8 bits as there are 256 sectors (2^8). Total number bits needed to specify a particular sector = 4+6+8 = 18 bits.

Solution

In horizontal microprogramming we use one bit for each control signal.

So the no. of bits required for the control signal in horizontal case = 4+8+10+20+3 = 45

But in vertical case we can represent the encoded version of the control signals:

So the no. of bits required for the control signal in vertical case = ceil(log 4)+ ceil(log 8)+ ceil(log 10) + ceil(log 20)+ ceil(log 3) = 2+3+4+5+2 = 16

Solution

The bus cycle is stolen 512 times as it is a cycle stealing mode.

Time to start + end DMA = 150 + 150 = 300 nsec.

Time for bus transfers (512 times) = 300*512 = 153600 nsec.

Time for 512 bus cycles = 512 * 80 = 40960 nsec.

Total = 153600 + 40960 = 194560 ns ~= 0.195ms

Solution

Disk rotation speed is given as 2400 RPM.

Means, 2400 rotations in 60 seconds.

Time for one rotation = 60 / 2400 seconds = 1/40 seconds = 25 ms

X = Average rotational latency = ½(Time for one rotation) = 25/2 = 12.5 ms

We know in one rotation one entire track can be read, that is in 25 ms we can read 32768 bits.

In one second we can read = 32768 / (25*10^-3) = 1.31 * 10^6 bits

Y = data transfer rate = 1.31 Mbits / sec

Solution

Number of words in control memory = 256 * 8 = 2^11 words

Address field = 11 bits

There are 12 flags. So no. of bits for the 12(next power of 2, 2^4) flags = 4

There are 120 control signals, no. of bits for the control signals = 7

Control word contains the following 3 fields:

Length of the control word = 4+7+11 = 22 bits.

Solution

Disk pack consist of 20 surfaces

256 tracks/surface,128 sectors/track and 256 bytes/sector.

Given formatted overhead = 32 bytes/sector

Total capacity of the disk pack = 20 surfaces * 256 tracks/surface * 128 sectors/track * 256 bytes/sector ⇒ 160 MB

Formatted overhead of the sectors= number of sectors * formatted overhead

= 20 surfaces * 256 tracks/surface * 128 sectors/track * 32 bytes/sector

= 20 MB

Formatted disk space= 160 MB - 20 MB= 140 MB

Solution

c⇒cylinder number

h⇒surface number

s⇒sector number

Hard disk has 128 sectors/track, 20 platters and each with 2 recording surfaces <500, 10, 84> means we need to cross the 500 cylinders, 10 surfaces in a particular cylinder and 84 sectors in a particular track to find the desired sector. So we need to find the how many sectors we are crossing to get the sector address <500, 10, 84>.

Number of sectors we need to cross = 500 cylinders(20 platters/cylinder * 2 surfaces/platter *1 tracks/surface *128 sector/track) + 10 surfaces(1 tracks/surface * 128 sectors/track) + 84 sectors ⇒ 2561364 sectors

Solution

The address is <20,15,33>

Number of sectors we need to cross=20 cylinders(20 platters/cylinder * 2 surfaces/platter * 1 track/surface * 128 sector/tack)+15 surfaces(1 track/surface * 128 sectors/track )+33 sectors=104353 sectors.