CPU: 1150118

Q1.a.

b.

Given instruction length is 15 bits, 5 bits are reserved for opcode, 4 bits are reserved for addressing mode. This leaves space for 15 – 5 – 4 = 6 bits for direct and absolute addressing. Giving range of 64 memory locations.

c. In base offset mode, the CPU can access +32 to -31 locations relative to address in base register. By changing base address CPU can access entire memory locations by altering the base address to required range

Q2.

Step	PC	MAR	MDR	IR	A
PC -> MAR	00101	X	X	X	X
	00101	00101	X	X	X
	00101	00101	10101011	10101011	X
	00101	01011	10111000	10101011	X
	00101	00101	10111000	10101011	10111000
	00110	00101	10111000	10101011	10111000
	00110	00110	10111000	10101011	10111000
	00110	00110	01001100	01001100	10111000
	00110	01100	01110100	01001100	10111000
	00110	01100	01110100	01001100	00101100
	00111	01100	01110100	01001100	00101100

Q3.

0x70000000 – 0x7FFFFFFF	Cache – 1MB
0x01C00000 – 0x6FFFFFFF	Empty Space – 483 MB
0x01400000 – 0x01BFFFFF	RAM – 8 MB
0x00400000 – 0x013FFFFF	Module 1 – 16 MB
0x00000000  – 0x003FFFFF	ROM  – 4MB

Q4.

As soon as Laser Printer needs CPU for completing its task. It generates Interrupt request that is forwarded to the INT pin on CPU.

CPU on receiving the interrupt request, completes the current instruction and initiates saving of CPU registers like Accumulator, PC and other status/data registers in to memory location specially reserved for the purpose. This memory area is called the Process Control Block or PCB. Each running process has it’s own PCB area.

Once all current register values are saved in a PCB, CPU switches to execute the Interrupt Serrvice Routine or ISR. Inside ISR, it starts checking on status of each connected device. This is called polling method. The device that generated interrupt would show its status to CPU as source of interrupt.

When CPU determines the printer as source, it branches to ISR of printer and starts executing the interrupt routine for serving the request. Once CPU is done serving the Laser printer, it returns to the previous job by reloading the registers from the saved PCB area and continues further execution where it was interrupted earlier.

Q5.

1. Capacity of each block = 512MB / Number of platters X number of sectors X number of cylinders

= 512 MB / 4 X 20 X 800 = 8KB

Cluster capacity = 4 x 8KB = 32 KB

• Transfer Rate = RPM x Number of Sectors x Capacity each sector / min

= 5400 x 20 x 8 KB / min

= 108000 * 8 KB KB / 60 sec

= 14400 KB / sec

= 14.4 MB /sec

• Average Latency time = Period of rotation / 2

           = ( (1 / 5400) x 60 ) / 2 sec

           = 1/180 sec = 5.55msec

Average Access Time = seek time  + Average Latency

= 36msec + 5.55msec

= 41.55msec