The hypothetical machine has two I/O instructions: 0011 = Load AC from I/o 0111 = Store AC to I/O In these cases the 12 bit address identifies a particular I/O device. 1.Load AC from device 5 2.Add contents of memory location 940 3.Store AC to device 6

Assume that the next value retrieved from device 5 is 3, and that location 940 contains a value of 2.

#1. Fetch: The instruction at memory location 300 (this is the content of PC) is loaded into IR. This instruction is  3005 (load AC from I/O device 5)

PC) is loaded into IR. This instruction is  3005 (load AC from I/O device 5)

Memory			CPU Registers
300	3005		300	PC
301	5940			AC
302	7006		3005	IR
940	0002		I/O Device
			005	0003

#2. Execute : AC is loaded with the content of I/O device 5. PC is incremented.

Memory			CPU Registers
300	3005		301	PC
301	5940		0003	AC
302	7006		3005	IR
940	0002		I/O Device
			005	0003

#3. Fetch: The instruction at memory location 301 (this is the content of PC) is loaded into IR.  This instruction is 5940 (add  AC to the content of location 940)

Memory			CPU Registers
300	3005		301	PC
301	5940		0003	AC
302	7006		5940	IR
940	0002		I/O Device
			005	0003

#4. Execute: The content of AC is added to the content of location 940. The result is in AC. PC is incremented

Memory			CPU Registers
300	3005		302	PC
301	5940		0005	AC
302	7006		5940	IR
940	0002		I/O Device
			005	0003

#5. Fetch: The instruction at memory location 302 (this is the content of PC) is loaded into IR.  This instruction is 7006 (store AC at I/O device 6)

Memory			CPU Registers
300	3005		302	PC
301	5940		0005	AC
302	7006		7006	IR
940	0002		I/O Device
			005	0003

#6. Execute: The content of AC is stored into I/O device 6. PC is incremented.

Memory			CPU Registers
300	3005		303	PC
301	5940		0005	AC
302	7006		7006	IR
940	0002		I/O Device
			005	0003
			006	0005