5. Consider a hypothetical machine below. There are 7 opcodes and the word struc

Question

5. Consider a hypothetical machine below. There are 7 opcodes and the word structure for an instruction is 16 bits and arranged such thatthe left most 4 bit s are the opcode and the right most 12 bits are memory address. Fill in the blanks for the remaining tables to show program execution. Begin program execution at address 620 hex. (4 pts) Opcode (Bits) Description 0001 Load AC from M(x) 0010 Store AC to MIX 0101 Add to AC from MXL Move ACH to AC 0011 0100 Multiply AC by M(x), 32 bit product, Bits (31:16 aced in ACH, Bits (15-0 aced in AC 1110 Load AC from I/O Address 1111 Subtract l/O Address from AC Opcode Bits 1512 Address, Bits (11:0) Data, Bits (15:0) Registers Memory Registers Memory 245 2785 620 pc 245 A 00 B 246 ACH 246 ACH 247 55 57 8 1 0 2 AC 247 AC /O 620 4245 /o 620 621 2247 1 040 301. 621 301 3301 2080 302 622 302 622 Registers Memory Memory Registers 245 245 PC 246 ACH 246 ACH 247 AC AC 247 /O 620 /o 620 621 301 621 301 622 302 622 302

Explanation / Answer

TABLE :

Address

Data

245

2785

246

A00B

247

5557

:

:

:

:

620

4245

621

2247

622

3301

Register

Data

PC

620

ACH

XXXX

AC

8102

IR

4245

I/O

ADDRESS

DATA

301

1040

302

2080

Starting from the instruction at address 620:

TABLE I

Register

Data

PC

621

ACH

13EA

AC

540A

IR

4245

I/O

ADDRESS

DATA

301

1040

302

2080

TABLE II

Register

Data

PC

622

ACH

13EA

AC

540A

IR

2247

I/O

ADDRESS

DATA

301

1040

302

2080

Address

Data

245

2785

246

A00B

247

540A

:

:

:

:

620

4245

621

2247

622

3301

TABLE III

Register

Data

PC

623

ACH

13EA

AC

13EA

IR

3301

I/O

ADDRESS

DATA

301

1040

302

2080

Address

Data

245

2785

246

A00B

247

540A

:

:

:

:

620

4245

621

2247

622

3301

Note: The instruction 3301 might be given incorrect because the opcode 3 has some operation to do and the address 301 pertains to I/O which is a mismatch. Had it been E301 instead of 3301, the first hex code E is “load AC from I/O address”, the contents would look like as below. The contents of the memory location 301 which is 1040 gets loaded in to the register AC.

TABLE IV

Register

Data

PC

623

ACH

13EA

AC

1040

IR

3301

I/O

ADDRESS

DATA

301

1040

302

2080

Address

Data

245

2785

246

A00B

247

5557

:

:

:

:

620

4245

621

2247

622

3301

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