Question

(a) (10 points) The hexadecimal form of a 3-byte instruction for SIC/XE is 010030. The opcode in the instruction is LDA. Indicate the contents of the A register in decimal after the above instruction is executed. Be sure to indicate how you arrived at your answer.

Answer 1

These are the instructions used in programming the Simplified Instructional Computer(SIC).

Here,
A stands for Accumulator
M stands for Memory
CC stands for Condition Code
PC stands for Program Counter
RMB stands for Right Most Byte
L stands for Linkage Register

a)

• Instruction set size – It tells the total number of instructions defined in the processor.
  

• Opcode size – It is the number of bits occupied by the opcode which is calculated by taking log of instruction set size.
  

• Operand size – It is the number of bits occupied by the operand.
  

• Instruction size – It is calculated as sum of bits occupied by opcode and operands.
  

MNEMONIC	OPERAND	OPCODE	EXPLANATION
ADDR	R1, R2	90	R2 = R2 + R1
CLEAR	R1	04	R1 = 0
COMPR	R1, R2	A0	compares R1 and R2
DIVR	R1, R2	9C	R2 = R2 / R1
LDB	M	68	B = M
LDS	M	6C	S = M
LDT	M	74	T = M
MULR	R1, R2	98	R2 = R2 * R1
RMO	R1, R2	AC	R2 = R1
SHIFTL	R1, n	A4	left shifts R1 n times
SHIFTR	R1, n	A8	right shifts R1 n times
STB	M	78	M = B
STS	M	7C	M = S
STT	M	84	M = T
SUBR	R1, R2	94	R2 = R2 – R1
TIXR	R1	B8	X = X + 1; compares X and R1

Here,
M stands for Memory
R1 and R2 are registers (A, B, S, T)

b)

JEQ

M

30

if CC set to =, PC = M

  JEQ m 3/4 30 PC <-- m if CC set to =

• For type-4 category having N instructions each having 1 floating point register operand (6 bits) will consume N* 2^6 = 2048 (calculated from previous step). Therefore, N = 32.

Answer 2

These are the instructions used in programming the Simplified Instructional Computer(SIC).

Here,
A stands for Accumulator
M stands for Memory
CC stands for Condition Code
PC stands for Program Counter
RMB stands for Right Most Byte
L stands for Linkage Register

a)

• Instruction set size – It tells the total number of instructions defined in the processor.
  
• Opcode size – It is the number of bits occupied by the opcode which is calculated by taking log of instruction set size.
  
• Operand size – It is the number of bits occupied by the operand.
  
• Instruction size – It is calculated as sum of bits occupied by opcode and operands.
  

MNEMONIC	OPERAND	OPCODE	EXPLANATION
ADDR	R1, R2	90	R2 = R2 + R1
CLEAR	R1	04	R1 = 0
COMPR	R1, R2	A0	compares R1 and R2
DIVR	R1, R2	9C	R2 = R2 / R1
LDB	M	68	B = M
LDS	M	6C	S = M
LDT	M	74	T = M
MULR	R1, R2	98	R2 = R2 * R1
RMO	R1, R2	AC	R2 = R1
SHIFTL	R1, n	A4	left shifts R1 n times
SHIFTR	R1, n	A8	right shifts R1 n times
STB	M	78	M = B
STS	M	7C	M = S
STT	M	84	M = T
SUBR	R1, R2	94	R2 = R2 – R1
TIXR	R1	B8	X = X + 1; compares X and R1

Here,
M stands for Memory
R1 and R2 are registers (A, B, S, T)

b)

JEQ

M

30

if CC set to =, PC = M

  JEQ m 3/4 30 PC <-- m if CC set to =

• For type-4 category having N instructions each having 1 floating point register operand (6 bits) will consume N* 2^6 = 2048 (calculated from previous step). Therefore, N = 32.