Question

Calculate the total memory traffic for the set of instructions that implements the expression A= (b/d) * (c - e) on 3, 2, 1, and 0 address machines. Assume opcodes occupy one byte, addresses occupy one byte, data values occupy two bytes and the device is byte addressable. Use Booth's multiplication algorithm to calculate m * n. (m is the sum of last four digits of your register number, n is the negative value of the largest digit in register number. For Eg. Reg no: 18bce0453, then m 0+4+5+3=12 and n- 5)

Answer 1

Answer-

lets assume we have following instruction sets in the given machine-

subtraction (SUB), multiplication (MUL), division (DIV) and data movement (MOV, LOAD, STORE, PUSH, and POP)

3-address machine has instructions like- SUB A, B, C means M [ A ]<----- M [ B ] -- M [ C ]

2-address machine has instructions like- SUB A,B means M [ A ] <------- M [A] -- M [B]

1-address machine has an accumulator to store the result.

0-address machine uses stack for both source and destination operands.

Given expression -   A = (b/d) * (c-e)

the following table shows the 4 programs of different type of machines-

3-address machine	2-address machine	1-address machine	0-address machine
Div R1, b, d	Mov R1, b	Load b	push b
Sub R2, c, e	Div R1, d	div d	push d
Mul A, R1, R2	Mov R2, c	store R	div
	Sub R2, e	load c	push c
	Mul R1, R2	sub e	push e
	Mov A, R1	mul R	sub
		store A	mul
			pop A

R1 and R2 are intermediate registers to hold the results temporarily.

we assume all instructions and the data values are in memory and must be fetched from memory.

Given that, opcode ----- 1 byte

address ------ 1 byte

data values ------ 2 bytes

so, instruction like SUB A, B, C requires

1 byte for the opcode, 1 byte for address A, 1 byte for address B, and 1 bytes for address C , in total of 4 bytes just to represent the instruction.

so 4 bytes require to fetch the instruction from the memory.

Also data at each address A, B and C requires 2 bytes, in total 6 bytes for the data value.

Therefore in total for the execution of SUB A,B,C we need 10 bytes.

So similar process is applied to other instructions also to calcute the number of bytes used.

total Memroy traffic for all types of instruction set:

3-address instruction	2-address instruction	1-address instruction	0-address instruction
Div R1, b, d 10 bytes	Mov R1, b 7 bytes	Load b 4 bytes	push b 4 bytes
Sub R2, c, e 10 bytes	Div R1, d 7 bytes	div d 4 bytes	push d 4 bytes
Mul A, R1, R2 10 bytes	Mov R2, c 7 bytes	store R 4 bytes	div 3 bytes
	Sub R2, e 7 bytes	load c 4 bytes	push c 4 bytes
	Mul R1, R2 5 bytes	sub e 4 bytes	push e 4 bytes
	Mov A, R1 5 bytes	mul R 4 bytes	sub 3 bytes
		store A 4 bytes	mul 3 bytes
			pop A 2 bytes
Memory Traffic--> 30 bytes	38 bytes	28 bytes	25 bytes

therefore , the total memory traffic is:

3-Address Machine: 30 bytes

2-Address Machine: 38 bytes

1-Address Machine: 28 bytes

0-Address Machine: 25 bytes