Single accumulator organisation

Question

The size of the instruction in a single accumulator is 16 bits. In order to evaluate the expression Y = A-B+C / E+F , how much memory space is required to store the program?

a) 18 bytes

b)24 bytes

c)22 bytes

d)20 bytes

Comments

Since each instruction is 16 bits, 16/8 = 2 bytes.

Instructions:

Load E

Add F

Store T

Load A

Sub B

Add C

Div T

Store Y

In this way we have 8 instructions.
Therefore, 8*2 = 16 bytes.

What is wrong with this answer guys?

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I think it will be C/E .Not (A-B+C) /( E+F)

Answer 1

Since each instruction is 16 bits, 16/8 = 2 bytes.

Instructions:

Load E

Add F

Store T

Load A

Sub B

Add C

store P

Div T

Store Y

In this way we have 8 instructions.
Therefore, 9*2 = 18 bytes.

Answer 2

Load C

Div E

Add F

Add A

Sub B

Store Y

there will be 2 Bytes

Ans will be 7*2 =14 Bytes

Answer 3

16 bit = 2byte

Instructions:

Load E

Add F

Store T

Load A

Sub B

Add C

store P

Div T

Store Y

total instruction 9

 

so 9*2 = 18  byte

option A is true