Practice Type | COA | Instruction Set |

Question

 

Consider a processor with 30 general purpose registers and an instruction set size of 28. Each instruction has 4 fields as shown below

 

Instruction Code

Operation | Source Reg | Source Reg | Destination Reg

 

Let the program length is 80 instructions and type of memory used to store the program is Byte Addressable. The amount of memory used to store the program is ____ Bytes.

(Note: The program is stored in Byte aligned fashion)

 

ANSWER- 240 bytes

Comments

These are belongs to COA subject, rt? - The why you selected study resources as the category ?

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yes sir, i have changed it now.

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30 general purpose reisters $\Rightarrow$ 5 bits required to identify the register

instruction set size of 28 $\Rightarrow$ 5 bits required to identify the operation.

Instruction Code : Operation | Source Reg | Source Reg | Destination Reg

Bits : 5 + 5 + 5+5 = 20 bits $\Rightarrow$ 3 Bytes for each instruction as byte addressable.

Total = $80 \times 3 \;Bytes = 240 B$