Instruction Cycle MCQ Quiz - Objective Question with Answer for Instruction Cycle - Download Free PDF

The correct answer is Option 4) Increment value in 32-bit register.

Key Points

• INC is an assembly language instruction that increases the value of a register or memory location by 1.
• EBX is a 32-bit general-purpose register in the x86 architecture.
• INC EBX increments the current 32-bit value stored in the EBX register by 1.
• This is commonly used in loop counters or arithmetic operations within low-level programming.

Additional Information

• Option 1 – Incorrect: There is no syntax error in INC EBX; it's a valid assembly instruction.
• Option 2 – Partially Correct: It’s true that it increments the value in EBX, but the option lacks clarity about register size.
• Option 3 – Incorrect: EBX is a 32-bit register, not an 8-bit one. The 8-bit part would be BL.
• Related Registers:
  • AX, BX, CX, DX – 16-bit registers
  • AL, BL, CL, DL – 8-bit lower halves
  • EAX, EBX, ECX, EDX – 32-bit registers
  • RAX, RBX, RCX, RDX – 64-bit registers (x86-64 architecture)

Correct Answer: Option 4) Initiation of interrupt service

Key Points

Let's examine the steps one by one:

1. MBR ← PC:
   • The content of the Program Counter (PC), which holds the address of the next instruction to be executed, is copied to the Memory Buffer Register (MBR).
   • This is typically done to save the return address before jumping to an interrupt service routine (ISR).
2. MAR ← X:
   • The Memory Address Register (MAR) is loaded with the address X.
   • This is the location where the return address (in MBR) will be stored.
3. PC ← Y:
   • The Program Counter is updated to Y, which points to the start of the ISR.
   • This redirect of the control flow to a new memory location is characteristic of interrupt handling.
4. MEMORY ← MBR:
   • The value in the MBR (the original PC value) is stored in memory at the address specified by MAR (which is X).
   • This effectively saves the return address so that the CPU can resume the original program after servicing the interrupt.

Therefore, this sequence of micro-operations matches the typical behavior during the initiation of an interrupt service routine, where:

• The current execution state (PC) is saved,
• The PC is redirected to the ISR location.

Final Answer: Option 4) Initiation of interrupt service

The correct answer is 2·24 

Explanation:

The formula for finding the total CPI is given below

CPI = ∑ Frequency_i * CPI_i

Frequencyi = IC_i / Instruction count

CPI = (0.60 * 1) + (0.18 * 2) + (0.12 * 4) + (0.10 * 8)

CPI = 2.24

The correct answer is 8*10^–10 sec

Explanation:

To determine the clock period of a processor, we use the formula: 

Given Data:

• Clock Speed = 1250 million cycles per second
• Convert the clock speed to standard units (Hz):

Step-by-Step Calculation:

The clock period (in seconds) is:

Substitute the clock speed:

                

Simplify:

                 

Rewriting:

                 

Final Answer: 4) 

• The Steps required by the CPU to fetch and execute an Instruction is called an instruction cycle. It consists of fetch and executes cycle.
• Instruction cycle (IC) = Fetch cycle (FC) + Execution cycle (EC)
• The time required by the microprocessor to complete the operation of accessing memory or I/O devices is called a machine cycle.
• Clock time is a known time state. It is reciprocal of clock frequency.
• Instruction cycle > Machine cycle > Clock cycle (time state)

Steps in the instruction cycle:

• First of all, the opcode is fetched by the microprocessor from a stored memory location.
• Then it is decoded by the microprocessor to find out which operation it needs to perform.
• If an instruction contains data or operand address which is still in the memory, the CPU has to perform read operation to get the desired data.
• After receiving the data, it performs to execute the operation.

 

Correct sequence: fetch → decode → read effective address → execute

Machine Cycle: Time taken to execute one OPERATION is known as a machine cycle.  One instruction will contain 1 to 5 machine cycles.

T-State: The portion of a machine cycle executed in one internal clock pulse is known as T-state.

Steps in the instruction cycle:

• First of all, the opcode is fetched by the microprocessor from a stored memory location.
• Then it is decoded by the microprocessor to find out which operation it needs to perform.
• If an instruction contains data or operand address which is still in the memory, the CPU has to perform read operation to get the desired data.
• After receiving the data, it performs to execute the operation.

Correct sequence: fetch → decode → read effective address → execute

Total Instruction = 100

Instruction Fetch, Instruction Decode, Operand Fetch, and Writeback (WB) performed in 1 cycle.

PO stage:

40 instructions take 3 cycle

35 instructions take 2 cycles

25 instructions take 1 cycle

Average number of cycles  = (40*3+35*2+25*1)/100 = 2.15 cycles.

On an average first instruction completed in 1+1+1+1+2.15 cycles

Remaining 99 instruction will takes 99*2.15 = 212.85 cycle

Total number of cycles is 6.15+212.85 = 219 cycles.

Steps in the instruction cycle:

• First of all, the opcode is fetched by the microprocessor from a stored memory location.
• Then it is decoded by the microprocessor to find out which operation it needs to perform.
• If an instruction contains data or operand address which is still in the memory, the CPU has to perform read operation to get the desired data.
• After receiving the data, it performs to execute the operation.

Correct sequence: fetch → decode → read effective address → execute

The correct answer is "option 4".

CONCEPT:

Micro-operations are also known as micro-ops.

They are the atomic or functional operations of a processor.

It is an elementary Central Processing Unit (CPU) operations performed during one clock pulse.

Micro-operations are low-level instructions, used in some designs, to implement complex instructions.

Micro-operations generally perform data stored in one or more registers.

EXPLANATION:

1. Program counter holds the next instruction value to be executed.

Here, MBR means the value of the program counter will get stored in MBR.

2. MAR means some address value X is storing in MAR so to access memory location X.

3. PC means storing new instruction value Y to the program counter to access new instruction.

4. Memory means MBR register will store its value to Memory. This saves the previous value of PC to memory.

This sequence of instructions matches with Interrupt Service Routine (ISR) since the sequence of instructions saved the address of current instructions into memory.

Then started executing new address by loading new instruction value to Program counter.

Hence, the correct answer is “option 4”.

Additional Information

1. Instruction fetch means fetching the instruction.

2. Operand fetch means fetching the operands.

3. Conditional branch means the program will branch to another instruction after satisfying a condition.

• The time required for the fetching and execution of one simple machine instruction is the CPU cycle. It is also known as an instruction cycle. 
• It is composed of three main stages: the fetch stage, the decode stage, and the execute stage.
• In simpler CPUs, the instruction cycle is executed sequentially, each instruction being processed before the next one is started.
• In most modern CPUs, the instruction cycles are instead executed concurrently, and often in parallel, through an instruction pipeline

Concept:

While executing a micro-instruction such as R0 ← R1 + R2, the CPU performs various micro-operations. Each of this micro-operation is performed in one time cycle.

Execution:

Step 1:

Fetch the instruction. Initially, the address of the instruction to be executed in Program Counter (PC). It is moved from PC to Memory Address Register (MAR). This is done via micro-operation PC_r, MAR_w, MEM_r

Step 2:

Once the instruction has been fetched, in next single time cycle, it is placed into Memory Data Register (MDR) and then to Instruction Register (IR). This is done via micro-operation MDR_r, IR_w

Step 3:

Operand Fetching and Decoding of contents from Register R1 and place it into temporary register Temp₁ via micro- operation R1_r, TEMP1_w

Step 4:

Contents of Register R2 is decoded and ALU performs the addition of fetched content of Temp1 and R2 and place it into Temp2 via micro-operation R2_r, TEMP1_f, ALU_add, TEMP2_w

Step 5:

Finally, the content of Temp₂ is moved into the target register R₀ via micro-operation TEMP2_r, R0_w

Answer: 50 to 50 

Explanation: 

Consider following memory cells

After Executing till 3 instruction we have the following value in Registers

After 4th instruction, M[R₃] ( M[3000]) will be updated as 50 + 10 = 60 

R₃ => R₃ +1 => 3001

R₁ = 9 Hence it is Nonzero; It will Branch to 1004 (which is instruction 2)

- R₂ = 50

R₂= R₁+R₂ = 59

M[R3] ( M[3001]) will be updated as 59.

R₁  = R₁ - 1 => 9-1 =>8 Hence it is Nonzero; It will Branch to 1004 (which is instruction 1)

If you try to observe the Pattern

Here M[3001] is updated as 59

M[3002] will be updated as 58

M[3003] will be updated as 57

M[3004] will be updated as 56

M[3005] will be updated as 55

M[3006] will be updated as 54

M[3007] will be updated as 53

M[3008] will be updated as 52

M[3009] will be updated as 51

R1 = 0

M[3010] will be updated as 50

Concept:

In both the cases number of Instructions will be same

Calculation:

Total time ratio: P1: P2 = 100: 75 = 4:3

CPI ratio: CPI1: CPI2 = 100: 120 = 5:6

Instruction register:

• Each instruction to be executed is loaded into the instruction register.
• It also contains instruction that is currently being executed.

Index register:

• Index register used to determine the address of an operand.

Memory address register (MAR):

• Used to handle the address transferred to the memory unit.
• MAR is a simple microprocessor needs enough bits for the address.
• MAR holds the memory location of data that needs to be accessed.

Memory data register (MDR):

• Also known as Memory buffer register.
• CPU that stores the data being transferred to and from the immediate access storage.

Hence option 2 is the correct option.