Computer Architecture and the Fetch-Execute
4.3.1 Von Neumann Architecture
John Von Neumann introduced the idea of the stored program. Previously data and
programs were stored in separate memories. Von Neumann realised that data and
programs are indistinguishable and can, therefore, use the same memory. This led to the
introduction of compilers which accepted text as input and produced binary code as output.
The Von Neumann architecture uses a single processor which follows a linear sequence of
fetch-decode-execute. In order to do this, the processor has to use some special registers.
PC Program Counter
CIR Current Instruction Register
MAR Memory Address Register
MDR Memory Data Register
Accumulator Holds results
The program counter keeps track of where to find the next instruction so that a copy of the
instruction can be placed in the current instruction register. Sometimes the program counter
is called the Sequence Control Register (SCR) as it controls the sequence in which
instructions are executed.
The current instruction register holds the instruction that is to be executed.
The memory address register is used to hold the memory address that contains either the
next piece of data or an instruction that is to be used.
The memory data register acts like a buffer and holds anything that is copied from the
memory ready for the processor to use it.
The central processor contains the arithmetic-logic unit (also known as the arithmetic unit)
and the control unit. The arithmetic-logic unit (ALU) is where data is processed. This
involves arithmetic and logical operations. Arithmetic operations are those that add and
subtract numbers, and so on. Logical operations involve comparing binary patterns and
The control unit fetches instructions from memory, decodes them and synchronises the
operations before sending signals to other parts of the computer.
The accumulator is in the arithmetic unit, the program counter and the instruction registers
are in the control unit and the memory data register and memory address register are in
A typical layout is shown in Fig. 188.8.131.52 which also shows the data paths.
4.3.2 The Fetch-Decode-Execute-Reset Cycle
The following is an algorithm that shows the steps in the cycle. At the end the cycle is reset
and the algorithm repeated.
Load the address that is in the program counter (PC) into the memory address register
Increment the PC by 1.
Load the instruction that is in the memory address given by the MAR into the memory data
Load the instruction that is now in the MDR into the current instruction register (CIR).
Decode the instruction that is in the CIR.
If the instruction is a jump instruction then
a. Load the address part of the instruction into the PC
b. Reset by going to step 1.
Execute the instruction.
Reset by going to step 1.
Steps 1 to 4 are the fetch part of the cycle. Steps 5, 6a and 7 are the execute part of the
cycle and steps 6b and 8 are the reset part.
Step 1 simply places the address of the next instruction into the memory address register
so that the control unit can fetch the instruction from the right part of the memory. The
program counter is then incremented by 1 so that it contains the address of the next
instruction, assuming that the instructions are in consecutive locations.
The memory data register is used whenever anything is to go from the central processing
unit to main memory, or vice versa. Thus the next instruction is copied from memory into
the MDR and is then copied into the current instruction register.
Now that the instruction has been fetched the control unit can decode it and decide what has
to be done. This is the execute part of the cycle. If it is an arithmetic instruction, this can be
executed and the cycle restarted as the PC contains the address of the next instruction in
order. However, if the instruction involves jumping to an instruction that is not the next one
in order, the PC has to be loaded with the address of the instruction that is to be executed
next. This address is in the address part of the current instruction, hence the address part is
loaded into the PC before the cycle is reset and starts all over again.
4.3.3 Other Architectures
Using the above architecture for a microprocessor illustrates that basically an instruction can
be in one of three phases. It could be being fetched (from memory), decode (by the control
unit) or being executed (by the control unit). An alternative is to split the processor up into
three parts, each of which handles one of the three stages. This would result in the situation
shown in Fig. 184.108.40.206, which shows how this process, known as pipelining, works.
Instruction 2 Instruction 1
Instruction 3 Instruction 2 Instruction 1
Instruction 4 Instruction 3 Instruction 2
Instruction 5 Instruction 4 Instruction 3
This helps with the speed of throughput unless the next instruction in the pipe is not the next
one that is needed. Suppose Instruction 2 is a jump to Instruction 10. Then Instructions 3, 4
and 5 need to be removed from the pipe and Instruction 10 needs to be loaded into the
fetch part of the pipe. Thus, the pipe will have to be cleared and the cycle restarted in this
case. The result is shown in Fig. 220.127.116.11
Instruction 2 Instruction 1
Instruction 3 Instruction 2 Instruction 1
Instruction 4 Instruction 3 Instruction 2
Instruction 11 Instruction 10
Instruction 12 Instruction 11 Instruction 10
Another method that is used is to have array processors. This involves more than one
arithmetic-logic unit but still only one processor. This is particularly useful when processing
data held in a one-dimensional array when the same operation is to be applied to every
element of the array. An example is where all the values represent the costs of different
items in stock and it is required to find the selling prices which are calculated by adding the
same percentage to each value.
An extension of array processors is to use parallel processors. This system uses many
independent processors working in parallel on the same program. One of the difficulties with
this is that the programs running on these systems need to have been written specially for
them. If the programs have been written for standard architectures, then some instructions
cannot be completed until others have been completed. Thus, checks have to be made to
ensure that all prerequisites have been completed. However, these systems are in use
particularly when systems are receiving many inputs from sensors and the data need to be
processed in parallel. A simple example that shows how the use of parallel processors can
speed up a solution is the summing of a series of numbers. Consider finding the sum of n
numbers such as
2 + 4 + 23 + 21 + …. + 75 + 54 + 3
Using a single processor would involve (n – 1) additions one after the other. Using n/2
processors we could simultaneously add n/2 pairs of numbers in the same time it would
take a single processor to add one pair of numbers. This would leave only n/2 numbers to
be added and this could be done using n/4 processors. Continuing in this way the time to
add the series would be considerably reduced.
4.3.4 Example Questions
The questions in this section are meant to mirror the type and form of questions that a
candidate would expect to see in an exam paper. As before, the individual questions are
each followed up with comments from an examiner.
1. The Program Counter (Sequence Control Register) is a special register in the processor
of a computer.
a) Describe the function of the program counter. (2)
b) Describe two ways in which the program counter can change during the normal
execution of a program, explaining, in each case, how this change is initiated. (4)
c) Describe the initial state of the program counter before the running of the program. (2)
A. a) -The program counter stores the address…
-of the next instruction to be carried out in the sequence of the program. (2)
b) -P.C. is incremented…
-as part of the fetch execute cycle.
-P.C. is altered to the value being held in the address part of the instruction…
-When the instruction is one that alters the normal sequence of instructions in the program.
-This second type of command involves the P.C. being reset twice in the same cycle. (4)
c) -The P.C. will contain the address of the first instruction in the sequence to be run…
-this must have been placed in the register by some external agent, the program loader.
Notes: Part (a) is often poorly understood by students. The majority believing that the
program counter is used to keep track of the number of programs running, or the order in
which programs have been called. There is obviously a confusion with the idea of a stack
storing return addresses of modules when they have been called.
Part (b) illustrates a characteristic of true examination questions. Most genuine questions
will have more mark points available than there are marks for the question. This is not true
of these sample questions. It should also be remembered that these sample questions have
not been through the rigorous testing process that a genuine paper would have undergone,
so any problems with the content should not be repeated in the examination. Candidates
find difficulty in making the distinction between different types of instruction, it may be of
value to spend some time talking about arithmetic/logic/jump/ command type instructions
as they all affect the cycle in different ways.
Part (c) refers back to the AS work in the need to understand that the loader will initially set
the value of the P.C. so that the program can begin.
2. Explain what is meant by the term Von Neumann Architecture. (2)
A. -A way of looking at the relationships between the various pieces of hardware in a
-A single memory used to store program instructions and the data for use with those
-A single processor is used which follows a linear sequence of instructions. (2)
Notes: Many students will be content with the correct answer that VN architecture is the
ability to store the instructions and data in the same memory. However, a look at the mark
allocation shows that something else is required or only one mark would have been
available. Always look at the mark allocation and think of the examiner, is there enough in
the answer given to be able to award the full number of marks?
3. Describe the fetch/decode part of the fetch/decode/execute/reset cycle, explaining the
purpose of any special registers that you have mentioned. (7)
A. -Contents of PC loaded into MAR
-PC is incremented
-Contents of address stored in MAR loaded into MDR
-Contents of MDR loaded into CIR
-Instruction in CIR is decoded.
-PC (program counter) stores the address of the next instruction to be executed.
-MAR (memory address register) holds the address in memory that is currently being used
-MDR (memory data register) holds the data (or instruction) that is being stored in the
address accessed by the MAR.
-CIR (current instruction register) holds the instruction which is currently being executed.
Notes: The whole cycle may be asked for in some questions but it is more likely that it
would be split up in some way in order to make the question shorter and more accessible.
This is a difficult question because there is no splitting up of the points asked for, the
student must rely on their own interpretation of the requirements of the question. There is a
hint in the question because it asks for two parts of the cycle specifically, but students
should be aware that that becomes a part of the question, in other words the answer must
not contain any further information because it has been specifically ruled out in the
question. A candidate who describes the execution of particular types of instruction has
demonstrated that they cannot differentiate between the parts of the cycle and would
probably be penalised.
4. a) Describe how pipelining normally speeds up the processing done by a computer. (2)
b) State one type of instruction that would cause the pipeline system to be reset, explaining
why such a reset is necessary. (3)
A. a) -All instructions have three phases…
-which are treated separately, by different parts of the processor…
-so that more than one instruction can be being dealt with simultaneously. (2)
b) -Jump instruction
-The instructions in the pipeline are no longer the ones to be dealt with next…
-so the pipeline has to be reset. (3)