Computer Organizations and Design - Solved Problems and Illustrations

This article provides solved problems and illustrations for Computer Organizations and Design. It covers topics such as addressing modes, signed integer representations, and pipeline systems. It is perfect for students in computer science and engineering.

Running head: COMPUTER ORGANIZATIONS AND DESIGN 1
Computer Organizations and Design
Student’s Name
Institutional Affiliation

COMPUTER ORGANIZATIONS AND DESIGN 2
You need to type in your solution except for diagrams (130 marks).
1. Using n= 8 bits and all of following internal representations for signed integers (30 marks):
1. Signed magnitude
2. one's complement
3. two's complement
Give the codes for each of the following signed decimal integer values, fill in values directly into
the table below. Given each answer as a 2-digit hexadecimal value. It is not necessary to show
your work (base conversion, twos complementing operations.
Decimal SM One’s complement Two’s complement
-96 11100000 1100000 0010 0000
-85 11010101 1010101 00101011
-1 1000 0001 1111 0111 1111
3 0000 0011 0011 1111 1101
78 01001110 1001010 10110010
128 010000000 10000000 110000000
2. Given the 32-bit two's complement internal representation for each of the following
signed integers. Give each answer in hexadecimal (8 digits). Show your work (15 marks).

COMPUTER ORGANIZATIONS AND DESIGN 3
1. 1,234,567,090
2 1234567090
2 617283545 -0
2 308641772 -1
2 154320886 -0
2 77160443 -0
2 38580221 -1
2 19290110 -1
2 9645055 -0
2 4822527 -1
2 2411263 -1
2 1205631 -1
2 602815 -1
2 301407 -1
2 150703 -1
2 75351 -1
2 37675 -1
2 18837 -1
2 9418 -1
2 4709 - 0
2 2354 -1

COMPUTER ORGANIZATIONS AND DESIGN 4
2 1177 -0
2 588 -1
2 294 -0
2 147 -0
2 73 -1
2 36 -1
2 18 -0
2 9 -0
2 4 -1
2 2 0
2 1 0
0100 1001 1001 0101 1111 1111 1011 0010
= (4995FFD2)16
2. -2,777,888,099
2 2777 888099
2 1388944059 -1
2 694472024 -1
2 347236012 -0
2 173618006 -0
2 86809003 -0
2 43404501 - 1
2 21702250 -1

COMPUTER ORGANIZATIONS AND DESIGN 5
2 10851125 -0
2 5425562 -1
2 2712781 -0
2 1356390 -1
2 678195 -0
2 339097 -1
2 169548 -1
2 84774 -0
2 42387 -0
2 21193 -1
2 10596 -1
2 5298 -0
2 2649 -0
2 1324 -1
2 662 -0
2 331 -0
2 165 -1
2 82 -1
2 41 -0
2 20 -1
2 10 -0
2 5 -0
2 2 -1

COMPUTER ORGANIZATIONS AND DESIGN 6
1 0
= (1101 0010 1100 1001 1001 1010 1011 0001)16
= (D2C99AB1)16
3. What signed decimal integer is represented by each of the following 32-bit two's
complement representation? Show your work (15 marks).
1. 7E45E012
(7E45E012)16 = (0111 1111 0100 0101 1110 0000 0001 0010)2
1*230 +1*229 +1*228 + 1*227 +1*226+1*225 +1*224 + 0*223 +1*222+0*221 +0*220
+ 0*219 +1*218+0*217 +1*216 + 1*215+1*214+1*213 +0*212 + 0*211 +0*210+0*29 +0*28
+ 0*27 +0*26+0*25 +1*24 + 0*23 +0*22+1*21 +0*20
= 2,135,285,778
2. (EA66110B)16
= (1111 1010 0110 0110 0001 0001 0000 1011)2
The first letter has a negative (-ve)
1*230 +1*229 +1*228 + 1*227 +0*226+1*225 +0*224 + 0*223 +0*222+1*221 +0*220
+ 0*219 +1*218+1*217 +0*216 + 0*215+0*214+0*213 +1*212 + 0*211 +0*210+0*29 +1*28
+ 0*27 +0*26+0*25 +0*24 + 1*23 +0*22+1*21 +1*20
= -2,053,509,387
4. Using figures to illustrate the following addressing modes (30 marks):
 Immediate addressing
 Direct addressing
 Indirect addressing

COMPUTER ORGANIZATIONS AND DESIGN 7
 Register addressing
 Register indirect addressing
 Displacement addressing
 Relative addressing
 Base-register addressing
 index addressing
Solution
Immediate addressing mode involves data following instructions. As an illustration, the
data that has to be used is already in the instruction itself. For instance, MOV A, #25H (This
instruction moves data 25H to the accumulator. However, the # sign shows that the preceding
term is data not address)
Instruction
Direct addressing mode. Within this mode, the address of data to be read is directly
given in the instruction. For instance, MOV A, 46H (This instruction will move the contents of
memory location 46H to accumulator).
Instruction
Memory
A
Operand
Opcode operand

COMPUTER ORGANIZATIONS AND DESIGN 8
Indirect addressing mode: The data is stored in registers or memory location indirectly.
Conversely, the instruction set the memory location of the address of the data to be stored. E.g.,
MOV R2, 37H. The instruction moves the content of address present at 37H to register R2.
Instruction
Memory
Register addressing mode: The source of data or destination of the result is registered.
However, the name of the register is given in the instruction where the data to be read or result to
be stored (Patterson & Hennessy, 2013). For example Add, R3, R4. The instruction adds the
contents of register R4 to R3 and then store it to R3.
Instruction
Registers
A
Operand
R
Operand

COMPUTER ORGANIZATIONS AND DESIGN 9
Register indirect addressing mode: The data is read or stored in register indirectly. The
register is provided in the instruction where the address of the other register is stored or which
points to other registers where data is stored or to be stored (Patterson & Hennessy, 2013). For
instance, MOV A, @R0. This instruction will move the data to accumulator from the register
whose address is stored in register R0.
Instruction
Memory
Registers
Displacement is addressing mode: This is similar to the index mode except that instead
of an index register, a base register will be used. The base register contains a pointer to a
memory location.
R
Operand

COMPUTER ORGANIZATIONS AND DESIGN 10
Instruction
RRRRRRRRRRRRFFF
Memory
Registers
Relative addressing mode is also called PC-relative addressing. The implicitly
referenced register is the program counter (PC). The next instruction is added to the address field
to produce the EA. The address field is usually treated as a twos complement number for this
operation. Therefore, the effective address is a displacement relative to the address of the
instructions (Patterson & Hennessy, 2013). For instance, Move X (PC), R1. The contents at
address X+PC are moved to R1.X contains a constant value.
A version of displacement addressing
R = program counter, pc
EA = A + (PC)
Base-register addressing mode: This mode is similar to index addressing mode, but
instead of suing index register it utilizes the value of base register which is called offset. For
instance, MOV R2, BR + offset. This moves the content of effective address by adding BR and
Offset to the register R2
R A
Operand
+

COMPUTER ORGANIZATIONS AND DESIGN 11
A holds displacement
R holds pointer to base address
R may be explicit or implicit
Index addressing is the address of the operand which is obtained by adding to the
contents of the general register a constant value. However, the number of the index register along
with the constant value is included in the instruction code.
A = base
R = displacement
EA = A + R
This is therefore good for accessing arrays
5 An instruction is stored at location 200 with its address field at location 201. The address
field has the value 300. A processor register R1 contains the number 100. Evaluate the
effective address if the addressing mode of the instruction is a) direct; b) immediate; c)
relative; d) register indirect; e) index with R1 as the index register (25 marks)
The instruction is stored at location 200 with address field at 201. However, the address
field has the value 300. The processor register R1 has the number 100.
(a). Direct.
To evaluate our addressing mode as direct, we pull the number from the value of the address
field. In this case, the direct is 300.
(b). Immediate addressing mode
For the case of immediate addressing mode, we seek out the instruction stored location.
In our case is 200 which is the answer.

COMPUTER ORGANIZATIONS AND DESIGN 12
(c) Relative
Concerning relative location, the formula for calculation is as follows:
Addressing Field Location + Processor Register = Relative Location
In this case, it is 201+ 100 = 301.
(d). Register indirect
The effective address for indirect register mode is 100.
(e). Index with R1 as the index register
The formula for getting this is;
Address Field Value + Processor Register = Indexed Location
300 + 100 = 400
6. A nonpipleline system takes 40ns to process a task. The same task can be processed in a
six-segment pipeline with a clock cycle of 10 ns. Determine the speedup ratio of the pipeline
for 80 tasks. What is the maximum speedup that can be achieved? (15 marks
Solution
The speedup ratio is defined as the speedup of a pipeline processing concerning the
equivalent non-pipeline processing. The below is the formula for getting speed up ratio
S = ten/ (K+n-1) tp
The number of tasks n = 80
For nonpipeline:
Time is taken by pipeline to process a task in = 40 ns
Total time is taken by pipeline to process 80 task = ten
= 80*40
= 3200 ns

COMPUTER ORGANIZATIONS AND DESIGN 13
For pipeline:
Number of segment pipeline K = 6
Time period of 1 clock tp = 10 ns
Total time requires to complete n tasks in K segment pipeline with tp clock cycle time:
= (k+n-1) tp
= (6+ 80-1)10
= 850 ns
Speed up Ratio:
When the total time taken by the pipeline to process 80 is divided by the total time
required to complete n tasks in K segment pipeline with tp clock cycle time then speed up ratio is
obtained.
S = 3200/850
= 3.76
The speedup ratio of the pipeline for 80 tasks is therefore 3.76
Maximum speed up ration:
As the number of tasks increases, n becomes very small k+n-1 approaches to n, and
hence the speed up ration becomes Smax = nTN/tp
Hence it is;
Smax = 40/10
= 4
In this case, the maximum speedup ratio that can be achieved is 4

COMPUTER ORGANIZATIONS AND DESIGN 14
References
Patterson, D. A., & Hennessy, J. L. (2013). Computer Organization and Design MIPS Edition:
The Hardware/Software Interface. Newnes.
http://ac.aua.am/Arm/Public/2017-Spring-Computer-
Organization/Textbooks/ComputerOrganizationAndDesign5thEdition2014.pdf