Assessment Item 3: Computer Organization and Architecture

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Table of Contents
Assessment item 3.................................................................................................................................2
Question 1:............................................................................................................................................2
a) Determine base value:...............................................................................................................2
b) Convert the followings:..............................................................................................................3
c) Complements:...........................................................................................................................6
Question 2:............................................................................................................................................8
a) Marie simulator- Fibonacci series..............................................................................................8
b) Maximum integer of Fibonacci series..........................................................................................11
Question 3:..........................................................................................................................................12
a) ‘Interrupts’...............................................................................................................................12
B) Multiple Vs Single Bus Architecture:...........................................................................................13
Reference:...........................................................................................................................................14

Assessment item 3
Question 1:
a) Determine base value:
The value of base b if (152)b = 0x6A. Please show all steps. [3 marks]
Given is ( 152 ) b=¿ Hexadecimal number 0x6A
Therefore, first of all, base-16 should be converted to base-10
Hexa-decimal to Decimal Conversion
In
(6 A)10=6∗161 +10∗160=96+10
Therefore,
(6 A)10=106
So,
(152)b=106
To determine base ‘b’,
1∗x2 +5∗x1 +2∗x0=106
x2+ 5 x +2=106
x2+ 5 x−104=0
By using Quadratic Formula,
x=−b ± √b2−4 ac
2 a
Here, a=1, b=5 & c=-104
By substituting the values of a, b & c into the quadratic formula,
x=−5 ± √52−4 (1)(−104)
2∗1

x=−5 ± √ 25+ 416
2
x=−5 ± √ 441
2
x=−5 ±21
2
Therefore, the values of x will be
Either, x=−5+21
2
x=−16
2
x=−8
Or,
x=−5−21
2
x=−26
2
x=−13
Base cannot be negative, therefore,
According to base 8
1∗82+ 5∗81+2∗80 =106
82 +5∗8+2=106
64+40+2=106
Therefore, base of (152) will be 8
Ans :(152)8=0x6A
b) Convert the followings:
(Please show all steps; no marks will be awarded if no steps are shown) [1.5x4 = 6 marks]
i) 0xBAD into 3-base representation
Given is Hexadecimal number 0xBAD
Therefore, first of all, base-16 should be converted to base-10

Hexa-decimal to Decimal Conversion
BAD=11∗162 +10∗161 +13∗160
¿ 11∗256 +160+13
(BAD)10=2989
Now, 2989 is a base-10 digit which should be converted to base-3 digit
Divisor Dividen
d
Remainder
3 2989
3 996 1
3 332 0
3 110 2
3 36 2
3 12 0
3 4 0
1 1
Backward sequence of Remainder will be the conversion of ( 2989 )3
Therefore,
Ans : ( BAD )10= ( 2989 )10= ( 11002201 )3
ii) (321)7 into 2-base (binary) representation
To convert base-7 into base-2 or binary,
First we need to convert, base-7 to decimal number
(321)7=3∗72 +2∗71+1∗70
(321)7=147+14 +1
(321)7=( 162)10
Now, 169 will be divided by 2 and backward sequence of remainders will be binary
conversion

Divisor Dividend Remainder
2 162
2 81 0
2 40 1
2 20 0
2 10 0
2 5 0
2 2 1
2 1 0
Backward sequence of Remainder will be the conversion of ( 162 ) 10
Therefore,
( 162 ) 10= ( 10100010 ) 2
Ans : ( 321 )7 = ( 10100010 )2
iii) (123)5 into octal representation
To convert base-5 into base-8 or octal,
First we need to convert, base-5 to decimal number
(321)7=1∗52+2∗51 +3∗50
(321)7=25+10+3
(321)7=( 38)10
Now, 38 will be divided by 8 and backward sequence of remainders will be binary conversion
Divisor Dividend Remainder
8 38
4 6
Backward sequence of Remainder will be the conversion of ( 38 )10 into octal
Therefore,
( 38 )10= ( 46 )8
Ans : ( 123 )5= ( 46 )8

iv) (21.21)8 into decimal representation
To convert base-8 into decimal
As the base-8 digit has fraction part
Positional values will be calculated according to the given base-8 number.
(21.21)8=2∗81 +1∗80+2∗8−1+1∗8−2
(21.21)8=16+1+0.25+ 0.015
Ans :(21.21)8=17.265
c) Complements:
Given a (very) tiny computer that has a word size of 3 bits, what are the lowest value (negative
number) and the highest value (positive number) that this computer can represent in each of the
following representations? [3 marks]
Given is a tiny computer that can store word of 3 bit size only.
Therefore,
For a computer having 3-bit word size will have.
Lowest Bit Value = 0
Highest Bit Value = 111
Moving to the question:
i) One's complement
One’s complement for lowest value will be
0 → 1
One’s complement for highest value will be
111
↓ ↓↓
000
ii) Two's complement
First, one’s complement will be calculated for each lowest and highest value.
One’s complement for lowest value will be
0 → 1

One’s complement for highest value will be
111
↓ ↓↓
000 Now ,‘ 1’ will be added ¿ the one ’ s complement ¿ get two ’ s complement .
For Lowest value:
1
+1
0
For Highest value:
000
+1
001
iii) Signed Magnitude
For the lowest value:
Since ‘0’ represents positive
And
‘1’ represents negative
Sign-magnitude of lowest value – ‘0’ will be positive
For the Highest value:
Sign-magnitude of highest value – ‘111’ will be:
1 1 1
First bit represents sign, that is negative,
Magnitude will be:
¿ 1∗22 +1∗21+ 1∗20
¿ 4 +2+1
Therefore, sign magnitude for the highest value will be:
¿−7

Question 2:
a) Marie simulator- Fibonacci series
The Fibonacci numbers are the numbers in the following integer sequence, called the Fibonacci
sequence, and are characterised by the fact that every number after the first two is the sum of the
two preceding ones: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 114, … etc. [8 marks]
Source code for the Fibonacci series:
ORG 000
INPUT USER /Get a number from operator
SUBT TWO /substraction of 2 from gotten number
PRINT, STORE P /after processing of value P, store AC in P
CLEAR
ADD R1 /Addition of R1 into Ac
ADD R2 /Addition of R2 into AC
STORE TOTAL /Added value from R1 to R2
/where R1 = R2
LOAD R2 /Process R2 into AC
STORE R1 /R1 will store value
LOAD TOTAL /to Process total AC value
STORE R2 /AC will store value of N2
LOAD P /Process register P
SUBT ONE /subtracts 1 from P
SKIPCOND 000 /skip cond 000, if value AC is less than 0
JUMP PRINT /begins Loop and prints
LOAD TOTAL /AC will contain fetched and processed total value
OUTPUT /shows output
HALT /stop present process
/Initialize process

USER, DEC 0
ONE, DEC 1
TWO, DEC 2
P, DEC 0
R1, DEC 0
R2, DEC 1
TOTAL, DEC 0
After processing the above program in Marie simulator and running the program,
Simulator asked for input values. Here are the outputs for the input value, 7, 15, 20 respectively.
For input = 7, received output = 13

For input = 15, received output = 610
For input = 20, received output = 6765

b) Maximum integer of Fibonacci series
For some values of n, your program will not produce correct results. You can check this by gradually
increasing the values of n and checking for the correct outputs. What is the maximum value of n for
which your program produces a correct result? Why? Please comment on this [3 marks].
The last value on which the Marie simulator can produce the correct output will be:
n=2147483647
This is the last value to get correct output from Marie simulator.

Question 3:
a) ‘Interrupts’
What are the 'interrupts' in a computer system? Describe the approaches to deal with multiple
interrupts in a system. [4 marks]
When a program or a device that is attached to the computer system sends a signal to the
microprocessor or the operating system of the computer which make the Operating system stay for
a while to think about its next step is called ‘interrupt’. Generally these signals are sent to the CPU
by the Input/output devices.
Normally there are three types of interrupts are there which occurs in a computer system:
1. External Interrupt
2. Internal Interrupt
3. Software Interrupt
Explanation:
1. External Interrupt: when an output or input device request for the connection and CPU stops
all the workings and processes for a while and first executes that request of I/O device, then
that stop of CPU is called External Interrupt. For an example, when a mouse is attached to
computer, it pauses all other process and takes instruction first from input device, mouse.
2. Internal interrupt: these are those kind of interrupts that occur when the internal execution
of any process or operation contains any error and cause the system to stop for a while and
think about its next move. Basically Internal interrupt occurs in that position where user or
operator requests for an operation which is not possible to execute by the CPU.
3. Software Interrupts: these interrupts occurs when the user try to access multiple software at
a time. CPU already accessing a program and on that time the user request for another
program which makes the CPU to stop.