ITC544 Assignment 1: Computers, Data, and Programming

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ITC544 - IT FUNDAMENTALS
ASSESSMENT ITEM 2 ASSIGNMENT 1
COMPUTERS, DATA AND PROGRAMMING
Student Name:
Student ID:

Table of Contents
Question 1..................................................................................................................................1
a).............................................................................................................................................1
b)............................................................................................................................................3
i).........................................................................................................................................3
ii)........................................................................................................................................4
iii).......................................................................................................................................4
iv).......................................................................................................................................4
c).............................................................................................................................................5
i).........................................................................................................................................5
ii)........................................................................................................................................5
iii).......................................................................................................................................5
Question 2..................................................................................................................................6
a).............................................................................................................................................6
b)............................................................................................................................................8
Question 3..................................................................................................................................9
a).............................................................................................................................................9
b)..........................................................................................................................................10

Question 1
a)
Firstly, converting Left Side of the equation into its decimal number.
0x6A = 6 x 161 + A x 160 = 106 ---------- (a)
106 is the decimal number so, comparing Left side with Right side.
A = 10
Now, converting (152) b into its equal expression.
(152) b = 1 x b² + 5 x b¹ + 2 x b⁰
(152) b = b² + 5b + 2 ------------- (b)
Solving (a) & (b), we get
b² + 5b + 2 = 106
b² + 5b - 104 = 0
b² + 13b - 8b - 104 = 0
b (b+13) - 8(b + 13) = 0
(b-8) (b+13) = 0
b = 8 or b = -13
b = 8 is the final answer as base can’t have a negative value.

b)
i)
 Hex  Dec conversion
Conversion of BAD into decimal form.
(BAD) 16 = B x 162 + A x 161 + D x 160
(BAD) 16 = 11 x 162 + 10 x 161 + 13 x 160
(BAD) 16 = (2989)10 -------- (a)
 Dec  three base conversion
Dividing eq. (a) result with 3
Dividend / 3 Divisor Remainder
2989/3 996 1
996 /3 332 0
332 / 3 110 2
110 / 3 36 2
36 / 3 12 0
12/3 4 0
4/3 1 1
1/3 0 1
Now, taking the reverse order of the remainder.
11002201 is the 3-base representation

ii)
 Converting 3217 to binary
 Dividing 3217 by 2
Dividend / 2 Divisor Remainder
3217/2  1608 1
1608/2 804 0
804/2 402 0
402/2 201 0
201/2 100 1
100/2 50 0
50/2 25 0
25/2 12 1
12/2 6 0
6/2 3 0
3/2 1 1
1/2  1 1
Writing the remainder in reverse order
110010010001 is the binary value of 3217.

iii)
 Converting 1235 to octal
 Dividing 1235 by 8
Dividend / 8 Divisor Remainder
1235/8 154 3 
154/8 19 2 
19/8 2 3 
2/8 2 2 
 Writing remainder in reverse order
2323 which is the octal value of 1235.
iv)
The decimal value of 21.218 is
21218/1000

c)
i)
Max number = 3  2n-1 -1
Min number = -3  - (2n-1 -1)
ii)
Max number = 3  2n-1 -1
Min number = -4  - 2n-1
iii)
Max number = 3  2n-1 -1
Min number = -3  - (2n-1 -1)

Question 2
a)
Code below:
ORG 000
INPUT USER /user inputs the number
SUBT TWO /result - 2
PRINT, STORE Q /Storing AC in processing of M
CLEAR
ADD HIT1 /AC added to HIT1
ADD HIT2 /AC added HIT2
STORE TOTAL /HIT1 added to HIT2 gives result
/now HIT1 is equal HIT2
LOAD HIT2 /loading of HIT2 --> AC
STORE HIT1 /value stored --> HIT1
LOAD TOTAL /value loaded --> AC
STORE HIT2 /N2 --> AC
LOAD Q /Q --> register loading
SUBT ONE /Q minus 1
SKIPCOND 000 / AC<0, avoid cond 000
JUMP PRINT /starting the Loop and print
LOAD TOTAL /total value loaded in AC
OUTPUT /displayed output
HALT /pause
/Progression initialization

USER, DEC 0
ONE, DEC 1
TWO, DEC 2
Q, DEC 0
HIT1, DEC 0
HIT2, DEC 1
TOTAL, DEC 0

b)
For certain values of n, the program does not give sufficient results as it should. This anomaly
is caused by the threshold of the integer crossed by the system. There is a value limit that can
be gathered by data type.
The simulator shows that the maximum value is 2147483647 for which value of n = 24.
After n = 24 there will be a wrong value generated.

Question 3
a)
Interrupt handling is a key feature of real-time technology that includes interrupts and
interrupts managers. The system interrupt table is only those physical interrupts which have
sufficient priority. It is only a routine with an operation sequence. Each one can ask for input
& output during the run.
Approaches to deal with multiple interrupts:
 A non-nestled handler is the simplest interrupt handler. That means that all others are
disabled while managing an interrupt until the control is reinstated. Such a type can
only provide one interrupt at a time so that multiple interrupts with different levels of
priority cannot be serviced in complex embedded systems. The microprocessor
disables further interruptions once the "interrupt request" pin is raised. The controller
then points to the right entry on the function table and executes the command to
indicate the interrupt handler.
 An interrupt manager can be used to perform another interrupt in the current manager.
This is obtained by allowing interrupts again prior to full maintenance by the handler.
This function increases system complexity and therefore allows timing problems
which can cause failure in the system.