Great Deal! Get Instant $10 FREE in Account on First Order + 10% Cashback on Every Order Order Now

MAT1101 Assignment 1 2020 MAT1101 ASSIGNMENT 1 SEMESTER 1, 2020 WEIGHT: 30% TOTAL MARKS: 30 Due date: Monday 5th April XXXXXXXXXX:55pm AEST1 SUBMISSION INSTRUCTIONS The assignment is to be...

1 answer below »
MAT1101 Assignment 1 2020

MAT1101 ASSIGNMENT 1
SEMESTER 1, 2020
WEIGHT: 30% TOTAL MARKS: 30
Due date: Monday 5th April XXXXXXXXXX:55pm AEST1
SUBMISSION INSTRUCTIONS
The assignment is to be electronically submitted via Study Desk. If you
cannot submit electronically please contact the Examiner, as soon as
possible to make alternative a
angements.
You are to submit your assignment as a Portable Document Format
(PDF/A) file. PDF/A is an archival format that embeds all font glyphs used
in the document within the PDF file. This means PDF/A documents display
co
ectly on any computer system.
Word files will not be accepted by the assignment system. Instructions on
how to save a Word document in PDF/A format are included on page 8.
Hand-written and scanned assignments are perfectly acceptable, as long as
they are submitted as a PDF2 file. You just need to ensure that the resulting
scanned assignment is clearly legible. Some guidelines on scanning your
hand written document is included on page 8.
If you choose to typeset your assignment, you must ensure that all
mathematical notation etc. follow standard mathematical conventions. A
quick guide to typing Mathematics in Word can be found on the Study Desk
(if you really need to typeset your assignment). A guide to technical
communications is given in Appendix B of the Study Book, and available
via the Study Desk.
If you have trouble submitting your assignment via the Study Desk etc.,
please contact the Examiner, via Email, USQAssist or phone ASAP.

1 Australian Eastern Standard Time
2 These just need to be PDF, as these will not contain any computer fonts.
https:
usqassist.custhelp.com
MAT1101 S1 2020
2 Due date: Monday 5th April, XXXXXXXXXX:55pm AEST
ASSIGNMENT INSTRUCTIONS
• Show full working
easoning for each question. Give the marker every
opportunity to see how you obtained your answers. Your reasoning is just
as important as the final answer. Part marks for each question will be
awarded based on your reasoning.
• Clear communication and good presentation will make it easier for the
marker to give you marks for each question. Tips on mathematical
communication, see the USQ Learning and Teaching Support “Maths
QuickTips” and Appendix B: Technical Communication of the Study Book.
• Generally, in all your calculations, use as many decimal places as your
calculator will allow. Only round your final answers.
ASSIGNMENT QUESTIONS
Question 1 [18 marks]
Computer specification
Everything stored on a computer is expressed as a string of bits. However, different
types of data (for example, characters and numbers) may be represented by the
same string of bits.
For this question, we assume that text characters (or symbols) are stored in 8-bits.
Table 1 maps the ISO—LATIN—1 character set to the hexadecimal value
epresenting the state of these 8 bits. For example, from Table 1 the character ‘A’
has the hexadecimal value 41. Converting this hexadecimal value to binary gives
the state of the 8 bits (e.g XXXXXXXXXXstoring the character ‘A’.
Table 1: Hexadecimal map of the “Control, Basic and Supplemental Latin 1 Character set” to an 8-bit
encoding scheme. See http:
www.unicode.org/charts/PDF/U0000.pdf and http:
www.unicode.org/charts/PDF/U0080.pdf for control character definitions.
XXXXXXXXXX A B C D E F
0 NULL SOH STX ETX EOT ENQ ACK BELL BS HT LF VT FF CR SO SI
1 DLE DC1 DC2 DC3 DC4 NAK SYN ETB CAN EM SUB ESC FS GS RS US
2 SP ! " # $ % & ' ( ) * + , - . /
XXXXXXXXXX9 : ; < = > ?
4 @ A B C D E F G H I J K L M N O
5 P Q R S T U V W X Y Z [ \ ] ^ _
6 ` a b c d e f g h i j k l m n o
7 p q r s t u v w x y z { | } ~ DEL
8 XXX XXX BPH NBH IND NEL SSA ESA HTS HTJ VTS PLD PLU RI SS2 SS3
9 DCS PU1 PU2 STS CCH MW SPA EPA SOS XXX SCI CSI ST OSC PM APC
A NBSP ¡ ¢ £ ¤ ¥ ¦ § ¨ © ª « ¬ SH ® ¯
B ° ± ² ³ ´ µ ¶ · ¸ ¹ º » ¼ ½ ¾ ¿
C À Á Â Ã Ä Å Æ Ç È É Ê Ë Ì Í Î Ï
D Ð Ñ Ò Ó Ô Õ Ö × Ø Ù Ú Û Ü Ý Þ ß
E à á â ã ä å æ ç è é ê ë ì í î ï
F ð ñ ò ó ô õ ö ÷ ø ù ú û ü ý þ ÿ
https:
lor.usq.edu.au/usq/file/f391ad40-fd5e-4c3d-a1d2-add2ef63c6fc/1/OER_maths_comm.pdf
https:
lor.usq.edu.au/usq/file/f391ad40-fd5e-4c3d-a1d2-add2ef63c6fc/1/OER_maths_comm.pdf
http:
www.unicode.org/charts/PDF/U0000.pdf
http:
www.unicode.org/charts/PDF/U0080.pdf
http:
www.unicode.org/charts/PDF/U0080.pdf
http:
www.unicode.org/charts/PDF/U0080.pdf
S1 2020 MAT1101
Due date: Monday 5th April, XXXXXXXXXX:55pm AEST 3
In this computer, we will also assume that numbers (signed integers, unsigned
integers and single precision floating point real) numbers are stored in 12-bits. For
single precision floating point (real) numbers, 6-bits of these 12-bits are reserved
for the mantissa (or significand) with 2?−1 − 1 as the exponent bias (where ? is the
number of bits for the characteristic).
Information will to be passed within the computer using 24-bits. For example, the
string of 24-bits:
XXXXXXXXXX XXXXXXXXXX,
within our computer might represent:
• three ASCII/LATIN-1 characters ‘695’ (i.e. 3×8-bits encoded as per Table
1); or
• two numbers (2×12-bits). The interpretation of these 12-bits will be
different depending whether the numbers are to be interpreted as:
– signed integers (e.g XXXXXXXXXX → 867 and XXXXXXXXXX →
−1739),
– unsigned integers (e.g XXXXXXXXXX → 867 and XXXXXXXXXX
→ 2357), or
– as single precision floating point (real) numbers (e.g XXXXXXXXXXand
− XXXXXXXXXXMore precisely, any single precision floating point
(real) number between XXXXXXXXXXand XXXXXXXXXXwill have the same 12-bit
pattern, in this not a very accurate computer. Similarly, any floating
point number between − XXXXXXXXXXand − XXXXXXXXXXwill also have
the same 12-bit pattern.
2 marks i) Find the computer representation for the negative integer −515.
2 marks ii) Find the computer representation for the floating point number 51.25.
2 marks iii) Is the number stored in Question 1(ii) exact? If not what is the actual number
stored?
2 marks
iv) Find the bit pattern required to store the five characters:
§USQ©
The remaining parts (v–ix) refer to the following 24-bits:
0111 XXXXXXXXXX
2 marks v) Represent this string of bits as a single hexadecimal number.
2 marks vi) What characters according to Table 1 are represented by these 24-bits?
2 marks vii) What pair of unsigned integers is represented by these 24-bits?
2 marks viii) What pair of single precision floating point numbers could be represented
y these 24-bits?
MAT1101 S1 2020
4 Due date: Monday 5th April, XXXXXXXXXX:55pm AEST
2 marks ix) Double precision on this computer will use 24-bits, where 8-bits are used to store the
characteristic. Assuming this, answer the following questions.
• What will be the state of the 24-bits, if 51.25 is stored as a double
precision floating point number on this computer?
Question 2 [6 marks]
An Internet Protocol address (IP address) is a numerical bit pattern assigned to each
device (e.g., computer, printer) participating in a computer network that uses the
internet for communication. An IP address serves two purposes: 1) host
identification, and 2) location addressing.
IPv4 uses 32-bit (four-bytes) addresses. IPv4 addresses may be written in any
notation expressing a 32-bit integer value, but for human convenience, they are
most often written in the dot-decimal notation, which consists of four octets of the
address expressed individually in decimal and separated by periods. However,
IPv4 addresses can also be expressed in Octal, Decimal, Hexadecimal etc as shown
in the Table 2.
Table 2: The following table shows several representations of the IP Address XXXXXXXXXX
Format Value Notes
Dotted Decimal XXXXXXXXXXEach 8-bits is expressed as decimal
Dotted
Hexadecimal
?0.00.02.?? Each 8-bits expressed as
hexadecimal
Dotted Octal XXXXXXXXXX Each 8-bits expressed as octal
Hexadecimal ?00002?? 32-bit binary number in
hexadecimal
Decimal XXXXXXXXXXbit binary number in decimal
Octal XXXXXXXXXXbit binary number in octal.
3 marks i) Consider the following IP address given in dotted decimal format XXXXXXXXXXConvert it to
the following formats.
i) Decimal format. ii)
Binary format.
iii) Octal format.
3 marks ii) You are given an IP address (011101110011 XXXXXXXXXX) as a 32bit binary
number. Convert this number to:
i) Dotted Hexadecimal format.
ii) Octal format. iii) Dotted Decimal
format.
Note: Internet Protocol version 6 (IPv6) is the latest version of the Internet Protocol
(IP), and is intended to replace IPv4. IPv6 uses a 128-bit address, allowing 2128, or
more than 7.9 × 1028 times as many as IPv4, which uses 32-bit addresses which
gives 232 addresses.
S1 2020 MAT1101
Due date: Monday 5th April, XXXXXXXXXX:55pm AEST 5
Question 3 [6 marks]
Consider the following algorithm.
1. Input ???? a four digit year.
2. if (???? mod 4 = 0) { Divisible by 4 }
2.1 if (???? mod 100 = 0) { Divisible by 100 }
2.1.1 if (???? mod 400 = 0) {Divisible by 400}
XXXXXXXXXXisLeapYear ← TRUE;
2.1.2 Else
XXXXXXXXXXisLeapYear ← FALSE;
2.1.3 End if
2.2 Else { Not divisible by 100}
2.2.1 isLeapYear ← TRUE;
2.3 End if
3. Else { Not divisible by 4 }
3.1 isLeapYear ← FALSE;
4. End if
5. Output isLeapYear
3 marks i) Trace the algorithm starting with the input 2600.
3 marks ii) Document the changes that would need to be made
Answered Same Day Apr 04, 2021 MAT1101

Solution

Valupadasu answered on Apr 05 2021
146 Votes
Answers:
Question 1 answers start==================
1. The sign bit is 1.
Convert 515 to binary: 515 = 010 0000 0011
Find the 2’s complement: 101 1111 1101
The 12-bit representation for signed integer 515 is: 1101 1111 1101
2. The Sign bit is 0
Convert 51 to binary = 110011
Convert 0.25 to binary = 01
The representation for 51.25 is: 0 110011.01
Scientific notation is 1.100101 * 2^5(2 power 5)
Bias value is 31, since the exponent value is positive. The decimal value of exponent is 35.
Final value to be stored is
0 100011 100101
(As per the instruction provided in the question 6-bits of from the 12-bits are reserved for
the mantissa and 1 for sign so will left with 5 bits for exponent part so wrong value will be
saved)
3. No, the actual 12-bit floating point stored value number is 0 00011 100101
Again, this is a positive number (the first bit, the sign, is 0), the exponent is 100011 and
the mantissa is 1.100101.
Converting the exponent to decimal:
The conversion is a basic binary to decimal conversion.
1 + 2 = 3
Since the bias value is 15(as per formula 2 ̂k-1 − 1 as the exponent bias (k is 5). 15 must be
subtracted from the converted value:
3-15=-12
The resulting exponent is -12.
The whole mantissa can now be converted to decimal:
Converting the mantissa does not need the normalization to be undone . The whole number
can be calculated as follows:
1.100101* 2^-12(2 power -12)
either with negative exponents: (1+ 0.5 + 0.0625 + 0.015625) * 2^-12 = 0.000244
or fractions: (1 + 1/2 + 1/16 + 1/64) * 2^-12 = = 0.000244


4. Hexadecimal value of §USQ© is: A7 55 53 51...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here