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...

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MAT1101 Assignment 1 2020 MAT1101 ASSIGNMENT 1 SEMESTER 1, 2020 WEIGHT: 30% TOTAL MARKS: 30 Due date: Monday 5th April 2020 11:55pm 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 arrangements. 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 correctly 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, 2020 11:55pm AEST ASSIGNMENT INSTRUCTIONS • Show full working/reasoning 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 representing 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. 01000001) storing 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. 0 1 2 3 4 5 6 7 8 9 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 ! " # $ % & ' ( ) * + , - . / 3 0 1 2 3 4 5 6 7 8 9 : ; < ==""> ? 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, 2020 11:55pm 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: 0011 0110 0011 1001 0011 0101, 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. 0011 0110 0011 → 867 and 1001 0011 0101 → −1739), – unsigned integers (e.g. 0011 0110 0011 → 867 and 1001 0011 0101 → 2357), or – as single precision floating point (real) numbers (e.g. 0.13672 and −0.0004044). More precisely, any single precision floating point (real) number between 0.13672 and 0.1406 will have the same 12-bit pattern, in this not a very accurate computer. Similarly, any floating point number between −0.0004044 and −0.0004119 will 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: 011100100110010111110011 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 by these 24-bits? MAT1101 S1 2020 4 Due date: Monday 5th April, 2020 11:55pm 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 192.0.2.235 Format Value Notes Dotted Decimal 192.0.2.235 Each 8-bits is expressed as decimal Dotted Hexadecimal ?0.00.02.?? Each 8-bits expressed as hexadecimal Dotted Octal 0300.0000.0002.0353 Each 8-bits expressed as octal Hexadecimal ?00002?? 32-bit binary number in hexadecimal Decimal 3221226219 32-bit binary number in decimal Octal 30000001353 32-bit binary number in octal. 3 marks i) Consider the following IP address given in dotted decimal format 10.0.20.105. Convert it to the following formats. i) Decimal format. ii) Binary format. iii) Octal format. 3 marks ii) You are given an IP address (01110111001100001000100010001001) 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, 2020 11:55pm 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} 2.1.1.1 isLeapYear ← TRUE; 2.1.2 Else 2.1.2.1 isLeapYear ← 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 DayApr 04, 2021MAT1101

Answer To: MAT1101 Assignment 1 2020 MAT1101 ASSIGNMENT 1 SEMESTER 1, 2020 WEIGHT: 30% TOTAL MARKS: 30 Due...

Valupadasu answered on Apr 05 2021
144 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...
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