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Test 2 for Computer Architecture - Fall 2003 | CSCI 4717, Exams of Computer Architecture and Organization

Material Type: Exam; Professor: Tarnoff; Class: Comp Architecture; Subject: Computer & Information Science (CSCI); University: East Tennessee State University; Term: Fall 2003;

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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Download Test 2 for Computer Architecture - Fall 2003 | CSCI 4717 and more Exams Computer Architecture and Organization in PDF only on Docsity! Points missed: _____ Student's Name: __________________________________ Total score: _____ /100 points East Tennessee State University Department of Computer and Information Sciences CSCI 4717 – Computer Architecture TEST 2 for Fall Semester, 2003 Section 201 Read this before starting! • The total possible score for this test is 100 points. • This test is closed book and closed notes • You may use one sheet of scrap paper that you will turn in with your test. • When possible, indicate final answers by drawing a box around them. This is to aid the grader (who might not be me!) Failure to do so might result in no credit for answer. Example: • If you perform work on the back of a page in this test, indicate that you have done so in case the need arises for partial credit to be determined. Binary Hex Binary Hex Power of 2 Equals 0000 0 1000 8 24 16 0001 1 1001 9 25 32 0010 2 1010 A 26 64 0011 3 1011 B 27 128 0100 4 1100 C 28 256 0101 5 1101 D 29 512 0110 6 1110 E 210 1K 0111 7 1111 F 220 1M 230 1G “Fine print” Academic Misconduct: Section 5.7 "Academic Misconduct" of the East Tennessee State University Faculty Handbook, June 1, 2001: "Academic misconduct will be subject to disciplinary action. Any act of dishonesty in academic work constitutes academic misconduct. This includes plagiarism, the changing of falsifying of any academic documents or materials, cheating, and the giving or receiving of unauthorized aid in tests, examinations, or other assigned school work. Penalties for academic misconduct will vary with the seriousness of the offense and may include, but are not limited to: a grade of 'F' on the work in question, a grade of 'F' of the course, reprimand, probation, suspension, and expulsion. For a second academic offense the penalty is permanent expulsion." Error Checking and Correction 1. True or false: Regarding types of memory errors, a hard error refers to a permanent defect in memory whereas a soft error is a random event causing no permanent damage. (2 points) 2. An error checking code is generated both when data is stored and when it is retrieved. These two codes are compared by doing a bitwise ___________ to see if an error has occurred. (2 points) 3. The graphic to the right depicts the digits of a 4-bit Hamming code where a single bit error has occurred. Circle the bit that has flipped. (2 points) 4. True or false: The graphic to the right depicts the digits of a 4-bit Hamming code with parity where a double bit error has occurred. (2 points) 5. If the check code stored with a piece of data is 10011 and the check code calculated on the data retrieved from memory is 10010, what is the syndrome word? (2 points) 6. True or false: Assuming that a single bit error occurred for the syndrome word from problem 5, the syndrome word indicates that there was an error in the check codes, not the stored data. (2 points) 7. If it takes a minimum of 7 bits to correct a single error in a 64-bit data word, then how many bits will it take to do single-error correction with double-error detection? (2 points) a.) 7 b.) 8 c.) 9 d.) Cannot be determined e.) None of the above 8. What is the maximum number of data bits that can be monitored for single error correction using a 5 bit check code? (2 points) a.) 15 b.) 16 c.) 20 d.) 26 e.) 27 f.) 31 g.) 32 h.) 33 i.) 64 9. The table below describes the position number of the data and code bits of a single error correction code. Determine the equations to derive C8, C4, C2, and C1 from D8 through D1. (6 points) M+K bit position 12 11 10 9 8 7 6 5 4 3 2 1 Position number 1100 1011 1010 1001 1000 0111 0110 0101 0100 0011 0010 0001 Data bit D8 D7 D6 D5 D4 D3 D2 D1 Check code bit C8 C4 C2 C1 C8 = C4 = C2 = C1 = 10. For the error correcting system of the previous question, assume that the check code retrieved from memory was 1001, and the newly calculated check code on the data retrieved from memory was 1111. Assuming a single bit error has occurred, which bit was the one that flipped, D8, D7, D6, D5, D4, D3, D2, D1, C8, C4, C2, or C1? (2 points) 1 0 1 0 11 0 1 0 1 0 0 01 0
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