CS6303 Computer Architecture April May 2015

Anna University Questions - CS6303 Computer Architecture April May 2015, Computer Science and Engineering (CSE), Third Semester, Regulation 2013

Exam	B.E/B.Tech. (Full Time) DEGREE END SEMESTER EXAMINATIONS
Academic Year	April May 2015
Subject Code	CS6303
Subject Name	Computer Architecture
Branch	Computer Science and Engineering
Semester	Third Semester
Regulation	2013

B.E / B.Tech. (Full Time) DEGREE END SEMESTER EXAMINATIONS, APRIL / MAY 2015

Computer Science and Engineering

Third Semester

CS6303 COMPUTER ARCHITECTURE

(Common to Information Technology)

(Regulations 2013)

Time : 3 Hours                      Answer A L L Questions                Max. Marks 100

PART-A (10 x 2 = 20 Marks)

1. List the eight great ideas invented by computer architects.

2. Distinguish Pipelining from Parallelism.

3. How overflow occur in subtraction?

4. What do you mean by sub word parallelism?

5. What are R-Type instructions?

6. What is a branch prediction buffer?

7. Differentiate between Strong scaling and Weak scaling.

8. Compare UMA and NUMA multiprocessors.

9. What is the need to implement memory as a hierarchy?

10. Point out how DMA can improve I/O speed.

Part-B (5* 16 = 80 Marks)

11. (a) Discuss about the various techniques to represent instructions in a computer system. (16)

Or

(b) What is the need for addressing in a computer system? Explain the different addressing modes with suitable examples. (16)

12. (a) Explain the sequential version of multiplication algorithm and its hardware. (16)

Or

(b) Explain how floating point addition is carried out in a computer system. Give an example for a binary floating point addition.

13. (a) Explain the different types of pipeline hazards with suitable examples. (16)

Or

(b) Explain in detail how exceptions are handled in MIPS architecture? (16)

14. (a) Discuss about SISD, MIMD, SMD, SPMD and VECTOR systems. (16)

Or

(b) What is hardware multithreading? Compare and contrast Fine grained Multi-Threading and Coarse grained Multi-Threading. (16)

15. (a) Elaborate on the various memory technologies and its relevance. (16)

Or

(b) What is virtual memory? Explain the steps involved in virtual memory address translation. (16)

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Find the functional dependenccies of the given relation

How to check whether a functional dependency hold in a relation R, Check the functional dependencies of a relation, Functional dependencies satisfied or not?, Find all the functional dependencies or a table

Exercise

For the following Student table instance, find all the possible functional dependencies that are held. Do not include trivial FDs.

[ Schema - Student (Regno, Name, DOB, Phone, Gender, Course_ID, Course_Name, Instructor_ID, Instructor_Name, Instructor_Office)]

Regno	Name	DOB	Phone	Gender	CID	CName	Ins_ID	Ins_Name	Ins_Office
14M01	Kumar	12-Jan-1996	12345	M	C1	DBMS	I1	Kesav	G123
14M05	Mary	10-Jun-1995	12367	F	C1	DBMS	I1	Kesav	G123
14M07	Ram	10-May-1996	12898	M	C1	DBMS	I2	Ragav	G127
14M01	Kumar	12-Jan-1996	12345	M	C3	DS	I5	Mani	G125
14B01	Revathi	10-Dec-1995	23456	F	C3	DS	I5	Mani	G125
14M09	Steve	23-Oct-1995	34567	M	C4	OS	I5	Mani	G125
14B03	Ramya	20-Jul-1996	23456	F	C4	OS	I5	Mani	G125

Solution:

Functional Dependency:

X → Y

-         X may be one or more attributes, and Y may be one or more attributes.

-         In this FD, X is the determinant set and Y is the dependent set.

-         We would read this as, “Y is dependent on X”, or “X determines the value of Y uniquely”

The following functional dependencies are held in Student;

Regno → Name

– Names are uniquely identified by a regno. In other words, for a given register number, there is exactly one name.

Regno → DOB

-         For any given register number in Student, there is exactly one DOB value.

Regno → Phone

Regno → Gender

You can write the above FDs collectively as follows;

Regno → Name, DOB, Phone, Gender

-         For any given register number of Student table, you would get unique name, DOB, phone, and gender values.

CID → CName

CName → CID

-         For the given instance, this FD is true. This FD may not be correct in some cases. For example, if suppose DBMS is offered for two different programs, say B.Tech and M.Tech with two different course IDs C1 and C10, then this FD does not hold in Student. You need to verify the semantics and the permitted values for a column carefully.

Ins_ID → Ins_Name, Ins_Office

Regno CID → Name, DOB, Phone, Gender, CName

-         Regno and CID together, i.e., any combination of these two values can uniquely identify the values of name, dob, phone, gender, and cname.

Regno CID Ins_ID → Name, DOB, Phone, Gender, CName, Ins_Name, Ins_Office

-         The combination of regno, cid and ins_id values can uniquely identify all the other attributes of the table Student. Hence, the combination (Regno, CID, Ins_ID) is the key for Student.

Example of the functional dependencies (Not all) that are not holding on Student:

Phone → Regno

-         As per the given instance of Student, for a given phone value 23456, there are two students with register numbers 14B01, and 14B03. Hence, this FD does not hold on Student.

CID → Ins_ID

-         CID C1 is taught by two instructors I1 and I2. Hence, this FD does not hold.

DOB → Regno, Name, Phone, Gender

-         Our Student instance does not contain duplicate values for the column DOB. But in reality, there may be more than one student with the same DOB. Hence, we cannot use DOB as the determiner.

The other set of FDs (if any), you are requested to try.

Go back to other Normalization Process Exercises