Part-2A Key Q1) 1. (A + B)(A + C) = AA + AC + AB + BC = A + AC + AB + BC = A(1 + C + B) + BC = A. 1 + BC = A + BC =2Marks

Transcription

1 Part-2A Key Q1) 1. (A + B)(A + C) = AA + AC + AB + BC = A + AC + AB + BC = A(1 + C + B) + BC = A. 1 + BC = A + BC =2Marks 2.What is the Golden rule of information? தகவல ன கக ல டன வ த என ற ல என ன? Value of Information Time 2 marks s add mark mark total =2Marks 5. 4) Q= (A+B)+B.C+A.C =2Marks A B C A+B A+B B.C A A.C (A+B)+B.C+A.C Q =2Marks Total 10 Marks

2 Q2) 1. What is the output of this Python tuple program? (gpd;tuk; Python tuple program apd; tpiyt vd;d?) a = ['Kamala', 'Apples', 150, ] a) print a =['Kamala', 'Apples', 150, ] b) print a[0] = Kamala c) print a[2] = 150 d) print a[-3] = Apples 2. >>> a + ['eggs'] 4x 0.5=2 a) Print a[0:2] =. ['Kamala', 'Apples'] b) Print a = =['Kamala', 'Apples', 150, ] 3. >>> a [0:2]=[ ] 2x 1=2 a) Print a[0:2] = [150, ] b) Print a[] = [150, ] 2x 1=2 4. >>> a=a*2 Print a[0:2] = [150, , 150, ] Print a[-1:-2] = [] 2x 1=2 5. >>> a. Print a[2] +100 = 250 b) Print a = [150, , 150, ] 2x 1=2 Total 10 Marks

3 Q3) Subject (, Subject _Name, Course Name,) Teachers (,, Teacher_name, Faculty name) This two tables are in relationship 1. Table Primary Key Foreign Key Subject Subject_No Teacher_ID Teachers Teacher_ID Subject_no 2x 1=2 2. M:M 2x 1=2 3. Create table Teachers (Tech _ID Varchar(5) unsigned Not Null Primary Key, Subject_no Varchar(6) Not Null Foreign Key, Teacher _name varchar(25), Faculty name Varchar(25)) 2x 1=2 4. Create table Subject (Subject_No Varchar(5) unsigned Not Null Primary Key, Subject_name Varchar(25), course _name varchar(25), Teacher _ID Varchar(5) Not Null Foreign Key) 2x 1=2 5. Select * from Teachers. 2x 1=2 Marks Total 10 Marks Q4) 1.clas B 2x 1=2 Marks x 1=2 Marks 3. many military applications, flight systems, ATM bank 2x 1=2 Marks 4. Credit card payment, on line exam. 2x 1=2 Marks 5. Von Neumann architecture 2x 1=2 Marks Total=10 marks Part 2 A total= =40

4 Part 2-b Q1 1 A B C F (A,B,C) m m 0 A.B.C m 1 A.B.C m 2 A.B.C m 3 A.B.C m 4 A.B.C m 5 A.B.C m 6 A.B.C m 7 A.B.C F (A,B,C)= A.B.C + A.B.C +A.B.C+ A.B.C + A.B.C 1 MARKS A B C D P (A,B,C,D P (A,B,C,D m 0 1 A.B.C.D m 1 1 A.B.C.D m 2 0 A.B.C.D m 3 1 A.B.C.D m 4 0 A.B.C.D m 5 0 A.B.C.D m 6 1 A.B.C.D m 7 1 A.B.C.D m 8 0 A.B.C.D m 9 0 A.B.C.D m 10 0 A.B.C.D m 11 0 A.B.C.D m 12 1 A.B.C.D m 13 0 A.B.C.D m 14 1 A.B.C.D m 15 1 A.B.C.D P (A,B,C,D= A.B.C.D +A.B.C.D+ A.B.C.D +A.B.C.D +A.B.C.D+ A.B.C.D + A.B.C.D + A.B.C.D 1 MARKS 2. F (A,B,C)= A.B.C + A.B.C +A.B.C+ A.B.C + A.B.C F (A,B,C)=( A.B.C ).( A.B.C).(A.B.C).( A.B.C ).( A.B.C) POS F (A,B,C)=( A+B+C).( A+B+C ).(A +B+C ).( A +B +C).( A +B +C ) 2 MARKS P (A,B,C,D= A.B.C.D +A.B.C.D+ A.B.C.D +A.B.C.D +A.B.C.D+ A.B.C.D + A.B.C.D + A.B.C.D P (A,B,C,D= (A.B.C.D ).(A.B.C.D). (A.B.C.D).(A.B.C.D ).(A.B.C.D). (A.B.C.D ).( A.B.C.D ).( A.B.C.D)

5 POS P (A,B,C,D= ( A+B+C+D). ( A+B+C+D ). (A+B+C +D ).(A+B +C +D).(A+B +C +D ). (A +B +C+D). (A +B +C +D). (A +B +C +D ) 3. 2 MARKS F (A,B,C)= A.B.C + A.B.C +A.B.C+ A.B.C + A.B.C 1 MARKS

6 P (A,B,C,D= A.B.C.D +A.B.C.D+ A.B.C.D +A.B.C.D +A.B.C.D+ A.B.C.D + A.B.C.D + A.B.C.D 1 MARKS 4. F (A,B,C)= A.B.C + A.B.C +A.B.C+ A.B.C + A.B.C= =0 P (A,B,C,D= A.B.C.D +A.B.C.D+ A.B.C.D +A.B.C.D +A.B.C.D+ A.B.C.D + A.B.C.D + A.B.C.D=0 5. = 3 marks = 2x2=4 marks Total=15 marks

7 Q2) 1 What is a Network Databases Some data were more naturally modeled with more than one parent per child. So, the network model permitted the modeling of many-to-many relationships in data. In 1971, The basic data modeling construct in the network model is the set construct. A set consists of an owner record type, a set name, and a member record type. A member record type can have that role in more than one set, hence the multiparent concept is supported. An owner record type can also be a member or owner in another set. = 1 marks 2. Relational Databases Relational Model (RDBMS - relational database management system) A database based on the relational model developed by E.F. Codd. A relational database allows the definition of data structures, storage and retrieval operations and integrity constraints. In such a database the data and relations between them are organised in tables. A table is a collection of records and each record in a table contains the same fields. Properties of Relational Tables: Values Are Atomic Each Row is Unique Column Values Are of the Same Kind The Sequence of Columns is Insignificant The Sequence of Rows is Insignificant Each Column Has a Unique Name = 1 marks 3. 2x 1=2 4. member member_no name address phone borrow isbn member_no checkout_date borrow_period

8 book isbn book_title authors 3x 1=3 5. Table Primary Key Foreign Key member member_no borrow isbn,member_no book isbn 1*3=3 marks 6. Table Table Relation ship member borrow M:M borrow book 1:1 book member M:M 1*2=2 marks 7. Q3) Create table member (member_no Varchar(5) unsigned Not Null Primary Key, name varchar(25), address Varchar(25), phone Varchar(10), Varchar(25)) Create table borrow (member_no Varchar(5) unsigned Not Null Foreign Key, isbn Varchar(5) unsigned Not Null Foreign Key) Create table book (isbn Varchar(5) unsigned Not Null Foreign Key, book title varchar(25), authors Varchar(25)) 1*3=3 Total=15 marks 1. Python has five standard data types: Numbers String List Tuple Dictionary = 3 marks 2. count=0 For i in range (0,25,2): Print sum=,sum

9 print "Good bye!" = 4 marks 3. def sumofthree( arg1, arg2, arg3 ): total = arg1 +arg2+arg3 print "Inside the function : ", total return total; total = sumofthree( 10, 20,50 ) print "Outside the function : ", total Inside the function : 80 Outside the function : 80 = 2*2=4 marks 4. X=open( teacher.txt, w ) x.write( TeacherName\t\tAge\nMrs.Mime\t\t20\nMr.Nimal\t\t40\n ) x.close X=open( teacher.txt ) = 2 marks For y in x: Print y.rstrip() Q4) = 2 marks Total=15 1. X Hub/Switch Y Router Z Hub/Switch = 2 Marks 2. A router is a device that forwards data packets between computer networks, creating an overlay... Routers perform the "traffic directing" functions on the Internet. routers join two networks together at a given protocol. A network switch is a small hardware device that joins multiple computers together within one local area network (LAN). 3. LAN1=Class A and LAN2=Class B 4. A= ,B= ,C= ,D= ,P= ,Q = ,,R= ,and S= Note: <255.<255 is correct <255.<255 is correct =.5*8 4Marks 5. Star 6. Advantages It is very easy to install and manage star network topology as it is the simplest of the lot when it comes to functionality. It is easy to troubleshoot this network type as all computers are dependent on the central hub which invariably means that any problem which leaves the network inoperable can be traced to the central hub. In star topology, the data packets don't have to make their way through various nodes which makes sure that the data transfer is fast.

10 At the same time, the fact that the data packets only make it through three different points ensures that the data is safe. As the nodes are not connected to each other, any problem in one node doesn't hamper the performance of other nodes in the network. Adding new machines or replacing the old ones is a lot easy in this network topology, as disruption of the entire network is not required to facilitate the same. Disadvantages The foremost problem with star network topology is the fact that it is highly dependent on the functioning of central hub. The size of the network is dependent on how many connections can be made to the hub. This network type requires more cable as compared to linear bus topology which means the expenses incurred would be relatively high. The performance of the entire network is directly dependent on the performance of the hub. If the server is slow, it will cause the entire network to slow down. If one of the numerous nodes utilizes a significant portion of the central hub's processing capability, it will reflect on the performance of other nodes. 7. =3*1 3 Marks Total= 15 Marks Q5) 1) What is the open system? Write 2 examples of that? a. An open system is a system which constantly interacts with its environment b. School system c. Computer system..ect 2) What is the close system? Write 2 examples of that? a. An isolated system that has no interaction with its external environment. b. Blood circulation system c. engine..ect 3) We consider following bus? a. Write a system? bus b. Write a Scope? peoples travel one place to other c. Write a Resource? seats, driver, engine,..ect

11 4) Write down the 3 scopes of the Bus? a. peoples travel one place to other b. Student travel to school c. Tourists travel one place to other d. Goods travel one place to other..ect 5) Write 2 functional requirement of this Bus? a. peoples travel one place to other b. Goods travel one place to other c. Very fast travel one place to other d. Relaxed travelling e. Easy travelling f. 6) Write 2 Nonfunctional requirement of this Bus? a. Can t go very fast (>250 km/h) b. Can t travel 500 people at a time. c. 7) Write 2 example of Natural System? a. Digestive system b. Nerves system..ect..ect 8) Write 2 example of Artificial System? a. School System b. Computer Q6)..ect =2*.5 1 Marks Total=15 marks 1. <table border="1"> < tr> < td> No</td> < td>name</td> < td>age</td> < /tr> < tr> < td>001</td> < td> Mr.Raju </td>< td> 15</td> < /tr> < /table> 2. <html> <body> <h4>> Travel Time</h4> <ul> <li>morning 7.00 am</li> <li> Morning 8.00 am</li> <li> Morning 9.00 am</li> </ul></body></html>

12 3. <img src="train.jpg" alt="train" width="150" height="50"> 4. Following file is save in name as Trainmain.html <html> <frameset rows="50%,50%"> <frame src=" heading.html "> <frameset cols="25%,75%"> <frame src=" Manu.html.htm"> <frame src=" Booking.html "> </frameset> </frameset> </html> 5. Following file is save in name as Booking.html <html> <body> <table border="1"> < tr> < td> No</td> < td>name</td> < td>age</td> < /tr> < tr> < td>001</td> < td> Mr.Raju </td>< td> 15</td> < /tr> < /table> <h4>> Travel Time</h4> <ul> <li>morning 7.00 am</li> =3*1 3 Marks <li> Morning 8.00 am</li> <li> Morning 9.00 am</li> </ul></body></html> 6. Following file is save in name as Booking.html <html> <body> <ul> <li> Train Booking</li> =3*1 3 Marks <li>train Time Table</li> <li> Contact</li></ul></body></html> =3*1 3 Marks Final=part2A(40)+part2B(60) =100 Total=15 marks