TUTORIAL -1(SOLUTIONS)

COMPUTER ORGANISATION IT-IV SEM (TIT-402)

Prepared by: Mr.Dilip Kumar gangwar Faculty,CS/IT Deptt(GEHU),Dehradun

Q1.(43)10 = ( ? )2’s (GATE 2000) a)01010101 b)11010101 c)00101011 d)10101011

Q2.The 2’s complement representation of (-539)10 in hexadecimal is (GATE

2001) a)ABE B)DBC c)DE5 d) 9E7

Q3. The decimal value 0.25 (GATE 2002)

(a) is equivalent to the binary value 0.1 (b) is equivalent to the binary value 0.01 (c) is equivalent to the binary value 0.00111…

(d) cannot be represented precisely in binary

Q4. The 2’s complement representation of the decimal value –15 is (GATE 2002)

(a) 1111 (b) 11111 (c) 111111 (d) 10001

Q5. In 2’s complement addition, overflow (GATE 2002) (a) is flagged whenever there is carry from sign bit addition

(b) cannot occur when a positive value is added to a negative value (c) is flagged when the carries from sign bit and previous bit match

(d) None of the above Q6. Assuming all numbers are in 2’s complement representation, which of the

following numbers is divisible by 11111011? (A)11100111 (B) 11100100 (C) 11010111 (D) 11011011

Q7. Find the following differences using twos complement arithmetic:

a. 110011 - 111000

All numbers are in 2’s complement form (110011)2s= -(001100+1)= -(12+1)= -13

(111000)2s = -(000111+1)=-(7+1)= -8 i.e we are given -13 - -8 which is equal to -13+8=-5

now

110011 + (-111000) convert 111000 in 2s complement form=-(000111+1)=-(001000)

110011 + 001000

11101 1 =now it is in 2s form so convert it (111011)2s=-(000100+1)=-(000101)=-5

b.

101110 - 11001100=00101110-11001100 (11001100)2s=-(00110011+1)=-(00110100)=-52

00101110 =46 46 - -52=46+52=98

Now 00101110 =46 00110100 =52

0110001 0 =98

c. 110011110011 - 111100001111

110011110011=-(001100001100+1)=-(001100001101)=-781 111100001111=-(000011110000+1)=-(000011110001)=-241

-781 - -241=-781 + 241=-540

110011110011 000011110001

110111100100 = -(001000011011+1)=-(001000011100)=-540 d.

11101000- 11000011 11000011=-(00111101)

11101000 00111101

00100101 =37 Q8. Is the following a valid alternative definition of overflow in twos

complementarithmetic? If the exclusive-OR of the carry bits into and out of the leftmost column is 1,

then there is an overflow condition. Otherwise, there is not. Ans:Yes,because on different bits ,xor gate gives result as 1 so if carry out and

carr in are different that it can be checked by their xoring .If result=0 then no overflow otherwise there is overflow

Q9. Represent the following twos complement values in decimal:

1101011=-21 0101101=47

Q10. Represent the following decimal numbers in both binary sign/magnitude and twos complement using 16 bits 1) +401 2) -14

1)401 sign/magnitude form=0000 0001 1001 0001

twos complement form=1111 1110 0110 1111 2)-14

sign/magnitude form=1000 0000 0000 1110 twos complement form=1111 1111 1111 0010

Q11.the unsigned (1011011)2 =( ? )10 a)63 b)91 c) 92 d)13

Q12.Which has the largest value? A)(110)10 b) ( 10011011)2 c)(1111)2 d)(9A)16 e) (222)8

Q13.What is the weight of digit 3 in base 7 number 12345?

a)3 b)5 c)7 d)14 e)49

Q14.if (321)4=(57)10 What is the decimal equivalent of (32100000)4 a)57 x 10

5 b) 57 x 10

4 c) 57 x 4

5 d) 57 x 4

10

ans: (32100000)4 =(321)4 x 45

=57 x 45

Q15. The unsigned binary number (110001)2 is equal to ( ? )8

a)49 b)61 c)31 d)15 e)None of the above Q16.In 6 bit 1’s complement binary no system ,what is the decimal value

represented by (010100)1s a)-11 b)43 c)-43 d)20 e)-20

Q17.In 6 bit 2’s complement binary no system ,what is the decimal value represented by

(100100)2s a)-4 b)36 c)-36 d)-27 e)-28

Q18.For 2’s complement binary number,the range of values for 5 bit numbers is a)0 to 31 b) -8

to +7 c)-8 to -8 d) -15 to -15 e) -16 to +15 Q19.For 4 bit 2s complement scheme,what is the result of this

(1011)2s + (1001)2s 1011 =-5

1001 = -7 1010 0 =4

a.0100 b. 0010 c. 1100 d.1001 e.overflow

Q20. (1217)8 is equivalent to (GATE 2009) (A) (1217)16

(B) (028F)16 (C) (2297)10

(D) (0B17)16 Q21. If 73x (in base x number system)=54y, (in base y number system) ,then

possible values of x and y are a)8,16 b)10 ,12 c) 9,13 d)8,11 (GATE 2004) Ans:

7 x x1 +3 x x

0 = 75 x y

1 +4 x y

0

7 x +3 = 75 y +4

Putting values of x and y from options,by hit and trial method We get x=8,y=11

Q22.Let A=1111 1010 and B=0000 1010 be two 8 bit 2’s complement no’s.Their product in 2’s

complement is a)1100 0100 b)1001 1100 c)1010 0101 d)1101 0101 (GATE 2004) Q23.The hexadecimal representation of (657)8 is (GATE 2005)

a)1AF b) D78 c)D71 d)32F Q24. how many kilobytes are accessible with a 23-bit address space?

A)2^23 KB B)2^23 Kb C)2^13 KB D)2^13Kb Ans:addresses locations=2^23 bytes= 2

13 x 2

10 bytes= 2

13 Kbytes=2

13 KB

Q25. Represent each of the following using the 8-bit two's-complement integer

representation. a. 10(10)=0000 1010

b. −60(10) = 1100 0100 c. −104(10) = 1001 10000

Q26. For the following, assume a six-bit two's-complement representation of integers. a. What numeric value does 110110 represent? = -10

b. What numeric value does 010101 represent?= 21 c. What bit pattern is used to represent −12(10)? = 110100

Q27. a. What is the smallest (most negative) number you can represent in seven bits using sign magnitude representation? Give both the bit pattern of the

number and its base-10 translation. Ans:smallest (most negative) number in 7 bit in sign magnitude format:

1 11 1111=-63

b. Answer the same question for a seven-bit two's-complement representation. 100 0000=-(64)10

Q28. What would be the bias value for a. A base-2 exponent in a 6-bit field?=2

6-1 -1=31

b. A base-8 exponent in a 7-bit field? = 27-1

-1=63

Q29. It is said that a 32-bit format can represent a maximum of different

numbers. How many different numbers can be represented in the IEEE 32-bit format?

Explain. Done in class

Q30.Express the following numbers in IEEE 32-bit floating-point format: a)-5 b) -6 c) -1.5 d) 384 e) 1/16 f)-1/32 G)1.0 H)-1.0 I) (438F0000)16

a)-5 = -(101)= -(1.01) x 2^2

exponent bits=2+127=129

1 10000001 0100000…00000

b)-6=-(110)=-(1.10) x 2^2

exp bits=2+127=129

1 10000001 1000000……000

c)-1.5 =- (1.1)2 x 20

exp=0+127=127

1 01111111 100…..0000

d)384 = 110000000 =(1.10000000) x 2^8

exp bits=8+127=135

0 1000 0111 10000000……000

e)1/16 =0.0625 = 0.00010000 =1.0000 x 2^-4

exp bits=-4+127 = 123

0 01111011 00000…..000

f)-1/32 =0.03125 =0.00001000 =1.000 x 2^-5

exp bits =-5+127 =122

0 01111010 0000…….000

g)1.0 =(1.0)2 = 1.0 x 2^0

exp bits=0+127=127

0 01111111 000……000

h)-1.0 = 1 01111111 00000000…….000000

i)(438F0000)16

convert in binary

(0100 0011 1000 1111 0000 0000 0000 0000)2

1.00 0011 1000 1111 0000 0000 0000 0000 x 2^30

Exp bits 30+127=157= (10011101)2

0 10011101 00 0011 1000 1111 0000 0000 0

Q31. The following numbers use the IEEE 32-bit floating-point format.What is the equivalent

decimal value? a. 1 10000011 11000000000000000000000

b. 0 01111110 10100000000000000000000 c. 0 10000000 00000000000000000000000

ans: 1 10000011 11000000000000000000000

10000011=(131)10

Now subtract 127 from 131= 131-127=4

-(1. 11000000000000000000000 )2 x 2^4

-(11100.0000000000000000000)2 x 2^4 x 2-4

=-(11100.0000000000000000000) = (-28.0)10

b) 0 01111110 10100000000000000000000

01111110=126

Now 126-127=-1

=1.10100000000000000000000 x 2^-1

= 0.110100000000000000000000 x 2^-1 x 2 ^1

=0.110100000000000000000000=0.5+0.25+0.0625=0.8125

c) 0 10000000 00000000000000000000000

10000000=128

128-127=1

=1. 00000000000000000000000 x 2^1

=1.00000000000000000000000 x 2^1 x2^-1

=10.0000000000000000000000 = (2.0)10

Q32. Express the following numbers in IBM’s 32-bit floating-point format,

which uses a 7-bitexponent with an implied base of 16 and an exponent bias of 64 (40 hexadecimal).A normalized floating-point number requires that the

leftmost hexadecimal digit be nonzero; the implied radix point is to the left of that digit.

Ans.LEAVE THIS QUESTIONS RIGHT NOW AS IBM HAS DIFFERENT METHOD TO REPRESENT BINARY NUMBERS

Q33.A hypothetical computer stores floationg point numbers in 7 bits.Th e first

bit is used forsign of number,the next three for the biased exponent and the next three for the magnitude of the mantissa.The number (0010110)2 represented in

base 10 is a)0.375 b)0.875 c) 1.5 d) 3.5

Q34. A hypothetical computer stores floationg point numbers in 7 bits.Th e first bit is used for sign of number,the next three for the biased exponent and the next

three for the magnitude of the mantissa.You are asked to represent 33.35 in the above word.The error in this case would be

a)underflow b)overflow c)NaN d) No error

Q35. Consider the following 32-bit floating-point representation scheme as shown in the formal below. A value is specified by 3 fields, a one bit sign field (with 0 for

positive and 1 for negative values), a 24 bit fraction field (with the binary point being at the left end of the fraction bits), and a 7 bit exponent field (in excess-64

signed integer representation, with 16 being the base of exponentiation). The sign bit is the most significant bit.

(a) It is required to represent the decimal value –7.5 as a normalized floating point number in the given format. Derive the values of the various fields.

Express your final answer in the hexadecimal.

(b) What is the largest values that can be represented using this format? Express your answer as the nearest power of 10. (GATE 2002)

ANS 35(A)

=1 01111000000000000000000 0100 0001 =1011 1100 0000 0000 0000 0000 0100 0001

=AC000081 B)LARGEST NO:

0 111………111 1111110 We take exp bits as 1111110 =(126)10 , as 1111111 is generally not valid but if

you take 1111111 then also here answer will not differ much.(as here nothing is

given whether to take 1111 111 or not.But if question is explicitly about IEEE 754 then you cannot take 1111 110)

=1.111……..11 x 16126-64

(1 is used for normalisation ,in (B) part nothing is given about to do normalisation or not,so it is optional to do normalisation) =

(1.111.11 )2x 1662

=(1.999)10 x 16

62

Converting 1662

in power of 10 We get 16

62 = 10

75 (nearly)

=>(1.999)10 x 1075

Q36. The following is a scheme for floating point number representation using

16 bits (GATE2003) Let s, e, and m be the numbers represented in binary in the sign, exponent, and mantissa fields respectively. Then the floating point number represented is:

What is the maximum difference between two successive real numbers

representable in this system? Ans 36:The diagram in the question paper tutorial was wrong .Above is the

correct diagram

Q37.Explain why, in a computer, floating point mathematics may not be associative or distributive, i.e. (A+B)+C may not equal A+(B+C).

Ans: While floating-point addition and multiplication are both commutative (a + b = b + a and a×b = b×a), they are not necessarily associative. That is, (a + b) + c is

not necessarily equal to a + (b + c). Using 7-digit decimal arithmetic:

Due to the fixed number of bits in the mantissa, floating point

arithmetic can lose precision when small numbers are added to or subtracted from big numbers. In the worst case, the small number

has no effect at all on the larger number. If you arrange your mathematics so that the small numbers are added to each other

first, they stand a better chance of affecting the bigger number. Order matters.

Q38.We use negative numbers ,positive numbers in computer,then it seems that

unsigned numbers are of no use in computer So What is the application of unsigned numbers?

ANS: Unsigned integers are used when we know that the value that we are storing will always be non-negative (zero or positive). One example is:

suppose the OS want to keep the information of no of total sectors in hard disk .Now No of sectors cannot be a negative quantity .so a programmer can use a unsigned variable to store this information. General Questions Q39. If a CPU has a clock frequency of 3.2 GHz, how long (in ns) does one

access cycle take? Ans:

Time period=1/frequency=1/3.2GHz =(1/3.2 x 10

9)

=0.3125 x 10-9

seconds=0.3125 nanoseconds Q40.Are hardware and software equivalent? Can you really do anything in

hardware that you can do in software, and vice-versa? Examples? Ans:NO .

Q41.What are the three basic-level pieces of a digital computer? Ans: 1- processor to interpret and execute programs

2- memory to store both data and programs 3- mechanism for transferring data to and from the outside world.

Q42.Give two reasons why you can't store 12GB of system memory on a hard

drive with an advertised capacity of 12GB.

Ans: Discussed in class

The operating system needs to store name and directory information on the disk, information about sectors etc.These information are called meta data