computer organization [e & tc] solution manual

8/16/2019 Computer Organization [E & TC] Solution Manual

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Important Solved Examples

Example 2.2.3

Solution :

Example 2.2.4

Solution : (49)10 = (0110001)2

Example 2.2.5

Solution : (A43)16 = (0101001000011)2

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2

Data Representation

and Arithmetic Algorithms

1 0 1 0 1 1 0 1

0 1 0 1 0 0 1 0

Number

NOT operation

1's complement of number

0 1 1 0 0 0 1

1 0 0 1 1 1 0

Number

NOT operation


Number

NOT operation


0

1

1

0

0

1

1

0

0

1

0

1

1

0

0

1

0

1

0

1

0

1

1

0

1

0

2 Arithmetic Unit



Example 2.2.6

Solution :

Example 2.2.7

Solution :

Example 2.2.8

Solution :

Example 2.2.11

Solution : We have

( )0111002

( )2810

and

( )01111 2 ( )15 10

( )10000 2 1's complement of 15

( )100011 2 1's complement of 28

Computer Organization 2 - 2 Arithmetic Unit

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0 1 1 1

0 1 0 0

0 0 1 1

0011

1

10100101

Number

1's complement

Add 1

2's complement

+

0 1 0 01 1 0

1

Binary equivalent of (50)10

1's complement of (50)10

Add 1

2's complement of (50)10

1011001+

1 0 0 1 1 1 0

Carry1

1 0 0 11 1 0

1

1

Number

1's complement

Add 1

2's complement

1001010

+

0 1 0 1 0 1 0

Carry

0

1

1

0



Addition of (– 28) and (– 15) :

Sign-extension

+

1 1 0 0 0 1 1

Sign-extension 1 1 1 0 0 0 0

Carry 1 1 0 1 0 0 1 1

1 Add end-around carry

1 0 1 0 1 0 0 (– 43) Result is in 1'scomplement form

Verification : 1 0 1 0 1 0 0

0 1 0 1 0 1 1 ( )43 10

Note

Here, the magnitude of greater number is 5-bit; however, the magnitude of the result

is 6-bit. Therefore, the numbers are sign-extended to 7-bits.

For proper result we suggest to use 1 sign-bit extension to the number having greater

magnitude and represent the number having smaller magnitude with extended

number of bits.

Example 2.5.8

Solution :

Multiplication :



1 1 1 10 Multiplicand

1

10100

00100 1's complement of multiplicand

Add 1

2's complement of multiplicand

+

Multiplicand =

Multiplier =

Recoded multiplier =

(11011) = (– 5)

(0 0 1 1 0 ) = (3)

0 + 1 0 – 1

2 10

2 10

Implied zero

1

0

00000

0

+

0

1 0 1 1

– 10+ 1×

1010 0 0

0 0

1 1 1 0 1 1

0 0 0 0 0

1 1 1 1 0 100

+

+

(– 15)10




Note Shaded region indicate sign extension.

Example 2.5.9

Solution :

Multiplication :

Note Shaded region indicate sign extension

Example 2.5.10

Solution :



1

0

1 0 1 1

001

+

0

1

0 0 1 0 1

Multiplicand


Add 1


Multiplicand =

Multiplier =

11011 (– 5)

0 0 1 1 1 0

Implied zero(7)

Recoded multiplier = 0 +1 0 0 – 1

1

+1

1 0 1 1

–1000

1

0 0 0 0

×

0 0 0 0 0 0 1 0

0000

0 0 0 0 0 0 0

1 1 1 0 1 1

0 0 0 0 0

1 1 1 0 1 1 1 0 1

+

+

+


(– 35)

+

1

0

0 1 0 1

0101

+ 1

1 0 0 1 1

Multiplicand


Add 1


Multiplicand =

Multiplier =

Recoded multiplier =

(+13) = (0 1 1 0 1)

(–5) = (1 1 0 1 1)

1 1 0 1 1 0

0 – 1 +1 0 – 1

10 2

10 2 Implied zero



Multiplication :

Note Shaded region indicate sign extension

Example 2.5.11

Solution :

Example 2.6.7

Solution :

+ 13 = 0 1 1 0 1– 6 = 1 0 1 0 2's complement of 6

Bit pair recoding of – 6



1

– 1

0 1 0 1

– 10+ 10

1 1 1 1 1 0 0 1 1

00000000

0 0 0 1 1 0 1

1 1 0 0 1 1

0 0 0 0 0

1 1 1 0 1 1 1 1 1 1



(– 65)

+

+

+

+

×

1

0

0 1 1 1

–100+1

Multiplier

Multiplicand

Recoded multiplier

Recoded multiplier

2's complement of – 13

(– 195)

11

0

0 0 1 1

–100+1

00 0 0 0 0 1 1 0 1

000000000

00 0 0 0 0 0 0

00 0 0 0 0 0

11 0 0 1 1

1 1 0 0 1 1 1 1 0 1

+

+

+

+

×

11 0 1Signextension

0 0

–1 –2

Implied 0 toright of LSB



Multiplication

( 13) ( 6) = (1 0 1 1 0 0 1 0)2 = ( )78 10

Example 2.7.13

Solution :



11 0 0

+

11 0 1

–1 –2

1 1 0

01 0 1 1

01 1 1 0 0 1 0

2's complement of 13 2

2's complement of 13

2's complement of 78

0 1 1 0 1

0 1 1 0

1 0 0 1 1 0

01

13

13 2 by shift left 1-bit

2's complement of 13 2

1

0

0 0 0

111

1

0011

+

Divisor (B)

1's complement

Add 1

2's complement

Initially

Shift left A, Q

Shift left A, Q

Shift left A, Q

Shift left A, Q

Subtract B

Subtract B

Subtract B

Subtract B

Set Q0

Set Q0

Set Q0

Set Q0

Restore (A+B)

Restore (A+B)

A Register

0 0 0 0 0

0 0 0 0 1

0 0 0 1 1

0 0 1 1 0

0 0 1 0 0

1 1 1 0 0

1 1 1 0 0

1 1 1 0 0

1 1 1 0 0

1 1 1 0 1

1 1 1 1 1

0 0 0 1 0

0 0 0 0 0

0 0 1 0 0

0 0 1 0 0

0 0 0 0 1

0 0 0 1 1

1 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

0 0 1 0

1 0 0 0

0 0 0 0

0 0 0 1

0 0 1 1

Q Register

First cycle

Second cycle

Third cycle

Fourth cycle

Dividend

Remainder

Quotient



Example 2.7.15

Solution :

Example 2.7.16

Solution :



A Register Q Register

0 0 0 0 1 1 0 0

0 0 0 0 0 1 1 0 0

1 1 1 0 1

1 1 1 0 1

0 0 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 0

0 0 0 0 0

1 1 1 0 1

1 1 1 0 1

0 0 0 0 0

0 0 0 1 1

1 1 1 0 1

0 0 0 1 1

0 0 0 1 1

1 0 0 0

0 0 0

0 0 1

0 1 0

0 0 0 1

0 0 1 0

0 1 0 0

Initially

Shift

Shift

Shift

Shift

Dividend

First cycle

Second cycle

Third cycle

Fourth cycle

Quotient

Restoreremainder

Subtract B

Subtract B

Add

Add

Add divisor

0 1 1

1 0 0

0

1

1

1 1 0 1

+

Divisor (B)

1's complement

Add 1

2's complement



Example 2.7.17

Solution :



Initially

Shift left A, Q

Shift left A, Q

Subtract B

Add B

A Register

0 0 0 0 0

0 0 0 0 1

1 1 1 0 0

1 1 1 0 1

0 0 0 1 1

1 1 1 1 0

1 1 1 1 1

0 0 0

Q Register

First cycle

Second cycle

Third cycle

Fourth cycle

Dividend

Quotient

1 0 0 0

0 0 0

Remainder

0 0 0 0

0 0 0 0

Shift left A, Q

Add B

1 1 1 0

0 0 0 0 1

0 0 0

0 0 0 1

1

0 0 1 10

Shift left A, Q

Subtract B

0 0 1 0

1 1 1 1 1

0 0 1

0 0 1 0

0

1 1 0 11

Add B

1 1 1 1

0 0 0 1 0

1

0 0 1 10 Restore remainder

Divisor

1's complement

Add 1

2's complement

0 0 1 1

1 1 0 0

1 01 1

+ 1



Example 2.8.9

Solution :

Step 1 : Convert decimal number in binary format.

Integer part :



Initially

Shift left A, Q

Shift left A, Q

Shift left A, Q

Shift left A, Q

Subtract B

Add B

Subtract B

A Register

0 0 0 0 0

0 0 0 0 1

1 1 1 0 0

1 1 1 1 1

0 0 1 0 1

1 1 1 0 1

0 0 0 1 1

0 0 0 1 1

1 1 1 0 1

1 1 1 1 0

1 1 1 1 1

0 0 0 1 0

0 0 0 1 0

1 0 1 1

0 1 1

1 1 0 0

1 0 0

0 0 1

1 1 0

1 0 0

0 0 1

Q Register

First cycle

Second cycle

Third cycle

Fourth cycle

Dividend

Remainder Quotient

0 1 1 00

Add B

1

1

16 309

19

1

0

1

3

5 LSD

MSD

RQ

16

16

(309)10 = (135)16



(309)10

= (135)16

= (000100110101)2

Fractional part :

(0.1875)10 = (0.001)2

(309.1875)10 = (100110101.001)2

Step 2 : Normalize the number

100110101.001 = 1.00110101001 28

Step 3 : Single precision representation for a given number

S = 0, E = 8, M = 0011 0101 0010

Bias for single precision format is = 127

E = E + 127 = 8 + 127 = (135)10 = (10000111)2

Number in single precision format is given as

Step 4 : Double precision representation

For a given number

S = 0, E = 8, M = 0011 0101 0010



0 0

1

1

Hex number

0 0 1

3

10 1 0

5

10 Binary number

0.1875 2 = 0 375.

0.375 2 = 0 75.

0.75 2 = 1 5.

MSD

LSD

0

0

1

Fraction Base Product

×

×

×

0 1 0 0 0 0 1 1 1 0 0 1 1 0 1 0

31 30 23 22

0

0

1 0 0 1 0

Sign Exponent Mantissa



Bias for double precision format is = 1023

E = E + 1023 = 8 + 1023 = (1031)10 = (10000000111)2

Number in double precision format is given as

Example 2.8.10Solution :

Step 1 : Convert the decimal number in binary format.

Fractional part :

(0.0625)10 = (0.0001)2


0.0001 = 1.0000 2– 4



Sign Exponent Mantissa (52 - bits)

0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 0 0

0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0

63 62 52 51

0 0

1 0

1 0 1 0 0 1 0


0.0625 2 = 0 125.

0.125 2 = 0 25.

0.25 2 = 0 5.

MSD

0

0

0


×

×

×

0.5 2 = 1 0.

LSD

1×



Step 3 : Single precision representation

For a given number

S = 0, E = – 4, M = 0000

Bias for a single precision format is = 127

E = E + 127 = – 4 + 127 = (123)10 = (01111011)2


Step 4 : Double precision representation

For a given number

S = 0, E = – 4, M = 0000


E = E + 1023 = (1019)10 = (01111111011)2


Example 2.8.11

Solution : i) 32.75

Step 1 : Convert the decimal number in binary format.

Integer part :



0 0 1 1 1 1 0 1 1 0 0 0

31 30 23 22

0


0 0

0

0 0 1 1 1 1 1 1 1 0 1 1

63 62

0

Sign Exponent Mantissa (52 bits)

00 0 0

52 51 1 0

16

16

32

2

0

0

2

Q R

MSD

LSD

(32)10 = (020)16

0 0 0 0 0 0 1 0 0 0 0 0

0 2 0 Hex number

Binary number



(32)10 = (020)16 = (0000 0010 0000)2

Fractional part :

(0.75)10 = (0.110)2

(32.75) = (010000.110)2


0100000.110 = 1.00000110 25


For a given number

S = 0, E = 5, M = 00000110


E = E + 127 = 5 + 127 = (132)10 = (10000100)2


ii) 18.125

Step 1 : Convert the decimal number into binary format

Integer part :



0.75 2 = 1 5.

0.5 2 = 1 0.

0.0 2 = 0 0.

MSD

LSD

1

1

0


×

×

×

0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0

31 30 23 22 0


16

16

18

1

0

2

1

Q R

MSD

LSD

(18)10 = (12)16



(18)10 = (12)16 = (000 100 10)2

Fractional part :

(0.125)10 = (0.001)2

(18.125)10 = (010010.001)2


10010.001 = 1.00100010 24


For a given number

S = 0, E = 4, M = 00100010


E = E + 127 = 4 + 127 = (131)10 = (10000011)2


Example 2.8.12

Solution : Refer section 2.3.1



0 0 0 1 0 0 1 0

1 2 Hex number

Binary number

0.125 2 = 0 25.

0.25 2 = 0 5.

0.5 2 = 1 0.

MSD

LSD

0

0

1


×

×

×

0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 1 0 0 0

31 30 23 22 0




Step 1 : Convert decimal number in binary format integer part.

Integer part :

2 84 0 LSD

2 42 0

2 21 1

2 10 0

2 5 1

2 2 0

2 1 1 MSD

0

(84)10 = (1010100)2

Fractional part :

0.25 2 = 0.50 = 0.50 with carry 0 MSD

0.50 2 = 1.00 = 0.00 with carry 1 LSD

(0.25)10 = (0.01)2

– (84.25)10 = (1010100.01)2


– 1010100.01 = – 1.01010001 26


For a given number

S = 1, E = 6 and M = 0101 0001

Bias for single precision format is = 127

E = E + 127 = 6 + 127 = (1 3 3 )10

= (1 0 0 0 0 1 0 1) 2


1

ign

1 0 0 0 0 1 0 1

Exponent

0 1 0 1 0

S

0 0 1 0 0 . .. . 0

Mantissa (23-bits)

Step 4 : Double precision representation for a given number

S = 1, E = 6 and M = 0101 0001






E = E + 1023 = 6 + 1023 = (1 0 2 9 )10

= (10000000101)2


1

ign

1 0 0 0 0 0 1 0 1

Exponent

0 1 0 1

S

0 0 0 0 0 1 0 0 .... 0

Mantissa (52-bits)





Important Solved Examples

Example 5.3.7

Solution : Word bits : Each block consists of 128 words.

Word bits are : 7 (27 = 128)

Block bits : Block in cache = 1128

K = 1024128

= 8

Block bits are : 3 (23 = 8)

Address bits : 16 address bits. Since 216 = 64 K

Tag bits : Address bits – (word bits + block bits)

= 16 – (7 + 3) = 6 bits

Main memory address =

Tag Block Word

6 3 7

Example 5.4.1

Solution : The Fig. 5.1 shows the address space and memory space split into groups of

1K words. At any given time, up to four pages of address space may reside in main

memory in any one of the four blocks.

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4 Memory Organization5 Memory Organization

Block 0

Block 1

Block 2

Block 3

Memory space

M = 4 K = 212

Page 2

Page 1

Page 0

Page 3

Page 4

Page 5

Page 6

Page 7

Address space

N = 8 K = 213

Fig. 5.1 Address space and memory space



i) LRU : Least Recently Used

ii) LFU : Least Frequently Used

iii) FIFO : First in First Out

Example 5.4.2

Solution :

Cache trace for LRU policy

Cache trace for FIFO policy

Computer Organization 5 - 2 Memory Organization

TM

Initial 4 2 0

0

1

1

2

2

6

6

1 4 0 1 0 2 3 5 7

4 4 4 4

222

0 01

6

2

41

6

0

41

2 2 2 2

0 0 0 7

4 3331 1 5 5

Initial 4 2 0

0

1

1

2

2

6

6

1 4 0 1 0 2 3 5 7

4 4 4 4

222

0 0

1

6

0

4

1

4

1

2

3

2

5

2

7

6 6

4

5

4

2

3 3

Initial 4 2 0

0

1

1

2

2

6

6

1 4 0 1 0 2 3 5 7

4 4 4 4

222

0 0

1

6

2

4

1

0

4

1

2 2

0 0

3

1

5

2

1

0

2

1

7

Initial 1 6 4

4 4 4 44 4 4

5 1 4 3

3 3 6 63 5 3

2

2 2 2 76 6 6

1 2 1 4 6 7 4

1 1 1 11 1 1

Fig. 5.2 (a)

Initial 1 6 4

4 3 33 74 4 4 3

5 1 4 3

3 1 61 63 3 1 1

2

2 5 44 42 5 5 5

1 2 1 4 6 7 4

1 2 22 26 6 6 6

Fig. 5.2 (b)

computer organization [e & tc] solution manual

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