Digital Equipment Corporation: 1132980

 Problem 1.

Ans.

A, 2.

B, register windowing and window overlapping.

 Problem 2.

50 %thoughoutput.

 Problem 3.

Ans.

a) 1.2

b) 1.4

Problem 4.

Ans.

1.  

Programmed I/O.

Interrupted driven I/O.

Direct Memory Access I/O.

Channel I/O.

b)

Programmed I/O: work well for control applications such as sensors. It works according to instruction programmed of specific register.

Interrupted driven I/O: Used in personal systems. They are used to signaled bits in CPU flags register.

Direct Memory Access I/O: Used in medium small systems. They are used when CPU offloads execution for tedious input and output instructions.

Channel I/O: Used in mainframes handles multiple users and handles huge tasks. They are used to control one or more I/O processors of different I/O pathways.

Problem 5.

Ans.

a) .Radia Joy Perlman

b) .Digital Equipment Corporation and invented in 1984.

Problem 6.

Ans

1. IPv6  cur=115.1,AVG=118.2,Max=146.6,min=99.2
2. 83.99%

 Problem 7.

 Ans

a) 62%

b) Normal

Problem 8,

Ans

.Data Link.

 Problem 9,

    Ans

       1

 Problem 10,

      Ans.

1. 2³²/2²⁵ = 2²⁷
2. 32 bit addresses with 17 bits in the tag field, 10 in the block field, and 5 in the word field.
3.  000063FA = 00000000000000000 1100011111 11010, which implies Block 799

Problem 11.

Ans.

The Effective Access Time.

EAT=0.05(8ns)+0.95(180ns)=171.4ns

Problem 12

Ans

B07H.

Problem 13,

              Ans

                S=1

                   (1-f)+f/k

                S= speed

                F=fraction

               K =speed of new component

               S=1

                   (1-0.4)+0.4/1.5

              Speedup=0.87

Problem 14,

                     Ans.

                       A

Problem 15,

Ans.

45 00 00 34 fb 24 40 00 40 06 0a 65 c0 a8 04 06 17 db 58 b1

Let us first find the value of header fields before answering the questions:

VER = 0x4 = 4 HLEN =0x5 = 5 ® 5 × 4 = 20 bytes Service =0x00 = 0 (Normal/routine)

Total Length = 0x0034 = 84 bytes Identification = 0x000002 = 2 Flags and Fragmentation = 0x5850 ® D = 1 M= 0 offset = 6224 Time to live = 0x20 = 32 Protocol = 0x06 = 6 Checksum = 0x0000?? Source Address: 0x7C4E0302 = 124.78.3.2 Destination Address: 0xB40E0F02 = 180.14.15.2

 Problem 16,

                    Ans

                    Garbage Collection

Problem 17,

Ans

                      RAID 4

 Problem 18,

                     Ans

                     2^128

Problem 19,

                  Ans

                   32

 Problem 20,

               Ans.

              101.01

Problem 22,

               Ans.

              Bridge

Problem 23

Ans

343.90625

Problem 24

Ans

101111 

Problem 25,

Ans.

F=1

Problem 26

Ans.

MIMD

Problem 27.

                     Ans.

                    ICANN

Problem 28.

HEX4DCE.

Problem 28.

Ans.

REGISTERS

Problem 29

 Ans

MAC address

Problem 30,

Ans.

3.

Problem 31,

Ans

RAID 0

Problem 32,

ANS

The computer has 16-byte cache and 256 bytes of main memory

Reference.

Null, L and Lobur,J..(2003) .The essentials of Computer Organization and Architecture.

       Jones and Bartlett   Publishers Inc., Canada.

 Stallings,W.(2019) .Computer Organization and Architecture 

         Designing for Performance.11th Ed., Pearson Education, Inc., Hoboken, New Jersey.

Stallings,W.(2016) .Computer Organization and Architecture 

         Designing for Performance.10th Ed., Pearson Education, Inc., Hoboken, New Jersey.