COA - Assignment 2 with Solutions, Study Guides, Projects, Research of Computer Fundamentals

The second assignment of this course along with solution to the problems.

Typology: Study Guides, Projects, Research

2016/2017

Uploaded on 09/17/2017

ammarnadeeem
ammarnadeeem 🇵🇰

4.4

(9)

11 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Question - 1
How many 8K x 1 RAM chips are needed to construct a memory capacity of 512K x 16? Draw
and properly label a complete block diagram of the above memory organization.
32x32 x (8Kx1) RAM chips
Question - 2
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number
of tracks on one side of platter: 30,000; number of sectors per track: 600; seek time: one ms for
every hundred tracks traversed. Let the disk receive a request to access a random sector on a
random track and assume the disk head starts at track 0.
a. What is the average seek time?
b. What is the average rotational latency?
c. What is the transfer time for a sector?
d. What is the total average time to satisfy a request?
(a)Average seek time :
100 tracks = 1ms ,
299,999 tracks = 1m* 29,999 /100 = 0.299 , average time = 0.299/2 =m 149.99ms
(b)
Rotational speed = 7200 rev per minute = 120 rev per second
Average rotational latency = 1/2r = 4.167ms
(c) Transfer time for a sector:
Transfer time for a complete track = 1/r = 8.33msec
Transfer time for a sector = 8.33m/600 = 13.89usec
(d) total = 154.17msec
Question - 3
Consider a 4-drive, 200GB-per-drive RAID array. What is the available data storage capacity for each of
the RAID levels, 0, 1, 3, 4, 5, and 6?
RAID 0: 800 GB
RAID 1: 400 GB
RAID 3: 600 GB
RAID 4: 600 GB
RAID 5: 600 GB
RAID 6: 400 GB
Question - 4
Divide 189 by 19 in binary twos complement notation, using 10-bit words, using the algorithm
described in section 10.3
Q = 0010 1111 01 , M =00 0001 0011, -M = 11 1110 1101
A Q Count
00 0000 0000 00 1011 1101 10
pf3
pf4
pf5

Partial preview of the text

Download COA - Assignment 2 with Solutions and more Study Guides, Projects, Research Computer Fundamentals in PDF only on Docsity!

Question - 1 How many 8K x 1 RAM chips are needed to construct a memory capacity of 512K x 16? Draw and properly label a complete block diagram of the above memory organization. 32x32 x (8Kx1) RAM chips Question - 2 Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side of platter: 30,000; number of sectors per track: 600; seek time: one ms for every hundred tracks traversed. Let the disk receive a request to access a random sector on a random track and assume the disk head starts at track 0.

a. What is the average seek time? b. What is the average rotational latency? c. What is the transfer time for a sector? d. What is the total average time to satisfy a request?

(a)Average seek time : 100 tracks = 1ms , 299,999 tracks = 1m* 29,999 /100 = 0.299 , average time = 0.299/2 =m 149.99ms (b) Rotational speed = 7200 rev per minute = 120 rev per second Average rotational latency = 1/2r = 4.167ms (c) Transfer time for a sector: Transfer time for a complete track = 1/r = 8.33msec Transfer time for a sector = 8.33m/600 = 13.89usec (d) total = 154.17msec Question - 3 Consider a 4-drive, 200GB-per-drive RAID array. What is the available data storage capacity for each of the RAID levels, 0, 1, 3, 4, 5, and 6? RAID 0: 800 GB RAID 1: 400 GB RAID 3: 600 GB RAID 4: 600 GB RAID 5: 600 GB RAID 6: 400 GB Question - 4 Divide 189 by 19 in binary twos complement notation, using 10-bit words, using the algorithm described in section 10.

Q = 0010 1111 01 , M =00 0001 0011, -M = 11 1110 1101 A Q Count 00 0000 0000 00 1011 1101 10

00 0000 0100 (A-M)

0000010010 (A-M)

Remainder = 0000010010, Quotient = 0000001001

National University of Computer & Emerging Sciences, Lahore Department of Electrical Engineering (Spring 2017)

Express the following numbers in IEEE 32-bit floating-point format:

National University of Computer & Emerging Sciences, Lahore

  • Question -
    • a. -1.5 b. 384 c. 1/64 d. -1/
    • a.
      • b. 384 = 110000000 = 1.1x
        • 127 + 8 = 135 = Change binary exponent to biased exponent:
        • Format:
      • c. 1/64 = 0.000001 = 1 x 2127 - 6 = 121 = 01111001 - - Format :
      • d. –1/32 = –0.00001 = -1 x 2- - 127 – 5 = 122 = - Format:
  • Question -
    • a. The following numbers use the IEEE 32-bit floating-point format. What is the equivalent decimal value?
    • b.
    • c.
  • a. exponent = 10000011=131=> 131-127=4=> -1.11x2^4 = –
  • b. exponent= 01111110 =126 => 126-127 = -1 => 1.101x2 -1 =0.
    • c. exponent = 128 => 128-127 = 1 => 1.00 x 2 1 =>

Question - 8 Compare zero-, one-, two-, and three-address machines by writing programs to compute for each X = (A + B * C) / (D - E * F) of the four machines. The instructions available for use are as follows:

0 address 1 address 2 address 3 address PUSH A PUSH B PUSH C MUL ADD PUSH D PUSH E PUSH F MUL SUB DIV POP X

LOAD E

MUL F

STORE T

LOAD D

SUB T

STORE T

LOAD B

MUL C

ADD A

DIV T

STORE X

MOV R0, E

MUL RO, F

MOV R1, D

SUB R1, R

MOV R0, B

MOV R0, C

ADD R0, A

DIV R0, R

MOV X, R

MUL R0, E, F

SUB R0, D, R

MUL R1, B, C

ADD R1, A, R

DIV X, R0, R