Scribd
Upload
232 views1 page

Quiz4 - Solution PDF

This document contains two quiz questions about computer architecture and organization. The first question asks to represent decimal numbers in binary sign/magnitude and twos complement formats using 16 bits. The second question asks to compute the product of 23 and 29 using Booth's multiplication algorithm with 6-bit words. It provides the full steps and calculation using Booth's algorithm shown in a table with the answer being 001010 011011 = 667.

Uploaded by

b2uty77_593619754
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
232 views1 page

Quiz4 - Solution PDF

This document contains two quiz questions about computer architecture and organization. The first question asks to represent decimal numbers in binary sign/magnitude and twos complement formats using 16 bits. The second question asks to compute the product of 23 and 29 using Booth's multiplication algorithm with 6-bit words. It provides the full steps and calculation using Booth's algorithm shown in a table with the answer being 001010 011011 = 667.

Uploaded by

b2uty77_593619754
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 1

KT14203- Computer Architecture and Organization

Sekolah Kejuruteraan dan Teknologi Maklumat

QUIZ #4

1. Represent the following decimal numbers in both binary sign/magnitude and


twos complement using 16 bits: +512; -29 [4 marks]

Answer:

Sign Magnitude: 512 = 0000 0010 0000 0000 [1 mark]


29 = 1000 0000 0001 1101 [1 mark]
Twos Complement: 512 = 0000 0010 0000 0000 [1 mark]
29 = 1111 1111 1110 0011 [1 mark]

2. Given x = 23 and y = 29. Compute the product X x Y with Booths


Algorithm using 6-bit words. [12 marks]

Answer:

For table, each row [1 mark]. Total = 11 marks

A Q Q-1 M
(multiplier) (multiplicand)
000000 011101 0 010111 [1]
101001 011101 0 A <- A-M [1]
110100 101110 1 Shift [1]
001011 101110 1 A <- A+M [1]
000101 110111 0 SHIFT [1]
101110 110111 0 A <- A-M [1]
110111 011011 1 SHIFT [1]
111011 101101 1 SHIFT [1]
111101 110110 1 SHIFT [1]
010100 110110 1 A <- A+M [1]
001010 011011 0 SHIFT [1]

A = 001010 , Q = 011011
001010 011011 = 667 [1]

Semester 2 2013/2014 Page 1

You might also like