Page 288 - FUNDAMENTALS OF COMPUTER

288 Fundamentals of Computers NPP

Solution: hc:
(a) (+32) + (–14)
+32 ⇒ 0 0 10 0 0 0 0
–14 ⇒ + 11110 0 10

end 1 0 0 0 1 0 0 1 0
carry
Neglect end carry; the result is (+18). A§{V_ hm{gc H$mo N>mo‹S>Zo na n[aUm_ (+18) AmVm h¡Ÿ&
Thus, (+32) + (–14) = (+18)
(b) (–47) + (–31)
+47 ⇒ 1 1 01 0 0 0 1
+ 1 1 1 0 0 0 0 1
-31 ⇒
end carry 1 1 0 1 1 0 0 1 0
(Neglected)
Result ⇒ 10110010 = (–78)
Thus, NPP (–47) + (–31) = (–78)
Problem 3.83 àíZ 3.83
Do the following using complement H$m°påßb_|Q> H$s {d{Y go {H«$`m H$amo…
method :
(–1110) + (–1011) 2
2
Solution: hc:
(a) Convert into 7-bits _mZ H$mo 7- {~Q> _| ~XbZo na
(–0001110) + (–0001011) 2
2
put a 1 for (–) sign and take 2's complement (-) Ho$ ñWmZ na 1 aIH$a n[a_mU H$s {~Q>m| H$m
of the magnitude to
2's H$m°påßb_|Q> boZo na…
1 1 11 0 0 1 0
+1 1 1 10 1 0 1

1 1 1 1 0 0 1 1 1

Thus, the result is AV… n[aUm_ h¡ …
11100111
Since MSB is 1, thus result is negative and My±{H$ MSB 1 h¡, AV… n[aUm_ F$UmË_H$ h¡ Am¡a
is in 2's complement form. Therefore actual result `h 2's H$m°påßb_|Q> _| h¡ & AV… dmñV{dH$ _mZ BgH$m
can be obtained by taking 2's complement of the nwZ… 2's H$m°påßb_|Q> boZo na hr kmV hmoJmŸ& `h Bg àH$ma
sum. Thus the result is –00011001 or –11001 AmEJm- 00011001 `m –11001.

(–1110) + (–1011) = (–11001) 2
2
2