ࡱ> q` JbjbjqPqP .::B 0 then ? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths restoring division algorithm, after performing operations (1) left shift operation on A,Q and (2) A=A-M, if sign of A is positive? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths restoring division algorithm, after performing operations (1) left shift operation on A,Q and (2) A=A-M, if sign of A is negative? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths restoring division algorithm, after performing operations (1) left shift operation on A,Q and (2) A=A-M, if MSB of A is 0? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths restoring division algorithm, after performing operations (1) left shift operation on A,Q and (2) A=A-M, if MSB of A is 1? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths restoring division algorithm, for Dividend=1000 and Divisor=100. How many numbers of cycles are required to get the correct division result? 4 5 3 6 In Booths restoring division algorithm, for Dividend=10000 and Divisor=100. How many numbers of cycles are required to get the correct division result? 4 5 3 6 In Booths restoring division algorithm, for Dividend=1000 and Divisor=0011. What size of divisor will give you correct answer? 4 5 3 6 In Booths restoring division algorithm, which register holds the quotient and remainder? Q=Quotient and A=Remainder A=Quotient and Q=Remainder M=Quotient and A=Remainder Q=Quotient and M=Remainder In Booths non-restoring division algorithm, after performing left shift operation on A,Q registers, if magnitude of A < 0 then ? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths non-restoring division algorithm, after performing left shift operation on A,Q registers, if magnitude of A > 0 then ? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths non-restoring division algorithm after performing left shift operation on A,Q register, if sign of A is positive? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths non restoring division algorithm, after performing left shift operation on A,Q register, if sign of A is negative? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths non restoring division algorithm, after performing left shift operation on A,Q register, if MSB of A is 0? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths non restoring division algorithm, after performing left shift operation on A,Q registers, if MSB of A is 1? Q0=0, A=A+M A=A+M Q0=1 A=A-M In Booths non restoring division algorithm, for Dividend=1000 and Divisor=100. How many numbers of cycles are required to get the correct division result? 4 5 3 6 In Booths non restoring division algorithm, for Dividend=10000 and Divisor=100. How many numbers of cycles are required to get the correct division result? 4 5 3 6 In Booths non restoring division algorithm, for Dividend=1000 and Divisor=0011. What size of divisor will give you correct answer? 4 5 3 6 In Booths non restoring division algorithm, which register holds the quotient and remainder? Q=Quotient and A=Remainder A=Quotient and Q=Remainder M=Quotient and A=Remainder Q=Quotient and M=Remainder In Booths non restoring division algorithm, at the end of last cycle, if magnitude of A < 0? A=A+M A=A-M M=A+M End of algorithm In Booths non restoring division algorithm, at the end of last cycle, if magnitude of A > 0? A=A+M A=A-M M=A+M End of algorithm How many bits are needed to represent floating point number in a single precision form? 32 16 64 50 How many bits are needed to represent floating point number in a double precision form? 32 16 64 50 What is the size of bias exponent in single precision representation of floating point number? 8 10 23 52 What is the size of bias exponent in double precision representation of floating point number? 10 11 25 21 What is the size of mantissa in single precision representation of floating point number? 23 32 16 24 What is the size of mantissa in double precision representation of floating point number? 52 11 23 16 What is the range of bias exponent in single precision representation of floating point number? 0 to 255 0 to 1023 0 to 256 0 to 127 What is the range of bias exponent in double precision representation of floating point number? 0 to 255 0 to 1023 0 to 256 0 to 127 What is the range of bias exponent in double precision representation of floating point number? 0 to 255 0 to 1023 0 to 256 0 to 127 What are the values of bias exponent to represent single precision floating point values of exact 0 and infinity respectively? 0 and 255 255 and 0 0 and 127 127 and 0 What are the values of bias exponent to represent double precision floating point values of exact 0 and infinity respectively? 0 and 1023 255 and 0 0 and 127 127 and 0 What is the relation in between in sign exponent E and bias exponent E in single precision floating point number? E=E-127 E=E+127 E=127-E E=E-1023 What is the relation in between in sign exponent E and bias exponent E in double precision floating point number? E=E-127 E=E+127 E=1023-E E=E-1023 What is the representation for IEEE single precision floating point number? +/- 1.M * eE-127 +/- 1.M * e127-E +/- 1.M * eE-1023 +/- 1.M * e1023-E What is the representation for IEEE single precision floating point number? +/- 1.M * eE-127 +/- 1.M * e127-E +/- 1.M * eE-1023 +/- 1.M * e1023-E What is the result of normalization on (10011101011.001)2 ? 1.0011101011001*210 1.0011101011001*211 1.0011101011001*29 1.0011101011001*28 What is the result of IEEE doulbe precision representation for (1259.125)10? 0100000010010011101011001. 1100000010010011101011001. 1111000010010011101011001. 0100001110010011101011001. What is the result of IEEE single precision representation for (1259.125)10? 0100010010011101011001. 1100000010010011101011001. 1111000010010011101011001. 0100001110010011101011001. What is the result of IEEE single precision representation for (-127.1075)10? 110000101111111000110111. 1100000010010011101011001. 1111000010010011101011001. 0100001110010011101011001. What is the result of IEEE double precision representation for (-127.1075)10? 1100000001011111111000110111. 1100000010010011101011001. 1111000010010011101011001. 0100001110010011101011001. g " ^ k  ) d 3 ? [ c j    4 5 S h p H]bchinotниаааШРМММh&?h&?h&?5h&?h[vT5hxh[vT5 hx5 hUh[vTh[vTh[vT5h$d*h[vT hU5hUhU5 hUhUhdJ=hU5hUhdJ=hdJ=5hdJ==JKg " ) - 3 ^ k v   gddJ= & FgddJ= & FgddJ=J ) 7 E d + 3 ? M [  & FgdU & FgdU & F h^hgdU & F ```gdU & FgdU & FgddJ= & FgddJ=   . [ c j n t    5  & Fgd[vT & Fgd[vT & FgdU & FgdU5 S p  C\zHch^hgd[vT & Fgd[vT & F $ `gd[vT]ciou ,TWZ]` & Fgdx & Fgdx & Fgd[vT & F $ `gd&? & F $ `gd[vT & Fgd[vT ,OSTW`"/06cl [`eo34ȼȴh} hshshshs5h53bhs hInhInhInhIn5hInh=hr] 5hr] h=h?5h? hxhxhdJ=hx5hxh[vTh[vT5h[vTh&?hxhx5hxh&?51"/:MS06_clty & Fgd? & Fgd? & Fgdx & Fgdx):G [`ejo & Fgds & Fgds & FgdIn & FgdIn & Fgdr]  & Fgdr] 4m-Sw 0] & Fgd; & Fgd_1l:XɽŹű~zvnzhXWhXW5h^hXWhXWh_}1DQ_ & Fgd[m3 & Fgd[m3 & Fgd$\ & Fgd0_ & FgdSm & FgdSm & Fgdp & Fgdp & Fgd;_l:X,9FS` & Fgd  & Fgd  & FgdXW & FgdXW & Fgd. & Fgd~ & Fgd~ & FgdV< & FgdV<---5.9.>.B.C........;/?/C/D/H/I/N//////B0F0J0K0P0U00000000001"1#1(1)1,1-12111111111ƲƲƮ h$hn.h=hhn.5 hd5hHchn. h$hc|{hhc|{5 hc|{5hc|{ h$hwYDh EhhwYD5 hwYD5hwYDh~h~5 h~h~ h;h~h~9>.C.H......;/?/D/I/N//////B0F0K0P0U00000 & FgdwYD & FgdwYD & Fgd~ & Fgd~001#1)1-1211111142@2F2K2Q222222 & FgdS9 & FgdS9 & Fgd= & Fgd= & Fgdn. & Fgdn. & Fgdc|{ & Fgdc|{ & FgdwYD1111222232425262?2@2F2G2H2J2K2P2Q2222222222333333333ȹȹzrhra h0h,/hJ@h,/5H*hJ@h,/5hJ@h,/H* hJ@h,/h,/ h0hS9hS9hS95H*hS9hS95hS9hS9H* hS9hS9hS9 h0h=h0h0H* h0h0h0h=5h0h05H*h0h05h=h0hq h$hn.hn.&23333384D4J4O4U444444~5555536567696 & Fgd-) & Fgd-) & Fgd^F & Fgd^F & Fgd,/ & Fgd,/34648494:4D4J4K4L4U4444444444~555555555553646567696;6u6v66ӼӣӫӍyrkgh"j hMBh-) h"jh-)h"jh-)5h"jhk5h=h-)h0h^FH*h0h^F5H*h0h^F5 h0h^Fh^Fh^F5H*h^Fh^F5h^Fh^FH* h^Fh^Fh^Fh0h,/H* h0h,/h0h,/5H*h0h,/5hT.h,/'96;666666\7^7`7b7d77778*88888 & Fgd%N & Fgd%N & Fgd> & FgdvXu & FgdvXu & Fgd^I & Fgd^I & Fgd= & Fgd= & Fgd-)666667)7Z7]7^7`7b7d7777777788*888888888888888889"9{s{lhh# hh%Nh0h%NH* h0h%Nhh%N5h#h%NH* h#h%Nh/|hh%N h$h>h`hvXu5h`h>5h>hvXu hMBh^Ih=h^I5 h$h^IhX{hsh^Ih= hMBh=h=h=5 h$h=(88K9W9]9b9h99999::::::;;%;*;0;;;;;;`< & Fgd%N & Fgd%N"9,9-9K9L9M9W9]9^9_9b9h9s9999999999: ::[:e::::::::::::;;;;;%;&;';ǿ􊂊{sh3GLh%NH* h3GLh%Nh^Fh%NH* h^Fh%Nh0h%NH* h0h%NhDyh3GLhDyh%N5hDyh%NH* hDyh%NhJ@h%NH* hJ@h%Nh qh%N5h qh%NH* h qh%NhS9h%NH* hS9h%Nh%Nh#h-';*;0;;;?;m;n;;;;;;;;;;;;;`<b<d<f<s<w<== = =========>,>->G>H>c>>>>>>ƿ}h`h 5 hC`5h&h  h$htZ_h`htZ_5htZ_h5(h=h%N5 h$h%N hMBh%N h"jh%Nh"jh%N5h7Dh0h%NH* h0h%Nh3GLh%NH*h3GLh%Nh3GLh%N5 h3GLh%N/`<b<d<f<h<== = = =======>->H>c>>>>>> & Fgd  & Fgd  & FgdtZ_ & FgdtZ_ & Fgd%N & Fgd%N>>>>>>>B?H?M?N?S?T?d?e????????!@$@&@'@*@-@@@@@@@@@@@@@@@@@οοθοƸ~vo h7h PhE*+h P5hE*+hE*+5 h Ph PhE*+h P h7hq& h$hq&hw h Phq&5 hw 5hq& h*ih*i h$h*ih*ih*i5h*i h9y`5h9y`h9y`5 h$hC` hghC`h h9y` h$h hC`+>B?H?N?T?e??????!@$@'@*@-@@@@@@@ & Fgd P & Fgdq& & Fgdq& & Fgd*i & Fgd*i & Fgd9y` & FgdC` & FgdC`@@@@A\A_AbAeAhAiAAAAAA/B8BBBKBTBB & Fgd2S & Fgde\ & Fgde\ & FgdUJ & FgdUJgd P & Fgd3KK & Fgd3KK & Fgd P@AA\A^A_AaAbAdAeAgAhAiAAAAAAAAAAAAA/B7B8BABBBJBKBSBTBBBBBBBBƾ˨zsl h7h2S h$h2S hcAh2Sh2S h7he\ h$he\h=lh Phe\5 h=l5he\haE h7hUJ h$hUJh$h PhUJ5 h%5hUJ h*ih3KK h7h3KK h$h3KKh Ph3KK5 h3KK5h3KK h*ih PhE*+(BBBBB9CBCLCUC^CCCCCDDDDDDD!E & FgdL-gd& & Fgdh & Fgdh & Fgd& & Fgd& & Fgdg & Fgdg & Fgd2SB^CCCCCCCCCCCCCCDDD=DDDDD!E*E2E3E;E_}1DQ_l:X,9FS`B#JP%+1X^(*,.0  PZdlv9G lpuz~1:CLZ/8AJX   t x } ! !!!!!!! " """""""""## #%#*######%$)$.$3$8$$$$$$.%2%7%<%A%%%%%%5&9&>&C&H&&&&&&;'?'D'I'N''''''B(F(K(P(U(((((()#)))-)2))))))4*@*F*K*Q******+++++8,D,J,O,U,,,,,,~-----3.5.7.9.;......\/^/`/b/d////0*000000K1W1]1b1h1111122222233%3*30333333`4b4d4f4h455 5 5 55555556-6H6c666666B7H7N7T7e777777!8$8'8*8-88888888889\9_9b9e9h9i999999/:8:B:K:T::::::9;B;L;U;^;;;;;<<<<<<<!=*=3=<=F======+>=>O>b>u>>>>> ? ?H?\?p?????@7@S@@@@@ A[AwAAAAAB:BVBrBBBBBBBBBBBBBBBBBBBBBB 0 00 0 0K 0 0 0 0 0 0 0 0 0 0 0 03 03 03 03 0 0 0 0 0 0 0 0 0 0 0 0E 0 0 0 0 0 0 0 0 0 0 0 0[ 0[ 0[ 0[ 0 0 0 0 0 0 0. 0. 0. 0. 0 0t 0t 0t 0t 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0  0C 0C 0C 0C 0  0 0 0 00 0  0 0 0 0 0  0 0 0 0 0  0u 0u 0u 0u 0 0,  0,  0,  0,  0 0`  0 0  0  0  0  0 0  0  0  0  0 0S  0S  0S  0S  0 0  0  0  0  0 06  06  06  06  0 0y  0y  0y  0y  0 0  0  0  0  0 0G  0G  0G  0G  0 0  0  0  0  0 0  0  0  0  0 0o  0o  0o  0o  0 0  0  0  0  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0] 0] 0] 0] 0  0 0 0 0 0! 0 0 0 0 0" 0 0 0 0 0# 0 0 0 0 0$ 0_ 0_ 0_ 0_ 0% 0 0 0 0 0& 0X 0X 0X 0X 0' 0 0 0 0 0( 0` 0` 0` 0`00 0) 0* 0B 0B 0B 0B 0+ 0 0 0 0 0, 0P 0P 0P 0P 0- 0 0 0 0 0. 0^ 0^ 0^ 0^ 0/ 0 0 0 0 00 0 0 0 0 01 00 00 00 000 02 0 0 0 0 03 0 0 0 0 04 0v 0v 0v 0v 05 0G 0G 0G 0G0 06 0 0 0 0 07 0 0 0 0 08 0~ 0~ 0~ 0~ 09 0 0 0 0 0: 0Z 0Z 0Z 0Z 0; 0 0 0 0 0< 0X 0X 0X 0X0 0= 0  0  0  0 0 0> 0  0  0  0  0? 0  0  0  0  0@ 0! 0! 0! 0! 0A 0! 0! 0! 0! 0B 0" 0" 0" 0" 0C 0" 0" 0" 0" 0D 0*# 0*# 0*# 0*# 0E 0# 0# 0# 0# 0F 08$ 08$ 08$ 08$ 0G 0$ 0$ 0$ 0$ 0H 0A% 0A% 0A% 0A% 0I 0% 0% 0% 0% 0J 0H& 0H& 0H& 0H& 0K 0& 0& 0& 0& 0L 0N' 0N' 0N' 0N' 0M 0' 0' 0' 0' 0N 0U( 0U( 0U( 0U( 0O 0( 0( 0( 0( 0P 02) 02) 02) 02) 0Q 0) 0) 0) 0) 0R 0Q* 0Q* 0Q* 0Q* 0S 0* 0* 0* 0* 0T 0+ 0+ 0+ 0+ 0U 0U, 0U, 0U, 0U, 0V 0, 0, 0, 0, 0W 0- 0- 0- 0- 0X 0;. 0;. 0;. 0;. 0Y 0. 0. 0. 0. 0Z 0d/ 0d/ 0d/ 0d/ 0[ 0*0 0*0 0*0 0*0 0\ 00 00 00 00 0] 0h1 0h1 0h1 0h1 0^ 02 02 02 02 0_ 02 02 02 02 0` 003 003 003 003 0a 03 03 03 03 0b 0h4 0h4 0h4 0h4 0c 0 5 0 5 0 5 0 5 0d 05 05 05 05 0e 0c6 0c6 0c6 0c6 0f 06 06 06 06 0g 0e7 0e7 0e7 0e7 0h 07 07 07 07 0i 0-8 0-8 0-8 0-8 0j 08 08 08 08 0k 09 09 09 090 0l 0i9 0i9 0i9 0i9 0m 09 09 09 09 0n 0T: 0T: 0T: 0T: 0o 0: 0: 0: 0: 0p 0^; 0^; 0^; 0^; 0q 0< 0< 0< 0<0 0r 0< 0< 0< 0< 0s 0F= 0F= 0F= 0F= 0t 0= 0= 0= 0= 0u 0u> 0u> 0u> 0u>0 0v 0 ? 0 ? 0 ? 0 ? 0w 0? 0? 0? 0? 0x 0S@ 0S@ 0S@ 0S@ 0y 0 A 0 A 0 A 0 A0 0z 0A 0A 0A 0A000000000000000000004 p$|(-136"9';>@B+F[GZIJJ&-147:<?BDFHIKNPRTUW 5 _^ "$8&(+>.02968`<>@B!EHGJJJ')*+,./0235689;=>@ACEGJLMOQSVXJ(8@0(  B S  ?=j=j=jlddqBmxxB9*urn:schemas-microsoft-com:office:smarttagsplace=*urn:schemas-microsoft-com:office:smarttags PlaceName=*urn:schemas-microsoft-com:office:smarttags PlaceType d?? AABV]^cUZ/4W[  ^q<G00 1"111Y2[22233D?G? AZAB3333333333333333333333333333??S@ A(A.A[AvAAAABBB AABB}*HI2. ^`o(.h ^`o(hH.$ $ ^$ `o(. @ @ ^@ `hH. ^`hH. L^`LhH. ^`hH. ^`hH. PLP^P`LhH.^`o(.hh^h`o(. 8L8^8`LhH. ^`hH.   ^ `hH.  L ^ `LhH. xx^x`hH. HH^H`hH. L^`LhH.2B}*  v|      Ppp       0~7>Z$w r] ! qg d?In'5(-)$d*E*+.2-T.?@ABCDEFGHIJKLMNOPQRSTUVWXY[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root Entry Fpa89m1TableZiWordDocument.SummaryInformation(DocumentSummaryInformation8CompObjq  FMicrosoft Office Word Document MSWordDocWord.Document.89q