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It depends Simple instructions do less work than complex instructions do Simple instructions execute faster than complex instructions Simple instructions -> big code size -> more cache miss, possibly DNZZZN"Homework 4  cont` RProblem 4: Two different implementations, P1 and P2, of the same instruction set. There are five classes of instructions. P1 s clock rate = 4GHz, P2 s clock rate = 6Ghz PHomework 4 - cont`  -Clearly, for P1, the ideal instruction sequence is the one composed entirely of class A instructions. Then, the peak performance of P1 is (4G cycles/sec) / (1 cycle/instn) = 4000 MIPS Similarly, the peak performance of P2 is 3000 MIPS, with the instruction sequence composed of class A, B, and/or C. N.ud.dud&    Homework 5 `  Problem 1 Speedup = (EX time b4 Imp)/(EX time aft Imp) It takes 100 seconds to complete program P1. Of this time, 15% is used for division, 40% for memory access. If you improve only division, what s the maximal possible speedup you can achieve? 1/(1  15%) = 117.65%@  ``d  Homework 5 - cont`  Problem 2: Explain how the instruction  add $t1, $t2, $t3 is being executed in the singlecycle datapath, using the figure 5.19 in your textbook. Four steps: Instruction fetch (IF) Instruction decode and reading registers (ID) ALU operation (EX) Write the result into the register file (WB) You should also be familiar with the multicycle case, pipelined case.  PPPFPP bb ffFfbfU           Homework 5 - cont`  8Problem 3 Add necessary datapaths the singlecycle datapath shown in the figure 5.17 in the textbook for a new instruction jr (jump register). Modification: The datapath to allow the new PC to come from a register (Read data 1 port) A new control signal (e.g., JumpReg) to control the new PC through a multiplexor~ ZZZ `p```dd        @  $  \  "    Homework 5 - cont`   Homework 5 - cont`  fProblem 4 Find the hazard and reorder the instructions to avoid pipeline stall S1: lw $t0, 0($t1) S2: lw $t2, 4($t1) S3: sw $t2, 0($t1) S4: sw $t0, 4($t1) RAW hazard between S2 and S3: the content of $t2 is not available when S2 tries to read $t2. We have to stall the pipeline, even with the help of forwarding. However, we can reorder the code to avoid it.t ZEZLZZ bEbLbftS         Homework 5 - cont`  The reordered code: S1: lw $t0, 0($t1) S2: lw $t2, 4($t1) S4: sw $t0, 4($t1) S3: sw $t2, 0($t1) Hazard is solved by a clever arrangement of the instructions, while it still guarantees the correctness. Instruction reorder/scheduling is a common compiler technique. HZLZZ`PHomework 5 - cont` Problem 5: Executing the following code on the pipelined datapath, what registers are being read and written at the end of the fifth cycle of the execution? S1: add $2, $3, $1 S2: sub $4, $3, $5 S3: add $5, $3, $7 S4: add $7, $6, $1 S5: add $8, $2, $6 J PPaP a:Homework 5 - cont`  Homework 5 - cont` Problem 6: How many cycles will it take to execute the code below on the pipelined datapath? S1: lw $4, 100($2) S2: sub $6, $4, $3 S3: add $2, $3, $5 S2 tries to read a register $4 right following S1, a load instruction that writes the same register Forwarding cannot help this time` R: R:,THomework 5 - cont`   Homework 6 `  2Problem 1: How many bits are required to implement a direct-mapped cache with 64KB of data and 4-B blocks, assuming a 32-bit address? Cache size = 2^16 bytes (64KB) Block size = 2^2 bytes (4-B) Number of cache blocks = 2^(16-2) = 2^14 Each block has 32 bits of data plus a tag, which is (32 - 14  2) = 16 bits, plus a valid bit. Thus, the total cache size is 2^14 * (32 + 16 + 1) = 2^14 * 49 = 784 KbitsL PP zHomework 6 - cont` Problem 3: Given a direct-mapped cache with 16-word data and 4-word blocks, what s the cache misses and hits when having a series of memory access by the addresses: 2,4,8,20,18,11,43,17? First, construct the cacheP ZZ Homework 6 - cont`  Homework 6 - cont` !Homework 6 - cont` Problem 4 Find the hazards in the code and reorder it to avoid pipeline stall lw $t0, 0($t1) addi $t3, $t0, 4 sw $t3, 0($t1) lw $t2, 4($t1) addi $t4, $t2, 4 sw $t4, 4($t1) The same thing as the problem 4, homework#5p ZDZ^Z-Z D]-tN     :"Homework 6 - cont` Problem 6 Explain how the instruction  lw $t1, 8($t2)  is being executed in the pipelined datapath, using the figure 6.17 in the textbook Five stages: Instruction fetch (IF); Instruction decode and register file fetch (ID); Address calculation (EX); Memory access (MEM); Write back (WB); It is important that you should be able to explain the details of the execution of a given instruction. For example, for  load , what should be stored in each pipeline registers?v ZZZZ  >'1  2Good Luck on your quiz! 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S  0\ "`P  B*   0 "`p   D*   0( "`0  D* H  0޽h ? fff80___PPT10. @F$ Shimmer9 1 0 9(   T ~  "~~b ~ # "~  c B@CW DEF"d@ W @W @W W @`".~  c B,CW DEF"d@ W ,W ,@`".  c B CDEF"2@  @`"(4[  c BC DEF"@ @`"  c BC DEF"d@ @`"_b \~  # "\~   c B CDEF"@   @`"(4   c B C DEF"d@ @`"(W4   c BtC DEF"d@t t tt@`"~   c B CDEF"@ @`"([4W  c B CDEF"d@ @`"(4  c BC DEF"@  @`"\  H ?"p  T Click to edit Master title style! !  0 " `   W#Click to edit Master subtitle style$ $  0A "`P  B*   04F "`@`   D*   0XK "`0  D* H  0޽h ? fff80___PPT10. @F$2 0  0(  x  c $Vp  x  c $W `   H  0޽h ? 33___PPT10i.$'+D=' = @B +} 1 0 $(  r  S 0F   r  S ̞   H  0޽h ? fff___PPT10i.%,+D=' = @B +R  1 0 8P(  8r 8 S ȭ0F    8 S Ȯ0<$@ 0  H 8 0޽h ? fff:2___PPT10.p_+ED' = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*8NY%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*8Y%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*8%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*8%(+{$ 1 0 H @ $2<(  <x < c $ང0F   ~ < s *@   @ f -< #"*   < BxɄ ?"6@ NNN?N0 mf I4 @` < Bx҄ ?"6@ NNN?Nm0 f I3 @` < BTۄ ?"6@ NNN?N@mf IE @` < B݄ ?"6@ NNN?N0 t m I4 @` < BX ?"6@ NNN?Nt 0 m I4 @` < B` ?"6@ NNN?N@t m ID @` < BD ?"6@ NNN?N0 { t  I2 @` < B  ?"6@ NNN?N{ 0 t  I3 @` < B ?"6@ NNN?N@{ t  IC @`  < B ?"6@ NNN?N0 {  I2 @`  < B$ ?"6@ NNN?N 0 {  I2 @`  < B- ?"6@ NNN?N@ {  IB @`  < B6 ?"6@ NNN?N0   I2 @`  < B? ?"6@ NNN?N 0  I1 @` < BH ?"6@ NNN?N@   IA @`  < BQ ?"6@ NNN?N0   Q CPI on P2   @`  < BL[ ?"6@ NNN?N 0  Q CPI on PI   @` < B d ?"6@ NNN?N@   MClass @`B < <o ?"0@NNN?N@  B < 61 ?"0@NNN?N@  B < 61 ?"0@NNN?N@  B < 61 ?"0@NNN?N@{ { B < 61 ?"0@NNN?N@t t B < 61 ?"0@NNN?N@mmB < <o ?"0@NNN?N@ffB < <o ?"0@NNN?N@ @fB < 61 ?"0@NNN?N fB  < 61 ?"0@NNN?N0 0 fB !< <o ?"0@NNN?N f# .< Hh ?"6@ NNN?N`UG m7The average number of cycles for each instruction class8 8T /< <n ?"6@ NNN?N   XPeak performance = the fastest rate that a computer can execute any instruction sequence6Y @fk 2< HlsGH ?"0 NNN?N 0,$D 0 Peak performances of PI and P2?0 (c H < 0޽h ?2< fff___PPT10.p_+P`G DO' = @B D ' = @BA?%,( < +O%,( < +DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*2<%(+8+0+2<0 +} 1 0 $(  r  S 0F   r  S    H  0޽h ? fff___PPT10i.&p+D=' = @B +} 1 0 $(  r  S 0F   r  S p  H  0޽h ? fff___PPT10i.'@ԇ+D=' = @B +r  1 0 %P(  r  S 0F     S 0@<$@ 0  H  0޽h ? fffR J ___PPT10* .(p+ED ' = @B D ' = @BA?%,( < +O%,( < +D ' =%(D ' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*$%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*$j%(+ 1 0 P(  r  S 0F     S 0<$@ 0  H  0޽h ? fff___PPT10.+@+EDZ' = @B D' = @BA?%,( < +O%,( < +DL' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*9%(+y 1 0 D (  Dr D S 00F   n D c &@A C"`H D 0޽h ? fff___PPT10i.Ц2+D=' = @B +: 1 0  P(  r  S 0F     S 00<$@ 0  H  0޽h ? fff"___PPT10.- +ED' = @B D' = @BA?%,( < +O%,( < +D' =%(Dh' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*9%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*9g%(+} 1 0  H$(  Hr H S `0F   r H S @  H H 0޽h ? fff___PPT10i.p+D=' = @B + 1 0 0L0(  Lx L c $|0F   x L c $T@  H L 0޽h ? fff___PPT10i.p+D=' = @B +5 1 0 M/E/@37P.(  Pr P S 0F    P B ?"6@ NNN?N_8   P`@  P N4 ?"6@`NNN?N :IF( P Nx ?"6@`NNN?NP   :EX( P ND( ?"6@`NNN?NP  :ID(  P N( ?"6@`NNN?N  ;MEM(  P N|( ?"6@`NNN?N :WB(mF   P    P NT ( ?"6@`NNN?N :IF( P N( ?"6@`NNN?NP   :EX( P N(( ?"6@`NNN?NP  :ID( P N( ?"6@`NNN?N  ;MEM( P N( ?"6@`NNN?N :WB(mF  P  P N( ?"6@`NNN?N :IF( P N"( ?"6@`NNN?NP   :EX( P N|&( ?"6@`NNN?NP  :ID( P NX*( ?"6@`NNN?N  ;MEM( P N)( ?"6@`NNN?N :WB(mF  P  @`  P N1( ?"6@`NNN?N :IF( P Nl5( ?"6@`NNN?NP   :EX( P N9( ?"6@`NNN?NP  :ID( P NX=( ?"6@`NNN?N  ;MEM( P N?( ?"6@`NNN?N :WB(mF  P ` @   P N8D( ?"6@`NNN?N :IF(  P NH( ?"6@`NNN?NP   :EX( !P N`L( ?"6@`NNN?NP  :ID( "P NN( ?"6@`NNN?N  ;MEM( #P NQ( ?"6@`NNN?N :WB( $P HTV( ?"6@ NNN?NK2 8S1  %P H,Z( ?"6@ NNN?N) 8S2  &P H^( ?"6@ NNN?N 8S3  'P H_( ?"6@ NNN?N  8S4  (P Hc( ?"6@ NNN?N` G  8S5 qF  )P @@  *P Ng( ?"6@`NNN?N ;CC1( +P Ndk( ?"6@`NNN?NP   ;CC3( ,P No( ?"6@`NNN?NP  ;CC2( -P NPr( ?"6@`NNN?N  ;CC4( .P Nv( ?"6@`NNN?N ;CC5( 0P Tz( ?"6@`NNN?N@  ;CC6( 1P Tl}( ?"6@`NNN?N@@ ;CC8( 2P T( ?"6@`NNN?N@ ;CC7( 3P T( ?"6@`NNN?N@@ ;CC9( 5P HL?"6@`NNN?N@@ ,$D 0L 6P H(( ?"6@ NNN?N@ 97 `S1: add $2, $3, $1 S2: sub $4, $3, $5 S3: add $5, $3, $7 S4: add $7, $6, $1 S5: add $8, $2, $6 a a 7P <( ?"6@ NNN?Nt ,$ 0 So, at the end of the fifth cycle of execution, registers $6 and $1 (of S4) are being read and register $2 (of S1) will be written. H P 0޽h ? fff___PPT10. +u-D' = @B DF' = @BA?%,( < +O%,( < +D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*5P%(DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*7P%(+8+0+7P0 +F 1 0 PT\(  Tx T c $(0F  (  T c $8(@<$@ 0 ( H T 0޽h ? fff"___PPT10.p+ED' = @B D' = @BA?%,( < +O%,( < +D' =%(Dh' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*T%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*T%(+E 1 0 \T`X(  Xx X c $(0F  ( 4 X # A H X 0޽h ? fff___PPT10i.p+D=' = @B +  1 0 p\\(  \x \ c $(0F  (  \ c $D(@<$@ 0 ( H \ 0޽h ? fff  ___PPT10 .p+EDr ' = @B D- ' = @BA?%,( < +O%,( < +Dd' =%(D ' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*\%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*\%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*\%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*\k%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*\k%(+; 1 0 443a"4(  `x ` c $0(0F  (  ` s *0( <$@ 0 ( u1 0 a# #"& <$@ 0 ( a B( ?"6@ NNN?NF0 @ @` a B( ?"6@ NNN?NVF @ @` a B( ?"6@ NNN?N V @ @` a B( ?"6@ NNN?N   @ @` a B( ?"6@ NNN?N  @ @` a B( ?"6@ NNN?Nz0___PPT10:___PPT9 UN8 " d## a BT( ?"6@ NNN?Nz0___PPT10:___PPT9 V118 " d## a B* ?"6@ NNN?NF0 @ @` a B * ?"6@ NNN?NVF @ @` a B* ?"6@ NNN?N V @ @` a BD!* ?"6@ NNN?N   @ @` a B** ?"6@ NNN?N  @ @` a B$,* ?"6@ NNN?Nz0___PPT10:___PPT9 UN8 " d## a Bt$* ?"6@ NNN?Nz0___PPT10:___PPT9 V108 " d## a B3* ?"6@ NNN?NF^0 @ @` a B88* ?"6@ NNN?NV^F @ @` a BHG* ?"6@ NNN?N ^V @ @` a BW* ?"6@ NNN?N ^  @ @` a BY* ?"6@ NNN?N^  @ @` a Bb* ?"6@ NNN?N^z0___PPT10:___PPT9 UN8 " d## a B0g* ?"6@ NNN?N^z0___PPT10:___PPT9 V018 " d## a Bs* ?"6@ NNN?NF0^ @ @` a BX|* ?"6@ NNN?NVF^ @ @` a Bh* ?"6@ NNN?N V^ @ @` a B* ?"6@ NNN?N  ^ @ @` a B* ?"6@ NNN?N ^ @ @` a Bؙ* ?"6@ NNN?N^z0___PPT10:___PPT9 UN8 " d## a B* ?"6@ NNN?N^z0___PPT10:___PPT9 V008 " d## a B* ?"6@ NNN?N 0z0___PPT10:___PPT9 ZData:(" d## a B`* ?"6@ NNN?N z0___PPT10:___PPT9 WTag8 " d## a Bd* ?"6@ NNN?Nz0___PPT10:___PPT9 UV8 " d## a B`* ?"6@ NNN?Nz0___PPT10:___PPT9 YIndex8 " d##B a 61 ?"0@NNN?N0B a 61 ?"0@NNN?NB a 61 ?"0@NNN?NB a 61 ?"0@NNN?N  B a 61 ?"0@NNN?N^0^B a 61 ?"0@NNN?N  B a 61 ?"0@NNN?NVVB a 61 ?"0@NNN?NFFB a 61 ?"0@NNN?N0B a 61 ?"0@NNN?N0B a <1 ?"0@NNN?N0B a <1 ?"0@NNN?NB a <1 ?"0@NNN?N00B a <1 ?"0@NNN?N0; a H* ?"6@ NNN?NFX ,$ 0 QThis is a 4-block cache H ` 0޽h ? fff___PPT10.p+{Dg' = @B D"' = @BA?%,( < +O%,( < +DY' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*`%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*a%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*a%(+8+0+a0 +5 1 0 4424v4(  tr u S  *0F  * 1 0 u #"& 0 * &u B$* ?"6@ NNN?NF0 @ @` %u B* ?"6@ NNN?NVF @ @` $u B* ?"6@ NNN?N V @ @` #u B* ?"6@ NNN?N   @ @` "u BD* ?"6@ NNN?N  @ @` !u B+ ?"6@ NNN?Nz0___PPT10:___PPT9 UN8 " d##  u B+ ?"6@ NNN?Nz0___PPT10:___PPT9 V118 " d## u B + ?"6@ NNN?NF0 @ @` u B+ ?"6@ NNN?NVF @ @` u B + ?"6@ NNN?N V @ @` u B)+ ?"6@ NNN?N   @ @` u B2+ ?"6@ NNN?N  @ @` u Bd4+ ?"6@ NNN?Nz0___PPT10:___PPT9 UN8 " d## u B9+ ?"6@ NNN?Nz0___PPT10:___PPT9 V108 " d## u BE+ ?"6@ NNN?NF^0 @ @` u BX@+ ?"6@ NNN?NV^F @ @` u BxI+ ?"6@ NNN?N ^V @ @` u B\R+ ?"6@ NNN?N ^  @ @` u B@[+ ?"6@ NNN?N^  @ @` u Bj+ ?"6@ NNN?N^z0___PPT10:___PPT9 UN8 " d## u Bo+ ?"6@ NNN?N^z0___PPT10:___PPT9 V018 " d##  u B{+ ?"6@ NNN?NF0^ SMem(03) @`  u B+ ?"6@ NNN?NVF^ SMem(02) @`  u B+ ?"6@ NNN?N V^ SMem(01) @`  u B+ ?"6@ NNN?N  ^ SMem(00) @`  u B+ ?"6@ NNN?N ^ P0000 @`  u B+ ?"6@ NNN?N^z0___PPT10:___PPT9 ]Y@ " d''  u BL+ ?"6@ NNN?N^z0___PPT10:___PPT9 V008 " d## u B+ ?"6@ NNN?N 0z0___PPT10:___PPT9 ZData:(" d## u B+ ?"6@ NNN?N z0___PPT10:___PPT9 WTag8 " d## u B+ ?"6@ NNN?Nz0___PPT10:___PPT9 UV8 " d## u B+ ?"6@ NNN?Nz0___PPT10:___PPT9 YIndex8 " d##B -u 61 ?"0@NNN?N0B /u 61 ?"0@NNN?NB 2u 61 ?"0@NNN?NB 5u 61 ?"0@NNN?N  B ?u 61 ?"0@NNN?N^0^B Pu 61 ?"0@NNN?N  B Tu 61 ?"0@NNN?NVVB Xu 61 ?"0@NNN?NFFB ]u 61 ?"0@NNN?N0B ~u 61 ?"0@NNN?N0B 'u <1 ?"0@NNN?N0B )u <1 ?"0@NNN?NB *u <1 ?"0@NNN?N00B (u <1 ?"0@NNN?N0. v <t+ ?"6@ NNN?N[  Read memory by address 2 Which block to look at? The memory block number: word address DIV word per block 2/4 = 0 which maps to cache block number: memory block number MOD # of cache blocks 0 module 4 = 0; Here, we suppose the memory space is 2^8 bytesj2(9 (L (0 H t 0޽h ? fff___PPT10i.h+D=' = @B +5 1 0 44224(  x  c $ص+0F  + u2 0 # #"&`  +  BЧ+ ?"6@ NNN?NF0 @ @`  Bp+ ?"6@ NNN?NVF @ @`  Bc+ ?"6@ NNN?N V @ @`  B, + ?"6@ NNN?N   @ @`  BD- ?"6@ NNN?N  @ @`   B- ?"6@ NNN?Nz0___PPT10:___PPT9 UN8 " d##   B- ?"6@ NNN?Nz0___PPT10:___PPT9 V118 " d##    B- ?"6@`NNN?NF0 SMem(2b) @`    B- ?"6@`NNN?NVF SMem(2a) @`    B$- ?"6@`NNN?N V SMem(29) @`   B-- ?"6@`NNN?N   SMem(28) @`  B6- ?"6@`NNN?N  P0010 @`  B8- ?"6@`NNN?Nz0___PPT10:___PPT9 ]Y@ " d''  B;- ?"6@ NNN?Nz0___PPT10:___PPT9 V108 " d##   BI- ?"6@`NNN?NF^0 SMem(17) @`   BR- ?"6@`NNN?NV^F SMem(16) @`   B$[- ?"6@`NNN?N ^V SMem(15) @`   B,d- ?"6@`NNN?N ^  SMem(14) @`  B4m- ?"6@`NNN?N^  P0001 @`  B+B#style.visibility<*%(+ 1 0 0\(  x  c $0F     c $0<$@ 0  H  0޽h ? fff  ___PPT10f .<+ED: ' = @B D ' = @BA?%,( < +O%,( < +D ' =%(D ' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(+ 1 0 ((  (r ( S $-   - H ( 0޽h ? fff___PPT10i.>@+D=' = @B +rp?jUg>==--v--$>??>>--w--$?@@??--x--$@AA@@--y--$ABBAA--z--$BCCBB--{--$CDDCC--|--$DEEDD--}--$EFFEE--~--$FGGFF----$GHHGG----$HJJHH----$JKKJJ----$KLLKK----$LMMLL----$MNNMM----$NOONN----$OPPOO----$PQQPP----$QRRQQ----$RSSRR----$SUUSS----$UVVUU----$VWWVV----$WYYWW----$YZZYY----$Z[[ZZ----$[]][[----$]^^]]----$^aa^^----$abbaa----$beebb----$ehhee----$hjjhh----$joojj----$ottoo----$twwtt---'-$m--$--2p--$--p--$----$----$----$----$  ----$ !!  ----$!""!!----$"##""----$#$$##--T~--$$%%$$---'-e--$  --d--$  --_--$  --V--$  --G--$  --4--$  --"--$  ---'---$  --$--$  --1--$  --@--$  --K--$  --V--$  --]--$  --b--$  --e--$ --f--$   ---'-g--$--h--$--j--$--m--$--o--$--s--$--w--$--{--$----$  ----$     ----$     ----$     ----$     ----$   ----$----$----$----$----$---'---$,00,,----$04400----$48844----$8::88----$:<<::----$<??<<----$?@@??----$@BB@@----$BCCBB----$CEECC----$EFFEE----$FGGFF----$GHHGG----$HIIHH----$IJJII----$JLLJJ----$LMMLL----$MNNMM----$NOONN----$OPPOO----$PQQPP----$QRRQQ----$RSSRR----$STTSS----$TUUTT----$UVVUU--}--$VWWVV--|--$WXXWW--{--$XYYXX--z--$YZZYY--y--$Z[[ZZ--x--$[\\[[--w--$\]]\\--u--$]^^]]--t--$^``^^--s--$`aa``--r--$abbaa--q--$bccbb--p--$cddcc--o--$deedd--n--$eggee--m--$ghhgg--l--$hjjhh--k--$jkkjj--j--$knnkk--i--$nppnn--h--$prrpp--g--$rvvrr--f--$vwwvv---'---$ ) )----$)) 3 3)----$33 < <3----$<< B B<----$BB F FB----$FF J JF----$JJ N NJ----$NN R RN----$RR V VR----$VV X XV----$XX \ \X----$\\ _ _\----$__ c c_----$cc e ec--}--$ee i ie--{--$ii l li--y--$ll o ol--w--$oo r ro--u--$rr u ur--s--$uu y yu--q--$yy ~ ~y--o--$~~ ~--m--$ --k--$ --i--$ --f--$ ---'-f--$%&&%%--d--$&''&&--]--$'((''--S--$())((--C--$)**))--0--$*++**----$+,,++---'---$----$--'--$--:--$--K--$--Y--$--a--$---'-l--$   --b--$   ----$   ----$   ----$    ----$  ----$  ----$  ----$  ----$  --`--$  --k--$  ---'@"Tahoma-. 2 .CPEG323 .. 2 -CPEG323 ."System:-@"Tahoma-. "2 :Homework Review II   .. "2 9Homework Review II   .-@"Tahoma-. 2 X Long Chen .. 2 W Long Chen .-@"Tahoma-. 2 b November, 30.. 2 a November, 30.-@"Tahoma-.  2 `Bth..  2 _Ath.-@"Tahoma-. 2 bF, 2005.. 2 aE, 2005.-՜.+,0    IOn-screen Show ] ArialTahomaTimes New Roman Wingdings宋体ShimmerCPEG323 Homework Review II Homework 4Homework 4 – contHomework 4 – contHomework 4 - cont Homework 5Homework 5 - contHomework 5 - contHomework 5 - contHomework 5 - contHomework 5 - contHomework 5 - contHomework 5 - contHomework 5 - contHomework 5 - cont Homework 6Homework 6 - contHomework 6 - contHomework 6 - contHomework 6 - contHomework 6 - contGood Luck on your quiz!   Fonts UsedDesign Template Slide Titles_0LongLong  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntrydO)PicturesoCurrent UserSummaryInformation(WQPowerPoint Document(nDocumentSummaryInformation8