Alex. Gerbessiotis  01/22/2026  SolKeyHW1 332S26
(c) All rights reserved. JANUARY   2026.
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Q1. TRUE,    FALSE,   TRUE,    18,    900s (ACD-B),  
    Von-N,   SharedM, 500,     -15,   0,       
    50,      1,       Both,    9ns,   99%,     
    7,       456,     9993,    12bit, 16bit

Q2. (1) 1ns,   (2)  2ns,   (3)  8000000000B/s
1ns, 2ns, 
 % f=1GHz => T= 1s/10**9 =10**9ns/10**9c  => 1ns  
      % 2 cycles is                       => 2ns
      % 2ns==1transfer==128-bit=16B=>                
 % 2ns/16B => 16B/2*10**-9s =  8x10**9 B/s=> 8000000000B/s

Q3.  50028ns
      %0.90 *20 =                     18.00ns
      %0.10*0.95*(80+20)=              9.50ns
      %0.10*0.05*(10000000+80+20)= 50000.50ns
      % 18+9.50+50000.50         = 50028.00ns i.e. 50028ns
0.90  i.e. 90% reference is in the cache  
      then (20ns to get it )
0.095 i.e.9.5% reference is in MM         
      then (20ns + 80ns to get it)
0.005 i.e.0.5% reference in in HDD        
      then (10ms HDD->MM, 80ns MM->cache)
      and then reissue request and
      100% reference is in the cache  (20ns) 
      for a grand total of 10ms+80ns+20ns)

Q4
a. Tr(J)=40s     AlsoCPU(J)=3/8=15/40 Programmed I/O (Blocked I/O)

         Job number indicated
                  1         2         3         4
Timeline 1234567890123456789012345678901234567890
     CPU 1     2     3     4     5     6666666666
 Printer  11111 22222 33333 44444 55555


 b.  Tr(J)=31s  Also   CPU(J)=15/31 Interrupt I/O

         Job number indicated
Timeline 1234567890123456789012345678901
     CPU 12345616666626666 3     4     5
Interrupt      ^     ^     ^     ^     ^ InterruptHandling
 Printer  11111 22222 33333 44444 55555


Q5.  Answ:(2)  512,  (4) 9bit, (5)  6bit
 
Q6.  (a) PC contains  0x00a    (b) AC contains  0x000c

                       1st instr  2nd instr  3nd instr
006 0006   PC  007     PC  008    PC  009    PC  00A
007 1006   AC 0077 --> AC 0006    AC  000C   AC  000C
008 5006   IR ____     IR 1006    IR  5006   IR  0000
009 0000
                       AC=M[006]  AC+=M[006] HALT

Q7. 6 : 15+0.4*15 = 15xH +(1-H)(15+100)=> H=0.94=> M=0.06=>M=6%

Q8.  a.Tr(A)= 20s   b. Tr(B)=11s  c. Tr(B)=15s.
     d. R(B) =  1s  e. R(C) = 5s  Moreover, CPUtil=100 
   A letter is 1s
  A                  B     C
  01234567890123456789012345678901234567890
   AAAAAAAAAAAAAAAAAAAABBBBBBBBBBCCCCCCCCCC

Q9 a. Tr(A)= 50s  
      Tr(B)= 21s     Tr(C)= 20s    CPUtil=80%
  A                  B     C
  012345678901234567890123456789012345678901234567890
   AAAAA----------AAAAABBBBBCCCCCAAAAABBBBBCCCCCAAAAA

Q10.  Trc: Tr(C)=15s , Rc: R(C)=0s
      a. Tr(A)=20s   b. Tr(B)= 16s  d. Tr(C)= 15s
      d.  R(B)= 1s   e. R(C)=0s     f. CPUtil=100
  A                  B     C
  01234567890123456789012345678901234567890
   AAAAAAAAAAAAAAAAAAAABBBBBCCCCCBBBBBCCCCC

Q11.  200   (as in 200ms)

                2ts
   CPU util = --------- >= 0.80 (80%)
              2ts + 2c
   For c=50ms, solve for ts >= 200ms

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Date created:     01/22/2026 @ 07:15
Date last edited: 01/22/2026 @ 07:45
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