Alex. Gerbessiotis 3/31/2026 HW3Ss26 (c) All rights reserved. April 2026. -------------------------------------- ----------------------------------- HW 3 CS 332 Solution key -------------------------------------- Q1.(6pt) 1.1. NO 1.2. 33 1.3. 128 1.4. 96 1.5. 257 1.6. 256 Needs 33+128+96 = 257 page > 256*4096B=1MiB Q2.(6pt) 2.1. 4096B 2.2. 36bit 2.3. 28bit 2.4. NO POINT :1024GiB GIVEN (2**40B) 2.1. 12bit for frame offset--> 2**12B framesize=pagesize --> S=4KiB=4096B 2.2. 12bit for frame offset == 12bit for page offset Logical address is 48bit-->page number #bit is 48-12=36bit 2.3. 12bit for frame offset == 12bit for page offset Physical address is 40bit-->frame number #bit is 40-12=28bit No question. #FrameNumbers x FrameSize = (2**28) x 4096B/frame = 2**(28+12)B=1TiB=1024GiB Q3.(8pt) 3.1. THRASHING 3.2. 5 3.3. 1 3.4. 1 3.5. 5 3.6. 560ns 3.7. 114ns 3.8. Kernel Mode Q4.(6pt) bit(T)=lg(16*1024)=14bit 4.1. 12bit bit(P1)=bit(P2)=bit(P)/2=(38-14)/2=12bit 4.2. 12bit bit(P2)=bit(P1) 4.3. 14bit 4.4. 18bit bit(F)+bit(T)=32bit=> bit(F)=32-14=18bit Q5.(8pt) 5.1. Running 5.2. Blocked 5.3. 0 5.4. 5 5.5. CORRECT 5.6. CORRECT 5.7. SWAPPER 5.8. PAGER Q6.(6pt) 6.1. 3, 6.2. 5, 6.3. 7 ----------------- 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 Time horizon : 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Page references : 5 4 7 2 4 5 4 1 5 2 5 3 5 2 3 Fault : F F F Total = 3 Frame-1 OPT : 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 [OPT] Frame-2 OPT : 4 4 4 4 4 4 1 1 1 1 3 3 3 3 Frame-3 OPT : 7 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 Time horizon : 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Page references : 5 4 7 2 4 5 4 1 5 2 5 3 5 2 3 Fault : F F F F F Total = 5 Frame-1 LRU : 5 5 5 2 2 2 2 1 1 1 1 3 3 3 3 [LRU] Frame-2 LRU : 4 4 4 4 4 4 4 4 2 2 2 2 2 2 Frame-3 LRU : 7 7 7 5 5 5 5 5 5 5 5 5 5 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 Time horizon : 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 Page references : 5 4 7 2 4 5 4 1 5 2 5 3 5 2 3 Fault : F F F F F F F Total = 7 Frame-1 FIFO : 5 5 5 2 2 2 2 1 1 1 1 3 3 3 3 [FIFO] Frame-2 FIFO : 4 4 4 4 5 5 5 5 2 2 2 2 2 2 Frame-3 FIFO : 7 7 7 7 4 4 4 4 5 5 5 5 5 [Per textbook Figure 8.14 definition of a fault] Q7.(3pt) 7.1. 52bit 7.2. 8B 7.3. 512 Q8.(5pt) 8.1. 131072B 8.2. 17bit 8.3. 28bit 8.4. 16384 8.5. 14bit Q9. (3pt) Answer : Average seek length is 50 SCAN @ 100 140 40 (=140-100) 160 20 (=160-140) 30 130 (=160-30) 20 10 (=30-30) -------------------------------------------------------------- Date created: 03/31/2026 @ 11:30 Date last edited: 04/13/2026 @ 10:54 --------------------------------------------------------------