My question is : in part a, the answer says that the read result is obtained from block offsets 3 and 2. Where are they getting those offsets from?

PROBLEM 3 (total of 11 marks) Consider a computer in which data words and memory addresses are 16 bits in width. There are separate L1 instruction and data caches, but no L2 cache. There is no virtual memory Below is the state of the writeback D-cache after a program has been running for some time. All numbers given are hexadecimal, except for the U (use), D (dirty) and V (valid) bits, which are binary. The U bit indicates which of the two blocks within a set has been accessed more recently; an LRU policy is used for replacement of blocks. index U DVta DV ta data (and block offsets) f e d c b a 98 7 6 S 4 3 2 1 0 data lock offsets f e d c b a 9 8 7 6 5 4 3 2 1 0 0 1 11 def7 5e 32dc 41fd cbeb 34 56a7 fd7e 5c 31 01 00 25 bdfe baf543 147c c3 36f2 f4af 9f 2c 72 1 1 01b 817cfb d9 13 551e 70 d1 b4 10 16 1e 8d 41 28 11 ad b2 b4 ec 05 e41f fc 1347 64 1b fa 88 d3 30 69 2 | 0 || 01 76 | 43 c1 e2 9abe 82 e2 50 fe 67 41 fc 3e 63 cf 18 01 6a | 60 8b 9a 06 2c 80 a9 30 24 cc c4 f 0 b9 ac 70 98 01 f561 21 86 3d da c408 1f 1f 3c 845f 5250 ba 36 01 2f a32d 8b 30 8e da 49 c166 c46c c3 f3 56 a001 01 86 | 85 73 87 81 3d cc 99 8e 58 ce 196a 55 26 7d 98 01 bf | 2d a8 2b7f 8ec9 66 e7dd a4 07 f50f al 28 1a 5 || 01 | a 1 | 66 e6 50 b2 48 2b bf 2e 11 c2 d4 ae Ob 8f 24 42 1 1 | c0 | e5 bd 93 4d 50 ec 40 98 b8 d5 52 77 a4 e6 47 27 6 101 c9 b3 64 27 e9 57 59 39 91 e5 57 df bd bb 29 9b1 6488 ad 06 13 ea a5 ae d27b cd cd 46 13 2f 8a 24 7001 42 cbdf c5 6a 36 56 92 bb f2f17ef4 a874 a7 c1 4a5a0 5470 27 cb 98 78 98 547a 7ef6 2a 380b 8 | 0 || 01 | ao lef 49 60 a5 33 94 52 25 e3 1e8c d9 f0 1614 ab | 11 d8 | a229 34 ao 84 3d bc 1978 92 ca 4 1 71 b9 c7b8 9 | 1 || 11 | 9a | ceOf Obd7 4d d9 17 c8 f4 4e f839 80 24 89 77 01 9f | d3 80 5 1 e8cc8039 ld 54 24 cf 90 9a166598 a 0 cd9d0e 4a 31904a 85 6b cf 74 2a41 34 57 1a e9 11c3 f918 87 20 4955 e3 fo7d 1c 4e 39 85 c491 ae 11 | c3 | 87 63 eS 09 a3 3e ac d4 5e 86 7a7? 23 01 ebab 11 a8 | d2 8 1 4e e9 2a f 6 b2 290f OS 52 3b0d 46 10 37 11 f9 | e8 de ad 91 ec 2a 63 15f3 89 2e a7 bd c9 dc 5a1 01 59 | a9 05 5e 44 5bdb de a5 5d 6c 7b 50 d27c 1c11 d 1 07a 1c6e c3 eb ce 36 141f e7 a40c bc 7e dd 64 10 01 7b 15 38 36 45 401e a499 db ef 117459 9b 8c bc 01 14 | 47 2b 01 86 2b ce 01 92 f2 do doaa e0 04 74 38 11 cb | 9377 f03a ce 37 56 4f7a 3d do 2f 37 c70704 f01 0cOd 45 ce 73 4a a224 22 5c 63 44c b753 3a 3201 90ef402 ac 42 be 9a b25c 69 84 27 7627 Answer parts b through d, using the answer to part a as a model. Part a. (Ezample, no marks). Explain briefly but precisely why an attempt to read a word at address 0x4b72 would hit in the cache. What data would be the result of the read attempt? The inder is Oa7 and tag is 0a4b. There is a tag match in way 0 with V-1 The read result is obtained using block offsets 3 and 2: 0zf62a