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For these problems, words are assumed to be 4 bytes, and the references are word-addresses. Thus, the words in memory are loc
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wood = 4bytes hiven ond Caches Cache ine of two-cwoads have and to tal size of 4 cods (bytes) block 22 Loods) 4 13 5 3 2 dock16 th byte t i oe miss Next f C>e fetch in first bloCk (o to bytes) in Cache Since and in 2nd block (s to 1s bytes)ase pdesen32lo 22 I0 Ref 6 11 3 2 yN NYYy yyN N N N HIT yN final Cootents o Cache - tog L4 3 2 O tog s 14 3 12

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