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Consider the sequence of word address references in hexadecimal (eight in total) shown on the left side of the following table. Let us assume that the system has a direct mapped cache with four blocks and two words per block. Fill out the table properly. Hint: first, convert the hexadecimal addresses into binary. Notice: - The byte-offset field has been removed from these addresses. In other words, each address specifies one word. - Initially, the cache is empty (entries are invalid). - When a miss happens, you must bring two words (not one) into the cache. the memory address shown in the leftmost column (of that row) has been referenced and the possible miss has been taken care of..

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- 1. Consider the sequence of word address references in hexadecimal (eight in total) shown on the left side of the following table. Let us assume that the system has a direct mapped cache with four blocks and two words per block. Fill out the table properly. Hint: first, convert the hexadecimal addresses into binary. Notice: - The byte-offset field has been removed from these addresses. In other words, each address specifies one word. - Initially, the cache is empty (entries are invalid). - When a miss happens, you must bring two words (not one) into the cache. the memory address shown in the leftmost column (of that row) has been referenced and the possible miss has been taken care of.