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Numerical: direct mapped cache and byte addressable memory

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Given, a byte addressable main memory of 2^16 bytes and block size of 8 bytes, using direct mapped cache consisting of 32 lines.

a. How many total memory bytes can be stored in the cache?
b. How is a 16 bit memory address divided into tag, line number and byte number?
c. Into what line would byte with the following address be stored?
0001 0001 0001 1011.

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a. total memory bytes that can be stored in the cache = number of cache lines * size of each cache line
= 32 * 8 (because each cache line holds one memory block)
= 256 ...

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