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If we represent physical addresses for the Megatron 747 disk by allocating a separate byte or...

If we represent physical addresses for the Megatron 747 disk
by allocating a separate byte or bytes to each of the cylinder, track within
a cylinder, and block within a track, how many bytes do we need? Make a
reasonable assumption about the maximum number of blocks on each track;
recall that the Megatron 747 has a variable number of sectors/track.

Please explain your solutions

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Answer #1

there are 4 cylinder providing 8 surfaces. Each surface has 2^13 or 8192 tracks. Each track has 256 sectors and each sector has 512 byte. The capacity of Megatron 747 is the product of 8 surface, times 8192 tracks, times 256 sectors, times 512 byte, Which is equals to 2^33 byte . Thus megatron 747 has 8 gigabyte. A single track hold 256* 512 byte = 128k byte.

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