Direct Mapped Cache – Problem 3

In this problem we analyze a Direct Mapped Cache configuration and determine the RAM size and the tag directory size. Throughout the analysis we assume the system is byte-addressable, meaning every byte in memory has its own address.

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Given

• Cache Size: 256 KB
• Block Size: 1 KB
• Tag Size: 8 bits

First convert the values to powers of two:

• Cache Size = 256 KB = 2¹⁸ bytes
• Block Size = 1 KB = 2¹⁰ bytes

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Step 1 – Determine the Offset Bits

Since the block size is 2¹⁰ bytes, we need 10 bits to address each byte inside the block.

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Step 2 – Determine Line Bits

The number of cache lines is calculated using:

Number of Lines = Cache Size / Block Size

2¹⁸ / 2¹⁰ = 2⁸

Therefore the cache contains:

To index 256 lines we need:

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Step 3 – Address Structure

The physical address in a direct mapped cache consists of:

• Tag
• Line Index
• Offset

From the problem statement:

• Tag = 8 bits
• Line Index = 8 bits
• Offset = 10 bits

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Step 4 – Calculate RAM Size

With a 26-bit physical address, the number of possible addresses is:

2²⁶ addresses

Because the system is byte-addressable, the RAM size is:

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Step 5 – Tag Directory Size

Every cache line stores a tag. Therefore the total tag storage (tag directory) is:

Tag Directory Size = Number of Lines × Tag Bits

= 2⁸ × 8 bits

Convert bits to bytes by dividing by 8:

(2⁸ × 8) / 8 = 2⁸

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Shortcut Method

Another quick way to solve the problem is:

1. Compute number of lines:
   2¹⁸ / 2¹⁰ = 2⁸
2. Line bits = 8
3. Offset bits = 10 (from block size)
4. Tag bits = 8 (given)
5. Total physical address bits = 8 + 8 + 10 = 26
6. RAM size = 2²⁶ bytes = 64 MB