Given

• Cache size $C =16$ KB $=16384$ B $=2^{14}$ B
• Associativity $k=4$ (4-way set associative)
• Block size = $8$ words, Word = $32$ bits $=4$ B
  $\Rightarrow$ Block = $8\times4=32$ B $=2^{5}$ B
• Physical address space $4$ GB $=2^{32}$ B $\Rightarrow$ Physical address $32$ bits

Calculations

1. Number of blocks in cache:
$N_{blocks}= C / Block =16384/32=512$ blocks $=2^{9}$

2. Number of sets:
$N_{sets}= N_{blocks}/k =512/4=128$ sets $=2^{7}$
$\Rightarrow$ SET field $s = \log_2 N_{sets}=7$ bits

3. WORD / Offset field:
Block = $32$ B. Byte offset within block = $\log_2 32 =5$ bits.
This $5$ bits comprises:

• Word offset within block: $\log_2 8 =3$ bits (select 1 of 8 words)
• Byte offset within word: $\log_2 4 =2$ bits (select 1 of 4 bytes)

If address is word-addressable, WORD field = $3$ bits.
If byte-addressable (standard), combined offset = $5$ bits.

Common naming in question: WORD field = $5$ bits (or split $3+2$).

4. TAG field:
$TAG = Physical\_bits - SET - OFFSET$

• Byte-addressable: $TAG =32 -7 -5 =20$ bits
• Word-addressable with byte split: $TAG =32 -7 -3 -2 =20$ bits same.

Answer

Physical Address (32 bits)
+-----------+---------+---------------+
| TAG 20b   | SET 7b  | WORD/OFFSET 5b|
|           |         | (3b word+2b byte)|
+-----------+---------+---------------+

• TAG = 20 bits
• SET = 7 bits
• WORD/Offset = 5 bits (3 bits word + 2 bits byte)

If counting WORD only (word index): WORD=3 bits, BYTE=2 bits, TAG=20, SET=7.

Check: $20+7+5=32$.

Alternative: If cache size given as $16$ KB words (unlikely), then $C=16K\times4=64KB$ $\Rightarrow$ sets $=512$ $\Rightarrow$ SET=9, TAG=18. But standard is bytes → answer above.