$\underline{\textbf{Answer:}\Rightarrow}\;\mathbf{1.5Mb}$
$\underline{\textbf{Explanation:}\Rightarrow}$
Data cache $=128\;\text{KB}$
Block Size $= 1\;\mathrm W = 4\;\text{Bytes} = 32 \;\text{bits}$
Main Memory address $=32$
Tag-bits (15) |
Index-bits(15) |
Block Offset(2) |
$\therefore $ Number of cache lines $=\dfrac{\text{Cache Memory Size}}{\text{Block Size}} = \dfrac{128\;\mathbf{ KB}}{4\;\text{Bytes}} = \dfrac{2^{17}}{2^2}= 2^{15}$
Now,
$\text{Tag Memory Size} = \text{Number of lines in cache memory} \times \text{Tag space in the line} =2^{15}\times 15 \;\text{bit}$
$\text{Total Cache Size} = \text{Tag Memory size + Data Memory size } =2^{15} \times \left (15+32\;\text{}\right ) = 1.5 \text{Mb}$