Step 1: Determine the Page Size
Let the page size be $2^p$ bytes.
Since T1 occupies exactly one page and each entry is 4 bytes (32 bits), the number of entries in T1 is:
\[
\frac{2^p}{4} = 2^{p-2}
\]
Thus, the number of bits needed to index T1 is $ p - 2 $. Similarly, since T2 and T3 are also paged and use the same entry size, each level uses $ p - 2 $ bits of the virtual page number.
The virtual address is 46 bits. The page offset consumes $ p $ bits, leaving $ 46 - p $ bits for the virtual page number. Therefore:
\[
3(p - 2) = 46 - p
\]
Solving:
$$\displaylines{3p - 6 = 46 - p \\4p = 52
p = 13}$$
Thus, the page size is $ 2^{13} $ bytes, or 8 KB.
Step 2: Cache Parameters
Cache size = $ 2^{20} $ bytes
Block size = $ 2^6 $ bytes
Number of blocks = $ \frac{2^{20}}{2^6} = 2^{14} $
Associativity = 16 = $ 2^4 $
Number of sets = $ \frac{2^{14}}{2^4} = 2^{10} = 1024 $
Hence, the cache requires 10 bits to select a set. Since the cache is virtually indexed, these 10 bits are derived from the virtual address.
Step 3: Virtual Address Bit Allocation
With page size $ 2^{13} $, the page offset is 13 bits.
With block size $ 2^6 $, the block offset is 6 bits.
Thus, the number of bits within the page used to select a cache block (i.e., block index within page) is:
\[
13 - 6 = 7 \text{ bits}
\]
In a virtually indexed cache, the 10 index bits are typically taken from virtual address bits [15:6], which includes:
Bits [12:6]: 7 bits from the page offset (same in physical address for synonyms) i.e safe bits ( meaning 6-12 bit no lie with boundary of page size of VA )
Bits [15:13]: 3 bits from the virtual page number (may differ across synonyms) i.e dangerous bits
Step 4: Synonym Problem and Page Coloring
Two virtual addresses mapping to the same physical page are called synonyms. In a VIPT cache, if synonyms differ in the virtual address bits used for indexing, they may be directed to different cache sets, leading to coherence or performance issues.
To prevent this, page coloring is employed: physical pages are assigned colors corresponding to specific values of the dangerous bits. The operating system ensures that a physical page is mapped only to virtual pages whose virtual addresses have matching dangerous bits.
The number of distinct values for the dangerous bits determines the number of colors required.
Step 5 : Compute Minimum Number of Colors
The dangerous bits are bits [15:13] of the virtual address 3 bits in total.
Number of distinct patterns = $ 2^3 = 8 $
Thus, the minimum number of page colors required is 8.
Final Answer
\[
\boxed{8}
\]
