Answer:
1. 2588672 bits
2. 4308992 bits
3. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Explanation:
1. Number of bits in the first cache
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits
2. Number of bits in the Cache with 16 word blocks
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^13(1 +13+(32*2^4)) = 4308992 bits
3. Caches are used to help achieve good performance with slow main memories. However, due to architectural limitations of cache, larger data size of cache are not as effective than the smaller data size. A larger cache will have a lower miss rate and a higher delay. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Digital data<span>. Biometric </span>data<span>. Digital </span><span>data</span>
Answer:
it would be amhv i think i hope it answered u'er question
Explanation:
Answer:
When an instruction is sent to the CPU in a binary pattern, how does the CPU know what instruction the pattern means
Explanation:
When the CPU executes the instructions, it interprets the opcode part of the instruction into individual microprograms, containing their microcode equivalents. Just so you know, a full assembly instruction consists of an opcode and any applicable data that goes with it, if required (register names, memory addresses).
The assembly instructions are assembled (turned into their binary equivalent 0s and 1s, or from now on, logic signals). These logic signals are in-turn interpreted by the CPU, and turned into more low-level logic signals which direct the flow of the CPU to execute the particular instruction.
