Now, we consider a 16-byte, four-way, fully-associative cache. Since the capacity of the cache is 16 bytes, the array "a" in our example (does/does not) fit inside the cache. We can deduce that the block size for this cache is ( ？ ) bytes per block. So the block index size b=2 bits. For a memory trace record such as: L 1fff000116,2 the 2-bit block offset is (0b01/0b10/0b11/0b00). The tag bits are all the rest of the bits not part of the block offset.

Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes address has 8 bits, to which cache set would the hexadecimal address OxBE map if the computer mapping? each. Assuming that each memory uses 2-way set associative

please no chatgpt answer . Consider a demand-paging system with a paging disk that has an average access and transfer time of 20 milliseconds. Addresses are translated through a page table in main memory, with an access time of 1 microsecond per memory access. Thus, each memory reference through the page table takes two accesses. To improve this time, we have added an associative memory that reduces access time to one memory reference, if the page-table entry is in the associative memory. Assume that 80 percent of the accesses are in the associative memory and that, of those remaining, 10 percent (or 2 percent of the total) cause page faults. What is the effective memory access time? Consider the following page reference string: 1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6. Assuming demand paging with four frames, Show which pages are resident under the LRU, FIFO, and Optimal replacement algorithms by filling out the following tables. How many page faults would occur…

cache.xls (or it can be a google sheet that you share with me in a cache.txt with the url in it )= a spreadsheet showing the cycle counts of a cached and non-cached machine. Also, you need to show the cache as the memory accesses progress in your program marking the hits and misses (misses in RED). You should have cells to show how many hits and how many misses your program has, and details on what is the cache replacement policy you implemented.   Here is code:   void c_cache_for_badge() {    int na = 8, nb = -3, nc = 10;    // pointers    int *npa;      npa = (int*)malloc(sizeof(int)*84);      // init integer pointers    for(int i = 0; i < 84; i++)    {        npa[i] = i;    }        // integer pointers    for(int i = 9; i < 70; i= i + 3)    {        nb += npa[i % 5] + npa[i % 3] + npa[i % 6];    }     }   void main() {    c_cache_for_badge(); }

correct answer is 13. Can you explain by sperately please?

Take into account the following scenario: we have a byte-addressable computer with 2-way set associative mapping, 16-bit main memory addresses, and 32 blocks of cache memory. Since there are 8 bytes in a block, you can use that information to calculate how big the offset field has to be.

Let's say p contains a memory address and the 8 bytes in memory beginning at that address are (in hex): 01 12 23 34 45 56 67 78. If p's type is "uint32_t *", what is the value of x on a little-endian computer after executing: uint32_t x = p[0]; Ox01122334 Ox0112233445566778 Ox34231201 Ox7867564534231201 O O O O

Consider a demand-paging system with a paging disk that has an average access/transfer time of 50 ms. Addresses are translated through a page table in main memory, with an access time of 500 ns per memory access. Thus, each memory reference through the page table takes two accesses. To improve this time, we have added a TLB that reduces access time to one memory reference if the page-table entry is in the TLB. Assume that 80% of the accesses are in the TLB and that, of those remaining, 15% (or 3% of the total) cause page faults. We assume that the TLB access time is 20 ns. What is the effective access time?