Final answer:
The total cost prior to adding cache is 25.6 cents. The total cost after adding cache is 32 cents. The percentage decrease in time due to the inclusion of cache is 90%.
Step-by-step explanation:
The total cost prior to the addition of cache can be calculated by multiplying the cost per byte of main memory (C_m) by the size of the main memory (S_m). In this case, the total cost prior to adding cache is 256 MB x 0.0001 cents/bytes = 25.6 cents.
The total cost after the addition of cache can be calculated by multiplying the cost per byte of cache memory (C_c) by the size of the cache memory (S_c). In this case, the total cost after adding cache is 32 KB x 0.1 cents/bytes = 32 cents.
The percentage decrease in time due to the inclusion of cache can be calculated using the formula: (T_m - T_c) / T_m * 100. Substituting the given values, the percentage decrease in time is (100 ns - 10 ns) / 100 ns * 100 = 90%.
Complete question:
Consider a memory system with cache having the following parameters: S_c = 32 KB C_c = 0.1 cents/bytes T_c = 10 ns S_m = 256 MB C_m = 0.0001 cents/bytes T_m = 100 ns What was the total cost prior to addition of cache? What is the total cost after addition of cache? What is the percentage decrease in time due to inclusion of cache with respect to a system without cache memory considering a cache hit ratio of 0.85?