Here is a series of address references given as word addresses: 2, 3, 11, 16, 21, 13, 64, 48, 19, 11, 3, 22, 4, 27, 6, and 11. A
ssuming a direct-mapped cache with 16 one-word blocks that is initially empty, label each reference in the list as a hit or a miss and show the final contents of the cache. Cache Cl is direct mapped with 16 one-word blocks. Cache C2 is direct mapped with 4 four-word blocks. Assume that the miss penalty for Cl is 8 memory bus clock cycles and the miss penalty for C2 is 11 memory bus clock cycles. Assuming that the caches are initially empty, find a reference string for which C2 has a lower miss rate but spends more memory bus clock cycles on cache misses than Cl. Use word addresses.
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Explanation:
Algorithm for solving flood condition:
We suggest an algorithm to resolve the flood condition by creating a flow network graph.
Let us assume for every patient "p" there is a node "2" and for every hospital "h" there is a node "uh" and there is an edge ()T, uh) exist between patient "p" and hospital "h" with flow capacity of 1 iff patient "p" is reachable to hospital "h" within a half-hour.
Then source node "s" is made between all the patient-nodes by an edge with flow capacity of 1 and then the sink "t" is made by linking all the hospital nodes by an edge with capacity "[n/k]".
There is an approach to send patients to hospitals: when there is a source "s" to sink "t" flow of "n". We can send 1 flow-unit from source "s" to sink "t" along the paths (s, yp, uh, t) whenever a probable approach is available to send patients.
This approach of sending patients to hospitals doesn't break the capacity limitation of edges. Hence we can send patient "p" to hospital "h" with 1 flow- unit if edge(m uh) permits at least 1 flow- unit.
The running-time of this algorithm is found by finding the time needed to solve max-flow graph with nodes O(n+k) and edges O() edges.