The need for effective policy management has led to the emergence of a class of software tools that supports policy development, implementation, and decentralization.

Write a function so that the main0 code below can be replaced by the simpler code that calls function mphandminutes tomiles0. original main0 int main) l double milesperhour-70.0; double minutestraveled = 100.0; double hourstraveled; double milestraveled; hourstraveled = minutestraveled / 60.0; milestraveled = hourstraveled * milesperhour; cout < "miles" 2 using namespace std; 4 /* your solution goes here/ 6 int maino 1 test passed 7 double milesperhour 70.0 all tests passed 8 double minutestraveled 100.0; 10 cout < < "miles: " < < mphandminutestomiles(milesper-hour, minutestraveled) < < endl; 12 return 0; 13

When making changes to optimize part of a processor, it is often the case that speeding up one type of instruction comes at the cost of slowing down something else. for example, if we put in a complicated fast floating-point unit, that takes space, and something might have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay to reach that unit. the basic amdahl's law equation does not take into account this trade-off. a. if the new fast floating-point unit speeds up floating-point operations by, on average, 2ă—, and floating-point operations take 20% of the original program's execution time, what is the overall speedup (ignoring the penalty to any other instructions)? b. now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5ă— slowdown (or 2/3 speedup). data cache accesses consume 10% of the execution time. what is the overall speedup now? c. after implementing the new floating-point operations, what percentage of execution time is spent on floating-point operations? what percentage is spent on data cache accesses?

Use a physical stopwatch to record the length of time it takes to run the program. calculate the difference obtained by calls to the method system.currenttimemillis() just before the start of the algorithm and just after the end of the algorithm. calculate the difference obtained by calls to the method system.currenttimemillis() at the start of the program and at the end of the program so that the elapsed time includes the display of the result. use the value returned by the method system.currenttimemillis() just after the end of the algorithm as the elapsed time.