Unverified Commit 6c7a7a27 authored by Bennett Piater's avatar Bennett Piater
Browse files

add a (bad) heuristic for machines to use.

What we could do is also implement a different strategy in that function.
Then, iterate HEFT until makespan * money rounded up is below the limit.
If it is not, use the other strategy to select a cheaper combination of
machines.
parent c73c9883
#!/usr/bin/env python3
from typing import List
from parser import parse_tasks, parse_machine_prices
from preprocessing import *
from preprocessing import parallel_width_assuming_mapreduce, best_performance_under_limit
def main():
def main(limit = 6):
# parse
tasks = parse_tasks()
prices = parse_machine_prices()
# preprocess
parallel_width_estimate = parallel_width_assuming_mapreduce(tasks)
# print(parallel_width_estimate)
cost, machine_count = best_performance_under_limit(parallel_width_estimate, prices, limit)
print(cost, machine_count)
if __name__ == "__main__":
main()
......@@ -5,7 +5,7 @@ from taskgraph import Task
from typing import List
def parse_machine_prices(fname: str) -> List[float]:
def parse_machine_prices(fname: str = "price.csv") -> List[float]:
with open(fname, "r") as f:
f.readline() # ignore header
lines = f.readlines()
......
#!/usr/bin/env python3
from typing import List
from typing import List, Tuple
from taskgraph import *
......@@ -23,6 +23,21 @@ def parallel_width_assuming_mapreduce(taskgraph: List[Task]) -> int:
return max_width
def best_performance_under_limit(taskgraph: List[Task], prices: List[float],
limit: float) -> List[float]:
pass
def best_performance_under_limit(
max_width: int,
prices: List[float],
limit: float,
strategy: str = "Biggest") -> Tuple[float, List[int]]:
current_solution = [0 for machine in prices]
current_price = 0.0
# TODO: this assumes more expensive is better and simply fills up with the largest possible
# TODO: it also completely ignores that we may need the machines for MORE than one hour
if strategy == "Biggest":
for machine, price in enumerate(reversed(prices)):
while current_price + price < limit and sum(
current_solution) < max_width:
current_price += price
current_solution[-machine - 1] += 1
return current_price, current_solution
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