started with TDA
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@ -19,9 +19,9 @@ def test(tasks):
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U_lub = n * ((2 ** (1 / n)) - 1)
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# for fewer tasks than 10, we use the exact computed least upper bound
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if n < 10:
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return U <= U_lub
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# if n < 10:
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# return U <= U_lub
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# from 10 tasks up unlimited, we use the limes of n(2 ** 1/n - 1)
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return U <= np.log(2) # round to 0.7 ?
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return U <= U_lub # np.log(2)
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@ -13,15 +13,42 @@ import include.TasksHelper as TH
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# C_i is accessed as: tasks[i][2]
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# The number of tasks can be accessed as: tasks.shape[0]
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#The Time Demand Analysis Test
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# The Time Demand Analysis Test
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def test(tasks):
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#Sorting Taskset by Period/Deadline
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#This makes implementing TDA a lot easier
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# Sorting Taskset by Period/Deadline
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# This makes implementing TDA a lot easier
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shape = tasks.shape
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sortedtasks = tasks[tasks[:, 0].argsort()]
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#####################
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#YOUR CODE GOES HERE#
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#####################
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print("\n======= TASK SET =======")
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# For each tasks in the ordered set
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for i in range(len(sortedtasks)):
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return False
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print(f'Task #{i} {tasks[i]}:')
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# calculate the time points for the demand function
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# t = j * P_k for k = 1, 2,...i and j = 1, 2,...,math.ceil(P_i / P_k)
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list_of_t = [TH.P_i(sortedtasks, k) * j for k in range(1, i)
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for j in range(1, int(np.ceil(TH.D_i(sortedtasks, i) / TH.P_i(sortedtasks, k))))]
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# print(f'\t list of ts: {list_of_t}')
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# at any time t between 0 and and TH.P_i
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for t in np.sort(list_of_t):
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# for t in range(TH.C_i(sortedtasks, i), TH.P_i(sortedtasks, i)+1, TH.P_i(sortedtasks, i)):
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print(f'\tTime-Demand for t ({t}): {time_demand_func(sortedtasks, i, t)} ')
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# if the demand for CPU time of task i exceeds the available time t
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if time_demand_func(sortedtasks, i, t) > t:
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return False # then the task i will not meet its deadline, hence taskset not schedulable
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return True
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def time_demand_func(tasks, i, delta=0):
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sum = 0
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for k in range(1, i-1):
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sum += math.ceil(delta / TH.P_i(tasks, k)) * TH.C_i(tasks, k)
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return TH.C_i(tasks, i) + sum
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