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6 Commits
Author | SHA1 | Date |
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Mukendi Mputu | 71de5bf9b5 | |
Mukendi Mputu | 317b8a1a9c | |
Mukendi Mputu | 49c1a64e52 | |
Mukendi Mputu | e398f3a492 | |
Mukendi Mputu | ae039e75be | |
Mukendi Mputu | 4a595665dc |
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@ -12,11 +12,15 @@ 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 sufficient Test for the Hyperbolic bound
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def test(tasks):
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#####################
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#YOUR CODE GOES HERE#
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#####################
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return False
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""" The Hyperbolic bound is given as Π(U + 1) of all tasks: """
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# compute the hyperbolic bound as product of the U_factor of each task + 1
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hb = 1
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for i in range(tasks.shape[0]): # len(tasks) is 10
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hb *= (TH.C_i(tasks, i) / TH.P_i(tasks, i)) + 1
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# compare this computed bound to 2.0
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# if greater then no guaranty of schedulability
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# otherwise task set is schedulable
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return hb <= 2.0
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@ -14,9 +14,14 @@ import include.TasksHelper as TH
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#The necessary Test for the Liu and Layland Bound
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def test(tasks):
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n = tasks.shape[0]
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U = TH.getTotalUtilization(tasks=tasks, NrTasks=n)
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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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#####################
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#YOUR CODE GOES HERE#
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#####################
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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 <= U_lub # np.log(2)
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return False
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@ -13,15 +13,39 @@ 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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set_num = 0
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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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# For each tasks in the ordered set
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#####################
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#YOUR CODE GOES HERE#
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#####################
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# calculate the time points for the demand function
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t_old = 10**-3
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i = 0
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return False
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while True:
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t_new = workload_func(sortedtasks, i, t_old)
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# if the workload of task i exceeds the deadline
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if t_new > TH.D_i(sortedtasks, i):
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return False # task not schedulable
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if t_new == t_old:
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i += 1
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t_old = 10**-3
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# chech array out of bounds
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if i == len(sortedtasks):
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return True
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t_old = t_new
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def workload_func(tasks, i, t):
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sum = 0
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for k in range(i):
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sum += math.ceil(t / 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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