Completed exercise M8 for embedded

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+= Opgaver til gang 8
+
+== Opgave 1
+
+Given a periodic taskset : T1=4,c1=3,T2=12,c2=2 with relative deadlines d1=4,d2=7
+
+. Examine the taskset w.r.t. schedulability using the exact criterion for DMA scheduling.  Is the taskset EDF schedulable ?
+
+=== Løsning
+
+____
+_Examine the taskset w.r.t. schedulability using the exact criterion for DMA scheduling._
+____
+
+Her kan man se at følgende priority scheme skal bruges.
+
+(T1, T2)
+
+Nu giver det mening at undersøge om dette er feasable.
+For T1 er completion time 3, hvilket er under dens deadline.
+
+For T2.
+
+image::M8f1.png[]
+
+Her kan man se at completion time er 8 hvilket er efter T2's deadline.
+Derfor er DMA ikke feasable.
+
+Dette kan man også se hvis man kører det.
+
+----
+0           3   2   4   7   
+1   |       2   2   3   6   
+2   |       1   2   2   5   
+3   |       0   2   1   4   
+4       |   3   1   4   3   
+5   |       2   1   3   2   
+6   |       1   1   2   1   
+Deadline missed in task 1
+----
+
+____
+Is the taskset EDF schedulable?
+____
+
+Når man kører med EDF kan man se at der ikke er mere arbejde efter LCM.
+
+----
+0           3   2   4   7   
+1   |       2   2   3   6   
+2   |       1   2   2   5   
+3   |       0   2   1   4   
+4       |   3   1   4   3   
+5       |   3   0   3   2   
+6   |       2   0   2   1   
+7   |       1   0   1   0   
+8   |       2   0   4   -1  
+9   |       1   0   3   -2  
+10  |       0   0   2   -3  
+11          0   0   1   -4
+----
+
+== Opgave 2
+
+Given a periodic taskset T1=40,c1=10,d1=20,T2=60,c2=20,d2=30,T3=80,c3=20,d3=80,(milliseconds) where task1 and task2 share data protected by semaphore S.
+Accessing the shared data lasts no more than 10 (time units-mill sec).
+Priority ceiling or immediate ceiling is assumed.
+
+. Is the taskset DMA schedulable ?
+. If the data above are instead shared between task1 and task3, is the taskset then DMA schedulable.
+
+=== Løsning
+
+____
+_Is the taskset DMA schedulable?_
+____
+
+Her bruger jeg den nye formel for at finde completion time.
+
+stem:[C_i = c_i + \sum_{j=1}{i-1}ceil(C_i/T_j) * c_j + B_i]
+
+Ved DMA bliver prioriteten (T1, T2, T3).
+Task 1 kan godt vente på T2, men T2 skal ikke vente på nogen.
+
+----
+C1 = 10 + 10 = 20
+C2 = 20 + ceil(C2/40)*10
+C3 = 20 + ceil(C3/60)*20 + ceil(C3/40)*10
+----
+
+T1 bliver færdig lige inden dens deadline.
+Først plottes C2.
+
+image::M8f2.png[]
+
+Her kan man se at den completer inden dens deadline 30.
+
+Under kan man se at T3 også bliver færdig, når man plotter C3.
+
+image::M8f3.png[]
+
+____
+_If the data above are instead shared between task1 and task3, is the taskset then DMA schedulable._
+____
+
+Her ændres C1 til C3 sig overhovedet ikke, da T1 alligevel skal vente 10.
+
+----
+C1 = 10 + 10 = 20
+C2 = 20 + ceil(C2/40)*10
+C3 = 20 + ceil(C3/60)*20 + ceil(C3/40)*10
+----
+
+== Opgave 3
+
+Implement the example of ex. 2) on JDN's kernel and measure worst case completion times.