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Commit 0991e740 authored by Per's avatar Per
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exam updated

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...@@ -599,6 +599,7 @@ gdb.execute("continue") ...@@ -599,6 +599,7 @@ gdb.execute("continue")
# #
# Assignment 1. # Assignment 1.
# Run the example and study the output. # Run the example and study the output.
# you may need to run xargo clean first
# #
# It generates `output data`, a list of list, something like: # It generates `output data`, a list of list, something like:
# Finished all ktest files! # Finished all ktest files!
...@@ -647,7 +648,8 @@ gdb.execute("continue") ...@@ -647,7 +648,8 @@ gdb.execute("continue")
# of the gdb integration regarding the return behavior. # of the gdb integration regarding the return behavior.
# #
# Verify that you can repeat the experiment. # Verify that you can repeat the experiment.
# The order of tasks/test may differ but it should look similar. # The order of tasks/test and cycles may differ but it should look similar.
# Two tests for EXTI1 and one for EXTI2 and one for EXTI3
# #
# Try follow what is going on in the test bed. # Try follow what is going on in the test bed.
# #
...@@ -678,7 +680,7 @@ gdb.execute("continue") ...@@ -678,7 +680,7 @@ gdb.execute("continue")
# (It may be a good idea to make first pass and extract wcet per task) # (It may be a good idea to make first pass and extract wcet per task)
## ##
# The total utilisation bound allows us to discard task sets that are # The total utilisation bound allows us to discard task sets that are
# obviously illegal.debug (not the case here though) # obviously illegal (not the case here though)
# #
# Assignment 3. # Assignment 3.
# #
...@@ -711,8 +713,8 @@ gdb.execute("continue") ...@@ -711,8 +713,8 @@ gdb.execute("continue")
# so bound priority inversion # so bound priority inversion
# #
# Notice 2, `output_data` does not hold info on WHAT exact resource is held # Notice 2, `output_data` does not hold info on WHAT exact resource is held
# (merely each 'claim time at a specific ceiling'). # merely timestamp information on each Enter/Exit and associated level.
# However this is sufficient for the analysis. # This is however sufficient for the analysis.
# #
# so # so
# R_EXTI2 = 11 + 10 = 21, well below our 30 cycle margin # R_EXTI2 = 11 + 10 = 21, well below our 30 cycle margin
...@@ -722,7 +724,7 @@ gdb.execute("continue") ...@@ -722,7 +724,7 @@ gdb.execute("continue")
# where I_EXTI3 is the interference (preemptions) # where I_EXTI3 is the interference (preemptions)
# #
# Here we can undertake a simple approach to start out. # Here we can undertake a simple approach to start out.
# Assume a deadline equal to our interarrival (50) # Assuming a deadline equal to our interarrival (40)
# I_EXTI3 is the sum of ALL preemptions until its deadline. # I_EXTI3 is the sum of ALL preemptions until its deadline.
# in this case EXTI2 can preempt us 2 times (40/30 *rounded upwards*) # in this case EXTI2 can preempt us 2 times (40/30 *rounded upwards*)
# I_EXTI3 = 2 * 11 # I_EXTI3 = 2 * 11
...@@ -730,7 +732,7 @@ gdb.execute("continue") ...@@ -730,7 +732,7 @@ gdb.execute("continue")
# The worst case blocking time is 15 # The worst case blocking time is 15
# (caused by the lower prio task EXTI1 holding X) # (caused by the lower prio task EXTI1 holding X)
# R_EXTI3 = 8 + 2 * 11 + 15 = 45, already here we see that # R_EXTI3 = 8 + 2 * 11 + 15 = 45, already here we see that
# EXTI2 may miss its deadline # EXTI2 may miss our deadline (40)
# #
# EXTI1 (our lowest prio task) # EXTI1 (our lowest prio task)
# R_EXTI1 = C_EXTI1 + B_EXTI1 + I_EXTI1 # R_EXTI1 = C_EXTI1 + B_EXTI1 + I_EXTI1
...@@ -746,6 +748,9 @@ gdb.execute("continue") ...@@ -746,6 +748,9 @@ gdb.execute("continue")
# Ouch, even though we had only a WCET of 37 we might miss our deadline. # Ouch, even though we had only a WCET of 37 we might miss our deadline.
# However we might have overestimated the problem. # However we might have overestimated the problem.
# #
# Implement the algorithm in a generic manner
# Verify that that the results are correct by hand computation (or make an Excel)
#
# Assignment 4. # Assignment 4.
# Looking closer at 7.22 we see that its a recurrent equation. # Looking closer at 7.22 we see that its a recurrent equation.
# Ri(0) indicating the initial value # Ri(0) indicating the initial value
......
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