1. Modifying Theorem 2
Theorem 4 (The synchronous completion time theorem)
In the above theorem, a task i will meet its completion deadline Di if it satisfies the shown relation.

2. One Final Example (Taken from “Software Engineering Fundamentals” by A
One Final Example (Taken from “Software Engineering Fundamentals” by A. Behforooz)
Consider the following 3 dependent and periodic tasks
Task 1: C=30; T=100; D=100; E=0; B=0
Task 2: C=70; T=200; D=200; E=0; B=30
Task 3: C=30; T=200; D=150; E=50; B=6
Using theorem 3: 1 = 30/100 = 0.3  schedulable
1,2 = (30/100)+(70+30)/200 = 0.8  schedulable
1,2,3 = (30/100)+(70/200)+( )/200 = 1.08  NOT schedulable
Using theorem 4: W3(t)=t0= = 136 ms
N1,1 = 1+( )/100 = 2
 t1 = 136+(2-1)30 = 166 ms (beyond 3 deadline)
Changing task priority from (1-2-3) to (1-3-2) results in:
W3(t)=t0= = 166 ms

3. Priority Inversion
Meaning: A lower priority task executing before one with a higher priority
Execution time = 5 ms
Period = 20 ms
Priority = 1
Execution time = 15 ms
Period = 30 ms
Priority = 2
Periodic Task 1
Periodic Task 2
Execution time = 25 ms
Period = 140 ms
Priority = 3
Periodic Task 3
Usage:
1(5 ms), 3(40 ms)
Server Task
Lockable resource

4. Timeline of previous slide
Time (ms) 80 20 30 60 40 90
100
120
150
140
180
210
240
160
Task 1 5 15
Task 2 15 10 5
Task 3 5 10 30
Server

5. Expanded Portion of Previous Timeline80 60 40 90
100
Time (ms)
Task 1 5 15
Task 2 15 5 10
Task 3 10 5
??...
Server

6. Modified and Augmented Portion from Previous Slide80 60 40 90
100
Time (ms)
120
140
150
160
1 queued
1 queued
Task 1 15 5
Task 2 15 5 10 20
Task 3 10 15 5 55
Server

7. A Solution to Priority Inversion
Use of Priority Ceiling Protocol (PCP) which is made up of 3 rules:
Pre-emption rule
Regular priority pre-emption
Inheritance rule
If a lower priority task blocks a higher priority one, the lower inherits the priority of the higher priority task
Priority ceiling rule
A task cannot get at a shared resource if it’s priority is less than any of the ceilings of the shared resources currently locked by other tasks

8. Bad PCP usage Example
Assume tasks A, B, C, and D have priorities as A highest and D lowest. The tasks use 2 server tasks as follows: S1 used by A and C; S2 used by tasks B and D.
Time (ms)
Task A 1
Task B 2
Task C 1 3
Task D A C B S1 S2

9. McCabe’s R-T System Analysis
Uses a modified interpretation of standard DFDs
Transform  Markov process state (a probabilistic queuing model)
Data flow  Transitions
Therefore the following reasoning may be applied:
0 < pij ] 1.0
(p: transitional probability)
True for each flow path pi pj
pij

10. Example of McCabe’s R-T System Analysis
Due to the lengthiness of this example it was thought better to hand it out as an MS-Word97 document. Therefore, it is available online and in hard-copy form separately. Please refer to it while it is explained during lecture sessions.