Embedded Systems Exercise 7: Models and Architecture …

1Swiss FederalInstitute of Technology

Computer Engineeringand Networks Laboratory

Embedded Systems

Exercise 7:Models and Architecture Synthesis

Yun [email protected]

9/11 Dec 2020



2./4.12.20202

Exercise Structure

Goal of today’s exercise:• Understand the interaction of energy input, storage and use in energy harvesting systems,

maximum power point tracking and optimal application control

Agenda:§ Wednesday 16:15 – 17:00 Clicker quiz (recorded)

Lecture summary (recorded)Solving a sample task (recorded)Tasks of exercise 7 (recorded)Q & A

§ Friday 16:15 – 17:00 Solution discussion (recorded)Q & A

Available Assistants:§ Yun Cheng - TA§ Julian Keller - SA

Naomi Stricker



Overview

• Summary of the Lecture

• Start solving the Exercise

• Discussion of the Solution



• Specification Models:– DG/MG/space exploration/timeing inequations

• Scheduling without resource constraints– ASAP/ALAP → Exercise 8

• Scheduling with resource constraints– List scheduling– Pipeline resources → Exercise 8– Integer Linear Programming (ILP) → Exercise 8

Scheduling



Dependance Graph (DG)

• A dependence graph describes order relations for the execution of single operations or tasks.

• It represents parallelism in a program but no branches in control flow.

• W= a+b;• X= a*b;• Y= X+c;• Z=X-W;



Scheduling : timing inequations Task 1



Scheduling : timing inequations

Assume all operations consume 1 time unit.Based on the dependence graph: we start formulating timing constraints represented by inequations:

𝜏(𝑊) >= 𝜏(𝑡0)

𝜏(𝑋) >= 𝜏(𝑡0)

𝜏(𝑌) >= 𝜏(𝑋) + 1

𝜏(𝑍) >= 𝜏(𝑋) + 1

𝜏(𝑍) >= 𝜏(𝑊) + 1

Task 1



Design space exploration

• For schedule, we try to minimize latency and cost (multi-objective optimization)

• Its goal is finding a balance between the number of resources used and the latency for executing the program, and choosing an implementation according to the available resources and any constraints on latency

Task 2

costs

Latency

bestbest

pareto optimal point



Pareto Points

Pareto Points : points that are not dominated by any other nodes



Pareto Set



Design space exploration Example

• No of resources =1• Latency =4


• No of resources =3• Latency =2 Pareto Points ?



Marked Graphs Task 3

Activated node consumes one token from input and produce one to the output



Marked Graph Example

Given the 2 nodes in the marked graph, How many times can node B fire before Node A has to fire to enable node B firing again?



List Scheduling

• Consider resource constraints– Assignment of priority to the tasks is required

• For every point in time, not just at events like ALAP

• Algorithm is heuristic– Does not yield optimal solution (min. latency)

Task 4



List Scheduling Example

x

+

-

<

nop

x

nop

x

+

1 multiplier, duration: 2

1 ALU, duration: 1

Allocation of resources

Determine Schedule for this resource constrained sequence graph:1. Resource constrained2. Priority : #follow-up nodes

x

+

-

<

nop

x

nop

1 3 3

2

1



List Scheduling

Allocation: resource type #instancesBinding: operation resource type

Ready operations

Running operations

Starting operations

Step-1 : ready nodesStep-2 : running nodesStep-3 : choose nodes from ready

Step-4 : iterate to next time




x

+

-

<

nop

x

nop




x

+

-

<

nop

x

nop

x

+

1 multiplier, duration: 2

1 ALU, duration: 1

Allocation of resources




Static priority assignment:# follow-up nodes

x

+

-

<

nop

x

nop

1 3 3

2

1

Priority Assignment



List Scheduling

x

+

-

<

nop

x

nop

1 3 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5

Step 1: Determine candidates U



x

+

-

<

x

nop

1 3 3

2

1

nopnop time 0

time 1

time 2

time 3

time 4

time 5

Step 2: Determine running operations TList Scheduling



x

+

-

<

x

nop

1 3 3

2

1

nopnop time 0

time 1

time 2

time 3

time 4

time 5

Step 3: Choose S Ì U with max P and |S|+|T| £ aList Scheduling



List Scheduling Step 4: Schedule and iterate

x

+

-

<

x

nop

1 3 3

2

1

nop

-x

nop time 0

time 1

time 2

time 3

time 4

time 5



List Scheduling

x

+

-

<

x

nop

1 3 3

2

1

nop

-x

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

x

+

-

<

x

nop

1 3 3

2

1

nop

-x

nop time 0

time 1

time 2

time 3

time 4

time 5

Step 2: Determine running operations T



List Scheduling

x

+

-

<

x

nop

1 3 3

2

1

nop

-x

nop time 0

time 1

time 2

time 3

time 4

time 5

Step 3: Choose S Ì U with max P and |S|+|T| £ a



List Scheduling

x

+

-

<

x

nop

1 3 3

2

1

nop

-x

nop time 0

time 1

time 2

time 3

time 4

time 5

Step 4: Schedule and iterate



List Scheduling Step 1: Determine candidates U

x

+

-

<

x

nop

1 3 3

2

1

nop

-x

nop time 0

time 1

time 2

time 3

time 4

time 5



List SchedulingStep 2: Determine running

operations T

x

+

-

<

x

nop

1 3 3

2

1

nop

-x

nop time 0

time 1

time 2

time 3

time 4

time 5



List Scheduling

x

+

-

<

x

nop

1 3 3

2

1

nop

-x

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-x

x

+

-

<

x

nop

1 3 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-x

x

+

-

<

x

nop

1 3 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-x

x

+

-

<

x

nop

1 3 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-x

x

+

-

<

x

nop

1 3 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-

<

xx

+

-

<

x

nop

1 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-

<

xx

+

-

<

x

nop

1 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-

<

xx

+

-

<

x

nop

1 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-

<

xx

+

-

<

x

nop

1 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




List Scheduling

nop

x+

-

<

x

nop

x

+

-

<

x

nop

1 3

2

1

nop time 0

time 1

time 2

time 3

time 4

time 5




Exercise 7 – Solution



Task 1 (1)



Task 1 (1)



Task1 (1)



Task 1 (2,3)



Task 2




Task 2




Task 2




Task 2




Task 2



Task 3 (I)



Question 3 (II)



Task 4



Task 4 - Solution

1a) & 1b)

L = 8 - 0 = 8L = 8 - 0 = 8



Task 4 - Solution

1c) Critical path is delayed by a multiplier → A multiplier should be added

1d) New latency = 71e) L = 7

Critical Path

Documents

Embedded Systems Exercise 7: Models and Architecture …