Instructions: Test II, Results Task 1 : Bluetooth Task 2

Institut für Technische Informatik
und Kommunikationsnetze
Computer Engineering and Networks Laboratory
Spring 2016
Embedded Systems–Sample Solution
First & Last Name:
Page 1
Legi-Number:
Instructions:
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Put your Legi-Card on the Table.
Write your Name and your Legi-Number on top of this page.
Accurately read each question before solving it.
There is at least one correct solution.
You will get points only for correct and complete answers.
Supporting Materials: Printouts, handwritten notes, and calculators are allowed.
Devices that can be used for communication (laptops, phones, tablets, mp3 players,
etc.) are NOT allowed.
• After test duration, leave your filled test and Legi-Card on the table infront of you.
Do Not collect/aggregate tests in your row.
• Test duration: 10 minutes. Good luck!
Test II, Results
Points:
Grade:
Task 1 : Bluetooth
(a) (1 Point) Consider a set of Bluetooth-enabled devices arranged in a scatternet containing two overlapping
piconets A and B, as illustrated in the figure below. An edge between two devices in the figure represents a
one-to-one Bluetooth communication link. The label of each device in the figure represents the communication
role as either a master M, or a slave S.
Piconet B
M
S
x
S
y
S
S
Piconet A
Mark all the correct statements for the above scatternet:
Role x is a slave in piconet A, and role y is a slave in piconet B.
⊠ Role x is a slave in piconet A, and role y is a master in piconet B.
Role x is a master in piconet A, and role y is a master in piconet B.
⊠ Each piconet supports a maximum of 7 active slave nodes.
Task 2 : Low Power
(a) (2 Points) Mark all the correct statements below.
⊠ The break-even time depends on the energy overheads when switching between active and sleep modes.
The YDS algorithm always minimizes the total energy consumption (both static and dynamic).
Reducing task operating frequencies always helps in reducing the overall energy dissipation.
⊠ With non-zero break-even time, work conserving scheduling is not necessarily optimal in minimizing energy.
Institut für Technische Informatik
und Kommunikationsnetze
Computer Engineering and Networks Laboratory
Spring 2016
Embedded Systems–Sample Solution
Page 2
(b) (3 Points) Consider two jobs J1 and J2 to be scheduled on a platform with two cores C1 and C2 . Each core has
an active and a sleep mode and is able to switch between the two modes independently. Dynamic voltage and
frequency scaling is not allowed on the platform. Once a job starts executing on a core, it runs to completion
on the same core. Mark all the correct statements.
Operating Frequency
Active Power
Sleep Power
C1
20MHz
40mW
0mW
C2
40MHz
100mW
0mW
Arrival Time
Deadline
Clock Cycles
J1
0s
3s
108
J2
0s
10 s
2 × 108
⊠ To meet job deadlines, two solutions exist: execute J1 on C2 and J2 on C1 , or execute both jobs on C2 .
To meet job deadlines and minimize energy, we must execute both J1 and J2 on C2 .
C2 is more energy efficient than C1 when executing any job with a fixed number of clock cycles.
Task 3 : Communication
Consider a network of three nodes, as shown in the figure. Nodes X and Y are separated by a distance of L,
while Y and Z are 2L apart. The signal propagation speed is σ, and messages sent propagate in all directions. The
CSMA/CD protocol is used for communication, and the minimum packet length is enough for every node to detect
a collision.
L
X
2L
Y
Z
(a) (2 Points) Mark all correct statements for this network.
X and Z can successfully send data to Y at the same time.
⊠ Y starts transmitting at time t = 0. The latest time Y can detect a collision is at time 4L/σ.
⊠ Y and Z both start transmitting at time t = 0. X will detect a collision at time 3L/σ.
X and Y both start transmitting at time t = 0. Z will detect a transmission at time 3L/σ.
(b) (2 Points) Upon detecting a collision, the nodes implement an exponential back-off algorithm wherein a node
draws a random number K from the set {1 . . . 2m }, where m = min{n, 10} and n is the number of subsequent
collisions that have already been detected. Each node then waits for K × tw time, where tw is the wait time.
All three nodes, X, Y and Z, transmit at time t = 0, therefore a collision is detected shortly after. What is the
probability that another collision happens when the next transmission is attempted?
0
1/8
1/4
3/8
1/2
⊠ 5/8
3/4
7/8
1

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