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The examination paper for the module ctec9012, networking embedded systems, of the meng in embedded systems engineering program at the cork institute of technology. The paper includes questions on various topics such as wireless sensor networks, data transmission systems, topology control, power aware routing, minimum spanning tree, maximum flow, and queuing systems.
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Semester 1 Examinations 2011/
Module Code: CTEC
School: Electronic and Electrical Engineering
Programme Title: MEng in Embedded Systems Engineering
Programme Code: EMBED_9_Y
External Examiner(s): Dr D. Heffernan, Mr M. Kelleher, Mr D. O’Connell Internal Examiner(s): Dr Oliver Gough
Instructions: Answer four questions
Duration: 2 Hours
Sitting: Winter 2011
Requirements for this examination:
Note to Candidates: Please check the Programme Title and the Module Title to ensure that you have received the correct examination paper. If in doubt please contact an Invigilator.
(a) Consider a single hop wireless sensor network with a large number of nodes each running a time critical data reporting application and a single aggregation server. The server receives 1200 packets per hour during normal operation according to a Poisson process. The application must operate with a mean latency below 250 ms. (i) Assuming exponentially distributed service times, specify the required server performance to achieve this bound. (ii) If sleeping schedules are introduced to save energy, how much will the mean service time (including sleep) be permitted to increase if a doubling of delay is tolerable (assume the service times remain exponentially distributed)? [10 marks]
(b) A transmission system, is shared among 25 nodes. Data packets of a fixed size of 300 bytes are generated by each computer system according to a Poisson process with rate = 1.2/sec. Due to variability in the communications environment the effective data rate has the following distribution. R 1 = 18 kbit/s,is available 10% of the time, R 2 = 36 kbit/s, 20% of the time, R 3 = 72 kbit/s, 30% of the time, R 4 = 144 kbit/s, 40% of the time. (i) Calculate the throughput of the system. (ii) Calculate the average waiting time and number of packets waiting for transmission. [15 marks] [Total: 25 marks]
(a) Greedy Perimeter Stateless Routing (GPSR) is a position based routing approach for wireless networks. Explain the use of graph-theoretic constructs such as the Relative Neighbourhood Graph (RNG) and Gabriel Graph (GG) to provide topology control that enable efficient operation of GPSR. [10 marks] (b) Consider the situation in figure Q3. The transmission links are labelled with the energy cost of transmitting a standard packet. Nodes are labelled with the current battery energy reserve. Node I is sending a message to node B. Consider the route(s) selected by the Min-Max Battery Cost Routing algorithm and the route(s) selected using the Min Total Transmit Power Routing. Explain the differences in these approaches to power aware routing. [15 marks]
2
2
2
2
(^3 )
1 1
3 1 1 2
2
2
2
14
20
7
4
9 20
6
15
12
[Total: 25 marks]
(a) Using Prim’s algorithm, develop a minimum spanning tree routed at node s = 1 for the network of 9 nodes shown in Figure Q4(a). The link costs are assumed bidirectional with equal costs in both directions. In your answer tabulate the sequence of edge additions and the total cost of the tree. [10 marks]
2
2
2
6
3 3
1 1
3 3 3 2
2
1
(b) Using the Ford-Fulkerson algorithm, calculate the maximum flow f between source node s = 1, and destination node d = 7 for the network of 7 nodes shown in Figure Q4(b). Link capacities and flow direction are shown with each link.
20
10 12
7
8
(^3 )
(^15 )
20
7
[Total: 25 marks]
The following formulae may be of use
2
M/M/m/ Queuing system :
A P and
A m P
(^0) m! 1 P p m^ m (^) Q , where
1 0
m i
i m m
m i
p m
and
m
Prob(Delay > T) = PQexp{(mμ-λ)T}
Pollazcek-Khinchine formula for an M/G/1 queuing system:
2 and T X W , X is service time random variable
k 0
k 0
2 k
, for discrete probability distributions
X 2 x^2 p(x)dx; for continuous probability distributions
(^) m
k
k
m
k
s
m
s
Ems
0
k s k
s k
s
Jackson’s Theorem:
K
i
1
j
j j