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Exercise I, 1a) – initiate rule
What is the likelihood of the
threat scenario to the left?
If the vertices t and e are
related by initiate, we have: I.e.:
Servers infected
by malicious code[1 per year]
Malicious
code on computer
spreads via LANEmployee
0.1
Malicious
code on computer
spreads via LAN
[1 per year]
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Exercise I, 1b) – leads-to rule
What is the likelihood of the
threat scenario to the right?
If the vertices e1 and e2 are
related by leads-to, we have: 1 per 1 year × 0.1 = 1 per 10
years
Servers infected
by malicious code[1 per year]
Malicious
code on computer
spreads via LANEmployee
0.1
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2a) Leads-to + statistically independent vertices
1 per 10 years × 0.2 = 1 per 50 years
1 per 1 year × 0.1 = 1 per 10 years
If the vertices e1 and e2 are
statistically independent and
separate, we
have:
(1/50 + 1/10) =
6 per 50 years
Servers malfunctioning
Servers infected
by malicious code
[1 per 10 years]
0.1
0.2
Malicious code
traffic jams network
[1 per year]
?
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2b) Leads-to + mutually exclusive vertices
If the vertices e1 and e2 are
mutually exclusive, we have:
1 or 5 per 50 years;
alternatively between 1 and 5
per 50 years
Servers malfunctioning
Servers infected
by malicious code
[1 per 10 years]
0.1
0.2
Malicious code
traffic jams network
[1 per year]
?
INF
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2c) Consistency check
Both 6 per 50 years and 5 per 50 years are higher values
than 1 per 20 years
The diagram is inconsistent
Servers malfunctioning
[1 per 20 years]
Servers infected
by malicious code
[1 per 10 years]
0.1
0.2
Malicious code
traffic jams network
[1 per year]
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Exercise II – computing likelihood intervals
Scale Value
1 Rarely <= 1 per 10 years
2 Seldom > 1 per 10 years & <= 1 per 5 years
3 Sometimes > 1 per 5 years & <= 1 per 1 year
4 Often > 1 per 1 year
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Exercise I, 1a) – initiate rule
What is the likelihood of the
threat scenario to the left?
If the vertices t and e are
related by initiate, we have: I.e.:
Servers infected
by malicious code[sometimes]
Malicious
code on computer
spreads via LANEmployee
0.1
Malicious
code on computer
spreads via LAN
[sometimes]
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Exercise I, 1b) – leads-to rule
What is the likelihood of the
threat scenario to the right?
If the vertices e1 and e2 are
related by leads-to, we have: sometimes = <1/5,1/1]
1 per 5 years × 0.1 = 1 per 50
years
1 per 1 year × 0.1 = 1 per 10
years
<1/50,1/10] = rarely
Servers infected
by malicious code[sometimes]
Malicious
code on computer
spreads via LANEmployee
0.1
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2a) Leads-to + statistical independence
[0,1/10] × 0.2 = [0,1/50]
<1/5,1/1] × 0.1 = <1/50,1/10]
If the vertices e1 and e2 are
statistically independent and
separate, we have:
([0,1/50] + <1/50,1/10] ) =
<1/50,6/50]
1.2 per 10 years ∈ seldom
we interpret this as seldom
Servers malfunctioning
Servers infected
by malicious code
[rarely]
0.1
0.2
Malicious code
traffic jams network
[sometimes]
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2b) Leads-to + mutually exclusive vertices
If the vertices e1 and e2 are
mutually exclusive, we have:
rarely multiplied with 0.2 gives
rarely
sometimes multiplied with 0.2
gives rarely
In which case, we may deduce
rarely
?
Servers malfunctioning
Servers infected
by malicious code
[rarely]
0.1
0.2
Malicious code
traffic jams network
[sometimes]
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2c) Consistency check
seldom > rarely
The diagram is inconsistent wrt 2a)
Servers malfunctioning
[rarely]
Servers infected
by malicious code
[rarely]
0.1
0.2
Malicious code
traffic jams network
[sometimes]
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3a) Consistency check
0.5
Internal
infrastructure
Virus attack
Shared infrastructure
resources
Low
robustness
External
resource failure
U2: Unauthorized
modification of
users’ personal
information
[seldom]
1.0
U4: Unavailability
of service due to
infrastructure
failure
[sometimes]
1.0
0.2
External
resources
Internal hardware
or software failure U5: Unavailability
of service due to
malicious code
[seldom]0.2
High traffic on the
service may cause the
server to crash
[sometimes]
All active user
sessions are deleted
leaving the information
partly incorrect
Users’ personal
information
Availability of
service
moderate
major
moderate
External sources of
information fails or
malfunctions
[sometimes]
Internal parts of
the service fails or
malfunctions
[seldom]
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Consistency check
We look at max values:
1 per 1 year × 0.2 = 1 per 5 years
1 per 5 years × 1.0 = 1 per 5 years
(1 per 5 years + 1 per 5 years)
= 2 per 5 years ∈ sometimes OK
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Consistency check
1 per 1 year × 0.2 = 1 per 5 years ∈ seldom OK
1 per 1 year × 0.5 = 1 per 2 years × 1.0 = 1 per 2 years ∉
seldom Not OK
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Exercise III – associativity
Statistical independence, separate events and
frequencies: Associativity follows trivially since + is
associative
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Exercise III - associativity
Statistical independence and probabilities
Again we have associativity. We get the same expression
using the above rule twice independent of where we put
the brackets
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Exercise III - associativity
Mutual exclusion and frequencies: Associativity follows
trivially if all events have the same frequency
If they have different frequencies f_1, f_2, f_3 we get
– f_1 or f_2 or f_3
independent of where we put the brackets; hence, we
also have associativity in the general case
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