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› Matthias Galster, University of Groningen, NL

› Armin Eberlein, American University of Sharjah, UAE

Identifying Potential Core Assets in Service- based Systems to Support the Transition to Service-oriented Product Lines

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Research problem

› Different instances of a SOA system can be treated as product line• To handle different instances• To support variability management

› But: how to make the transition from individual products to a product line?

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Research question

› What services should be included in what instance of a service-based system so that• value can be maximized• stability is maintained

S1

s1

s5

s7

s11

S2

s1

s11

s13

s15

s13

s15

s18

s18

s19

Co

re asset services

s1

s11

s13

s18

s17

s25

S1

s1

s5

s7

s11

S2

s1

s11

s13

s15

s13

s15

s18

s18

s19

s17

s25

Original services

Added services

Original services + added services

Identification of potential core asset services

Background and related work

› Product lines and service-oriented systems, e.g.,• Service-oriented PL architectures• Feature analysis to identify services for SO

systems

› Transition to PL and identifying core assets, e.g., • Core assets as foundation for new PL

architecture• Components reused on PL• Feature orientation

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Proposed approach

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Step 1: Determine available services, i.e., services which are not yet assigned to a service composition

Step 2: Prioritize available services

Step 3: Select services to be added to SOA instances

Step 4: Determine potential core asset services

Step 1: determine available services

› What services are still available?

› S’v with S’v = S \ Sv

› S: set of all services

› Sv : set of services already assigned to an instance v

Step 2: prioritize available services

› S’v is sorted in decreasing order

› If services have the same value order is chosen randomly

Step 3: select services (I)

› Trade-off between stability and accumulated added value

› stabilityv(i) = 1 – [i / (M + i)](1)

› AAVv(i) = k=1..i value(sk) (2)

Step 3: select services (II)Definition: I = number of services in S’v

Input: S’v(sorted), Sv, M = sizeOf(Sv)Output: S’’v(resulting list of services; initialized empty)copy Sv into S’’v

k = 0 /* counter for added services is 0 */for i = 1 to Iadd S’v(i) to S’’v

k = k + 1 /* service is added */calculate stabilityv of S’’v using (1)calculate AAVv of S’’v using (2)normalize stabilityv and AAVv

endfor

Determine i so that (i, AAVv(i)) and (i, stabilityv(i)) are closest to each other to represent the best compromise (trade-off)

Step 4: determine core asset services

› Sca = S’’1 S’’2 S’’3 … S’’V (3)

› Sca: services that could form basis for a product line

Case study

› “Exploratory” case study• No proposition

› Identification of web services that could form the foundation for a service-based product line• Web services of telecom company

Case study settings – services

Service Description Value

s1 Display name and number of caller 20

s2 Alerts incoming calls when on the phone 15

s3 Takes messages when call is not answered 17

s4 Forwards incoming calls to pager, cell phone, voice mail or any other phone number 12

s5 Intercepts calls from pre-selected numbers and routes them to a standard recording 5

s6 Provides a second phone number on existing line, with distinctive ring 12

s7 Prompts anonymous callers to unblock phone number or say their name 5

s8 Lets restrict selected outbound calls 10

s9 Shows names and numbers of incoming callers when on the phone 0

s10 Prevents own name / number from being shown when calling somebody with call display 0

s11 Allows to trace last incoming call in the event of threatening, harassing, or obscene phone calls 0

s12 Lets add a third party to call 4

s13 Lets monitor a busy line to get in touch with caller when line is free 4

Case study settings – initial SOA instances

Service Customer 1 (S1) Customer 2 (S2) Customer 3 (S3)

s1

s2

s3

s4

s5

s6

s7

s8

s9

s10

s11

s12

s13

Case study: step 1 – identify services

› S’1 = S \ S1 = S \ {s3, s9, s10, s11, s13} = {s1, s2, s4, s5, s6, s7, s8, s12} “inverse” of columns from previous slide

› S’2 = S \ S2 = S \ {s2, s3, s5, s9, s10, s11, s12} = {s1, s4, s6, s7, s8, s13}

› S’3 = S \ S3 = S \ {s1, s3, s9, s11} = {s2, s4, s5, s6, s7, s8, s10, s12, s13}

Case study: step 2 – prioritize services› Available services are ranked based on value

Rank S’1 S’2 S’3

1 s1 s1 s2

2 s2 s4 s4

3 s4 s6 s6

4 s6 s8 s8

5 s8 s7 s5

6 s5 s13 s7

7 s7 - s12

8 s12 - s13

9 - - s10

Case study: step 3 – select services (I)

› Trade-off analysis for instance 1

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1

1 2 3 4 5 6 7 8

Number of added services

Normalized

AAV

[a.u.]

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1

Normalized stability [a.u.]

AAV (normalized) stability (normalized)

Case study: step 3 – select services (II)› Illustration of the algorithm for instance from

previous slide

i (number of added services) Added service Increase in AAV (normalized) Decrease in stability

1 s1 norm(20) = 0 norm(0.84) = 1.00

2 s1, s2 norm(20 + 15) = 0.24 norm(0.71) = 0.74

3 s1, s2, s4 norm(20 + 15 + 12) = 0.43 norm(0.63) = 0.54

4 s1, s2, s4, s6 norm(20 + 15 + 12 + 12) = 0.62 norm(0.54) = 0.38

5 s1, s2, s4, s6, s8 norm(20 + 15 + 12 + 12 + 10) = 0.78 norm(0.50) = 0.26

6 s1, s2, s4, s6, s8, s5 norm(20 + 15 + 12 + 12 + 10 + 5) = 0.86 norm(45) = 0.16

7 s1, s2, s4, s6, s8, s5, s7 norm(20 + 15 + 12 + 12 + 10 + 5 + 5) = 0.94 norm(0.42) = 0.07

8 s1, s2, s4, s6, s8, s5, s7, s12 norm(20 + 15 + 12 + 12 + 10 + 5 + 5 + 4) = 1 norm(0.38) = 0.00

Case study: step 3 – select services (III)› Trade-off analyses for instances 2 and 3

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

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0,9

1

Number of added services

Normalized

AAV

[a.u.]

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

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0,9

1

Normalized stability [a.u.]

AAV (normalized) stability (normalized)

1 2 3 4 5 60

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1

Number of added services

Normalized

AAV

[a.u.]

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1

Normalized stability [a.u.]

AAV (normalized) stability (normalized)

1 2 3 4 5 6 7 8 9

Case study: step 4 – determine services› S’’1 = S1 {s1, s2, s4} = {s3, s9, s10, s11, s13}

{s1, s2, s4} = {s1, s2, s3, s4, s9, s10, s11, s13}

› S’’2 = S2 {s1, s4, s6} = {s2, s3, s5, s9, s10, s11, s12} {s1, s4, s6} = {s1, s2, s3, s4, s5, s6, s9, s10, s11, s12}

› S’’3 = S3 {s2, s4, s6} = {s1, s3, s9, s11} {s2, s4, s6}= {s1, s2, s3, s4, s6, s9, s11}

Case study: step 4 – determine services- Sca = S’’1 S’’2 S’’3

- Sca = {s1, s2, s3, s4, s9, s11} see services in bold on previous slide

Case study: discussion of results

› Investigating how SOA instances diverge from the PL is important

› Method is light-weight

› Trade-0ff between value and stability

› Threats to validity

Limitations

› No effort or resource constraints

› No interface problems

› Simplified trade-off algorithm

› Interpretation of stability

Conclusions and future work› SO PL help manage versions of SOA

› Trade-off analysis between value and stability to ensure homogenity between SOA systems / instances

› Future work• More comprehensive evaluation• Inclusion of types of variability• Inclusion of more complex definitions for

stability

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Thank you for your attention

Case study: threats to validity› External validity

• More evaluations are needed

› Internal validity• See limitations (simplifications made in

method)

› Assumptions• Price correlates with value• No value for services which are free of charge

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