SDF-GA: a service domain feature-oriented approach for manufacturing cloud service composition

Abstract

Cloud manufacturing (CMfg) is a new service-oriented manufacturing paradigm in which shared resources are integrated and encapsulated as manufacturing services. When a single service is not able to meet some manufacturing requirement, a composition of multiple services is then required via CMfg. Service composition and optimal selection (SCOS) is a key technique for creating an on-demand quality of service (QoS)-optimal efficient manufacturing service composition to satisfy various user requirements. Given the number of services with the same functionality and a similar level of QoS, SCOS has been seen as a key challenge in CMfg research. One effective approach to solving SCOS problems is to use service domain features (SDF) through investigating the probability of services being used for a specific requirement from multiple perspectives. The approach can result in a division of the service space and then help streamline the service space with large-scale candidate services. The approach can also search for optimal subspaces that most likely contribute to an overall optimal solution. Accordingly, this paper develops an SDF-oriented genetic algorithm to effectively create a manufacturing service composition with large-scale candidate services. Fine-grained SDF definitions are developed to divide the service space. SDF-based optimization strategies are adopted. The novelty of the proposed algorithm is presented based on Bayes’ theorem. The effectiveness of the proposed algorithm is validated by solving three real-world SCOS problems in a private CMfg.

Appendix

• 1. Proof of Lemma2. A set of values of $P_{T - GA} \left( {B/A_{h}^{\left( 0 \right)} } \right)$ can be derived based on the formula in Property 1 for calculating $P_{T - GA} \left( {B/A_{h}^{\left( 0 \right)} } \right)$ and sorted in ascending order $0 \le P_{T - GA} \left( {B/A_{1}^{\left( 0 \right)} } \right) \le P_{T - GA} \left( {B/A_{2}^{\left( 0 \right)} } \right) \le \cdots \le P_{T - GA} \left( {B/A_{\text{h}}^{\left( 0 \right)} } \right)$, and then we randomly split the set of values as $0 \le P_{T - GA} \left( {B/A_{1}^{\left( 0 \right)} } \right) \le P_{T - GA} \left( {B/A_{2}^{\left( 0 \right)} } \right) \le \cdots \le P_{T - GA} \left( {B/A_{h - u}^{\left( 0 \right)} } \right) \le P_{T - GA} \left( {B/A_{1}^{\left( 0 \right)} } \right) \le P_{T - GA} \left( {B/A_{2}^{\left( 0 \right)} } \right) \le \cdots \le P_{T - GA} \left( {B/A_{u}^{\left( 0 \right)} } \right)$. We assure that the optimal subspace is one of subspaces among u subspaces with bigger values of $P_{T - GA} \left( {B/A_{u}^{\left( 0 \right)} } \right)$ and use all solutions from u subspaces to initialize the population of SDFs-GA. Based on the Property 1, we can have

  $$ \begin{aligned} P_{SDF - GA} \left( {A^{\left( 0 \right)} } \right) = & \mathop \sum \limits_{j = 1}^{u} P_{SDF - GA} \left( {A_{j}^{\left( 0 \right)} } \right) = \frac{{\mathop \sum \nolimits_{j = 1}^{u} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right) \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}}{{\mathop \sum \nolimits_{i = 1}^{h - u} P_{T - GA} \left( {B/A_{i}^{\left( 0 \right)} } \right) \times P_{T - GA} \left( {A_{i}^{\left( 0 \right)} } \right) + \mathop \sum \nolimits_{j = 1}^{u} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right) \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}} \\ = & \frac{{\mathop \sum \nolimits_{j = 1}^{u} \frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}}{{\mathop \sum \nolimits_{i = 1}^{h - u} \frac{{P_{T - GA} \left( {B/A_{i}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} \times P_{T - GA} \left( {A_{i}^{\left( 0 \right)} } \right) + \mathop \sum \nolimits_{j = 1}^{u} \frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}} \\ \end{aligned} $$

Because of that the value of $P_{T - GA} \left( {B/A_{i}^{\left( 0 \right)} } \right)$ is smaller than the value of $\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)$, so we can have

$$ \begin{aligned} P_{SDF - GA} \left( {A^{\left( 0 \right)} } \right) \ge & \frac{{\mathop \sum \nolimits_{j = 1}^{u} \frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}}{{\mathop \sum \nolimits_{i = 1}^{h - u} P_{T - GA} \left( {A_{i}^{\left( 0 \right)} } \right) + \mathop \sum \nolimits_{j = 1}^{u} \frac{{P_{GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}} \\ = & \frac{{\mathop \sum \nolimits_{j = 1}^{u} \frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}}{{1 - \mathop \sum \nolimits_{j = 1}^{u} P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right) + \mathop \sum \nolimits_{j = 1}^{u} \frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}} \\ = & \frac{{\mathop \sum \nolimits_{j = 1}^{u} P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right) + \mathop \sum \nolimits_{j = 1}^{u} \left( {\frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} - 1} \right) \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}}{{1 + \mathop \sum \nolimits_{j = 1}^{u} \left( {\frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} - 1} \right) \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}} \\ \end{aligned} $$

On account of $\sum\nolimits_{i = 1}^{h - u} {P_{T - GA} } \left( {A_{i}^{\left( t \right)} } \right) + \sum\nolimits_{j = 1}^{u} {P_{T - GA} } \left( {A_{j}^{\left( t \right)} } \right) = 1$, we can have

$$ \begin{aligned} P_{SDF - GA} \left( {A^{\left( 0 \right)} } \right) = & \frac{{1 + \mathop \sum \nolimits_{j = 1}^{u} \left( {\frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} - 1} \right) \times P_{GA} \left( {A_{j}^{\left( 0 \right)} } \right) - \mathop \sum \nolimits_{i = 1}^{h - u} P_{T - GA} \left( {A_{i}^{\left( 0 \right)} } \right)}}{{1 + \mathop \sum \nolimits_{j = 1}^{u} \left( {\frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} - 1} \right) \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}} \\ = & 1 - \frac{{\mathop \sum \nolimits_{i = 1}^{h - u} P_{T - GA} \left( {A_{i}^{\left( 0 \right)} } \right)}}{{1 + \mathop \sum \nolimits_{j = 1}^{u} \left( {\frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} - 1} \right) \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right)}} \\ \end{aligned} $$

Because of $\sum\nolimits_{j = 1}^{u} {\left( {\frac{{P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}}{{\hbox{min} P_{T - GA} \left( {B/A_{j}^{\left( 0 \right)} } \right)}} - 1} \right)} \times P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right) \ge 0$, so we can have

$$ P_{SDF - GA} \left( {A^{\left( 0 \right)} } \right) \ge 1 - \mathop \sum \limits_{i = 1}^{h - u} P_{GT - A} \left( {A_{i}^{\left( 0 \right)} } \right) = \mathop \sum \limits_{j = 1}^{u} P_{T - GA} \left( {A_{j}^{\left( 0 \right)} } \right) = P_{T - GA} \left( {A^{\left( 0 \right)} } \right) $$

• 2. Proof of Lemma3. Based on the Property 1, we can have

  $$ \begin{aligned} P_{T - GA} \left( {A^{\left( 1 \right)} } \right) = & \frac{{\mathop \sum \nolimits_{j = 1}^{u} P\left( {B/A_{j}^{\left( 0 \right)} } \right)P\left( {A_{j}^{\left( 0 \right)} } \right)}}{{\mathop \sum \nolimits_{i = 1}^{h - u} P\left( {B/A_{i}^{\left( 0 \right)} } \right)P\left( {A_{i}^{\left( 0 \right)} } \right) + \mathop \sum \nolimits_{j = 1}^{u} P\left( {B/A_{j}^{\left( 0 \right)} } \right)P\left( {A_{j}^{\left( 0 \right)} } \right)}} \\ = & \frac{1}{{\frac{{\mathop \sum \nolimits_{i = 1}^{h - u} P\left( {B/A_{i}^{\left( 0 \right)} } \right)P\left( {A_{i}^{\left( 0 \right)} } \right)}}{{\mathop \sum \nolimits_{j = 1}^{u} P\left( {B/A_{j}^{\left( 0 \right)} } \right)P\left( {A_{j}^{\left( 0 \right)} } \right)}} + 1}} \\ \end{aligned} $$

Because that the operators of SDFs-GA use feature sets generated from $A_{GA}^{\left( 1 \right)}$, we can have

$$ \begin{aligned} P_{SDF - GA} \left( {A^{\left( 1 \right)} } \right) = & \frac{{\mathop \sum \nolimits_{j = 1}^{u} P_{T - GA} \left( {B/A_{j}^{\left( 1 \right)} } \right)P\left( {A_{j}^{\left( 0 \right)} } \right)}}{{\mathop \sum \nolimits_{i = 1}^{h - u} P_{T - GA} \left( {B/A_{i}^{\left( 1 \right)} } \right)P\left( {A_{i}^{\left( 1 \right)} } \right) + \mathop \sum \nolimits_{j = 1}^{u} P_{T - GA} \left( {B/A_{j}^{\left( 1 \right)} } \right)P_{GA} \left( {A_{j}^{\left( 1 \right)} } \right)}} \\ = & \frac{1}{{\frac{{\mathop \sum \nolimits_{i = 1}^{h - u} P_{T - GA} \left( {B/A_{i}^{\left( 1 \right)} } \right)P\left( {A_{i}^{\left( 1 \right)} } \right)}}{{\mathop \sum \nolimits_{j = 1}^{u} P_{T - GA} \left( {B/A_{j}^{\left( 1 \right)} } \right)P\left( {A_{j}^{\left( 1 \right)} } \right)}} + 1}} \\ \end{aligned} $$

Based on the Lemma 2 we can get that $\sum\nolimits_{j = 1}^{u} {P_{GA} } \left( {B/A_{j}^{\left( 1 \right)} } \right)P\left( {A_{j}^{\left( 1 \right)} } \right) \ge \sum\nolimits_{j = 1}^{u} {P_{GA} } \left( {B/A_{j}^{\left( 0 \right)} } \right)P\left( {A_{j}^{\left( 0 \right)} } \right)$, so we can have

$$ P_{SDF - GA} \left( {A^{\left( 1 \right)} } \right) \ge P_{GA} \left( {A^{\left( 1 \right)} } \right) $$