Component ColdCost MTTR MTBF No.of Resources Exhaustive Iteration Optimal Solution Calculation x86Server $2400 S2640 5300 3600sec 75days 1000 56 POWERServer S85000 S93500 S1500 3600sec 150days 10 171 LinuxOS 5S0 S0 $0 120sec 45days 1ò 1010 295 WindowsOS S200 120sec 30days 1 105 1516 AIXOS 0 S400 240sec 100days 20 1020 2021 WASServer $0 $100 100sec 30days 30 1030 3011 THSServer $0 S45 $0 50sec 25days DB2Server $0 S60 $0 80sec 30days TABLE V TABLE II COMPUTATIONAL COMPLEXITY CHARACTERISTICS COMPONENT FAILURE BEHAVIOR AND COSTS hardware is powered on,or the software license cost if an Finally we have the optimal solution (X1,X2)=(0.0128, application software component has a usage-based licensing 0.0042),so we have (n1,n2)=([1.3],[1.17)as the near scheme.RepairCost specifies the annual cost to repair the component when the component is down.The annual cost optimal solution,that is (2,2)in this example,and Fig.4 depicts the HA enhanced topology. of a component is the sum of the annual cost to operate it in its operating mode and the initial cost of the component IV.EXPERIMENTAL EVALUATION annualized by dividing by its useful lifetime in years.The A.Experiment Settings annual cost of a resource is the sum of the annual cost of each component. We illustrate the efficiency of our weak-point analysis In this experimental scenario,we create four resources to framework using a simple experimental scenario.For the host relevant web services.Table III depicts the components fundamental parameters for resource availability and cost,we of the resources in our scenario.We specifiy two BPEL specify them as Table II shows.In this table we allocate workflows over the application and resource topologies,as the failure behavior (MTTR.MTBF)and cost (ColdCost, table IV shows. ActiveCost,RepairCost)for each component comprising the three-tier stack of the specified resources. B.Performance Evaluation Based on the above settings for the experimental scenario, Resource Server OS hardware maxSize we can form the utility functions and the workflow-resource resourcel IHSServer LinuxOS x86Server 10 resource2 WASServer LinuxOS x86Server 10 mapping matrix.We get the high availability solution through resource3 WASServer AIXOS POWERServer 10 two approaches for comparison:exhaustive iteration and op- resource4 DB2Server WindowsOS x86Server 10 timal solution calculation methods.Through iteration,we TABLE III generate the optimal solution by simply iterating all possible RESOURCE COMPONENTS solutions for availability requirement and getting the solution for minimum overall cost according to various cost modes. For the optimal solution calculation,we use MATLAB [13]to perform the calculation,leveraging the fimincon algorithm with BPEL Workflow Application Resource workflowl EAR Componentl medium-scale optimization [9]in the MATLAB optimization resourcel EARComponent2 resource2 toolbox.Therefore,we can compare the above two different EARComponent3 resource3 approaches in the following two aspects:solution efficiency DatabaseComponentl resource4 workflow2 EAR Componentl and computational complexity. resourcel DatabaseComponentl resource2 For the solution efficiency,we compare the overall cost for resource4 the solutions found by the two different methods,as Fig.6 TABLE IV shows.From Fig.6 we note that the overall cost raises as WORKFLOW TO RESOURCE MAPPING the availability requirement increases:exhaustive iteration can always achieve the optimal cost when the upper bound for cluster size is set large enough.Our optimal solution achieves For the availability parameters,MTTR specifies the mean the minimum cost of the iteration method most of the time. time to recover after each failure,and MTBF specifies the The reason for any disparity is that we leverage the Lagrange mean time between failures.The single component availability multiplier method,which yields as a solution a vector with can be calculated as MTTR/(MTTR+MTBF) fractional values;however,the solution must be a vector of The cost parameters specify various costs associated with integer values (one cannot deploy 1.2 application servers!). the components:ColdCost specifies the cost when the com- Therefore,the solution we achieve may not be the actual ponent is powered off(e.g.,as a cold spare),ActiveCost optimal solution,but it is near-optimal and can achieve the specifies the cost when the component is powered on (e.g., minimum overall cost for most of the cases. as an active spare).The cost difference may account for For the computational complexity,we compare the number the electrical power costs that are incurred only when the of operations to compute the solutions in Table V.In thisComponent ColdCost ActiveCost RepairCost MTTR MTBF x86Server $2400 $2640 $300 3600sec 75days POWERServer $85000 $93500 $1500 3600sec 150days LinuxOS $0 $0 $0 120sec 45days WindowsOS $0 $200 $0 120sec 30days AIXOS $0 $400 $0 240sec 100days WASServer $0 $100 $0 100sec 30days IHSServer $0 $45 $0 50sec 25days DB2Server $0 $60 $0 80sec 30days TABLE II COMPONENT FAILURE BEHAVIOR AND COSTS Finally we have the optimal solution (X1, X2) = (0.0128, 0.0042), so we have (n1, n2) = (d1.3e, d1.1e) as the near optimal solution, that is (2, 2) in this example, and Fig.4 depicts the HA enhanced topology. IV. EXPERIMENTAL EVALUATION A. Experiment Settings We illustrate the efficiency of our weak-point analysis framework using a simple experimental scenario. For the fundamental parameters for resource availability and cost, we specify them as Table II shows. In this table we allocate the failure behavior (MTTR, MTBF) and cost (ColdCost, ActiveCost, RepairCost) for each component comprising the three-tier stack of the specified resources. Resource Server OS hardware maxSize resource1 IHSServer LinuxOS x86Server 10 resource2 WASServer LinuxOS x86Server 10 resource3 WASServer AIXOS POWERServer 10 resource4 DB2Server WindowsOS x86Server 10 TABLE III RESOURCE COMPONENTS BPEL Workflow Application Resource workflow1 EAR Component1 resource1 EARComponent2 resource2 EARComponent3 resource3 DatabaseComponent1 resource4 workflow2 EAR Component1 resource1 DatabaseComponent1 resource2 resource4 TABLE IV WORKFLOW TO RESOURCE MAPPING For the availability parameters, MTTR specifies the mean time to recover after each failure, and MTBF specifies the mean time between failures. The single component availability can be calculated as MTTR/(MTTR+MTBF). The cost parameters specify various costs associated with the components: ColdCost specifies the cost when the com￾ponent is powered off (e.g., as a cold spare), ActiveCost specifies the cost when the component is powered on (e.g., as an active spare). The cost difference may account for the electrical power costs that are incurred only when the No. of Resources Exhaustive Iteration Optimal Solution Calculation 3 1000 56 5 105 171 10 1010 295 15 1015 1516 20 1020 2021 30 1030 3011 TABLE V COMPUTATIONAL COMPLEXITY CHARACTERISTICS hardware is powered on, or the software license cost if an application software component has a usage-based licensing scheme. RepairCost specifies the annual cost to repair the component when the component is down. The annual cost of a component is the sum of the annual cost to operate it in its operating mode and the initial cost of the component annualized by dividing by its useful lifetime in years. The annual cost of a resource is the sum of the annual cost of each component. In this experimental scenario, we create four resources to host relevant web services. Table III depicts the components of the resources in our scenario. We specifiy two BPEL workflows over the application and resource topologies, as table IV shows. B. Performance Evaluation Based on the above settings for the experimental scenario, we can form the utility functions and the workflow-resource mapping matrix. We get the high availability solution through two approaches for comparison: exhaustive iteration and op￾timal solution calculation methods. Through iteration, we generate the optimal solution by simply iterating all possible solutions for availability requirement and getting the solution for minimum overall cost according to various cost modes. For the optimal solution calculation, we use MATLAB [13] to perform the calculation, leveraging the fmincon algorithm with medium-scale optimization [9] in the MATLAB optimization toolbox. Therefore, we can compare the above two different approaches in the following two aspects: solution efficiency and computational complexity. For the solution efficiency, we compare the overall cost for the solutions found by the two different methods, as Fig.6 shows. From Fig.6 we note that the overall cost raises as the availability requirement increases; exhaustive iteration can always achieve the optimal cost when the upper bound for cluster size is set large enough. Our optimal solution achieves the minimum cost of the iteration method most of the time. The reason for any disparity is that we leverage the Lagrange multiplier method, which yields as a solution a vector with fractional values; however, the solution must be a vector of integer values (one cannot deploy 1.2 application servers!). Therefore, the solution we achieve may not be the actual optimal solution, but it is near-optimal and can achieve the minimum overall cost for most of the cases. For the computational complexity, we compare the number of operations to compute the solutions in Table V. In this