The Grid Background and Architecture

1. Keys to success for IT technologies Infrastructure Open Standards

Infrastructure without, no one can use the technology financed  by governments, very first  by industry, after interest increases

Open Standards fast evolution efficiency will be increased more and more integrated

2. How Infrastructure is built railroads, telephones and power development was very complicated started in small regional areas connected to a bigger network to be successful:  acceptance by the users  support by the governments financial promotive

3. Scientific demands computational vs. observational = 1 to 10 remote access to  data  instrumentation data and computation intensive powerful management of resources

Problems of experimental science only few resources worldwide research must be done on site The Grid may allow corporative work between scientists team spread all over the world

4. Business Impact large corporations are global in extent The Grid may link suppliers, manufactures and customers unite a company into a single collaborative team

Infrastructure for the masses only accepted widely if it  becomes transparent to the user  doesn‘t need much knowledge  is highly reliable

The growth of technology development phase  technology itself is important  mainly experts mass adoption  applications, reliability and availability  control returns to experts sinks into background

5. History ARPANET started in the early 1970s experimental network developed important protocols  TCP/IP  notion packet switching

Evolution (1) 1997, GT2  usability and interoperability  solutions for authentication and  resource discovery and access  protocols, APIs and services  GT2 „standards“ not formular not for public review

Evolution (2) 2002, OGSA  extended GT2 concepts and technologies  service-oriented architectures  Web services  provides framework

6. Concepts (1) Analogies to Peer-to-Peer file sharing sharing in terms of The Grid  direct access to computers software data sensors all other resources

6. Concepts (2)  sharing under a certain set of rules  mechanisms for accounting payment (if needed)  in money  in access to user‘s local resources

Concepts (3) achieving various QoS decomposing of integrated infrastructure  into fragmanted systems different resources  shared under certain circumstances pool of resources  members can use under a certain set of rules

7. Architecture seperated into different layers  providing different levels of abstraction lowest level  first step for resources into the Grid core protocol  establishing secure connection between Grid members  shared access to local resources  base for many different applications

Fabric Layer (1) provides local resources  shared over the Grid  computational power  storage  access to sensors translating local protocols to Grid protocols components in this layer will act as proxy objects

Fabric Layer (2) Component provides access to one kind of resources implements resource specific operations general operations for concurrent access show higher-level protocols the resources‘  structure  state  capabilities

Connectivity and Resource Layer narrow neck (hourglass model) based on many maybe different fabric layer technologies base for many very highspread technologies small set of core abstractions and protocols local resources connected to those how ask for them

Communication protocols include  transport  routing  naming defined by the ISO/OSI model TCP/IP protocol stack

Security (Connectivity Layer) one base functionality of this layer secure exchange of data identity verifying  users  resources implementations should  base on existing standards  support single sign-on

Resource Access (Resource Layer) enabling user to interact with remote resources defines protocols for  secure negotiation  initiation  monitoring  control  accounting  payment information protocols managing protocols

Collective Layer protocols and services to provide interactions across collections of resources in many cases built inside the application for examples  weather forecast program  Netsolve/GridSolve 2.0

Application Layer comprises the user applications applications constructed by calling upon services of any layer may introduce different layers

8. Implementations Globus Toolkit Version 2 (GT2) Open Grid Service Architecture (OGSA)

GT2 (1) first standardized implementation Grid protocols at higher levels assumes suitable software on fabric elements  CPU scheduling  file system management  sytem monitoring some components for discovering information about common resource types

GT2 (2) connectivity layer defined by GSI protocols  single sign-on authentication  communication protection  restricted delegation of rights

GT2 (3) implements GRAM protocol provides secure, reliable creation and management of remote computation uses  „gate-keeper“ to initiate  „job manager“ to manage „GRAM reporter“ for local computations

GT2 (4) Monitoring and Discovery Service (MDS-2)  discovering and accessing configuration status information  data model  resource-level protocols  configurable local registry  collective registry

OGSA (1) standardization of core GT protocols use essential Grid functions in different settings service orientated uniform treatment of all network entries

OGSA (2) Grid service  implements standard interfaces behaviors conventions  services are defined by the OGSI

Resources Need some sort of refundment  Financial  other

P2P-Networks: eMule Refundment by Priority Modifier * Waitingtime => Queue Rank

Credit System in eMule Ratio1 = 2*Up / Down Ratio2 = SQRT(Up+2) Modifier = Min{Ratio1,Ratio2} 1<=Modifier<=10 If Up Modifier = 1 If Down = 0 => Modifier = 10

OurGrid CPU-Sharing Round based Problems:  Free Riders  ID Changers

r A (B) = v(B,A)−v(A,B) r A (B): reputation of B relative to A v(B,A): Value of favours B done to A v(A,B): Value of favours A done to B rho: probability of consumer in turn f: probability of freerider epsilon: probability of a freerider getting a resource

r A (B) = v(B,A)−v(A,B) f=0,5

r A (B) = v(B,A)−v(A,B) rho=0,5

r A (B) = v(B,A)−v(A,B) rho=0,5

r A (B) = max{0, v(B,A) - v(A,B)} rho=0,5

r A (B) = max{0, v(B,A) - v(A,B) + log(v(B,A))} rho=0,5

Thank You