In this chapter, of infinite-state we focus on the solution with a finite, but possibly specification. 4 (birth-death state space, once they have been generated CTMCs from a high-level The solution has been discussed in Chapter queueing models), Chapter 8 (quasi-birth-death queueing models) and Chapter models) and will be discussed further in Chapter 17. Finite haviour. | Performance of Computer Communication Systems A Model-Based Approach. Boudewijn R. Haverkort Copyright 1998 John Wiley Sons Ltd ISBNs 0-471-97228-2 Hardback 0-470-84192-3 Electronic Chapter 15 Numerical solution of Markov chains ALTHOUGH a CTMC is completely described by its state space Z. its generator matrix Q and its initial probability vector p 0 in most cases we will not directly specify the CTMC at the state level. Instead we use SPNs or other high-level specification techniques to specify the CTMCs. In this chapter we focus on the solution of CTMCs with a finite but possibly large state space once they have been generated from a high-level specification. The solution of infinite-state CTMCs has been discussed in Chapter 4 birth-death queueing models Chapter 8 quasi-birth-death queueing models and Chapter 10 open queueing network models and will be discussed further in Chapter 17. Finite CTMCs can be studied for their steady-state as well as for their transient behaviour. In the former case systems of linear equations have to be solved. How to do this using direct or iterative methods is the topic of Section . In the latter case linear systems of differential equations have to be solved which is addressed in Section . Computing steady-state probabilities As presented in Chapter 3 for obtaining the steady-state probabilities of a finite CTMC with N states numbered 1 through A we need to solve the following system of N linear equations pQ Q 5 1- 15-9 iel where the right part is added to assure that the obtained solution is a probability vector. We assume here that the CTMC is irreducible and aperiodic such that p does exist and is independent of p 0 . Notice that the left part of in fact does not uniquely define 330 15 Numerical solution of Markov chains the steady-state probabilities however together with the normalisation equation a unique solution is found. For the explanations that follow we will transpose the matrix Q and denote it as A.
