Dependability state model

<h2 id="example">Example</h2>

<p>A redundant computer system consist of identical two-compute nodes, which each fail with an intensity of 
  
    
      
        λ
      
    
    {\displaystyle \lambda }
  
. When failed, they are repaired one at the time by a single repairman with negative exponential distributed repair times with expectation 
  
    
      
        
          μ
          
            −
            1
          
        
      
    
    {\displaystyle \mu ^{-1}}
  
.
</p>
<ul><li>state 0: 0 failed units, normal state of the system.</li>
<li>state 1: 1 failed unit, system operational.</li>
<li>state 2: 2 failed units. system not operational.</li></ul>
<p>Intensities from state 0 and state 1 are 
  
    
      
        2
        λ
      
    
    {\displaystyle 2\lambda }
  
, since each compute node has a failure intensity of 
  
    
      
        λ
      
    
    {\displaystyle \lambda }
  
. Intensity from state 1 to state 2 is 
  
    
      
        λ
      
    
    {\displaystyle \lambda }
  
.
Transitions from state 2 to state 1 and state 1 to state 0 represent the repairs of the compute nodes and have the intensity 
  
    
      
        μ
      
    
    {\displaystyle \mu }
  
, since only a single unit is repaired at the time.
</p>
<h3>Availability</h3>
<p>The asymptotic <a href="/facts/Availability/l8SGSyyf">availability</a>, i.e. availability over a long period, of the system is equal to the probability that the model is in state 1 or state 2. 
</p><p>This is calculated by making a set of linear equations of the state transition and solving the linear system. 
</p><p>The matrix is constructed with a row for each state. In a row, the intensity into the state is set in the column with the same index, with a negative term.
</p>

A
            
              0
            
          
        
        =
        
          
            [
            
              
                
                  0
                
                
                  −
                  μ
                
                
                  0
                
              
              
                
                  −
                  λ
                
                
                  0
                
                
                  −
                  μ
                
              
              
                
                  0
                
                
                  λ
                
                
                  0
                
              
            
            ]
          
        
        .
      
    
    {\displaystyle \mathbf {A_{0}} ={\begin{bmatrix}0&-\mu &0\\-\lambda &0&-\mu \\0&\lambda &0\end{bmatrix}}.}

<p>The identities cells balance the sum of their column to 0:
</p>

A
            
              1
            
          
        
        =
        
          
            [
            
              
                
                  (
                  λ
                  )
                
                
                  −
                  μ
                
                
                  0
                
              
              
                
                  −
                  λ
                
                
                  (
                  λ
                  +
                  μ
                  )
                
                
                  −
                  μ
                
              
              
                
                  0
                
                
                  −
                  λ
                
                
                  (
                  μ
                  )
                
              
            
            ]
          
        
        .
      
    
    {\displaystyle \mathbf {A_{1}} ={\begin{bmatrix}(\lambda )&-\mu &0\\-\lambda &(\lambda +\mu )&-\mu \\0&-\lambda &(\mu )\\\end{bmatrix}}.}

<p>In addition the equality clause must be taken into account: 
</p>

∑
          
            n
          
        
        
          P
          
            n
          
        
        =
        1.
      
    
    {\displaystyle \sum _{n}P_{n}=1.}

<p>By solving this equation, the probability of being in state 1 or state 2 can be found, which 
is equal to the long-term availability of the service.
</p>
<h3>Reliability</h3>
<p>The reliability of the system is found by making the failure states absorbing, i.e. removing all outgoing state transitions.
</p><p>For this system the function is:
</p>

R
        (
        t
        )
        =
        
          e
          
            −
            λ
            t
          
        
        
      
    
    {\displaystyle R(t)=e^{-\lambda t}\,}

<h2 id="criticism">Criticism</h2>
<p>Finite state models of systems are subject to <a href="/facts/State_explosion_problem/TN8pUrtI">state explosion</a>. To create
a realistic model of a system one ends up with a model with so many states that it is infeasible to solve or draw the model.
</p>

<h2 id="references">References</h2>

<ol>
<li id="fn:1"><p>Bjarne E. Helvik (2007). Dependable Computing Systems and Communication Networks. Gnist Tapir. <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li>
</ol>

Dependability state model open-in-new

Dependability state model