First textbox is
"Probability that fault is permanent". This is defined as S in the formulas
shown below.. As we know failure may be either transient or permanent.
It is a fact that
Prob(failure is transient) + Prob(failure is permanent) = 1.
Second is : "Average Number of Instructions per program". This is defined as 'L'
Third is : "Required setup time, when a permanent failure happens." This is defined as delta1.
Fourth is : "Required Diagnosis and repair time when a permanent failure happens". This is defines as 'delta2'
Second ScrollPane is also an input field which consist of each instructions failure, recover rates. Also time and frequnce values. Key point here is total frequnce for all instructions should be 1.
Finally output :
Outputs are in an order way: P ^{( C ) }, P ^{(RB1 ) },
P ^{(RB2 ) }and P ^{(PF ) } , tau1, tau2 that
we tried to formulate below.
We proceed now to calculate these conditional probabilities. First:
P_{i} ^{( C ) }= P_{0}(lambda_{i }, T _{i }) = e ^{(- lambdai * Ti )}If we denote T¬ =^{ }( f_{1}*T_{1} + f_{2}* T_{2} +...+ f_{N}* T_{N})P_{i} ^{(RB1 ) }= P¯_{00}( (1-S) * lambda_{i}, µ_{i}, t_{i}, t_{i}) * P_{0}(S*lambda_{i }, T _{i }) * (P_{i}^{( C )} / M) * ( ( (1 -P^{ C })^{M })/(1 -P^{ C }) )
P_{i} ^{(RB2 ) }= P¯_{00}( (1-S) * lambda_{i}, µ_{i}, t_{i}, t_{i}) * P_{0}(S*lambda_{i }, T _{i }) * P_{i}^{(RB1 )}
P_{i} ^{(PF ) }= 1 - P_{i}^{( C ) }- P_{i} ^{( RB1 ) }- P_{i} ^{(RB2 )}
Mean time required to successfully execute an instruction with a one rollback will be :
tau1 = T¬ + P ^{(RB1 ) }* (delta1 + ((M+1)/2) * (T¬) ) + P ^{(PF )}* (delta1 + ((M+1)/2) * (T¬) + delta2 +(L+1)*(W/2) )
Mean time required to successfully execute an instruction with two
rollback will be :
tau2 = T¬ + P ^{(RB1 ) }* (delta1 + ((M+1)/2) * (T¬) )+ P ^{(RB2 )}* ( (2* delta1) + (M+1)* (T¬) )+ P ^{(PF )}* ( (2*delta1)+(M+1)* (T¬) + delta2 + (L+1)*(W/2) )