Fault Torroance

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  • Topic: Reliability theory, Fault model, The Mistake
  • Pages : 2 (423 words )
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  • Published : December 22, 2012
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1. 因為軟體用硬體的方式複製一模一樣的程式碼屆時發生邏輯上的錯誤無論是原本的程式碼或是複製的程式碼都會發生相同的錯誤,所以用硬體的方式多準備一份是沒有用的。若我要用軟體方法解決此問題,我會想用NVP(N-Version Programming)技術,也就是叫不同人開發不同軟體達到相同功能,最後在check point用voting的方式達到跟hardware voting相同的效果。 2. fsd’( X’) = [X’n+1f(X’) + Xn+1fd(X’)]’

= [X’n+1f(X’)]’[Xn+1fd(X’)]’
= [Xn+1+f’(X’)][X’n+1+fd’(X’)]
= Xn+1 X’n+1 + Xn+1 fd’(X’) +f’(X’)X’n+1+f’(X’) fd’(X’) = 0 + Xn+1 f(X) + f(X’)Xn+1 + fd(X) f(X) ………………. (1) a. if f(X) != fd(X)
(1) = 0 + Xn+1 f(X) + Xn+1 fd’(X) + 0= fsd(X) = fsd’( X’) b. if f(X) = fd(X)
fsd( X) = Xn+1 f(X) + X’n+1 fd(X) = (Xn+1 + X’n+1) f(X) = f(X) f’sd( X’) = f’(X’) = fd(X) = f(X) = fsd( X)
所以當X為任意值,都可證得fsd(X) = f’sd( X’),故其是self-dual 3. 有四種情形是failure,分別是(0,0)->1,(0,1)->0,(1,0)->0,(1,1)->0 a. (0,0)->1
代表兩個輸入中有發生stuck-at-1的情形
我們知道兩個input都無stuck-at-1的機率是(1-P1)^2
又輸入為(0,0)之機率為0.25
故此狀況發生機率為0.25(1-(1-P1)^2)
b. (0,1)->0
代表後面那個input發生stuck-at-0,而且前面的input無stuck-at-1。 機率為 0.25(P0*(1-P1))
c. (1,0)->0
機率同b
d. (1,1)->0
代表兩者皆stuck-at-0,機率為 0.25(P0^2)

總體機率為四者相加,整理後得機率為
0.25 (P1^2 + P0^2 – 2P1P0 + 2P1 + 2P0)
4.
a. Fault- is a physical defect, imperfection or flaw within the hardware or software component. Error- is the manifestation of a fault. Specifically, it is a deviation from accuracy or correctness. Failure- is the deviation from expected actions or services. Fault -> Error -> Failure

It is failure that really annoying users.
b. permanent fault是永久錯誤,一但出現就不會因重複而消失。例如RAM燒壞,重開機是沒用的。 intermittent fault 是偶發性錯誤,有時正常有時會發生錯誤。例如接觸不良。 transient fault 是暫態錯誤,不需理會就會回復穩定與正常狀態,是相對短暫的錯誤。例如剛按下電風扇時轉速過慢。 c. Reliability is the probability to get the right outcome, availability is the probability the service/function can be accessed, and performability is the probability the service do the right thing due to the user’s thought. d. True. Maybe the mistake is as follow:

number >> 3 //shall be 2, the programmer mistype it.
And...
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