Department of Civil Engineering-I.I.T. Delhi CEL 899: Environmental Risk Assessment HW5 Solution

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1 Departmet of Civil Egieerig-I.I.T. Delhi CEL 899: Evirometal Risk Assessmet HW5 Solutio Note: Assume missig data (if ay) ad metio the same. Q. Suppose X has a ormal distributio defied as N (mea=5, variace= ) (ote the otatio used here), aswer the followig (i) Calculate mea, media, stadard deviatio, variace (ii) Calculate P(X=3), P(X<0)? (iii) Calculate P (<X<8), i.e., probability of X obtaiig betwee ad 8? Hit: P(X<X<X) = P(X<X)-P(X<X)? (iv) Calculate 5 th, 50 th, 90 th, 95 th, ad 99 th percetile values. Also calculate 90% cofidece iterval value (= 95 th percetile value 5 th percetile)? Aswer: (i) Mea = 5; Stadard deviatio =; Variace = = 4 (Aswer) At media value, 50% of X values are lower tha this value ad 50% of X values are higher tha this. So, P(X< Media) = P(X> Media) =0.50 So if we determie 50 th percetile value, it will be media value for radom variable X with N (mea=5, variace= ). Give that P(X<media) = 0.5 or P([(X-5)/]< [(media-5)/]) =0.5 here Z=[(X-5)/] ad Z * =[(media-5)/] So we have P(Z<Z*)=0.5 where Z has a stadard ormal distributio with mea 0 ad stadard deviatio equal to. Now look at your stadard ormal distributio table to fid Ф (Z * )= P(Z<Z * ) = 0.5. The table gives Z * = 0 i.e., Z * =[(media-5)/]=0 => media = 5 (Aswer) Note: For ormal distributio, mea = media. But I showed you this approach for calculatig ay percetile value. Here we did the calculatio for 50 th percetile value (i.e, media). Similarly, you ca calculate all 5 th, 50 th, 95 th percetile values. (ii) P(X=3) =? Put X=3 i the followig formula of probability fuctio for ormal distributio. X µ f ( x, µ, σ ) = exp 0.5 σ π σ => 3 5 f (3;5,) = exp 0.5 π => P(X=3) = f(3; 5,) = 0. (Aswer) = (/5.0) * exp(-0.5) = (/5.0)*65 Now, P(X<0)=? First get Z value which has a stadard ormal distributio with mea 0 ad stadard deviatio equal to. or P([(X-5)/]< [(0-5)/]) =? here Z=[(X-5)/] ad Z * =[(0-5)/] =.5 So we have P(Z<.5)=? Now look at your stadard ormal distributio table to fid Ф (.5) as Z*=.5 is calculated ad kow here.

2 => P(Z<Z * ) = So, P(X<0) = P(Z<Z * ) = (Aswer) (iii) P(<X<8) = P(X<8)-P(X<) First get Z value which has a stadard ormal distributio with mea 0 ad stadard deviatio equal to. P(X<8)= P([(X-5)/]< [(8-5)/]) = P(Z <.5] = Ф (.5) =0.933 P(X<)= P([(X-5)/]< [(-5)/]) = P(Z < -.0] = Ф (-.0) (or P(Z<-)) Here, remember the formula: P(Z<Z * )=P(Z>-Z*) (because of symmetry of ormal distributio about mea). so, P(Z<-) = P(Z>)=-P(Z<) So, Ф (-.0) =- Ф (.0) = = 0.07 So, P(<X<8) = P(X<8)-P(X<) = = => (i.e., 0.905*00 = 9.05% of times X will lie betwee ad 8). (aswer) Note: Here 9.05% cofidece iterval is give by (, 8). (iv) Calculate all percetile values as we calculated for 50 th percetile values i part (i). Say 5 th percetile value = X* i.e., P(X<X * )=0.05 P(X<X*)= P([(X-5)/]< [(X*-5)/]) = P(Z < Z*) = 0.05 [here, Z*= (X*-5)/] As P (Z<Z*) = -P(Z>Z*) 0.05=-P(Z>Z*) P(Z>Z*)=0.95 Now, as P(Z<Z * )=P(Z>-Z*) (because of symmetry of ormal distributio about mea), it meas that P(Z>Z*)= P(Z<-Z*). As we kow from table that for P(Z>Z*)=0.95= P(Z<-Z*) Here, Ф (Z)= 0.95 happes for Z lyig betwee.64 ad.65. So usig liear iterpolatio, Z comes out to be.645. So, it meas that Z* =.645 Z*=-.645 Now as Z*= (X*-5)/ = X* = *(-.645)+5 =.7 So 5 th percetile value =.7 Q. Look at the followig failure data: Failure o Operatig time (days) (say X) Usig the operatig time data, aswer the followig (i) Calculate mea, media, stadard deviatio, variace, miimum, maximum operatig time values (ii) Develop a frequecy histogram for operatig time before failure (radom variable X) usig the biig approach (called as probability mass fuctio, if X is discrete or probability desity fuctio if X is cotiuous). Develop this histogram for both cases. (iii) Develop cumulative distributio fuctios for two cases: For X as a discrete radom variable) ad for X as a cotiuous variable. Plot. (iv) Usig the cumulative distributio fuctio for X as a discrete, calculate P(X=0 days), P(X< 0 days). Repeat this calculatio for the case whe X is a cotiuous variable. (v) Calculate operatig time without failure with 95 % cofidece (i.e., calculate 95 th percetile value) usig the developed cumulative desity fuctio whe X is a cotiuous variable case. Also calculate 90% cofidece iterval value (= 95 th percetile value 5 th percetile)? Note: 90% cofidece iterval value idicates that your operatig time lies betwee these two eds 90% of the observatio times.

3 Aswer: (i) First arrage the data i icreasig order. Arraged data: X (i days): 40, 98, 65, 35, 3, 48, 547, 70, 95, 340 (total N=0) Miimum = 40 days, maximum = 340 days Mea = N = 0 X = x i N = 0 =(/0) *[ ] => mea = 48 days As N is eve umber here, media = average of two middle terms = ½ (5 th term + 6 th term) Media =/*(3+48)=370 days Stadard deviatio (σ) σ = σ = N = 0 0 ( X X ) i N [( 40 48) + ( 98 48) +...( ) ] Table. Operatig time (days) (say X) (X-mea) Total sum of differet (X-mea) terms = So, stadard deviatio (σ) = (59406/9) 0.5 = (69934) 0.5 = 4. days Variace = σ = (4.) = days (ii) Table. Failure o. operatig frequecy (X/total sum) F(X)= P(X<x) time (days) [i.e. P(X=x)) 40 40/480 40/ /480 38/ / / / / / / /480 78/ /480 85/ / / / / /480.0 total sum 480 3

4 .00 Probability (or frequecy) histogram P(X=x) X (operatig time, days) Fig. Probability (or frequecy) histogram (see attached spreadsheet) (X=discrete; thus poits are ot coected here) Probability (or frequecy) histogram (X=cotiuous).00 P(X=x) X (operatig time, days) Fig. Probability (or frequecy) plot (see attached spreadsheet) (X=cotiuous; thus poits are coected here) (iii).00 Cumulative Probability Histogram P(X=x) X (operatig time, days) Fig.3 Cumulative Probability Histogram (see attached spreadsheet) (X=discrete; thus poits are ot coected here) 4

5 Cumulative Probability Histogram (X=cotiuous).00 P(X=x) X (operatig time, days) Fig.4 Cumulative Probability Plot (see attached spreadsheet) (X=cotiuous; thus poits are coected here) (iv) For X: discrete variable. P(X=0 days)=? Refer Table. As X is discrete ad as per the data, it does ot have ay value at X=0, so P( X=0 days) =0 (aswer) Now, P(X< 0 days)= P(X=98) +P(X=40) (as X is smaller tha 0 days) = (98/480)+(40/480)=38/480 (aswer) For X: cotiuous variable. P(X=0 days)=? Refer Table. As X is cotiuous variable, it ca take ay value icludig X=0. So X=0 lies betwee 98 days ad 65 days. Do liear iterpolatio to determie frequecy of gettig X=0 days (or P(X=0 days)): (0-98)/(65-98) = (P(X)-98/480)/(65/480-98/480)=> so, P(X=0) = 0/480 (aswer) Now, P(X< 0 days)=? Refer Table. As X is cotiuous variable, assume liearity holds for F(x) fuctio as well (see Figure 4). So F(X<=0) lies betwee F(X<=65) ad F(X<=98). Do liear iterpolatio to determie cumulative probability of gettig operatig time < = 0 days: (0-98)/(65-98) = (F(0)-38/480)/(303/480-38/480) ()/(67) = (F(0)-38/480)/(65/480) => (/67) *(65/480) = F-(38/480) => F (X=0)= P(X<=0) = 9.5/480 (aswer) (v) From table : Table 3. Failure o. operatig time (days) F(X)= P(X<x) 40 40/480 = /480 = /480= /480= /480= /480= /480= /480= /480= =.0 5

6 Assumig X as a cotiuous variable ad its cumulative probability desity fuctio is give by F(X) as per Table 3, first determie 5 th percetile value ad 95 th percetile ad the calculate 90% cofidece iterval. 5 th percetile value = that X * for which F(X * )=0.05 From table 3, it lies betwee F(98) =0.087 ad F(65)= Say after liear iterpolatio, 0.05 value of F(X*) comes for X * = 30 days (assumed for illustratio; calculate i homework ad exam). Similarly, F(X**) =0.95 comes out to be 00 days (say) (agai assumed; you should calculate it). So 90% cofidece iterval value for operatig time before failure (days) = 95 th percetile value -5 th percetile value =00 days-30 days = 070 days (aswer) 90% cofidece rage: (5 th percetile value, 95 th percetile value) = (30 days, 00 days) 6

7 Departmet of Civil Egieerig-I.I.T. Delhi CEL 899: Evirometal Risk Assessmet st Semester 0-3 HW 6 Solutio Q. Look at the followig failure data: Failure o Operatig time (days) For the operatig time data (Variable =X), fit a Weibull failure model for operatig time (i.e., time-tofailure) ad determie values of z(t), f(t), F(t), ad R(t)? [0 poits] solutio Aswer: Table. Ordered shifted data ad calculated media-based F(t) values usig give observed data (total umber =0) Failure No. Ordered Operatig time without failure (days) (say t * = t-0.5) Ordered Shifted Operatig time without failure (days) Order Cumulative failure distributio usig media based formula (here formula F(t) = (order-0.3)/(total umber+0.4) As we have shifted the data, we have effectively Z(t)= k(t * ) m So for this expressio, from otes, R(t*) = exp[-k(t*) m+ /(m+)] Thus, F(t*)=-R(t*) = - exp[-k(t*) m+ /(m+)] or -F(t*)= exp[-k(t*) m+ /(m+)] Now we eed to determie k ad m values. Take l o both sides=> l [-F(t*)]=-[k(t*) m+ /(m+)] or - l [-F(t*)]=[k(t*) m+ /(m+)] if we assume (m+) = β ad [k/(m+)]=α, the or - l [-F(t*)]= α (t*) β or l[l(/(-f(t*))] = l α + β l (t*) So if we have compare with Y= a+bx here, X=l (t*); Y= l[l(/(-f(t*))] ; a= l α ; b= β Now calculate values of X ad Y ad the use formula to calculate values of a ad b. 7

8 b = X iyi X i X i X i Y i a = Yi b X ad [ ] [ ] i Table. Order Cumulative failure distributio usig media based formula /[-F(t*)] Y X (=l(t*) D E F G H (XY) X sum [ (/0) *[ 5.3* 57.57] ] [ (/0) * (57.577) ] b = =0.8555/0.589= a=(/0)*[ *57.577]= b a alpha =exp(a)= beta =b= m =beta-= k =alpha*beta=6457 So, F(t*)= - exp[-k(t*) m+ /(m+)] =- exp[-6(t*).0308 /(.0308)] (Aswer) 8

9 F(t*) Shifted operatig time (days) F(t*) empirical F(t*) fitted Figure. Plots of empirical ad fitted CDF values with shifted operatig time i days Prob (t>=500 days) =-Prob (t<500 days) => - F( =499.5 days) So Prob (t>=500 days) = - (- exp[-6(499.5*).0308 /(.0308)]) = = (Aswer) Q3. Plot values of f (t), F (t), ad R (t) for time = 0 to 00 days for λ =0/day, /day, ad 90/day ad commet o effect of differet λ values o reliability of the system defied by expoetial failure fuctio. [5 poits] solutio f(t) Lambda=0/day Lambda=/day Lambda=90/day 9

10 f(t) Lambda=0/day Lambda=/day Lambda=90/day Figure. f (t) as a fuctio of time for differet λ values. Commet: Here, chaces of failure for a give operatig time are small, irrespective of λ values R(t) Lambda=0/day Lambda=/day Lambda=90/day 0

11 .0.00 R(t) Lambda=0/day Lambda=/day Lambda=90/day Figure. R (t) as a fuctio of time for differet λ values. Commet: Here, reliability is highest for λ =0/day ad lowest for λ =90/day as i earlier o failure per day ad o latter case, 90 failures per day. As operatig time icreases, R (t) decreases faster for λ =90/day compared to λ =/day as ca be see from part of Fig.. It also shows the effect of umber of failures per day o reliability as ca also be see by lookig at the formula ad calculatig value of d{r(t)}/d λ ad the see the slops values F(t) Lambda=0/day Lambda=/day Lambda=90/day

12 .0.00 F(t) Lambda=0/day Lambda=/day Lambda=90/day Figure 3. F(t) as a fuctio of time for differet λ values. Commet: Here, ureliability (or total umber of failures up to give time period) behaves complimetary to behavior of reliability curves. Additioal: F(t) or R(t) Lambda=0/day_F(t) Lambda=0/day_R(t) Lambda=90/day_F(t) Lambda=90/day_R(t) Figure 4. Relatioship betwee R (t) ad F (t). Commet: This figure shows iterrelatioship betwee reliability ad ureliability with operatig time for two extremes of λ values, idicatig the sesitivity of these parameters o λ values. Q8. If two similar equipmets are purchased from compay A ad compay B (deoted as product A ad product B, hereafter). As per the followig iformatio, calculate reliability of these two equipmets for 5 hours of operatio (i.e., 5 hours of operatio without failure)? [ =5 poits] Parameter Product A Product B

13 Distributio of times-to-failure for products (i.e., Failure desity fuctio) Mea life of product (i.e., mea time betwee two successive failures)(hours) Expoetial Normal 00 hours mea =0 hours; stadard deviatio = 40 hours Aswer: Product A: Reliability : R(t) A = exp(-λ A *t) Give that mea time betwee two successive failures =00 hours, i.e., oe failure i every 00 hours, thus λ A =/00 (/hour) For 5 hours of operatio, Reliability R(t) A = exp(-0.0*5) = (i.e., 86.07% reliable) Product B: Reliability : R(t) A =-F(t) here f(t) = distributio for time to failure is ormal, so first we eed to determie CDF value (i.e., F(5 hours) ad the we ca calculate R(t) value ad the F(t) = Prob(t<5 hours)=? Here, Z= (t-0 hours)/40 hours => Z=(5-0)/40=-.65 So, F(t) = Prob(t<5 hours)= Prob (Z<-.65) = Ф (-.65) Here, remember the formula: P(Z<Z * )=P(Z>-Z*) (because of symmetry of ormal distributio about mea). so, P(Z<-.65) = P(Z>.65)=-P(Z<.65) So, Ф (-.65) =- Ф (.65) =? Values from the Table are: Ф (.6) = ad Ф (.64) = Calculate Ф (.65)=(0.9956)+ [(/4)*( )]= So, Ф (-.65) = =433 = F(5 hours) Thus, R(5 hours) =-F(5hours) = -433= (i.e., % reliable) 3

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