╨╧рб▒с>■  &■   ■   %                                                                                                                                                                                                                                                                                                                                                                                                                                                 Г╠с┴┐└т\pWalter Wawruck BфЬ=X; +О8X@Н"╖┌1╚ РArial1╚ ╝Arial1╚ РArial1╚ ╝Arial1╚ РArial1╚ РTimes New Roman1Ё РTimes New Roman1Ё ╝Times New Roman1Ё РTimes New Roman"$"#,##0;\-"$"#,##0"$"#,##0;[Red]\-"$"#,##0"$"#,##0.00;\-"$"#,##0.00!"$"#,##0.00;[Red]\-"$"#,##0.003*0_-"$"* #,##0_-;\-"$"* #,##0_-;_-"$"* "-"_-;_-@_-*)'_-* #,##0_-;\-* #,##0_-;_-* "-"_-;_-@_-;,8_-"$"* #,##0.00_-;\-"$"* #,##0.00_-;_-"$"* "-"??_-;_-@_-2+/_-* #,##0.00_-;\-* #,##0.00_-;_-* "-"??_-;_-@_- д0.000000 е0.00000 ж0.0000з0.000и0.0рї  └ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ рї  Ї└ р └ р+ї  °└ р)ї  °└ р,ї  °└ р*ї  °└ р ї  °└ р+8└ @АIА@ р(└ р"8└ IА@ р (└ IА@ р └ р"8└ H@ р (└ H@ р"8└ @АH@ р (└ @АH@ р"└ р (└ H@ рж ,└ H@ р (└ р 8└ @АIА@ р ,└ IА@ рж ,└ IА@ р ,└ H@ р ,└ @АH@ рж ,└ @АH@ р!└ р!└ р!└ р"└ УА УА УА УА УА УА Тт8            ААААААААА└└└ААААА А `  └арр`А ААА└└└ А      ААААА ╠ i  ╠ ╠  Щж╩Ё╠Ь╠╠Щ ууу3f 3╠╠3Щ3ЩЩ3Щf3ЩffffЩЦЦЦ33╠3ff333f3Щ3f33ЩBBBЕ#Solution Г╠ =ч :ш(  d№йё╥MbP?_*+ВА% МБ┴QP&C&"Times New Roman,Bold"&12PMP Exam Preparation Solution to PERT Sample ProblemХФ&L&"Times New Roman,Regular"&8PERT Sample.xls&C&"Times New Roman,Bold"&8Walter A. Wawruck Management Consultant&R&"Times New Roman,Regular"&8Page &PГДMHP LaserJet 6Lman,Regular"&8PERPФИяъ odXXWalter A. Wawruck Management CoLPT1:22√0 SP51HP LaserJet 6Lб"dXXр?р?U} ╢+} $} ╢} } Т }  }  =  ■T0А ;b ;b ;b ;┴Б ; ; ;b ;b ;i ; 0 ; ; ; ¤hТHВ ;пC ;dТ ╧ ;М ;b ;Э0 ;ЄП ;╪Т╕В ;пМ ;Й ;Н( ;mm ;b ;┼0╟ ;b ;bМ ; ; Activity Name Best Case Most Likely Worst CaseExpected DurationStandard DeviationVarianceCritical DurationsCritical Variances  Probability A╜@"@5@$@ ╝!  L└¤L└■L└ @■╝  L└■L└№"@ ╝  L└ L└ $@ D└"@ D└  B╜8@>@N@A@ @ B@ ╛   C╜8@А@@K@АA@ @ 9@ ╛   D╜5@8@А@@9@ @ @ ╛   E╜"@(@;@,@¤@ "@ ╛   F╜5@>@АF@?@ @ 0@ ?@ D└0@ D└  G╜@(@2@(@ @ @ (@ D└@ D└  H╜;@B@АO@АC@ @ B@ АC@  D└B@  D└  I╜ @2@>@ 2@   @   0@   2@  D └ 0@  D └ ╛  ╛  Totals: "@  D └A ! А[@   %└ └w! @T@   %└ └w% ° ├   ▀? D └n A%╛  ╛ !!!! ╛"D Total Duration  "Probability of D or less╛~ #Y@!% $@ЗRyм┴? D└n A%(The General Solution: ╛~ %А[@!% ° ├   ▀? D└n A%╛#&а└жБ]@   D└ n A'!~  'ш?╛)))))***e)]The expected duration of each activity is calculated as (best + (4 x most likely) + worst)/6.╛)))))***╛))))))***i)aThe total of the expected durations is calculated for each path through the network which can be ╛)))))***i)atraced from start to finish, proceeding always from left to right. The three possible paths are: ╛)))))***f)^A-B-C-I, A-D-E-H-I, and A-F-G-H-I. The critical path is the path through the network with the╛)))))***N)Fgreatest total expected duration; in this case 110 days for A-F-G-H-I.╛)))))***╛))))))***o)gStandard deviations and variances need be calculated only for the activities which lie on the critical ╛)))))*** )path.╛)))))***╛))))))***l)dThe standard deviation of the duration of each critical activity is calculated as (worst - best)/6.╛)))))***k)cThe variance of the duration of each critical activity is calculated as the square of the standard ╛)))))***#)deviation for the activity.╛)))))***╫D╧l¤$ЦЦЦЦ╠╠╠╠├bYzWГЗЗДlН+КЙ ;T0! ;b" ;b# ;b$ ;┴Б% ;& ;' ;b( ;b) ;i* ; 0+ ;, ;- ;. ;hТHВ/ ;пC0 ;dТ ╧1 ;М2 ;b3 ;Э04 ;ЄП5 ;╪Т╕В6 ;пМ7 ;Й9  Н(:  mm;  b<  ┼0╟╛ ))))))***n!)fThe sum of the variances along the critical path is 81.The square root of 81, 9 days, is the standard ╛!)))))***2")*deviation of the total expected duration.╛")))))***╛#))))))***g$)_The Z value, to be used in reading probability values from the table for a normal distribution,╛$)))))***`%)Xis (Specified Total Duration - Expected Total Duration)/(9 days).The table for a normal ╛%)))))***╓f&)[distribution can be found in Meredith and Mantel, Project Management, A Managerial Approach1 ╛&)))))***╓R')Apage 734 in the Third Edition, or page 576 in the Fourth Edition. 1 @╛')))))***╛())))))***')(Solutions to Specific Questions╛))))))***╛*()))))***~ +)Ё?V+)NThe expected total duration is 110 days for path A-F-G-H-I as described above.╛+))))***╛,))))))***~ -)@e-)]For a specified duration of 100 days, the Z value is (100 - 110)/9 = -1.111. The probability ╛-*******.)a.)Yfrom the table of 0.8665 must be subtracted from 1.000 for a negative Z value, giving an ╛.*******/)!/)answer of close to 0.1335╛/*******╛0))*******~ 1)@e1)]For the Expected Total Duration, the Z value is zero, and the probability, from the table is ╛1*******2)a2)Y0.5000. This makes sense, since the Normal distribution is symmetrical , and the mean is ╛2*******3[3Slocated at the center of gravity of the area under the curve. The center of area is4.4&coincident with the center of gravity.╛ 5~ 6)@c6[For a probability of 0.750, the table indicates a Z value of about 0.675. The corresponding7K7CSpecified Total Duration is 110days + 9days(.675) = 11 6.075 days.9!9*Revised to April 27, 2005g:_Thank you to Diane Ehman of Regina for pointing out an error in the description of the formula =;5for calculating the expected duration of an activity.#<The error is now corrected.╫<f МPЕ~ДpEАПЗGПЗi<uY/kA=X; +О8X> ╢л" p№                               ■  рЕЯЄ∙OhлС+'│┘0░@H`x Р ЬифWalter WawruckWalter WawruckMicrosoft Excel@pP╜Щ┐@+░сc┐■  ╒═╒Ь.УЧ+,∙о0Ф8@ LT \ qфicr  Solution  Worksheets ■   ■    !"#$■   ¤   ■                                                                                                                                                                                                                                                                                                                                                                       Root Entry        └F■   Book             Й)SummaryInformation(    DocumentSummaryInformation8