# Mba h2040 quantitative techniques for managers 238 a

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MBA-H2040Quantitative Techniquesfor Managers238A0B12C12D12E12F19G23H29I32J45Backward pass:Calculation of Latest Allowable Time of Occurrence of EventsNodeLatest Allowable Time of Occurrence of Node8Maximum time of a path in the network= 617Time for Node 8 - Time for Activity J = 61 -16 = 456Time for Node 7 - Time for Activity I = 45 -13 = 325Time for Node 7 - Time for Activity H = 45 -14 = 314Time for Node 6 - Time for Activity G = 32 - 9 = 233Time for Node 5 - Time for Activity F= 31- 10 = 212Min {Time for Node 3 - Time for Activity B,Time for Node 4 - Time for Activity C,Time for Node 5 - Time for Activity D,Time for Node 6 - Time for Activity E}= Min {21 - 7,23 - 11, 31 - 8,32 - 6}= Min {14, 12,23,26} = 121Time for Node 2 - Time for Activity A= 12- 12 = 0Usingthe above values, we obtain the LatestFinish Times of the activities as follows:ActivityLatestFinish Time(Weeks)J61I45H45G32F31
MBA-H2040Quantitative Techniquesfor Managers239E32D31C23B21A12Calculation of Total Float for each activity:ActivityDuration(Weeks)Earliest StartTimeEarliestFinish TimeLatestStart TimeLatestFinishTimeTotal Float = LatestFinish Time - EarliestFinish TimeA120120120B7121914212C11122312230D81220233111E61218263214F10192921312G9233223320H14294331452I13324532450J16456145610The activities with total float = 0 are A, C, G, I and J. They are the critical activities.Project completion time = 61 weeks.Problem 2:The following are the details of the activities in a project:ActivityPredecessorActivityDuration (Weeks)A-15BA17CA21DB19EB22
MBA-H2040Quantitative Techniquesfor Managers240FC, D18GE, F15Calculate the earliestand latest times, the total float for each activity and the project completion time.Solution:The following network diagram is obtained for the given project.E22BA17D19G1515C21F18Consider the paths, beginning with the start node and stopping with the end node. There arethree such paths for the given project. They are as follows:Path IABEG15172215Time ofthe path = 15 + 17 + 22 + 15= 69 weeks.Path IIABDFG1517191815Time ofthe path = 15 + 17+ 19+ 18 + 15 = 84 weeks.Path IIIACFG15211815Time ofthe path = 15 + 21+ 18+ 15 = 69 weeks.Compare the times for the three paths. Maximum of {69, 84, 69} = 84. We see that the maximum time of a pathis 84 weeks.1654216532154326214563
MBA-H2040Quantitative Techniquesfor Managers241Forward pass:Calculation of Earliest Time of Occurrence of EventsNodeEarliest Time of Occurrence of Node102Time for Node 1 + Time for Activity A =0 + 15 = 153Time for Node 2 + Time for Activity B = 15 + 17 =324Max {Time for Node 2 + Time for Activity C,Time for Node 3 + Time for Activity D}= Max {15 + 21, 32 + 19} = Max {36, 51} = 515Max {Time for Node 3 + Time for Activity E,Time for Node 4 + Time for Activity F}= Max {32 + 22, 51 + 18} = Max {54, 69} = 696Time for Node 5 + Time for Activity G= 69 + 15 = 84Calculation of Earliest Time forActivitiesActivityEarliestStart Time(Weeks)A0B15C15D32E32F51G69Backward pass:Calculation of the Latest Allowable Time of Occurrence of EventsNodeLatest Allowable Time of Occurrence of Node6Maximum time of a path in the network= 845Time for Node 6 - Time for Activity G = 84 -15 = 694Time for Node 5 - Time for Activity F = 69 - 18 = 513Min {Time for Node 4 - Time for Activity D,Time for Node 5 - Time for Activity E}= Min {51 - 19,69 - 22} = Min {32, 47} = 322Min {Time for Node 3 - Time for Activity B,Time for Node 4 - Time for Activity C}= Min {32 - 17, 51 - 21} = Min {15, 30} = 15

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Term
Spring
Professor
brhanu
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