The following diagram shows a project with ten tasks. Tasks can only be done in the...

 

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The following diagram shows a project with ten tasks. Tasks can only be done in the order shown in the diagram. For example, Task A must be complete before Task D can begin. Task D must be complete before Task H can begin. Calculate the length of time to complete the project. For parts 2 and 3, this will be a distribution. 1) Find the time to complete this project assuming the given duration of each task is constant. Note, how many possible paths are there through this system? 2) Find the time to complete this project assuming the duration of each task is normally distributed with a mean equal to the given duration and a standard deviation equal to 20% of that duration. (Example: for Task J, mean = 3, standard deviation = 0.6) 3) Find the time to complete this project assuming given the duration of each task has a triangular with the given duration equal to the most likely value, the minimum value is 20% less and the maximum value is 50% more. (Example: for Task J, min = 2.4, most likely = 3, max = 4.5) A 1 day B 2 dys C 3 dys 4 dys E 5 dys F 4 dys 6 dys H 6 dys 2 dys 3 dys The following diagram shows a project with ten tasks. Tasks can only be done in the order shown in the diagram. For example, Task A must be complete before Task D can begin. Task D must be complete before Task H can begin. Calculate the length of time to complete the project. For parts 2 and 3, this will be a distribution. 1) Find the time to complete this project assuming the given duration of each task is constant. Note, how many possible paths are there through this system? 2) Find the time to complete this project assuming the duration of each task is normally distributed with a mean equal to the given duration and a standard deviation equal to 20% of that duration. (Example: for Task J, mean = 3, standard deviation = 0.6) 3) Find the time to complete this project assuming given the duration of each task has a triangular with the given duration equal to the most likely value, the minimum value is 20% less and the maximum value is 50% more. (Example: for Task J, min = 2.4, most likely = 3, max = 4.5) A 1 day B 2 dys C 3 dys 4 dys E 5 dys F 4 dys 6 dys H 6 dys 2 dys 3 dys

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ANSWER AND EXPLANATION Refer to image A B 4 4HETITIONS 1 G B ... View the full answer