GB
Gabin Burgun
Answers (2)
GB
Answered
Adam Smith's pin factory and Henry Ford's automobile assembly line are examples of
A) Product lines
B) Functionally organized firms
C) Inefficient processes
D) In-line production
A) Product lines
B) Functionally organized firms
C) Inefficient processes
D) In-line production
On Sep 27, 2024
B
GB
Answered
Consider the network described in the table below.
Activity Immediate Predecessor(s) Pessimistic Probable Optimistic J −−15108 K −−987 L J 1065 M J 333 N K, M 951 O K, M 1074 P L, N 1083\begin{array} { | c | c | c | c | c | } \hline \text { Activity } & \begin{array} { c } \text { Immediate } \\\text { Predecessor(s) }\end{array} & \text { Pessimistic } & \text { Probable } & \text { Optimistic } \\\hline \text { J } & - - & 15 & 10 & 8 \\\hline \text { K } & - - & 9 & 8 & 7 \\\hline \text { L } & \text { J } & 10 & 6 & 5 \\\hline \text { M } & \text { J } & 3 & 3 & 3 \\\hline \text { N } & \text { K, M } & 9 & 5 & 1 \\\hline \text { O } & \text { K, M } & 10 & 7 & 4 \\\hline \text { P } & \text { L, N } & 10 & 8 & 3 \\\hline\end{array} Activity J K L M N O P Immediate Predecessor(s) −−−− J J K, M K, M L, N Pessimistic 15910391010 Probable 10863578 Optimistic 8753143 a. Calculate the expected duration of each activity.
b. Calculate the expected duration and variance of the critical path.
c. Calculate the probability that the project will be completed in fewer than 30 time units.
Activity Immediate Predecessor(s) Pessimistic Probable Optimistic J −−15108 K −−987 L J 1065 M J 333 N K, M 951 O K, M 1074 P L, N 1083\begin{array} { | c | c | c | c | c | } \hline \text { Activity } & \begin{array} { c } \text { Immediate } \\\text { Predecessor(s) }\end{array} & \text { Pessimistic } & \text { Probable } & \text { Optimistic } \\\hline \text { J } & - - & 15 & 10 & 8 \\\hline \text { K } & - - & 9 & 8 & 7 \\\hline \text { L } & \text { J } & 10 & 6 & 5 \\\hline \text { M } & \text { J } & 3 & 3 & 3 \\\hline \text { N } & \text { K, M } & 9 & 5 & 1 \\\hline \text { O } & \text { K, M } & 10 & 7 & 4 \\\hline \text { P } & \text { L, N } & 10 & 8 & 3 \\\hline\end{array} Activity J K L M N O P Immediate Predecessor(s) −−−− J J K, M K, M L, N Pessimistic 15910391010 Probable 10863578 Optimistic 8753143 a. Calculate the expected duration of each activity.
b. Calculate the expected duration and variance of the critical path.
c. Calculate the probability that the project will be completed in fewer than 30 time units.
On Sep 23, 2024
(a) See table below.
(b) Tasks J-M-N-P are critical. The sum of their expected durations is 26.00; the sum of their variances is 4.50.
(c) The standard deviation along the path is = 2.12; the probability that Duration < 30 is the probability that z < (30 - 26.00)/2.12 = 1.89. The associated normal curve area is 0.97062.
Task Early Start Early Finish Late Start Late Finish Slack Mean Variance J010.5010.5010.51.361111K085.513.55.58L10.51719.52696.5M10.513.510.513.5030N13.518.513.518.5051.777778O13.520.519265.57P18.52618.52607.51.361111 Project 26 Project 4.5 Std.dev 2.12132\begin{array}{|c|c|c|c|c|c|c|c|}\hline\text { Task } & \begin{array}{c}\text { Early } \\\text { Start }\end{array} & \begin{array}{c}\text { Early } \\\text { Finish }\end{array} & \begin{array}{c}\text { Late } \\\text { Start }\end{array} & \begin{array}{c}\text { Late } \\\text { Finish }\end{array} & \text { Slack } & \text { Mean } & \text { Variance } \\\hline \mathrm{J} & 0 & 10.5 & 0 & 10.5 & 0 & 10.5 & 1.361111 \\\hline \mathrm{K} & 0 & 8 & 5.5 & 13.5 & 5.5 & 8 & \\\hline \mathrm{L} & 10.5 & 17 & 19.5 & 26 & 9 & 6.5 & \\\hline \mathrm{M} & 10.5 & 13.5 & 10.5 & 13.5 & 0 & 3 & 0 \\\hline \mathrm{N} & 13.5 & 18.5 & 13.5 & 18.5 & 0 & 5 & 1.777778 \\\hline \mathrm{O} & 13.5 & 20.5 & 19 & 26 & 5.5 & 7 & \\\hline \mathrm{P} & 18.5 & 26 & 18.5 & 26 & 0 & 7.5 & 1.361111\\\hline &\text { Project } & 26 & & & & \text { Project } & 4.5 \\\hline & & & & & & \text { Std.dev } & 2.12132\\\hline\end{array} Task JKLMNOP Early Start 0010.510.513.513.518.5 Project Early Finish 10.581713.518.520.52626 Late Start 05.519.510.513.51918.5 Late Finish 10.513.52613.518.52626 Slack 05.59005.50 Mean 10.586.53577.5 Project Std.dev Variance 1.36111101.7777781.3611114.52.12132
(b) Tasks J-M-N-P are critical. The sum of their expected durations is 26.00; the sum of their variances is 4.50.
(c) The standard deviation along the path is = 2.12; the probability that Duration < 30 is the probability that z < (30 - 26.00)/2.12 = 1.89. The associated normal curve area is 0.97062.
Task Early Start Early Finish Late Start Late Finish Slack Mean Variance J010.5010.5010.51.361111K085.513.55.58L10.51719.52696.5M10.513.510.513.5030N13.518.513.518.5051.777778O13.520.519265.57P18.52618.52607.51.361111 Project 26 Project 4.5 Std.dev 2.12132\begin{array}{|c|c|c|c|c|c|c|c|}\hline\text { Task } & \begin{array}{c}\text { Early } \\\text { Start }\end{array} & \begin{array}{c}\text { Early } \\\text { Finish }\end{array} & \begin{array}{c}\text { Late } \\\text { Start }\end{array} & \begin{array}{c}\text { Late } \\\text { Finish }\end{array} & \text { Slack } & \text { Mean } & \text { Variance } \\\hline \mathrm{J} & 0 & 10.5 & 0 & 10.5 & 0 & 10.5 & 1.361111 \\\hline \mathrm{K} & 0 & 8 & 5.5 & 13.5 & 5.5 & 8 & \\\hline \mathrm{L} & 10.5 & 17 & 19.5 & 26 & 9 & 6.5 & \\\hline \mathrm{M} & 10.5 & 13.5 & 10.5 & 13.5 & 0 & 3 & 0 \\\hline \mathrm{N} & 13.5 & 18.5 & 13.5 & 18.5 & 0 & 5 & 1.777778 \\\hline \mathrm{O} & 13.5 & 20.5 & 19 & 26 & 5.5 & 7 & \\\hline \mathrm{P} & 18.5 & 26 & 18.5 & 26 & 0 & 7.5 & 1.361111\\\hline &\text { Project } & 26 & & & & \text { Project } & 4.5 \\\hline & & & & & & \text { Std.dev } & 2.12132\\\hline\end{array} Task JKLMNOP Early Start 0010.510.513.513.518.5 Project Early Finish 10.581713.518.520.52626 Late Start 05.519.510.513.51918.5 Late Finish 10.513.52613.518.52626 Slack 05.59005.50 Mean 10.586.53577.5 Project Std.dev Variance 1.36111101.7777781.3611114.52.12132