NETWORK: PERT $$ Ex$.1$ SPRING $24-$ KEY Consider the following table:

ACTIVITY

A

B

C

D

E

F

G

PREC. ACT.

$---$

$---$

$---$

A

B

C

D$,$ E$,$ F

TIME ESTIMATES $($days$)$

To

$3$

$2$

$2$

$3$

$3$

$2$

$4$

Tml

$5$

$4$

$3$

$4$

$5$

$4$

$5$

Tml

$7$

$6$

$4$

$5$

$13$

$6$

$6$

Te

$5$

$4$

$3$

$4$

$6$

$4$

$5$

$\backslash$sigma

$(6-2)/6=(2)/(3)$ TO $4/9$

$(13-3)/6=10/6$ TO $25/9$

$(6-4)/6=(1)/(3)$ TO $1/9$

$1)$ Construct a PERT network.

$3223324$

$5746$

$34$

$45$

$513$

$46$

$56$

G $(10,15)5(10,15)$

$5(4-2)/6=1/3$

$4(6-2)/6=2/34/9$

$3(6-2)/6=2/3$

$4(5-3)/6=1/3$

$6(13-3)/6=10/625/9$

$4(6-2)/6=2/3$

$5(6-4)/6=1/31/9$

A $(0,5)$

$5(1,6)$

B $(0,4)4(0,4)$

C$(0,3)3(3,6)$

D $(5,4(6,10)$

E $(4,10)6(4,10)$

F $(3,7)4(6,10)$

$9)$

Te$=1*$t$0.+4*$tml $+1*$tp $6$

$*$ExpectedTimes:Ate $=5$; Bte $=4$; Cte $=3$; Dte $=4$; Ete $=6$; Fte $=4$; Gte $=52)$ By using ES$,$ EF$,$ LF$,$ LS$,$ find the slack of each activity and the critical path.

Slack $=$ LF $$ EF or LS $$ ES Activity A: Slack $=65=1$ Activity B: Slack $=44=0$ Activity C: Slack $=63=3$ Activity D: Slack $=109=1$ Activity E: Slack $=1010=0$ Activity F: Slack $=107=3$ Activity G: Slack $=1515=0$

$*$The critical path would be Activity B$,$ E$,$ G$.$ These activities must be fulfilled without delay, as shown by there being no slack or allowance associated with these activities.

$3)$ Find the completion time and the standard deviation $(\backslash$sigma $)$ for the project$$ round off the standard deviation to the nearest hundredth: two decimal places.

$\backslash$sigma $=($tp $$ to$)/6$ squared$$ of squared SUM

B: $(6-2)/6=2/3$ squared$4/9$ OR $0\mathrm{.}4444$ E: $(13-3)/6=10/6$ squared$100/36=25/9$ OR$.2\mathrm{.}7777$ G: $(6-4)/6=1/3$ squared$1/9.$ OR $0\mathrm{.}1111$

Xbar$=15\backslash$sigma $=30/9=1\mathrm{.}83$

$4)$ What is the completion time at $99\%$ confidence level? $99\%$ confidence $=($Xbar$)+/-3\backslash$sigma $=153(1\mathrm{.}83)9\mathrm{.}51.$

Similarly for: $95\%$ CI $=($Xbar$)+/-2\backslash$sigma

$*$

$15+3(1\mathrm{.}83)20\mathrm{.}49$

Therefore, we can say with $99\%$ confidence that it will take about $10$ and $2012$ days to complete the

project.

$5)$ What is the likelihood of this project being completed in more than $16$ days?

Z $=(1615)/1\mathrm{.}83=0\mathrm{.}56$ then, from ND table$\mathrm{.}7123....$ More than means :$1\mathrm{.}712328\mathrm{.}77\%$

$1\mathrm{.}71220\mathrm{.}28774.$ or about $28\mathrm{.}77\%$

$*$There is about an $29\%$ chance likelihood that the project will be completed in greater than $16$ days.

$6)$ Between $14$ and $17$ days?

Z$1=(1415)/1\mathrm{.}83=0\mathrm{.}550\mathrm{.}2912($using table provide$)$

Z$2=(1715)/1\mathrm{.}83=1\mathrm{.}090\mathrm{.}8621$

In between means subtract table answers!

$0\mathrm{.}86210\mathrm{.}2912=\mathrm{.}570957\mathrm{.}09\%$

$*$There is about a $57\%$ chance $($that the project will take between $14$ & $17$ days.

i$.$ GRAPH of Network with TIMES$$ See above

SUMMARY:

$($You should provide a Summary$)$ All work must be clearly shown!

ii Critical Path: B $$ E $$ G$.4+6+5=15$

iii. Xbar $=15$

v$.$ CI:$99\%[102012].$

Highlighted on Network

iv$.$ Sigma $=1\mathrm{.}83($work provided above$)$

vi$.$ND$1$: $($Px$>16)=29\%.$ vii.ND$2$: $($P$14>$x$>17)=57\%$