\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION

\begin{description}

\item[(a)]
A project consists of activities $A,B,\ldots,L$ whose
prerequisites are given in the table below.  Draw a network,
suitable for analysis by the critical path method, to represent
the project. You should avoid using dummy activities, where
possible. For each event, write the earliest and latest event
times on the network, and deduce the critical path.

\begin{center}

\begin{tabular}{ccc}
\hline Activity&Prerequisites&Duration (days)\\ \hline
A&-&8\\B&-&6\\ C&-&6\\ D&A&9\\ E&A,B&7\\ F&C&3\\ G&C&9\\ H&D&3\\
I&D,E,F&7\\ J&G&8\\ K&H,I&8\\ L&H,I&5\\ \hline
\end{tabular}

\end{center}

If the duration of activity $E$ is increased to 10 days, discuss
the effect on the overall project duration.


\item[(b)]
An electronics company manufactures measuring instruments which
have a reading device that must be correctly aligned.  In some
cases, the reading device gets out of alignment during delivery to
customers.  Such a fault occurs in 10\% of deliveries. If the
equipment is installed with an incorrect alignment, then the cost
of correcting this fault is \pounds5000.

The company has three possible courses of action.  First, it can
proceed with the installation of the equipment, and incurs no cost
if the reading device is correctly aligned.  Second, the company
can send an engineer, at a cost of \pounds400, to arrive just
before installation.  The engineer will check the alignment, and
correct it, if necessary, prior to installation.  Third, the
company can make a field check of the equipment just before
installation, at a cost of \pounds100, with a view to deciding on
whether to call an engineer before installation.    However, the
field check is not 100\% reliable.
 Specifically, if the reading device is correctly aligned, the field
check will indicate that adjustment is needed with probability
0.1; and if the reading device is
 not correctly aligned, the field check will indicate that no
adjustment is needed probability 0.2.  To call an engineer after
the field check, costs a further \pounds500.

Develop a decision tree to find out whether the field check should
be performed, and to decide in which situations an engineer should
be called.

\end{description}

ANSWER


\begin{description}

\item[(a)]

\setlength{\unitlength}{.4in}

\begin{picture}(12,6)
\put(1,3){\circle{.4}}\put(1,3){\makebox(0,0){1}}
\put(3,5){\circle{.4}}\put(3,5){\makebox(0,0){2}}
\put(3,3){\circle{.4}}\put(3,3){\makebox(0,0){3}}
\put(3,1){\circle{.4}}\put(3,1){\makebox(0,0){4}}
\put(5,5){\circle{.4}}\put(5,5){\makebox(0,0){5}}
\put(5,3){\circle{.4}}\put(5,3){\makebox(0,0){6}}
\put(7,5){\circle{.4}}\put(7,5){\makebox(0,0){7}}
\put(7,1){\circle{.4}}\put(7,1){\makebox(0,0){8}}
\put(9,5){\circle{.4}}\put(9,5){\makebox(0,0){9}}
\put(11,3){\circle{.4}}\put(11,3){\makebox(0,0){10}}

\put(1.2,3){\vector(1,1){1.8}}\put(1.2,3){\vector(1,0){1.6}}\put(1.2,3){\vector(1,-1){1.8}}
\put(3.2,5){\vector(1,0){1.6}}\put(3,4.8){\vector(0,-1){.4}}\put(3,4.){\vector(0,-1){.4}}
\put(3.2,3){\vector(1,0){1.6}}
\put(3.2,1){\vector(1,1){1.8}}\put(3.2,1){\vector(1,0){3.6}}
\put(5.2,5){\vector(1,0){1.6}}\put(5,4.8){\vector(0,-1){.4}}\put(5,4.){\vector(0,-1){.4}}
\put(5.2,3){\vector(1,1){1.8}}
\put(7.2,5){\vector(1,0){1.6}}\put(7.2,5){\vector(2,-1){3.6}}
\put(7.2,1){\vector(2,1){3.6}}
\put(9.2,5){\vector(1,-1){.4}}\put(10,4.2){\vector(1,-1){.4}}


\put(1,2.5){\makebox(0,0){(0,0)}}
\put(3,5.5){\makebox(0,0){(8,8)}}
\put(3,2.5){\makebox(0,0){(8,10)}}
\put(3,.5){\makebox(0,0){(6,14)}}
\put(5,5.5){\makebox(0,0){(17,17)}}
\put(5.5,2.5){\makebox(0,0){(17,17)}}
\put(7,5.5){\makebox(0,0){(24,24)}}
\put(7,.5){\makebox(0,0){(15,24)}}
\put(9,5.5){\makebox(0,0){(32,32)}}
\put(11.5,3.5){\makebox(0,0){(32,32)}}

\begin{small}

\put(1.5,4){A, 8} \put(2,2.5){B, 6} \put(1.8,1.5){C, 6}
\put(3.7,5.5){D, 9} \put(3.7,3.5){E, 7} \put(3.7,2.5){F, 3}
\put(5,.5){G, 9} \put(5.6,5.5){H, 3} \put(6,3.5){I, 7}
\put(8.8,2){J, 8} \put(7.6,5.5){K, 8} \put(9,3.5){L, 5}

\end{small}

\end{picture}

The critical path is 1 - 2 - 5 - 6 - 7 - 9 - 10,\ A - D - I - K

The total float for activity E is $TF_E=17-7-8=2$

The increase in duration of 3 for E exceeds $TF_E$ by 1 day, so
the project duration becomes 33 days.

\item[(b)]
Let $A$ be the event that the reading device is correctly aligned

$\overline{A}$ be the event that it is not

$C$ be the event that the field test indicates correct alignment

$\overline{C}$ be the event that it does not.

\begin{eqnarray*}
P(A)&=&0.9\\ P(\overline{A})&=&0.1\\ P(C|A)&=&0.9\\
P(\overline{C}|A)&=&0.1\\ P(C|\overline{A})&=&0.2\\
P(\overline{C}|\overline{A})&=&0.8\\
P(C)&=&P(C|A)P(A)+P(C|\overline{A})P(\overline{A})\\
&=&0.9\times0.9+0.2\times0.1=0.83\\
P(\overline{C})&=&P(\overline{C}|A)P(A)+P(\overline{C}|\overline{A})P(\overline{A})\\
&=&0.1\times0.9+0.8\times0.1=0.17\\
P(A|C)&=&\frac{P(C|A)P(A)}{P(C)}\\
&=&\frac{0.9\time0.9}{0.83}=0.9759\\
P(\overline{A}|C)&=&\frac{P(C|\overline{A})P(\overline{A})}{P(C)}\\
&=&\frac{0.2\times0.1}{0.83}=0.0241\\
P(A|\overline{C})&=&\frac{P(\overline{C}|A)P(A)}{P(\overline{C}}\\
&=&\frac{0.1\times0.9}{0.17}=0.5294
\end{eqnarray*}

\setlength{\unitlength}{1cm}

\begin{picture}(11,20)
\put(1,11){\framebox(.3,.3)}\put(6,18){\framebox(.3,.3)}
\put(6,14){\framebox(.3,.3)}\put(4,6){\framebox(.3,.3)}
\put(4,16){\circle{.3}}\put(6,4){\circle{.3}}
\put(8,17){\circle{.3}}\put(8,13){\circle{.3}} \put(8,5){0}

\put(6,8){400} \put(8,19){600} \put(8,15){600} \put(8,3){5000}
\put(10,18){100} \put(10,16){5100} \put(10,14){100}
\put(10,12){5100}

\put(1.2,11){\line(3,5){2.8}} \put(1.2,11){\line(3,-5){2.8}}

\put(4.2,16){\line(1,1){1.6}} \put(4.2,16){\line(1,-1){1.6}}
\put(4.2,6){\line(1,1){1.6}} \put(4.2,6){\line(1,-1){1.6}}

\put(6.2,18){\line(2,1){1.6}} \put(6.2,18){\line(2,-1){1.6}}
\put(6.2,14){\line(2,1){1.6}} \put(6.2,14){\line(2,-1){1.6}}
\put(6.2,4){\line(2,1){1.6}} \put(6.1,4){\line(2,-1){1.6}}

\put(8.2,17){\line(2,1){1.6}} \put(8.2,17){\line(2,-1){1.6}}
\put(8.2,13){\line(2,1){1.6}} \put(8.2,13){\line(2,-1){1.6}}

\put(.3,11.5){285.02} \put(3.5,16.5){285.02} \put(3.5,5.5){400}
\put(5.5,18.5){220.5} \put(5.5,13.5){600} \put(5.5,3.5){500}
\put(7.5,16.5){220.5} \put(7.5,12.5){2453}

\put(1,14){Field check} \put(1,8){No field}\put(1,7.5){ check}
\put(4.2,17.5){$C$ 0.83} \put(4.2,14.5){$\overline{C}$ 0.17}
\put(4.2,7.5){Send} \put(4.2,7){ Engineer}
\put(4.2,4.5){No}\put(4.2,4.){Engineer}
\put(6,19.5){Send}\put(5.8,19){Engineer}
\put(6,16.7){No}\put(6,16.2){ Engineer}
\put(6,15.3){Send}\put(6,14.8){ Engineer} \put(6,13.5){
No}\put(5.5,13){ Engineer}

\put(6.5,5){$A$ 0.9} \put(6.5,3){$\overline{A}$ 0.1}
\put(8.2,18){$A$ 0.9759} \put(8.2,16){$\overline{A}$ 0.0241}
\put(8.2,14){$A$ 0.5294} \put(8.2,12){$\overline{A}$ 0.4706}

\put(3.5,7.7){\line(0,-1){1}} \put(5.5,5.2){\line(0,-1){1}}
\put(7.5,19.2){\line(0,-1){1}} \put(7.5,13.8){\line(0,-1){1}}

\end{picture}

Thus, the field check should be undertaken. If it indicates that
no adjustment is necessary, then do not call the engineer;
otherwise call the engineer.

\end{description}




\end{document}