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\begin{document}

\parindent=0pt

QUESTION

\begin{description}

\item[(a)]
A project consists of activities $A,B,\ldots,J$ 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&amp;Prerequisites&amp;Duration(days)\\ \hline
$A$&amp;-&amp;8\\$B$&amp;-&amp;9\\$C$&amp;$A$&amp;7\\$D$&amp;$A$&amp;4\\$E$&amp;$B$&amp;6\\$F$&amp;$B,D$&amp;1\\$G$&amp;$B,D$&amp;1\\
$H$&amp;$C,F$&amp;7\\$I$&amp;$E,G$&amp;4\\$J$&amp;$E$&amp;8\\ \hline
\end{tabular}

\end{center}

If  it is possible to transfer resources from activity $E$ to $H$
so that the duration of $H$ is reduced by one day but the duration
of $E$ is increased by one day, decide, giving a reason, if the
project duration can be reduced.


\item[(b)] A doctor diagnoses that a patient has either disease $A$
or disease $B$, and that on the current evidence these two
alternatives are equally likely.

Although there is no test for disease $B$, it is possible to
undertake an expensive test for disease $A$.  A disadvantage of
this test is unreliability.  In particular, it gives a positive
indication of disease $A$ for only 70\% of patients who have
disease $A$, and also gives  a positive indication of disease $A$
for 10\% of patients who have disease $B$ rather than $A$.

There is no effective treatment for disease $B$, and patients with
this disease remain in poor health thereafter. Similarly, if
patients suffering from disease $A$ go untreated, then they also
remain in poor health.  However, there is an expensive treatment
for disease $A$.  A patient with disease $A$ receiving treatment
is returned to good health with probability 0.7, and remains in
poor health with probability 0.3. Using this treatment on a
patient with disease $B$ has no effect.

The utility of a patient in good health is 30, and of a patient in
poor health is 10.  Utilities are reduced by 2 if a patient
undergoes the test, and are reduced by 1 if a patient receives the
treatment.

Develop a decision tree to find out whether the patient should
undergo the test, and whether the treatment for disease $A$ should
be administered.

\end{description}

\bigskip

ANSWER


\begin{description}

\item[(a)]\ \\

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\put(11,3){\makebox(0,0){8}} \put(10.8,3.5){(24,24)}

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4} \put(5,.5){E, 6} \put(7,3.7){F, 5} \put(6.5,2.5){G, 1}
\put(10.2,4.){H, 7} \put(9.5,3.1){I, 4} \put(9.2,1.7){J, 8}

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The critical path is $A$ - $D$ - $F$ - $H$

$TF_E=16-9-6=1$

Since $H$ is not a successor of $E,$ there is a path containing
$E$ that becomes critical when the duration of $E$ is increased.
Therefore, the project duration is not reduced.

\item[(b)]
Let $T$ denote that the test is undertaken.

$\overline{T}$ denote that the test is not undertaken

$P$ denote a positive indication of disease $A$

$\overline{P}$ denote no positive indication of disease $A$

$H$ denote that the patient returns to good health

$\overline{H}$ denote that the patient remains in poor health.

We are told that $P(A)=P(B)=0.5$

\begin{tabular}{cc}
$P(P|A)=0.7$&amp;$P(\overline{P}|A)=0.3$\\
$P(P|B)=0.1$&amp;$P(\overline{P}|B)=0.9$
\end{tabular}

\begin{eqnarray*}
P(P)&amp;=&amp;P(P|A)P(A)+P(P|B)P(B)=0.4\\
P(\overline{P})&amp;=&amp;P(\overline{P}|A)P(A)+P(\overline{P}|B)P(B)=0.6
\end{eqnarray*}

The posterior probabilities are
\begin{eqnarray*}
P(A|P)&amp;=&amp;\frac{P(P|A)P(A)}{P(P)}=0.875\\
P(B|P)&amp;=&amp;\frac{P(P|B)P(B)}{P(P)}=0.125\\
P(A|\overline{P})&amp;=&amp;\frac{P(\overline{P}|A)P(A)}{P(\overline{P})}=0.25\\
P(B|\overline{P})&amp;=&amp;\frac{P(\overline{P}|B)P(B)}{P(\overline{P})}=0.75\\
\end{eqnarray*}

The decision tree is given below

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\begin{footnotesize}
\put(14,31){$H$, 0.7} \put(14,29.9){$\overline{H}$, 0.3}
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\put(7.6,1.4){B, 0.5}

\put(7.,29.5){Treatment} \put(7.,25.4){No}\put(7,25){Treatment}
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\end{footnotesize}

\put(1,20){$T$} \put(1,11){$\overline{T}$}

\put(5,26){$P$, 0.4} \put(5,21){$\overline{P}$, 0.6}

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It is recommended that the treatment should be administered, but
the test should not be performed.

\end{description}



\end{document}
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