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<p class=3DMsoNormal align=3Dcenter style=3D'text-align:center'><b style=3D=
'mso-bidi-font-weight:
normal'><i style=3D'mso-bidi-font-style:normal'><u><span style=3D'font-size=
:14.0pt'>Tips
for the Study of PERT<o:p></o:p></span></u></i></b></p>

<p class=3DMsoNormal align=3Dcenter style=3D'text-align:center'><u><o:p><sp=
an
 style=3D'text-decoration:none'>&nbsp;</span></o:p></u></p>

<p class=3DMsoNormal>I hope that the following suggestions will be helpful =
in
your study of the PERT topic.<span style=3D'mso-spacerun:yes'>&nbsp; </span=
>This
list is of course not exhaustive one.<span style=3D'mso-spacerun:yes'>&nbsp;
</span>In particular, <span style=3D'background:yellow;mso-highlight:yellow=
'>for
the stat portion see the special file on my home page.</span></p>

<p class=3DMsoNormal align=3Dcenter style=3D'text-align:center'><u><o:p><sp=
an
 style=3D'text-decoration:none'>&nbsp;</span></o:p></u></p>

<p class=3DMsoNormal>1)<span style=3D'mso-spacerun:yes'>&nbsp; </span>Solve=
 one
complex problem completely before turning to the next one.<span
style=3D'mso-spacerun:yes'>&nbsp; </span>It is a mistake to do one problem =
80
percent and then turn to the next and solve it too 80 percent.<span
style=3D'mso-spacerun:yes'>&nbsp; </span>PERT problems are essentially
conceptually similar to each other. <span
style=3D'mso-spacerun:yes'>&nbsp;</span>Thus mastering one completely will =
sharply
increase your chance to solve any other.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>2) The challenge in solving PERT problems is in the <s=
pan
class=3DGramE>internalizing<span style=3D'mso-spacerun:yes'>&nbsp; </span>o=
f</span>
the great number of details in the conceptual formal (notational and
definitional) and computation development.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>3). <span class=3DGramE>To</span> help mastering the
complexity, we break the PERT treatment to the following topics and within a
topic to some subtopics.<span style=3D'mso-spacerun:yes'>&nbsp;&nbsp; </spa=
n>The
topics are:</p>

<p class=3DMsoNormal>A) Set Up</p>

<p class=3DMsoNormal>B) Event Scheduling</p>

<p class=3DMsoNormal>C) Activity Scheduling</p>

<p class=3DMsoNormal>D) Statistics.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>So make sure to simply check if you have the full
understanding of each topic.</p>

<p class=3DMsoNormal><b style=3D'mso-bidi-font-weight:normal'><o:p>&nbsp;</=
o:p></b></p>

<p class=3DMsoNormal><b style=3D'mso-bidi-font-weight:normal'><u>Set Up<o:p=
></o:p></u></b></p>

<p class=3DMsoNormal>4)<span style=3D'mso-spacerun:yes'>&nbsp; </span>In a =
set up,
notice the following:</p>

<p class=3DMsoNormal><span
style=3D'mso-spacerun:yes'>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>a. There i=
s no
loose activity.<span style=3D'mso-spacerun:yes'>&nbsp; </span>Every activit=
y is
an arrow that must start at an event and end in <span style=3D'mso-tab-coun=
t:
1'>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>an event.</p>

<p class=3DMsoNormal><span
style=3D'mso-spacerun:yes'>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>b. O=
nly
the dummy activity arrow is marked as a dashed arrow.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>5). In a Set Up, remember that the greatest challenge =
is
that of dealing with dummy variables.</p>

<p class=3DMsoNormal><span
style=3D'mso-spacerun:yes'>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span><u><o:p></=
o:p></u></p>

<p class=3DMsoNormal><span class=3DGramE><u>Dummy Activities</u>.</span><sp=
an
style=3D'mso-spacerun:yes'>&nbsp; </span>The need for those arise only when=
 some
activity, x, requires only a subset of activities required by some other
activity, y. <span style=3D'mso-spacerun:yes'>&nbsp;&nbsp;</span><span
class=3DGramE>The</span> activities in this subset are obviously commonly
required by both x and y. </p>

<p class=3DMsoNormal><span style=3D'mso-spacerun:yes'>&nbsp;</span></p>

<p class=3DMsoNormal>These common activity (<span class=3DSpellE>ies</span>=
) in the
subset would be set up first, leading to some event (I<span class=3DGramE>)=
 <span
style=3D'mso-spacerun:yes'>&nbsp;</span>whose</span> declaration allows the
logical start of x. </p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>The uncommon activities in the set are leading to even=
t (II)
whose declaration would allow y to begin.</p>

<p class=3DMsoNormal><span style=3D'mso-spacerun:yes'>&nbsp;</span>Since st=
arting Y
requires also the inputs of the common inputs, a dummy directed from (I) to
(II) will be drawn.<span style=3D'mso-spacerun:yes'>&nbsp; </span>Declarati=
on of
event (II) would mean that all inputs have been completed and activity y can
begin.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>6) Please notice, the complexity can be by far larger,=
 for
example x requires as inputs a subset of activities that y requires as inpu=
ts,
while the inputs for y are only a subset of inputs required by z, etc.<span
style=3D'mso-spacerun:yes'>&nbsp;&nbsp; </span>While the principle of const=
ructing
the dummies is as described in 5), it is important to practice with differe=
nt
more complex cases, so <span style=3D'background:yellow;mso-highlight:yello=
w'>it
worthwhile to invent some complex cases and deal with them constructing the
proper dummy activities.</span></p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal><b style=3D'mso-bidi-font-weight:normal'><u>Event Sche=
duling<o:p></o:p></u></b></p>

<p class=3DMsoNormal><b style=3D'mso-bidi-font-weight:normal'><o:p>&nbsp;</=
o:p></b></p>

<p class=3DMsoNormal>7) Remember the notations a square for TE a circle for=
 TL a
triangle for event slack. <span class=3DGramE>TL-TE.</span></p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>8) Use different colors:<span
style=3D'mso-spacerun:yes'>&nbsp; </span><span style=3D'background:aqua;mso=
-highlight:
aqua'>Blue for TE</span>, <span style=3D'background:red;mso-highlight:red'>=
RED
for TL</span>, <span class=3DGramE><span style=3D'background:silver;mso-hig=
hlight:
silver'>Black</span></span><span style=3D'background:silver;mso-highlight:s=
ilver'>
for event slack</span>.<span style=3D'mso-spacerun:yes'>&nbsp; </span>When =
you
process red you need not pay attention to the numbers in blue and this will
reduce considerably the chance for computational mistake. </p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>9) Please remember:<span style=3D'mso-spacerun:yes'>&n=
bsp;
</span>Earlier time is smaller that later time. And there is no negative ti=
me.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>10)<span style=3D'mso-spacerun:yes'>&nbsp; </span>Unde=
rstand
conceptually the logic in setting up both TE (LEFT) and TL (ELST).</p>

<p class=3DMsoNormal>Pay particular attention to TL.<span
style=3D'mso-spacerun:yes'>&nbsp; </span>Most students face no difficulty i=
n computing
TE.<span style=3D'mso-spacerun:yes'>&nbsp; </span><span style=3D'background=
:yellow;
mso-highlight:yellow'>The difficulty is in TL.</span> <span
style=3D'mso-spacerun:yes'>&nbsp;</span></p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal><span class=3DSpellE>TE&#8217;s</span> computation dea=
ls with
the event as a summary point and considers only the activities leading to t=
he
event.<span style=3D'mso-spacerun:yes'>&nbsp; </span></p>

<p class=3DMsoNormal>TE assumes that <span style=3D'background:yellow;mso-h=
ighlight:
yellow'>all</span> activities leading to the event can be <span
style=3D'background:yellow;mso-highlight:yellow'>completed</span> by TE.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>TL deals with the event as initiation point and consid=
ers
the activities that are starting from the event.</p>

<p class=3DMsoNormal>TL assumes that <span style=3D'background:yellow;mso-h=
ighlight:
yellow'>all</span> activities starting from the event can <span
style=3D'background:yellow;mso-highlight:yellow'>start</span> by TL <span
style=3D'background:yellow;mso-highlight:yellow'>without delaying the proje=
ct.</span>
</p>

<p class=3DMsoNormal><b style=3D'mso-bidi-font-weight:normal'><u>Activity
Scheduling<span style=3D'mso-spacerun:yes'>&nbsp; </span><o:p></o:p></u></b=
></p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>11) Do not mix up the notation for start and finish of
activities ( ES, LS, EF, LF ) with the notations for earliest and latest
declaration of events (TE, TL).<span style=3D'mso-spacerun:yes'>&nbsp; </sp=
an></p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>12) Process the table of activity scheduling by column=
.<span
style=3D'mso-spacerun:yes'>&nbsp; </span>It is much easier.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>Consider any activity <i style=3D'mso-bidi-font-style:=
normal'>x
</i>starting at event (I) and targeted to event (II). </p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>ES, the earliest starting time for <i style=3D'mso-bid=
i-font-style:
normal'>x</i> is simply, TE for event (I), which is the number marked inside
the square by event (I).<span style=3D'mso-spacerun:yes'>&nbsp; </span><span
style=3D'mso-spacerun:yes'>&nbsp;</span></p>

<p class=3DMsoNormal>Similarly, LF, the latest finishing time for <i
style=3D'mso-bidi-font-style:normal'>x</i> is simply, TL for event (II), wh=
ich is
the number marked inside the circle by event (II). </p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>Compute the EF columns by adding (t), the activity time
column and the ES column</p>

<p class=3DMsoNormal>Compute the LS column by subtracting (t) from the LF c=
olumn.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>13) The activity slack is the gap between the LS and ES
column or alternatively equivalently the gap between the LF and LS
columns.<span style=3D'mso-spacerun:yes'>&nbsp; </span>This can be used to =
check
for possible mistakes in the columns computation.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal>14) Again, latest time is always greater than earliest=
 time
and there is no negative time.</p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

<p class=3DMsoNormal><span style=3D'mso-spacerun:yes'>&nbsp;</span></p>

<p class=3DMsoNormal><o:p>&nbsp;</o:p></p>

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