Time Cost Tradeoff

1
Time Cost Tradeoff
2
Activity-On-Arc
N the event (node) set representing the numbers
of the node A the activity set (arc set)
represented by a collection of pairs of nodes
enclosed in parentheses and separated by a comma
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Example
N 1, 2, .. , 7 and A (1,2), (1,3), (2,3),
(1,4), (2,5), (3,5), (3,6), (4,6), (5,7), (6,7).
Note that activity (2,3) is a "dummy" activity.
It merely helps represent the precedence
relationships. It takes 0 time units and consumes
no resources
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Notation
With each activity (i , j) ? A
dijc shortest achievable (crash) duration for
activity (i , j). dijn normal duration for
activity (i , j). NCij direct cost of activity
(i, j) if performed at normal
duration. CCij direct cost of activity (i, j)
if performed at crash duration. cij
additional direct cost of crashing activity (i,
j) by one time unit. Note CCij NCij
cij (dijn - dijc ) dij a decision variable,
representing the duration of activity (i,
j)
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Notation
With each event i ? N of the project
?i partial payment to be received when event i
is realized. Ti a decision variable,
representing the realization time of
event i. O Overhead (indirect) cost of the
project expressed as /unit time.
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Time-Cost Tradeoff Problem
To find the optimal project duration that will
yield the minimum total project cost. The total
project cost is the sum of total activity direct
costs plus the total indirect cost. That is,
TC TD TI
where, and TI O?Tn
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Linear Formulation

Min
Subject to Tj - Ti - dij ? 0 for all (i , j) ?
A (1) dijc ? dij ? dijn , for all (i ,
j) ? A (2) T1 0
(3)
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Payment Scheduling Problem

• In this section we assume that activity
• durations have already been determined
• as dij , for all (i , j) ? A.

• Given the realization time of two events Ti
• and Tj at two ends of an activity (i, j), it
  is
• to our advantage to schedule (i, j) as late
• as possible without violating the
• corresponding constraint in (1).

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Payment Scheduling Problem
Let Sij be the scheduled start time of activity
(i, j), then we must satisfy Sij dij ?
Tj Since dij is constant, the latest Sij that
will still satisfy the inequality is
Sij Tj - dij
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Payment Scheduling Problem
Assuming continuous compounding with the company
interest rate of r per period, we can calculate a
single payment equivalent of this equal payment
series to occur at the termination of the
activity as
Note that Fij ? 0 since NCij ? 0.
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Payment Scheduling Problem
Let us assume that with the realization of event
j at time Tj there will be a partial payment to
us by an amount, ?j ? 0. Therefore, it is to our
advantage to receive this amount as early as
possible. Also, let
be the net cash flow associated with node j
(event j).
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Formulation
The associated cash flow scheduling problem is
one of finding realization times, Tj, of each
event j 1, 2, , n such that the PW(T) is
maximized. That is
Subject to Tj - Ti - dij ? 0 for all (i , j)
? A T1 0 Tn ? D
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Transformation to LP
Let yj e-rTj (Tj -(ln yj)/r by taking
logarithms of the both sides) Kij erdij.
The new objective function is easy to
determine. For the constraints, Tj - Ti - dij ?
0 is equivalent to -(ln yi)/r(ln yj)/r-dij
?0. -ln yiln yj-rdij ?0. ln (yj/yi) ?rdij when
we take exponent of both sides eln (yj/yi) ?
erdij yj/yi ? Kij and this gives yjKij - yi ?
0, for all (i, j) ? A, i ? 1. Note that y11,
since T10.
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Linear Formulation
Max
Subject to yjK1j ? 1, for all (1,
j) ? A, yjKij - yi ? 0, for all (i,
j) ? A, i ? 1, -yn ? -Kn1 where,
Kn1 e-Dr Once the LP is solved for
(y2,y3, , yn) we obtain the optimal Tj
values by inverse transformation Tj
ln(yj)/(-r) Here, r is the nominal discount
rate per period.