Replacement Model

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```					Replacement Model

Dr Wang ShouQing
Department of Building
National University of Singapore
Need for Replacement

It is need to replace an existing machine/
equipment with a new one because the
existing one will result:
 Economic decline or obsolescence: loss of
efficiency
 Technical obsolescence

 Failure or impending failure

Dr Wang ShouQing, Department of Building, National University of Singapore
2
Replace Model
 Replacement model deals with the determination
of optimum time for replacement (Fig.1).
 Longer life gives:
 low average cost;
 high average
O&M cost.
 Failure: It is
economically
replace or repair
on a scheduled
basis before
failure occurs.
Dr Wang ShouQing, Department of Building, National University of Singapore
3
Aim & Methods of Analysis
 Aim: Determine the optimum
replacement interval/time for problems
that with increased O&M cost
 Tabular method vs. Mathematical
optimization method
 Tabular method is simple and permits one to
use discontinuous (discrete) data
 Mathematical optimization method permits
one to use continuous data

Dr Wang ShouQing, Department of Building, National University of Singapore
4
Tabular Method – Example 1
Question (Interest rate not considered):
Fleet cars have increased costs as they continue
in service due to increased direct operating (gas
& oil) and maintenance (repairs, tires, batteries,
etc.) cost as shown in Column 3. The initial
capital cost, P = \$3,500, and the trade-in value
drops as time passes until it reaches a constant
value of \$500 as shown in Column 2. Assume the
interest rate is 0, please determine the proper
length of service before cars should be
replaced.
Dr Wang ShouQing, Department of Building, National University of Singapore
5
Example 1 (con’t)
Solution: Determine the optimum replacement
interval by comparing the average annual cost.
Years Year-end Annual Capital Average          Average O&M Cost       Total
of   Trade-in O&M Investment Capital                               Average
Service Value    Cost (P=3500) Investment                           Annual Cost
(1)     (2)     (3) (4)=P-(2) (5)=(4)/(1)        (6)=(3)/(1)     (7)=(5)+(6)
1     1900    1800   1600       1600            1800/1=1800         3400
2     1050    2200   2450       1225        (1800+2200)/2=2000      3225
3      600    2700   2900       967      (1800+2200+2700)/3=2233    3200
4      500    3200   3000       750             9900/4=2475         3225
5      500    3700   3000       600            13600/5=2720         3320

Therefore, replace after 3 years of service for it
results in the lowest average annual cost.
Dr Wang ShouQing, Department of Building, National University of Singapore
6
Example 2 (Interest rate considered)
In Example 1, taking account of the interest rate, i=10%
Solution: To consider the time value of money, the
analysis must be based on an equivalent annual cost.
Years Present- Capital Present-          Cumulative Total Total Present-  Annual
of   worth of Invest- worth of           of O&M Cost worth of Capital Equivalent of
Service Trade-in ment O&M Cost                             & O&M Cost     Total Cost
(1)     (2) (3)=P-(2)    (4)                 (5)=(4)     (6)=(3)+(5)      (7)
1     1727     1773    1637                   1637           3410        3751
2      867     2632    1818            1637+1818=3455        6087        3507
3      450     3050    2028            3455+2028=5483        8533        3431
4      341     3159    2186            5483+2186=7669       10828        3416
5      310     3190    2297            7669+2297=9966       13156        3470
(Click above table for MS-Excel calculation sheet)
Therefore, replace after 4 years of service.
If replace after 3 years: increased cost = 3431 - 3416
= 17 (0.4% of minimum cost).
Dr Wang ShouQing, Department of Building, National University of Singapore
7
Effect of Interest Rate on
Optimum Replacement Interval
 Higher interest rate  Higher cost & Longer
service life (Fig.2)
 If i is large (>
20%) & applied
to large
investment
value, the                                                      3416
analysis should
3200
consider the
time value of
money.
Dr Wang ShouQing, Department of Building, National University of Singapore
8
Mathematical Method
 Assumption:
 Ignore interest rate over short life
 O&M cost increasing linearly.

 Formula derivation
Let: I = capital investment = (purchase price) – (trade-in)
n = years of service
Co = operating cost in the 1st year of service; \$
O = increase in operating cost per year; \$/year
Cm = maintenance cost in the 1st year of service; \$
M = increase in maintenance cost per year; \$/year

Dr Wang ShouQing, Department of Building, National University of Singapore
9
Mathematical Method (con’t)
 Formula derivation (con’t)
Then, the Average Total Cost (ATC):
ATC = average investment cost + average O&M cost
= I/n + (Co+Cm) + [(n-1)/2](O+M)                                         … (1)
Differentiation gives:
Optimum/Min ATC* = [2I(O+M)]1/2 - (O+M)/2 +(Co+Cm)
… (2)
and
Optimum life n* = [2I/(O+M)]1/2

Dr Wang ShouQing, Department of Building, National University of Singapore
10
Math Method – Example 3
 Question: A company maintains a pool of cars
to be used by its area representative. The
following cost data have been collected.
I = 3000, Co = 1500, Cm = 200, O = 225, M
=175. Find the optimum service life n* and the
minimum average total cost ATC*.

 Solution:
Substituting the data in equation (1) & (2) gives:
n* = 3.873 years, and ATC* = \$3049.18.
Dr Wang ShouQing, Department of Building, National University of Singapore
11
For Non-linear O&M Cost Function
 Assumption: The average O&M cost is a direct
product of the first year's cost and nk. Hence:
ATC* = I/n + (Co+Cm)nk
where k is selected to provide the best fit of the
estimated O&M cost.

 Differentiation gives the optimum life n*:
1/( k 1)
       I     
n*               
 k (Co  Cm) 
Dr Wang ShouQing, Department of Building, National University of Singapore
12
Effects of Values of k
 Values of k > 1
(or <1) are used
to represent O&M
cost which
increases over
time at an
increasing (or
decreasing) rate.
 This is depends
on the properties
of equipment &
its environment.
 Increasing k has the effect of decreasing n*.

Dr Wang ShouQing, Department of Building, National University of Singapore
13
Example 4
If given: I = 3000, Co = 1500, Cm = 200, k = 0.25,
Substituting the data in the formula gives:
n* = 4.76 and ATC* = 3141
Data for each year could also be obtained:
Life n   Average investment cost per        Average annual O&M           Total annual average
(year)            year (\$)                        cost (\$)                      cost (\$)
1                3,000                           1,700                         4,700
2                1,500                           2,022                         3,522
3                1,000                           2,237                         3,237
4                 750                            2,404                         3,154
5                 600                            2,542                         3,142
6                 500                            2,661                         3,161
7                 429                            2,765                         3,194
8                 375                            2,859                         3,234
Dr Wang ShouQing, Department of Building, National University of Singapore
14

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