capital equipment replacement decisions

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Capital Equipment Replacement Decisions

Ali Zuashkiani

Condition Based Maintenance Laboratory

University of Toronto

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Capital Equipment Replacement• Constant Annual Utilization

• Varying Annual Utilization

• Technological Improvement

• Tracking Individual Units

• Repair versus Replace

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www.ipamc.orgReplacement Age ( years)

Optimum replacement age

Total cost

Fixed cost

Ownership cost

Operations and maintenance costA

nnua

l Cos

t

Economic Life Problem

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Economic Life Model: [constant utilization]

Construction of model:

C1 C2 C3 A – Sn

1 2 3 n years

Note:Above assumes costs in year are paid at the end of year.

0Replacement Cycle

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Time Value of Money:Discounted Cash Flow

Analysis

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Concept of Economic LifeN = 1 year

0 1 2 3 4 5 6 Years

A – S1c1

A – S1c1

A – S1c1

A – S1c1

A – S1c1

A – S1c1

A – S1c1

……..

……..

N = 2 years

0 1 2 3 4 5 6 Years

A – S2c1

A – S2c1 c1

A – S2 A – S2c1c2 c2 c2

……..

……..

N = 3 years……..

……..

0 1 2 3 4 5 6 Years

A – S3c1

A – S3c1

A – S3c1c2 c3 c2 c3

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Time Value of Money

$100$100 + 10 = $110

$110 + 11= $121

i = 10% i = 10%0 1 2

The above concept is familiar to many, i.e. investments grow in value

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Example

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Problem: Assume we have a payment to make, 2 years from now, of $121. What is its value today? (i.e. its present value)

$121?

0 1 2

Solution: P = 121 1 (Assume i=10%)1 + 0.1

= $100

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Equivalent Annual Cost (EAC): Use of the Capital Recovery Factor (CRF)

EAC = Present Value x CRF

( ) 11)1(−+

+= n

n

iiiCRF

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Selecting an Alternative

In the following example, we will consider two (equivalent) criteria:

(i) Present value (ii) Equivalent annual cost

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Example: Statement of Problem

A contractor requires specialized equipment for a period of 3 years. Given the costs and salvage values in the following table, which is the best alternative?

321

100

400

100

80

300

100

3500501006000C

15002001003000B

30001001005000A

SalvageValue

Operating CostInstallationCost

PurchasePrice

Equipment

NOTE: Costs in $ x 100

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For Equipment A

• Present Value (PV) = ?• Assume that discount factor r = 0.9

Recall: r = 1 / 1+i where i = interest rate appropriate for discounting(in this case, assume i = 11%)

• PV = 5000 + 100 + 100 (0.9)1 + 100 (0.9)2 + 100 (0.9)3 –3000 (0.9)3

= $3157

$5000$100

$100 $100 $100 $3000

0 1 2 3

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For Equipment B

PV = 3000 + 100 + 200 (0.9)1 + 300 (0.9)2

+ 400 (0.9)3 – 1500 (0.9)3

= $2721

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For Equipment C

Therefore, the best alternative using the present value concept is B (since it has the minimum PV).

PV = $3731

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An Alternate Approach …• Dealing with the same example, rather than the

present result in terms of the Present Value of a stream of cash flows, we frequent convert this PV to an Equivalent Annual Cost (EAC) - sometimes referred to as Annual Equivalent Evaluation.

• To convert PV to EAC, multiply PV by the Capital Recovery Factor (CRF):

EAC = PV x CRFwhere CRF = i (1+i)n

(1+i)n - 1

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EAC = 3157 i (1+i)n

(1+i)n - 1 = 3157 0.11 (1 + 0.11)3

(1 + 0.11)3 - 1= 3157 x 0.4092= $1291.89

Recall: PV = $3157

Note: The PV of this is the same as the PV of the original stream of cash flows for Equipment A.

0 1 2 3

$1291.89 $1291.89$1291.89

Graphically, we have:

Example: Equipment A

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Construction of model:

C1(n) = C1r1 + C2r2 + C3r3 + -------- + Cnrn + ( A – Sn ) rn

PV of above cycle:

In general:

)(C(n)Cn

1i1 n

nii SArr −+= ∑

=

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C1 C2 C3 A – Sn

n years0

First Cycle

Construction of model:Consider the second cycle:

2n years

n years

PV of 2nd cycle discounted to start of 2nd cycle:

C1r1 + C2r2 + C3r3 + -------- + Cnrn + ( A – Sn ) rn

In general: )( nni

i SArr −+= ∑=

n

1iC(n)C2

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Optimal Replacement Interval for Capital Equipment: Minimization of Total Cost

0 1 2 … n-1 n 1 2 … n-1 n

Replace Replace ReplaceC1 C2 Cn C1 C2 Cn C1 C2 Cn C1

1 2 … n-1 n 1 …

Cycle 1 Cycle 2 Cycle 3

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Construction of model:Similarly, we can obtain: C3(n) , C4(n) , etc

Thus PV of this chain of replacement is:

C(n) = C1(n) + C2(n)rn + C3(n)r2n + C4(n)r3n + ……….

Summing to infinity then for the geometric progression we get:

C1(n)C(n) = -----------

1 - rn

Since C1(n) = C2(n) = C3(n) ……

n

nni

i

r

SArr

−

−+=∑=

1

)(CC(n)

n

1i

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ExampleHave A Ci, Si, r :

0

0.2

0.4

0.6

0.8

1

1.2

Thus the economic life of the asset is 2 years with an associated total discounted cost of $19,421

EAC = 19,421 * i Recall r = 1 / (1 + i) r = 0.9

Therefore i = 0.11 and EAC = $2,136

22500

19421 20790

21735

23701

Economic life2 3 4 51 years

C(n)

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Note :

C1 C2 C3Cn

Sn

0 1 2 3 n - 1 n

A

Replacement Cycle

USED IN AGE/CON

ANDPERDEC

SOFTWARE

NOTE : NEED TO BE CLEAR WHERE CASH FLOW OCCURS

A better assumption for cash flows may be:

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Fleet StatisticsS Fleet

Utilization: 60,000 km/year per tractor

Fleet size: 19

Tractor weight: 18,000 kg

Current policy: 5 years replacement cycle

K Fleet

Utilization: 110,000 km/year per tractor

Fleet size: 17

Tractor weight: 23,000

Current policy: 5 year replacement cycle

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Title: BEST ESTIMATE RESALE VALUES

Number of Years: 5

Acquisition Cost: 85000

65,78770,54172,45971,10168,234

6000040000250002000015000

10%10%10%10%10%

2935245246526265332442363

1 Year Old2 Year Old3 Year Old4 Year Old5 Year Old

EACBest Year

Highlighted

Trade-in Value

Rate for Cash Flow

Discounting

O&M Cost(In Today’s

Dollars)

Age of Vehicle

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60

65

70

75

80

1 2 3 4 5

Age of Trucks (years)

EAC - $$ (Thousands)

DATA ANALYSIS:Equivalent Annual Cost vs Age of Trucks

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Haul Truck

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Haul Truck ReplacementMD452

July 11997

$165,000 $165,000 $105,000 $246,000 $115,000

July 11998

July 11999

July 12000

Today

$187,500 $187,500 $28,500

Jan 1, 1998 Jan 1, 1999 Jan 1, 2000 Jan 1, 2001

$330,000 $369,000 $302,000 $216,000

Historical Costs

Year 1 Year 2 Year 3 Year 4

2002 Dollars (inflation = 5% P.A.)

$401,117 $427,164 $332,955 $226,800

Year 1 Year 2 Year 3 Year 4

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Haul Truck Replacement

TruckYear MD450 MD451 MD452 MD453 Average

1 $401,117 $396,2702 $427,164 $411,2463 $332,955 $329,9634 $226,800 $279,279

Continued…

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Why are O&M costs declining in years 3 and 4?

Or

Why are O&M costs so high in years 1 and 2?

Due to developing maintenance expertise?

? ?


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Average yr 1 O&M cost: $171,300

Average yr 2 O&M cost: $207,400

In today’s prices we have:

Yr 1: $188,858

Yr 2: $217,770

Examining newest group of 5 trucks [MD464-468] that came into service 2 years ago we get:


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Input Data: 100T Haul Truck

Acquisition Cost: $700,000

Interest Rate Appropriate for Discounting: 11%Year O&M Cost

($/year)Resale (or Trade-In) value at End of Year

Equivalent Annual Cost ($)

1 188,858 450,000 523,244

2 217,770 300,000 491,415

3 329,963 275,000 471,321

4 279,279 250,000 448,800

5 300,000 (est) 150,000 450,678

6 350,000 (est) 100,000 451,2246


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Remarks: Haul Truck Economic Life

When dealing with capital equipment replacement problems, there are frequently significant uncertainties associated with future costs, interest rates and possibly the demands that will be placed on the equipment. However, the availability of specially-designed pc software enables valuable sensitivity analyses to be undertaken.

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These “what if” analyses allow the engineer to examine the effect that various estimates of trade-in values, interest rates and so on, will have on replacement cycles. Since a high degree of confidence can be associated with final recommendations to senior management on equipment economic life, the chances of obtaining approval for major capital expenditures is generally increased significantly.

Remarks: Haul Truck Economic Life

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0 12 18 24 30 360.8

1.0

1.2

1.4

Replacement Age (Months)

Seam

erD

isco

unte

d To

tal C

ost

Dol

lars

* 1

06Seamer Replacement

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Feller Buncher

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120000

140000

160000

180000

200000

220000

1 2 3 4 5 6 7 8 9 10

Year

EA

CEAC (Before tax)

EAC (After tax)

Economic Life: before and after tax calculation (Feller Buncher data)

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Replacement Age Study: Alcao Vehicles

Determining the optimal replacement age for two floor sweepers, a fork-lift truck, and a GM Suburban

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Example of the Use of AGE/CON:

In this case, optimal replacement age is after 7 years of operation

Determining Optimal Replacement Age for Municipal Dump Trucks

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A Further Example:

In this case, optimal replacement age is after 5 years of operation

Determining Optimal Replacement Age for 4x4 Pick-up Trucks

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Analysis of Alcao Vehicle #1:

Conclusion: Continue using this vehicle (i.e. do not replace yet)

Floor Sweeper

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Results

• For each of the four vehicles analyzed, the Equivalent Annual Cost continually decreased as the replacement age increased

• As a result, the optimal replacement policy is to continue using the vehicles

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Recommendations• Although the EAC is still decreasing as the

vehicles get older, the marginal decrease from year to year is getting smaller

• This suggests that the EAC may increase in the near future

• Therefore, this very same cost analysis should be performed on a yearly basis to determine the year where the EAC begins to increase

• It is at this point that the vehicles should be replaced

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$$

Time

total cost

operations

depreciation

inventory

downtime

Economic Life Model

Source: H. Greene & R.E. Knorr, Managing Public Equipment, American Public Works Association, Kansas City, 1989.

maintenance

Economic Life

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Establishing The Economic Life of a Fleet of Mobile Equipment

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Kiruna trucks - INCO

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Objectives

• Determine the optimal replacement policy for the fleet of Kiruna Haulage Trucks

• Introduce capital equipment replacement in an overall maintenance strategy

• Present a framework to be applied to other capital equipment fleets at Inco

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An Optimal Replacement Policy

• Ensures the efficient use of capital equipment by minimizing the total costs/age of a mobile fleet

• Allows for better planning in vehicle replacement

• Employs a structured decision making technique for replacement cycles

• Allows for flexibility and practicality in replacement decisions

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Model Variables• Two Major Components:

1. Ownership Costs

• A Acquisition Cost of the Kiruna Truck

• CCA Capital Cost Allowance, the depreciation rate of the truck

• Un Undepreciated Capital Cost in year n

• Sn Salvage Value in year n

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Model Variables2. Operating and Maintenance Cost

• Cj Maintenance Cost in year j, comprised of labour, material and other costs

– Labour costs work hours associated with a work order, multiplied by a standard rate

– Material costs total of all part costs used in performing maintenance, as recorded on a work order

– Other Costs Warranty, Subcontracting, and miscellaneous charges

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Values for the Model Variables• Recall from the model development, the key

variables in the model:

Variable Value A $1.8 M i 12% r 0.89286 Sn 0 CT 33.7% CCA 30%

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Forecasted O&M Costs• Forecasts presented below (in red):

Age of Truck O&M Costs % Increase 1 134,258 - 2 341,271 154 3 444,073 30 4 669,650 51 5 803,682 20 6 767,651 -4 7 806,034 5 8 846,335 5 9 888,652 5

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Bus Economic LifeTerms of reference:

To determine the economic life for the range of GMCbuses operated by the Société de Transport de la CommunautéUrbaine de Montréal (STCUM)

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Optimum Replacement

Age Total Cost

Maintenance and Operation

Cost

Fixed Cost

Ownership Cost

Bus Replacement Age

Ann

ual C

ost

Economic Life Calculations: Transit Fleet Replacement

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Table A:“High” Trend in Resale Values

Table B:“Low” Trend in Resale Values


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0.00.10.20.30.40.50.60.70.80.91.01.11.2

0 200000 400000 600000 800000 1000000km

$/km

c(t)

c(t) = 0.302 + 0.723 (cum.km/106)2


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0

12000

24000

36000

48000

60000

72000

84000

96000

0 400 800 1200 1600 2000Bus Number

km/y

rBus Utilization Trend

y = [9.11574031 x 104] - [9.722352231 x 101]x- [1.475740136 x 10-2]x2 + [7.403868263 x 10-4]x3

- [1.575402140 x 10-6]x4 + [9.488071440 x 10-10]x5

+ [4.192383586 x 10-13]x6 - [7.567287679 x 10-16]x7

+ [3.266561486 x 10-19]x8 - [4.786922264 x 10-23]x9

xHighestUtilized

LeastUtilized

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1) Newest 100 buses will travel 8,636,059 km each bus travels 86,361 kmcost/bus = 86,361 (0.303) = $26,197 in 1st year

2) Similarly next 100 buses each travel on average 78,093 km/yr in 2nd year cost/bus = 78,093 (0.313) = $24,472 in 2nd year


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1 2 3 4 19 20

A = $96,300

$26,197 $24,472 S = $1000BusAge

EAC = $30,031


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Low Resale Value Trend Acquisition cost at start of replacement cycle

Economic Life Calculation

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* Using high resale value trend+ Using low resale value trend

Overall Fleet Savings Summary

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www.ipamc.orgBy: Cheung- Ki Derek Siu Monique Ka Yee Ho

Optimizing Economic Life Decisions for Bus Fleets of the Town of Markham

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Background

• The analysis will cover the whole bus fleet, which consists of conventional and specialized buses.

• The conventional bus fleet is composed of 54 coaches, which provide 11 service routes.

• The specialized bus fleet is composed of 6 vehicles.

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Existing Problems

• Increase in costs due to frequent repair– Mechanical problems due to engine wear-out

• Purchasing problem– Shortage of new buses from the bus manufacturer

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Primary Analysis

• The two bus fleets had different number of buses and different utilization patterns. Therefore, the analysis was divided into two main parts:– Conventional Bus Fleet– Specialized Bus Fleet

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Conventional Bus Fleet Analysis

• AGE/CON was employed and the results were deduced from its outputs.

• EAC for 18 years, for interest rates from 0% to 20%, was obtained. The interest rate selected was equal to the prime rate of 6.75% at the time; i = 6% and 8% were considered, thus the optimal economic life determined was 13 years.

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Assumptions

• O&M Cost was obtained by adding up the cost of repair, fuel, tires, and cleaning.

• Parameters used in AGE/CON:– “Low” trend of resale value.– Utilization trend of year 2000: fitted to the

9th degree polynomial function.– Linear Operations and Maintenance Cost

function.

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AGE/CON Example

i = 6%

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Utilization Trend CurveU

sage

(km

)

Bus Number

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Equivalent Annual Cost (EAC)Eq

uiva

lent

Ann

ual C

ost (

$/ye

ar)

Year of Service

Minimum EAC in year 13

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Interpretation of Results

• The optimal solution’s (13 years) corresponding EAC is between $116,000 and $125,000.

• The current replacement policy (18 years) has a corresponding EAC between $120,000 and $129,000.

• Therefore, the potential cost savings are about $216,000/year (= $4,000 × 54 vehicles)

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Fixed future operating time, n periods

Replace with technologically improved equipment

Ct,n-TCt,1 Ct,2Cp,1 Cp,2 Cp,3

0 1 2 3 . . . T T+1 T+2 . . . n-1 n

Optimal Replacement Policy for Capital Equipment Taking Into Account Technological Improvement

A-Sp,T

Finite Planning Horizon (Section 4.5)

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Noranda Equipment Replacement System

B. Buttimore, A. Lim

Source: B. Buttimore, A. Lim, Applied Systems and Cybernetics, Edited by G.E. Lasker, Vol. II: Systems Concepts, Models and Methodology, Pergamon Press Inc., 1981, pp. 1069-1073

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Introduction

The Noranda Equipment Replacement System is designed to analyze an existing fleet of equipment over a specified period to determine, using a total discounted cost method, if or when individual units in the fleet should be replaced by a proposed new unit.

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1 2 4 63 5 7 8

existing shovel new shovel

Years

Time diagram for example

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System Input Includes

• Expected rate of return.

• Depreciation rate.

• Investment tax credit, capital cost allowance, CCA depreciation type, federal, provincial and mining tax rates.

• Inflation rates.

• Unit purchase year, and price.

• Unit yearly total operating and maintenance cost.

• Unit yearly total production.

• Unit yearly salvage values.

• Proposed replacement unit data.

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Conclusion

The Noranda Equipment Replacement System can be used to do the following analyses:

• Compare the productivity of individual units with a fleet.

• Find the “lemon” in a fleet of equipment.

• Calculate the optimum year to replace a unit.

• Sensitivity testing to see what effect rate of return, taxes or production, for example, have on replacement timing.

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Cp,1 Cp,2 Cp,T Ct,1 Ct,2 Ct,n Ct, Ct, Ct,n

Total discounted costs C(T,n)

Replacement with technologically improved

Optimal Replacement Policy for Capital Equipment Taking Into Account Technological Improvement:

Infinite Planning Horizon

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Caterpillar 992D Wheel Loader

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(Years)

R A-Sp,T A-Sn A-Sn

0 1 2 … T-1 T 1 2 … n-1 n n+1 n+2 …n-1 2n

Cp,1 Cp,2 Cp,3 … Cp,T Ct,1 Ct,2 Ct,3 … Ct,n Ct,1 Ct,2 Ct,3 … Ct,n

Today

Cash Flows Associated with Acquiring New Equipment at Time ‘T’

Repair vs. Replace

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A = $ 1,083,233 (Unit operational)

(Years)

R A-Sp,T A-Sn A-Sn

0 1 2 … T-1 T 1 2 … n-1 n 1 2 . . . n-1 2n

Cp,1 Cp,2 Cp,3 …Cp,T Ct,1 Ct,2 Ct,3 … Ct,n Ct,1 Ct,2 Ct,3 … Ct,n

Today Cash Flows Associated with Acquiring New Equipment at Time T

R = $390,000 (includes arms + Z•bars) Excludes Front Frame, Rear Frame and Bucket components

Cp,1 = $ 138,592 Cp,2 = $ 238,033 Cp,3 = $ 282,033

Sp,0 = $ 300,000 Sp,1 = $ 400,000 Sp,2 = $ 350,000 Sp,3 = $ 325,000

Variable Maintenance Costs(excludes operator, fuel, wear parts)

Ct,1 = $ 38,188 Ct,2 = $ 218,583 Ct,3 = $ 443,593 Ct,4 = $ 238,830

S1 = $ 742,500 S2 = $ 624,000 S3 = $ 588,000 S4 = $ 450,000

••••

••••

Repair vs. Replace

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The Solution

(Years)

R A-Sp,T A-Sn A-Sn

0 1 2 … T-1 T 1 2 … n-1 n 1 2 . . . n-1 2n

Cp,1 Cp,2 Cp,3 …Cp,T Ct,1 Ct,2 Ct,3 … Ct,n Ct,1 Ct,2 Ct,3 … Ct,n

Today Cash Flows Associated with Acquiring New Equipment at Time T

Change-over time to new loader, TT = 0 T = 1 T = 2 T = 3

OverallEAC ($)

449,074 456,744 444,334 435,237

Note: n = 11 yrs

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The Bad News

Most likely additional component to be replaced at re-build: Front frame ($99,870)

∴R = $ 489,970 (was $ 390,000)

Change-over time to new loader, T

T = 0 T = 1 T = 2 T = 3

Overall

EAC ($)

449,074 471,725 459,319 450,217

Minimum

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The Life Cycle Cost Iceberg

Source: B.S. Blanchard and W.J. Fabrycky, Systems Engineering and Analysis, Prentice Hall, 1990

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Example: Nimrod Aircraft• BAE Systems signed a contract with the Ministry of

Defence to supply an upgraded fleet of Nimrod aircrafts for the year 2001

• Initial estimated cost of project was £2 billion• However, cost of project definition (initial stage of

system planning) was only £24 million• Consequence: Initial estimated cost of the project was

too optimistic, resulting in a lack of funding to complete the programme

• BAE Systems now uses a life-cycle management approach

Reference: Professional Engineering, May 2001

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Whole Life Costing: B- 52

From: Professional Engineer, p 59, May 2002

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Annual Maintenance Cost Limit

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‘As New’ Condition

Equi

pmen

t Co

nditi

on

Overhaul and Repair Consequences

Trend in Equipment Condition

O/H

Repair

O/H

O/H

Replace

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I

Age of vehicle

n years to end of planning horizon

JEstimated cost, (X) greater than AMCL, (LI) for a vehicle of age I

= 1Age of vehicle

JEstimated cost, (X) less than or equal to AMCL, (LI) for a vehicle of age I

= I + 1

Estimate, (X), of maintenance cost

in next year.Distribution of

this cost of fI(X)

Age of vehicle

n-1 years to end of planning horizon

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Consider a vehicle of age 5 years…

If buy new vehicle and sell 5 year old vehicle:

Cost = $2800 – 15% of $2000

= $2500 (± Bal. Adj.)

0

MEAN MAINTENANCE COST = $700

AMCL = L5 = $1055300 1500

17%

$

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I

5 YR OLD

Estimate

10 YRS TO GO

L5= $1055J

Cost = $2500

= 1 YR OLD

PROB. = 17%

L5= $1055J

Cost = $700

= 6 YR OLD

PROB. = 83%

EXP. COST = $2500 (0.17) + $700 (0.83) = $1006

0

MEAN MAINTENANCE COST = $700

AMCL = L5 = $1055300 1500

17%

$

If buy new vehicle and sell 5 year old vehicle:

Cost = $2800 – 15% of $2000

= $2500 (± Bal. Adj.)

Recall

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10 YR TO GO 9 YR 8 YR

76543210

AGE

… 2 YR 1 YR 0 YR

76543210

KEEP

REPLACE

K

R

K

R

K

R

K

R

R

For each vehicle age, there are two alternatives – keep or replaceTherefore, over 10 year period, there are:

210 = 1024 “Paths” – for each vehicle ageDecision on whether to keep or replace depends on AMCL – which can an almost infinite number of values.

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Asset Replacement(Optimizing Life Cycle Decisions)

Software:•AGE/CON (Mobile equipment)•PERDEC (Fixed equipment)

•www.banak-inc.com

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capital equipment replacement decisions

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