CHAPTER 7
Replacement and Retention Decisions
7.1 INTRODUCTION
Business organizations use physical assets in the form of machines and equipment to produce goods and services. After a particular period of service, there might be a need for the replacement of the existing assets and therefore, the firms must constantly monitor the performance of the assets so as to decide whether they should be continued in service or they should be replaced with new assets. Replacement study is carried out to make economic decision to retain or replace an existing asset. If the decision is to replace, the study is complete. However, if the decision is to retain, the cost estimates and decision will be reconsidered each year to ensure that the decision to retain is still economically correct.
A replacement study involves an application of the A.W. method of comparing alternatives having different useful lives. In situations, where replacement study is required to be performed for no specified study period, the A.W. values are determined by a technique of cost evaluation called the economic service life (ESL) analysis. However, there is a different procedure for conducting replacement study for cases where study period is specified. All these procedures are discussed in this chapter.
7.2 REASONS FOR REPLACEMENT
Replacement study can be required for the following reasons:
Reduced Performance: Due to physical impairment/deterioration or accident, the ability of the asset to perform its intended functions reliably reduces. It leads to increased costs of operation, higher scrap and rework costs, lost sales, reduced quality and higher maintenance costs.
Altered Requirements: In situations, where the existing asset can not meet the new requirements of accuracy, speed, and other specifications.
Obsolescence: In the presence of a technologically advanced machine/equipment, the existing asset becomes obsolete. The product development cycle time to bring new products to market is getting shorter and this necessitates the replacement study before the estimated useful life of the asset is reached.
7.3 TERMINOLOGIES USED IN REPLACEMENT STUDY
Before the procedure for replacement study is explained, it is important to understand the following terms:
Defender and Challenger: Defender and challenger are two mutually exclusive alternatives, and a replacement study compares these two alternatives. The existing asset is called the defender, whereas its potential replacement is known as a challenger.
A.W. Values: The defender and challenger are compared on the basis of annual worth (A.W.) values. Since often, only costs are included in the computation of A.W. values, the term equivalent uniform annual cost (EUAC) may be used in lieu of A.W.
Economic Service Life (ESL): The number of years at which the lowest A.W. of cost of an asset occurs is called its economic service life (ESL). The equivalency calculations to determine ESL establish the life n for the best challenger, and it also establishes the lowest cost life for the defender in a replacement study.
Defender First Cost: It represents the investment amount P used for the defender. In replacement study, the value of P is correctly estimated by the current market value (MV) of the defender. The estimated salvage value of the defender at the end of a particular year becomes its MV at the beginning of the next year, provided the estimates remain correct as the years pass. If the defender must be upgraded or augmented to make it equivalent to challenger, this cost is added to the MV to obtain the estimate of defender first cost.
Challenger First Cost: It represents the amount of capital that must be recovered when replacing a defender with a challenger. It is indeed, equal to P, the first cost of challenger. There may be occasions when high trade-in value (TIV) may be offered for defender. In this event, the correct amount to recover and use in economic analysis for the challenger is P – (TIV – MV). It is obvious that when the trade-in value and market value are the same, P is used as challenger first cost.
7.4 ECONOMIC SERVICE LIFE
The economic service life (ESL) is defined as the number of years n at which the equivalent annual worth (A.W.) of costs is the minimum, considering the most current cost estimates over all possible years during which the asset may provide the needed service. ESL indeed, represents the estimated life of an asset and this life should be used in engineering economy studies. When n years have passed, the ESL indicates that the asset should be replaced to minimize overall costs. In replacement study, it is required to determine ESL of both defender and challenger.
While an asset is in service, its ESL is determined by calculating the total A.W. of costs in 1 year, 2 years, 3 years, and so on of its service. Total A.W. of costs is the sum of capital recovery (CR), which is the A.W. of the initial investment and any salvage value, and the A.W. of the estimated annual operating cost (A.O.C.), i.e.,
Total A.W. = – Capital recovery – A.W. of annual operating costs
CR is calculated from the following formula:
Figure 7.1 shows the variation of CR, A.W. of A.O.C. and total A.W. of costs with the life of the asset. It can be seen from this figure that CR decreases with each year of life of the asset. However, A.W. of A.O.C. increases with the life of the asset. The figure also exhibits that the total A.W. of costs decreases in the early life of the asset, attains a minimum value and then increases as the asset gets older. The life of the asset at which total A.W. of costs is the minimum is its ESL. In order to calculate A.W. of A.O.C., determine the present worth of each A.O.C. value with the P/F factor, then redistribute this P value over the entire life of the asset using the A/P factor.
Fig. 7.1 A.W. curves of costs and ESL
The following equation is used to calculate A.W. of costs over k years:
where, P = Initial investment or current market value
SK = Salvage value or market value after k years
A.O.C.j = Annual operating cost for year j (j = 1 to k)
The current MV is used for P when the asset is the defender, and the estimated future MV values are substituted for the S values in years 1, 2, 3, … … …
The following example illustrates the determination of ESL.
Example 7.1 A small manufacturing asset which is 3 years old is being considered for early replacement. Its current market value is 
65,000. Estimated future market values and annual operating costs for the next 5 years are given below. What is the economic service life of this defender if the interest rate is 10% per year?
| Year, j | Market Value, MVj () |
Annual Operating Costs, A.O.C.j () |
|---|---|---|
| 1 | 45,000 | –12,500 |
| 2 | 40,000 | –13,500 |
| 3 | 30,000 | –15,000 |
| 4 | 10,000 | –17,500 |
| 5 | 0 | –22,500 |
Solution Equation (7.3) is used to calculate total A.W.k, for k = 1, 2, 3, 4 and 5.
Year 1
| Total A.W.1 | = | –65,000(A/P, 10%, 1) + 45,000(A/F, 10%, 1) –12,500 |
| = | –65,000(1.10000) + 45,000(1.0000) |
|
| = | –39,000 |
Year 2
| Total A.W.2 | = | –65,000(A/P, 10%, 2) + 40,000 (A/F, 10%, 2) – [12,500(P/F, 10%, 1) + 13,500(P/F, 10%, 2)](A/P, 10%, 2) |
| = | –65,000(0.57619) + 40,000(0.47619) – [12,500(0.9091) + 13,500(0.8264)]
(0.57619) |
|
| = | –31,380.6 |
Year 3
| Total A.W.3 | = | –65,000(A/P, 10%, 3) + 30,000 (A/F, 10%, 3) – [12,500(P/F, 10%, 1) + 13,500 (P/F, 10%, 2) + 15,000(P/F, 10, 3)] (A/P, 10%,3) |
| = | –65,000(0.40211) + 30,000(0.30211) – [12,500(0.9091) + 13,500(0.8264) + 15,000 (0.7513)](0.40211) |
|
| = | –30,661.01 |
Year 4
| Total A.W.4 | = | –65,000(A/P, 10%, 4) + 10,000(A/F, 10%, 4) – [12,500(P/F, 10%, 1) + 13,500 (P/F, 10%, 2) + 15,000(P/F, 10, 3) + 17,500(P/F, 10%, 4)] (A/P, 10%,4) |
| = | –65,000(0.31547) + 10,000(0.21547) – [12,500(0.9091) + 13,500(0.8264) + 15,000(0.7513) + 17,500(0.6830)](0.31547) |
|
| = | –32,781.13 |
Year 5
| Total A.W.5 | = | –65,000(A/P, 10%, 5) + 0(A/F, 10%, 5) – [12,500 (P/F, 10%, 1) + 13,500 (P/F, 10%, 2) + 15,000 (P/F, 10, 3) + 17,500(P/F, 10%, 4) + 22,500 (P/F, 10%, 5)] (A/P, 10%, 5) |
| = | –65,000(0.26380) – [12,500(0.9091) + 13,500(0.8264) + 15,000(0.7513) + 17,500 (0.6830) + 22,500(0.6209)](0.26380) |
|
| = | –32,899.13 |
Thus, the ESL of the asset is 3 years as the total A.W. is the minimum.
The problems discussed in the previous chapters had an estimated life of n years with other associated estimates such as first cost in year 0, possibly salvage value in year n, and an A.O.C. that remained constant or varied each year. For all such problems, the calculation of A.W. using these estimates determined the A.W. over n years. This is also the ESL when n is fixed. In all previous problems, there were no year-by-year MV estimates applicable over the years. Thus, we can conclude that when the expected life is known for the challenger or defender, determine its A.W. over n years, using the first cost or current market value, estimated salvage value after n years, and A.O.C. estimates. This A.W. is used in the replacement study.
The following points should be remembered about n and A.W. values to be used in a replacement study.
- In situations where year-by-year market value estimates are made, use them to perform an ESL analysis, and determine the n value with the lowest total A.W. of costs. These are the best n and A.W. values for the replacement study.
- In cases, where yearly market value estimates are not made, use it to calculate the A.W. over n years and these values of n and A.W. are used in the replacement study.
7.5 PROCEDURE FOR PERFORMING REPLACEMENT STUDY
Replacement studies are performed in one of the two ways: without a study period specified or, with a defined study period. This section describes the procedure for making replacement study when no study period (planning horizon) is specified. The procedure for making replacement study when study period is specified is discussed in the subsequent section.
A replacement study determines when a challenger replaces the in-use defender. If the challenger (C) is selected to replace the defender (D), now then the replacement study is finished. However, if the defender is retained now, the study may extend over a number of years equal to the life of the defender nD, after which a challenger replaces the defender. Use the annual worth and life values for C and D determined in the ESL analysis to apply the following replacement study procedure. It assumes that the services provided by the defender could be obtained at the A.W.D amount.
- Determine A.W.C and A.W.D values using the estimates of challenger and defender respectively. Based on the better A.W.C or A.W.D value, select the challenger or defender. When the challenger is selected, replace the defender now, and keep the challenger for nC years. This replacement study is complete. If the defender is selected, plan to retain it for up to nD more years. Next year, perform the following steps.
- After one year look at the estimates, especially first cost, market value, and A.O.C. of both challenger and defender. If these estimates are not current estimates, proceed to Step 3. However, if the estimates are current estimates and this is year nD, replace the defender. If this is not year nD, retain the defender for another year and repeat the same step. This step may be repeated several times.
- Whenever the estimates have changed, update them and determine new A.W.C and A.W.D values. Initiate a new replacement study (Step 1).
If the defender is selected initially (Step 1), estimates may need updating after 1 year of retention (Step 2). It may be possible that a new best challenger is available and therefore it should be compared with the defender. Either significant changes in defender estimates or availability of a new challenger indicates that a new replacement study is to be performed. In actual sense, replacement study is performed each year to take decision about replacing or retaining any defender, provided a competitive challenger is available.
The following example illustrates the procedure discussed above for performing replacement study.
Example 7.2 Five years ago, a company purchased a small welding machine for 2,25,000. It is expected to have the market values and annual operating costs shown for the rest of its useful life of upto 3 years. It could be traded now at an appraised market value of 40,000.
| Year | Market Value at the End of Year () |
A.O.C.() |
|---|---|---|
| 1 | 30,000 | –2,50,000 |
| 2 | 20,000 | –2,60,000 |
| 3 | 5,000 | –3,00,000 |
A replacement welding machine costing 6,25,000 has an estimated 50,000 salvage value after its 5-year life and an A.O.C. of 1,55,000 per year. At an interest rate of 15% per year, determine how many years the company should retain the present welding machine.
Solution Find ESL of the defender; compare with A.W.C over 5 years.
| For n = 1: A.W.D | = | –40,000(A/P, 15%, 1) –2,50,000 + 30,000 (A/F, 15%, 1) |
| = | –40,000(1.15) –2,50,000 + 30,000(1.0000) |
|
| = | –2,66,000 |
|
| For n = 2: A.W.D | = | –40,000(A/P, 15%, 2) –2,50,000 + (–15,000 + 20,000)(A/F, 15%, 2) |
| = | –40,000 (0.61512) – 2,50,000 + 5,000(0.46512) |
|
| = | –2,72,279 |
|
| For n = 3: A.W.D | = | –40,000(A/P, 15%,3) – [2,50,000(P/F, 15%, 1) + 2,65,000(P/F, 15%, 2)] (A/P, 15%, 3) + (–3,00,000 + 5,000)(A/F, 15%, 3) |
| = | –40,000 (0.43798) – [2,50,000(0.8696) + 2,65,000(0.7561)] (0.43798) –2,95,000(0.28798) |
|
| = | –2,85,447 |
The ESL is now 1 year with A.WD = –2,66,000
| A.W.C | = | –6,25,000 (A/P, 15%, 5) – 1,55,000 + 50,000 (A/F, 15%, 5) |
| = | –6,25,000(0.29832) – 1,55,000 + 50,000 (0.14832) |
|
| = | –3,34,034 |
Since the ESL, A.W. value is lower than the challenger A.W., the company should keep the defender now and replace it after 1 year.
7.6 REPLACEMENT STUDY OVER A SPECIFIED STUDY PERIOD
In cases, where replacement study is performed over a specified study period, for example, 5 years, the determination of A.W. values for the challenger and for the remaining life of the defender are usually not based on the economic service life. It is assumed that the services of the alternatives are not needed beyond the study period and therefore, what happens to them after the study period is not considered in the replacement analysis.
When performing a replacement study over a fixed study period, it is essentially important that the estimates used to determine the A.W. values be accurate and used in the study. When the defender's remaining-life is shorter than the study-period, the cost of providing the defender's services from the end of its expected remaining life to the end of the study period must be estimated as accurately as possible and included in the replacement study.
In order to perform replacement study over a specified study-period, the P.W. or A.W. for each option i.e. defender and challenger is calculated over the study-period. The option with the lowest cost or highest income if revenues are estimated is then selected.
The following examples illustrate the procedure for performing replacement study over a specified period:
Example 7.3 Three years ago, Delhi Fire Service purchased a new fire truck. Because of increase in fire cases, new fire-fighting capacity is needed once again. An additional fire truck of the same capacity can be purchased now, or a double capacity truck can replace the current fire truck. Estimates are given below. Compare the options at 10% per year using a 10-year study period.
Solution Let us identify Option 1 as retention of the presently owned truck and augmentation with a new truck of same capacity. Option 2 as replacement with the double-capacity truck.
For a full-life 10-year study period of Option 1
Retain the currently owned fire truck and perform the replacement study next year.
Example 7.4 An oil exploring company placed an equipment into service 5 years ago for which a replacement study is required. It has been decided that the current equipment will have to serve for either two, three, or four more years before replacement. The equipment has a current market value of 5,00,000 which is expected to decrease by 1,25,000 per year. The A.O.C. is expected to remain constant at 1,25,000 per year. The replacement challenger is a fixed-price contract to provide the same services at 3,00,000 per year for a minimum of 2 years and a maximum of 5 years. Use M.A.R.R. of 10% per year to perform replacement study over a 6-year period to determine when to sell the current equipment and purchase the contract services.
Solution Since the defender will be retained for two, three or four years, there are three feasible options (A, B, and C)
| Option | Defender Retained | Challenger Serves |
|---|---|---|
| A | 2 years | 4 years |
| B | 3 | 3 |
| C | 4 | 2 |
The A.W. values of the defender for two, three, and four years are
| A.W.D2 | = | –5,00,000(A/P, 10%, 2) + 2,50,000(A/F, 10%, 2) – 1,25,000 |
| = | –5,00,000(0.57619) + 2,50,000(0.47619) – 1,25,000 | |
| = | –2,94,048 |
|
| A.W.D3 | = | –5,00,000(A/P, 10%, 3) + 1,25,000(A/F, 10%, 3) – 1,25,000 |
| = | –5,00,000(0.40211) + 1,25,000(0.30211) – 1,25,000 | |
| = | –2,50,528 |
|
| A.W.D4 | = | –5,00,000(A/P, 10%, 4) – 1,25,000 |
| = | –5,00,000(0.31547) – 1,25,000 | |
| = | –2,82,735 |
For all options, the challenger has A.W. of
A.W.C = –3,00,000
Table on next page shows the cash flows and P. W values for each option over the 6-year study period.
P.W. computation for options A, B, and C are shown below:
| P.W.A | = | –2,94,048(P/A, 10%, 2) – 3,00,000(F/A, 10%, 4)(P/F, 10%, 6) |
| = | –2,94,048(1.7355) – 3,00,000(4.6410)(0.5645) | |
| = | –12,96,274 |
| P.W.B | = | –2,50,528(P/A, 10%, 3) – 3,00,000(F/A, 10%, 3)(P/F, 10%, 6) |
| = | –2,50,528(2.4869) – 3,00,000(3.3100)(0.5645) | |
| = | –11,83,587 |
| P.W.C | = | –2,82,735(P/A, 10%, 4) – 3,00,000(F/A, 10%, 2)(P/F, 10%, 6) |
| = | –2,82,735(3.1699) – 3,00,000(2.1000)(0.5645) | |
| = | –12,51,877 |
Option B has the lowest cost P.W. value (– 11,83,587). Keep the defender for 3 years, then replace it. It should be noted that the same answer will result if the annual worth, or future worth, of each option is calculated at the given M.A.R.R. as shown below:
PROBLEMS
- Anas Khan wishes to replace his Scoda car, which he bought for 8,00,000 two years back, with a new Toyota brand. The Toyota costs (MRP) 12,00,000. He expects 3,00,000 for his Scoda. He approaches a Toyota showroom which gives him a replacement offer of 3,85,000 for the scoda which is 25,000 more than the currently owned Scoda model and a discount of 1,50,000 on the MRP of the new Toyota brand. The showroom further offers free insurance and accessories with 65,000. In case Anas wishes to consider evaluating the replacement offer, what is the correct first cost for (a) the defender and (b) the challenger.
- Eveready Industries Ltd bought a testing and inspection machine for 80,00,000 two years back. At the time of its purchase, the company expected to use it for 5 years. The company also expected that the machine could be salvaged for 9,00,000 after 5 years. The fast developments in its business and high competition forced the company to procure a new machine for 1.0 cr. In case the company wishes to sell the machine now, it would fetch the company 42,00,000. The company, although if so wish may retain the machine for two more years at which time the salvage value of the machine is estimated at 80,000. If the company retains the machine for two more years, the company expects that it would have to spend 1,20,000 annually as maintenance and overhead cost on the machine (other than operating cost). Determine the values of P, n, S and A.O.C. for this defender if a replacement analysis were to be performed today.
- Zoya purchased a keyboard one year ago for 2,50,000 which consumes more battery than expected. She purchased the keyboard with a view to use it for 5 years with annual battery consumption worth 2,500 and a salvage value of 20,000. However, she had to spend 12,000 towards some maintenance during the previous year. She expects that the keyboard now requires periodic maintenance with the present year's maintenance cost of 13,500 and an increase of 1,500 each year there after. The salvage value is estimated to follow the relationship S = 20,000 – 3,500C where S is the salvage value and C is the number of years since the keyboard was purchased. It is now estimated that the keyboard would be useful for a maximum of 2 more years. Determine the values of P, A.O.C., n and S for a replacement study performed now.
- Toyama Electric commissioned a new SMT system in its manufacturing unit two years ago for 2.5 cr and an expected life of 5 years. At the time of installation, its market value was described as (2,10,00,000 – 15,00,000C1.3), where C was the number of years since purchase. It was estimated that O&M cost of the SMT system follows a relationship: 25,000+100C2.5. Determine the values of P, S and A.O.C. for this defender if a replacement analysis is performed (a) now with a study period of 3 years specified and (b) 2 years from now with no study period specified.
- Wockhardt Ltd wishes to replace one of the incubators installed in its plant with a new one at an estimated first cost of 45,00,000 and an annual operating cost of 6,40,000 and a maximum useful life of 5 years. The salvage value of the incubator any time it is sold, is expected to be 4,80,000. At an interest rate of 12% per year, determine its economic service life and corresponding A.W. value.
- A heavy mining equipment with a first cost of 9,00,00,000 is expected to have a useful life of 10 years and a market value that decreases by 9,00,000 every year. The annual operating cost of the equipment is to remain constant at 9,00,000 for 5 years and to increase 35% per year every year thereafter. Unitech Limited which owns the equipment, manages to finance at a low interest rate of 7% per year. (a) Verify that the ESL is 5 years. (b) During a replacement analysis, the company finds that equipment would have an ESL of 10 years when it is evaluated against any challenger. If the estimated A.O.C. series has proved to be correct, determine the minimum market value that will make ESL equal to 10 years.
- Emad consulting estimated the following A.W. values for a precision inspection system owned by its client Sameer Aerospace.
Number of Years (To be Retained) A.W. Value ( /Year)1 –6,40,000 2 –5,00,000 3 –4,40,000 4 –5,50,000 5 7,50,000 A challenger has ESL = 2 year and A.W.C = –
4,80,000 per year. If the consultant must recommend a replacement/retain decision today, should the company purchase the challenger? The M.A.R.R. is 15% per year. - Sachin Sethi Movers Pvt. Ltd. is considering to replace a heavy duty crane which was purchased two years back for 5,00,00,000 with a new one. The existing crane is valued at present at 10,00,000. If the company decides to upgrade the existing crane, it would cost 3,50,00,000 and would be worthy for use for another 3 years of its operation after which the crane would have to be sold for an estimated value of 8,50,000. The challenger can be purchased at a cost of 6,00,00,000, has an expected life of 10 years and has a salvage value of 10,00,000. Determine whether the company should upgrade or replace at a M.A.R.R. of 12% per year. Assume the A.O.C. estimates are the same for both the options.
- Kitply Ltd, manufactures different types of wooden board and plies. It is evaluating the replacement alternative for one of its chemical treatment facility. The cost estimates for this possible replacement is given in the following table.
Current System New Chemical Treatment Facility First cost 8 years ago ( )–2,00,00,000 First cost ( )–3,85,00,000 Remaining life (Years) 5 10 Current value ( )24,00,000 A.O.C. ( /year)–65,00,000 –57,00,000 Future salvage ( )0 24,00,000 For the said replacement proposition (a) perform the replacement analysis, and (b) find the minimum resale value required to make the challenger replacement choice now. Is this a reasonable amount to expect for the current system? Use an interest rate of 12% per year.
- Rishi Laser Cutting Ltd. purchased a CNC WireEDM for 6,50,000 five years ago and its A.W. data for rest of its useful life of upto 3 years is shown in the following table:
Year Year-end Market Value ( )A.O.C. ( )1 1,20,000 –87,500 2 96,000 –99,000 3 56,000 –1,25,000 It is known that the machine can now be traded for
45,000. A new EDM with latest software and pulse control power source costs 11,00,000 with a 3,60,000 salvage value after five years and an A.O.C. of 55,000 per year. Find out for how many more years, the company should retain the present machine if interest rate is 10% per year. - A replacement analysis is being performed for an in-situ machining centre with an equivalent annual worth of –6,00,000 for each year of its maximum remaining useful life of two years. A suitable replacement is determined to have equivalent annual worth values of –8,25,000, –7,15,000 and –7,65,000 per year is kept for 1, 2 or 3 years respectively. If company uses a fixed 3-year planning period, find out when should the company replace the machine? Assume an interest rate of 15%.
- Two years ago, Cipla Ltd installed an emission control system for its medicine plant for 4,50,00,000 with an estimated salvage value of 3,00,000 after 10 years. Currently the expected service life is 8 years with an A.O.C. of 4,50,000 per year. Thermax Ltd has recently developed a challenger which costs 9,60,00,000 with an A.O.C. of 2,50,000 and has an estimated useful life of 15 years with a salvage value of 2,80,000. If the M.A.R.R. is 15% find (a) the minimum trade in value necessary now to make the challenger economically viable, (b) number of years to retain the defender upto break even if the trade in offer is 4,50,00,000. Assume that the salvage value can be realized for all retention periods up to 8 years.
- Three years ago, Mahindra & Mahindra Ltd purchased a multi-axis CNC machining centre for 5,00,00,000. To meet the demand to catch-up with competition, the company finds that either the centre be upgraded with new accessories or be replaced with a new machining centre. The cost of upgradation is 2,00,00,000 which enables the machine to be good for operation for 3 more years. In case the company goes for replacement, the existing centre may be traded for 1,10,00,000. The annual operating cost of the centre is 2,20,00,000. The replacement machine would cost 7,50,00,000, if installed would satisfactorily serve for at least 8 more years. The salvage value of the replaced machine would be 1,50,00,000 for year 1 through 5, 60,00,000 after 6 years and 30,00,000 thereafter. It will have an estimated A.O.C. of 5,50,000 per year. Using a 5-year time horizons, perform an economic analysis at 20% per year. Also recommend if the company should replace the existing machine now or do so after 3 years from now? What are the A.W. values?
- Three years ago, the Heart Centre, Lajpat Nagar New Delhi installed a Japanese robotic heart surgery facility for 3,50,00,000. The system can be satisfactorily used for 3 more years. The hospital is considering to evaluate a replacement option. If replaced the hospital can realize 50,00,000. The market value and operating cost of the present system and replacement system are tabulated as follows:
The replacement option, a Taiwan make, is although cheaper at 27,500,000, but has a considerably higher O&M costs in later years of service. It has a maximum life of 7 years. It is also anticipated that if replaced, the new system would incur a recurring cost, after 4 years, for periodic inspection by a service personnel from Taiwan costing 1,250,000 per year. Using the tabulated financial data and i = 15%, estimate ESL and A.W. values for the defender and challenger and also find in what year the current system be replaced.