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2. Need for Replacement Study. Reduced Performance:Reduction in quality, reliability and productivity due to wear and tearAltered Requirements:New production needs, accuracy, speed, etc.Obsolescence:Not competitive. 3. Terminology. Defender Asset:Currently installed asset Challenger Asset:

Slide Set 7: Ch. 5 Replacement Analysis

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**1. **1 Slide Set 7: (Ch. 5) Replacement Analysis

**2. **2 Need for Replacement Study Reduced Performance:
Reduction in quality, reliability and productivity due to wear and tear
Altered Requirements:
New production needs, accuracy, speed, etc.
Obsolescence:
Not competitive

**3. **3 Terminology Defender Asset:
Currently installed asset
Challenger Asset:
Potential replacement, to the current defender (possibly “best” among many challengers)
AW values are used, often referred to as EUAC (equivalent uniform annual cost)
EUAC would be the negative of AW, as it treats costs as positive and revenues/salvage as negative

**4. **4 The Proper Viewpoint Consultant’s viewpoint
The analyst assumes that he/she is an outsider (a consultant)
Thus, the analyst owns neither asset and must assign reasonable value to the defender and challenger
Self-study: See Section 5.11.1 and e.g. 5.32 for an “insider viewpoint” cash-flow approach; compare the incremental cash-flows with those in e.g. 5.35 (tables 5.19, 5.22)

**5. **5 First Costs of Defender/Challenger The outsider viewpoint dictates that the first costs are estimated as follows:
The proper first cost to apply to keeping the defender in service for it’s remaining life- is its current Market Value. Note that market value can be different from an depreciated asset value.
The proper first cost to apply to a challenger asset that might replace the current defender asset – is its investment cost P, which may be reduced if defender could actually be traded-in for value higher than its estimated market value, in this case it is P – (Trade-in Value – Market Value)defender

**6. **6 First costs - summary Defender First Cost
MVdefender
Challenger first cost
Amount of capital used to replace the defender with the challenger
Usually equals challenger initial investment, P
Sometimes equal to P - (TIV - MV)defender
A meaningful BE question here: What minimum market value of the defender will make the current challenger economically attractive?

**7. **Outsider’s viewpoint- illustration Suppose a company is considering replacing an equipment (D) with a new and more sophisticated one (C). The undepreciated value is $12,600; however, due to its obsolescence, its market value is only $9000. The current equipment has a 5 year remaining life. The annual M&O costs of (D) are $7000. The challenger has a 10 year life, initial cost of $36,000, M&O 9000 and salvage value reducing by $3600 in each year. Assuming a 5 year planning horizon, either retaining the existing equipment or replacing it right now, develop the cash flows from an outsider’s viewpoint. 7

**8. **8

**9. **Fixed planning horizon approach Fix planning horizon and enumerate succession options.
Assume defender (current asset) will be replaced immediately with challenger, one year from now, two years from now etc.
This will result in mutually exclusive replacement options over the planning horizon and you can choose the replacement option with the lowest annual cost (we compute EUAC = -AW) 9

**10. **Example An oil field equipment placed into service 5 years ago is up for a replacement study. The current equipment can serve for 2, or 3, or 4 more years before it is replaced. AOC is $25,000 per year and constant. This equipment has a current MV of $100,000 and it decreases by $25,000 every year. The replacement challenger is a fixed price contract that provides the same service at $60,000 per year. Use 12% MARR to perform replacement study over a 6 year study period to determine when to sell the current equipment and purchase the new services. 10

**11. **Solution Enumerate the succession options over 6year study period:
(X) Defender replaced 2yrs from now;
(Y) Defender replaced 3yrs from now;
(Z) Defender replaced 4yrs from now;
X = {2D,4C}
NCF in years {0,1,2,3,4,5,6}
= {-100K, -25K,-25K+50K,-60K,-60K,-60K,-60K}
AWX = [-100K-25K(P/A,12%,2) +50K(P/F,12%,2)](A/P,12%,6) -60k(F/A,12%,4)(A/F,12%,6) = -60.24K
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**12. **12 X = {2D,4C}
EUAC = -AWX = 60.24K
Y = {3D,3C}
NCF in years {0,1,2,3,4,5,6}
= {-100K,-25K,-25K,-25K+25K,-60K,-60K,-60K}
EUACY = [100K+25K(P/A,12%,2)](A/P,12%,6) +60K(F/A,12%,3)(A/F,12%,6) = 59.55K
Z = {4D,2C}
NCF in years {0,1,2,3,4,5,6}
= {-100K,-25K,-25K,-25K,-25K,-60K,-60K}
EUACZ = [100K+25K(P/A,12%,4)](A/P,12%,6) +60K(F/A,12%,2)(A/F,12%,6) = 58.47K

**13. **Optimal service life approach For each alternative, defender and challenger, (with outsider’s viewpoint for the defender- that still holds!) do the following. Assume that alternative will be in service for K years, for different K values, 0,1,2,3 ...
Find EUAC(K), which is the equivalent uniform annual cost (-AW) over the K years the asset is in service.
Find out for which K, EUAC(K) is the smallest, that K is the optimal service life (OSL) of that alternative
You would have KD and EUAC(KD) for the defender; and KC and EUAC(KC) for the challenger
Note that they are independent of each other
13

**14. **OSL intuition 14

**15. **Optimal replacement time approach –contd. If EUAC(KD) is smaller, decide keep defender for KD more years
If EUAC(KC) is smaller, dispose defender now, and replace with challenger to be retained for KC more years
Note that there is an implicit assumption here that the planning horizon is an integer multiple of LCM and all the cash-flows are accurate forever
Ideally, replacement analysis should be done every year with the most accurate cost estimates available
Typically, people would rather not do this analysis “fearing” that they may have to actually replace!
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**16. **Example Challenger:
First cost: $50,000
Future market values decreasing by 20% per year
Estimated retention period: no more than 5 years
AOC: $5000 in year 1, increasing by $2000 per year
Defender:
Current MV: $15,000
Future market values decreasing by 20% per year
Estimated retention period no more than 3 years
AOC: $4000 next year, increasing by $4000 per year thereafter, plus the $16,000 retro-fit next year
MARR = 10%
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**17. **17

**18. **Sample calculations Defender, k = 1
EUAC(1) = 15,000(A/P,10%,1) + 20K – 12K = 24,500
Defender, k = 3
EUAC(3) = 15,000(A/P,10%,3) +20K(P/F,10%,1)(A/P,10%,3) +8K(P/F,10%,2)(A/P,10%,3) +12K(P/F,10%,3)(A/P,10%,3) -7680(A/F,10%,3) = 17,307
Challenger, k = 4
EUAC(4) = 50K(A/P,10%,4)-20480(A/F,10%,4) + [5K + 2K(A/G,10%,4)] = 19,123
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**19. **19 Conclusion Defender: OSL = 3, EUACD = 17,307
Challenger: OSL = 4, EUACC = 19,123
EUACC > EUACD = 17,307
Use Defender for 3 more years and replace with challenger
This conclusion assumes costs are not going to change and the estimates are perfect!
In reality, you should use defender for one more year and redo the analysis.

**20. **20 OSL analysis Summary Given: (C) and (D) find OSL (nC , nD), and the corresponding EUAC at OSL: EUAC(nC) , EUAC(nD)
EUAC(nC) vs. EUAC(nD)
Select the better alternative, i.e.,
Stay with the Defender for nD years or,
Go with the challenger for nC years.