   Replacement Analysis

Replacement Analysis

The analysis to determine whether an in-place item requires replacement due to:

• Obsolescence
• New requirements / Inadequacy
• Deterioration

Definitions

• Defender - Currently owned (in place) item.
• Challenger - The new possible replacement item /alternative.
• Outsider perspective - Analyst neither owns nor uses either the defender or challenger.
• Analysis uses EAC for comparison.
• First cost of  defender is :
• Current market value.
• Trade in value which is the actual value obtained. This is the actual first cost.
• The owner foregoes this amount of capital by not trading in the asset.
• Do not include "Sunk Costs" which are unrecoverable due to loss of value prior to beginning of study period.
• No past costs are used.

Bought 10 years ago for \$1,000,000
Trade in now                 \$    100,000 = First cost
Sunk costs                     \$    900,000

Example:

• Ex Machine bought 3 years ago for \$100,000.
• 8 years life remaining.
• Annual operating cost = \$23,000/year.
• Salvage value = \$10,000 (In 8 years)
• Sell existing machine for \$75,000
• Buy more efficient machine for \$150,000.
• New machine operating costs = \$10,000/yr
• New machine life = 8 years (with no salvage).

Keep old machine or replace with new? MARR = 10%

Defender      Challenger
P           \$75,000       \$150,000
AOC     \$23,000       \$10,000
S           \$10,000       \$0
N          8                  8

Note that defender first cost is the current price obtained by selling it.

Defender EACD = \$23,000 + \$75,000(A/P,10,8) -\$10,000(A/F,10,8)
= \$23,000 + \$75,000(0.18744) -\$10,000(0.08744)
= \$36,184

Challenger EACC = \$10,000 + \$150,000(A/P,10,8)
= \$10,000 + \$150,000(0.18744)
= \$38,116

The Defender is less cost. Keep the old machine.

Opportunity Cost vs. Cash Flow Approaches

We started by using P = Current market value for the first cost of the defender. However, there is another approach to use.

Cash Flow Approach

• Use defender current market or trade in value as a discount to the challenger first cost.
• This method works well if several vendors give different trade-in values.

Example: i = 10%

Defender                    Challenger 1        Challenger 2
First Cost         \$30,000 (2 years ago)         \$35,000               \$40,000
Trade-in           \$18,000 (Current value)      \$20,000               \$23,000
O & M            \$3,500                                \$1,500                 \$1,000
Salvage            \$1,000                                \$2,000                 \$5,000
Est Life          10 years                               10 years               10 years

EACD = \$3500 - \$1000(A/F,10,10) = \$3500 - \$1000(0.06275)
= \$3,437
Note that the  first cost was omitted from the EAC of the defender.

EACC1 = \$1500 + (35000 - 20000)(A/P,10,10) - \$2000 (A/F,10,10)
= \$1500 + (35000 - 20000)(0.16275) - \$2000 (0.06275)
= \$3,816
Note that the first cost of this challenger was reduced by the defender trade-in cost.

EACC2 = \$1000 + (40000 - 23000)(A/P,10,10) - \$5000(A/F,10,10)
= \$1000 + (40000 - 23000)(0.16275) - \$5000(0.06275)
= \$3,453
Again the trade-in cost was subtracted from the Challenger first cost.

Keep the Defender. Less cost.

Unequal Service Lives

If the asset will be needed indefinitely, assume "Repeat Replacement" and the defender will be replaced with the minimum cost replacement.
If asset replacement is a one-time occurrence, use a study period approach.

• Shorter study period approach with salvage life estimate on challenger.
* Use for projects with definitive end.
• Longer study period best to use on indefinite projects.
* Assume replacement can be found for defender at same cost.

Example: A quarry plans on opening a new pit that will last 5 years in production. An existing bulldozer with 5 years of service remaining can be used. The existing dozer has a book value of \$50,000, a salvage value of \$5,000 and O&M first year cost of \$10,000 which will increase at \$2,000 /year.
A new dozer can be purchased for \$100,000 with a 10 year service life. O&M costs for the first year are \$4,000 with \$500 /year increase. Estimated resale is \$40,000@ 5 years. Should the quarry buy the new dozer with i =10% ?

Existing             New
N           5 years           10 years
P           \$50,000         \$100,000
A'          \$10,000          \$4,000
G           \$2,000           \$500
S           \$5,000             \$40,000 @5 years

Choose shorter 5 year period. (This project has a definitive end. Why choose a longer N, just to buy a dozer that isn't needed ?)

EACD = -\$50,000(A/P,10,5) + \$5,000(A/F,10,5) - (\$10,000 + 2,000(A/G,10,5))
= -\$50,000(0.26380) +\$5,000(0.16380)- (\$10,000 + 2,000(1.81013))
= -\$25,991

EACC = -\$100,000(A/P,10,5) + \$40,000(A/F,10,5) - (\$4,000 + 500(A/G,10,5))
= -\$100,000(0.26380) + \$40,000(0.16380) - (\$4,000 + 500(1.81013))
= -\$24,733

If the \$40,000 salvage value is reliable, the quarry would be better suited to purchase the new dozer.
If a longer study period was used, a cost for a replacement would need to be estimated.

However, this would create the assumption of a required cost carrying on for 5 more years, but it wasn't needed due to project duration. Thus the longer study period would be inappropriate to use.

Cost of Capital Recovery - A Closer Look

Capital assets are purchased in the belief that they will earn more than they cost. The cost of these assets are "recovered" with the combination of :

1) Capital Recovery : Income from services rendered through use of capital expended.
2) Return : Interest on unrecovered capital that could have been invested elsewhere.

Consider the following example :

Current     : Manual operation costs = \$8,000 /year
Challenger : New machine cost = \$12,000
Annual operation cost = \$6,000 / year
Salvage  = \$3,000
i = 12%,   N = 6 year life Naive view

If we didn't know any better we might think that the :

Average cost = (12000 + 6(6000) - 3000) / 6 = \$7,500 / year

Hence the purchase looks advantageous when i = 0% or is not considered.
This is not Correct.

Actually we know that the average cost is :

EAUC = \$12,000(A/P,12,6) + \$6,000 - \$3,000(A/F,12,6)
= \$8549 / year

Thus, the equivalent cost of interest = \$8549 - \$7500 = \$1049 / year
Hence the current manual operation is economically more efficient when interest and capital recovery are considered.

Capital Recovery (CR)

The equivalent annual values of P and S are spread over N with i.

Cost of capital recovery CR = P(A/P, i, N) - S(A/F, i, N)

which translates to  CR = (P - S)(A/P, i, N) + Si

where Si is the annual amount of interest lost on future salvage cost and (P - S)(A/P, i, N) is the annual average amount of i weighted for cost of capital consumed.

For our example: CR = (12,000 - 3,000)(A/P,12,6) + 3000(0.12)
= \$2,549 / year (Interest weighted cost / year)

Average cost / year without interest = (12,000 - 3,000) / 6 = \$1,500 / year

Difference = \$2,549 - \$1,500 = \$1,049 = Average interest cost/year
= Dollar cost/year for opportunities forgone
= Cost of loss of use of money elsewhere.

Capital recovery can also be determined using the sinking fund equation where
CR = (P-S) (A/F,i,N) + Pi.
This equation will yield an identical result, but is sometimes used to determine the amount required to be saved to purchase a replacement in N years.

Example: A company buys a new machine for \$10,000. It can be salvaged for \$2,000 at the end of its 5 year life. How much money should be budgeted for replacement of the machine and compensation for lost interest of 6% ?

With Sinking fund :

CR = (10000 - 2000)(A/F,6,5) + 10000(6%)
= 8000(0.17740) + 10000(0.06)
= \$2019.20 /year

With Capital recovery :

CR = (10000-2000)(A/P,6,5)+2000(6%)
= 8000(0.2374)+2000(0.06)
= \$2019.20/year

Economic service life

How long should an asset be kept in service? The longer an asset is kept, the less the annual cost of capital recovery, but the higher the operation and maintenance cost. The economic service life occurs at the minimum total EAC (The sum of CR and O&M).  Thus, the EAC of ownership at any year j can be found by :

EACj = -P(A/P, i, j) - SV(A/F, i, j) + [(Sum from j =1 to N  of O&Mj)(P/F, i, j)](A/P, i, N)

Example: Determine economic service life for i = 8%, equipment purchase price = \$24,000, decline in salvage value = 20% / year, O&M cost = \$3,000 for the first year, O&M increase = \$1,000 /year.

Life           Salvage Value                  Capital recovery        EAC          Total
j, years              SVj               O&Mj            Cost                 O&M          EAC
1                \$19,200        \$3,000          \$6720                \$3000         \$9720
2                \$15,360         \$4,000          \$6074                \$3481         \$9555
3                \$12,288         \$5,000          \$5528                \$3949         \$9477
4                \$9,830           \$6,000          \$5065                \$4404         \$9469
5                \$7,864           \$7,000          \$4671                \$4847         \$9518
6                \$6,291           \$8,000          \$4334                \$5276         \$9610

The minimum total EAC is in year 4. Thus, the equipment should be sold after year 4 yielding a  four year economic service life.

As an example, let's see how the year 4 EAC is calculated.

Capital Recovery cost = \$24,000(A/P,8,4) - \$9,830(A/F,8,4)
= \$24,000(0.30192) - \$9,830(0.22192)
= \$5,065

EAC O&M (4) = [3000(P/F,8,1) + 4000(P/F,8,2) + 5000(P/F,8,3) + 6000(P/F,8,4)] (A/P,8,4)
= [3000(0.9259) + 4000(0.8573) + 5000(0.7938) + 6000(0.7350)] (0.30192)
= \$4,404
or in this case, since we have an arithmetic gradient

EAC O&M (4) = 3000 + 1000(A/G,10,4)
= 3000 + 1000(1.4090)
= \$4,404

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