Depreciation October 1, 2012

Observation: most things lose value over time… …though a few things do not…

Things lose value for several reasons: they wear out through the passing of time.. They wear out through use… …they become obsolete or out of fashion

We need to keep track of this wearing out for three reasons: 1.To keep track of the company’s value 2.To know when to replace them 3.We can deduct depreciation costs from company taxes. Note: we need not use the same formula for all three

Two main ways of calculating depreciation: (though there are others) Straight – Line Depreciation Declining balance depreciation

Straight-line depreciation is simple: If the asset has a life of N years, it loses 1/N of its original value every year N time Book Value

In some cases, the item depreciates to zero; In others, it depreciates to a salvage value, S N time P S

N Book Value S P n Annual Depreciation, D = (P-S)/N Book value, BV = P - nD BV

Declining-balance depreciation is slightly harder: The asset loses d% of its original value every year N time Book Value P

n time Book Value P Book value (BV) = P(1 – d) n BV D n = BV n-1 d

Sample Problem: A machine costs $10,000 when new. If it depreciates by straight-line depreciation over 10 years, and has a salvage value of $1,000 at the end of the tenth year, what is its book value after five years and what is its depreciation in the sixth year? If it instead depreciates 10% per year, what is its book value after five years and what is its depreciation in the sixth year?

Replacement Analysis I have a 2001 Honda CRV. This week I paid $1,400 in repairs. When should I replace it with a new car? (Based on a true story)

Background: the Bathtub Curve N (Years before replacement) EUAC Capitalised cost EUAC = P(A/P,i,N)

Background: the Bathtub Curve Years before replacement EUAC Capitalised cost Operating costs

Background: the Bathtub Curve Years before replacement EUAC Capitalised cost Operating costs Total

Background: the Bathtub Curve Years before replacement EUAC Total n n is the economic life of the asset

Terminology The current asset is the defender, its prospective replacement is the challenger. The economic life of a defender is the period over which its EUAC is the least. Method: calculate and compare the EUAC’s for all challengers, considering all possible dates for replacement.

Simplest Case: Retirement without Replacement This is the case where the asset generates some income, but is not vital to the continuation of the business. Example: video game in a corner store Method: calculate the PW’s of the asset and associated cash flows for all feasible lifetimes. If any of these PW’s are positive, keep it; otherwise, get rid of it.

Example: A supermarket has a mechanical horse that is nearing the end of its life. It won’t survive past the end of Year 2. It brings in $1,200/year in quarters, but this is expected to decline by $100/year as the neighbourhood kids grow up. Maintenance costs are expected to be $800 this year and $900 next year. Right now it has a salvage value of $600. After one year it will only be worth $300, and after that it’s worthless. If the supermarket’s MARR is 10%, how long should the horse be kept?

Second Case: Retirement with Identical Replacement If the technology of an asset changes slowly with time, we may be able to replace it with an identical item in the future (e.g., a wheelbarrow). In this case, we just look at the EUAC of the asset for various lifetimes and find its economic life. Is the defender younger than its economic life? If so, keep it. If it’s at the economic life, replace it. If it’s older than its economic life, calculate the cost of keeping it for 1, 2, … years and compare with the minimum EUAC

Example: A fork-life truck costing $10,000 has a maximum service life of 6 years. Its salvage value and operating costs are given by the following table: YearSalvage ($)Operating Costs ($) 17,0005,000 24,5005,400 32,5006,100 41,0007, , ,000

Solution: YearSnSn O. C.Capital RecoveryAv. Op. Cost.EUAC 17,0005,0004,0005,2469,246 24,5005,4003,6195,4469,065 32,5006,1003,2665,7349,000 41,0007,1002,9396,1049, ,4002,6386,5489, ,0002,2967,1959,491 Capital Recovery = 10,000(A/P,i,N) – S n (A/F,i,N) Average Op Cost * = (Σ PW(Op. Costs)) (A/P,i,N) EUAC = Cap. Recovery + Op. Costs * Make mid-period assumption

Third Case: One Challenger, Different from Defender

(This analysis assumes both provide the same service) 1.Calculate the economic life of the challenger and its EUAC for that life (`AC(C N *)’) 2. Compare that EUAC with the cost of keeping the defender one more year (`AC(D 1 )’) 3. If AC(D 1 ) <AC(C N *), do nothing for a year. 4. Otherwise, find AC(D n ), 1 < n < Physical life 5. If AC(D n ) >AC(C N *) for all n, replace

Fourth Case: Several Challengers This is a simple extension of the third case: Find the minimum EUAC for each challenger, Select the challenger with the lowest minimum to go up against the defender Then follow the procedure for the third case.

The General Case Your company wants to come up with a replacement strategy for the next N years, knowing that a different replacement may be used every time (`Chao’s problem’). Difficulty: to determine the economic life of the next-generation challenger, we need to know what the next-generation-plus-one challenger will be. Solution: Dynamic Programming

Dealing with Trade-Ins The present cost of a defender is its salvage value (since by keeping it, we’re losing the opportunity to sell it and invest the money). Suppose we have a defender and two challengers, A and B. The vendor of A offers us a $5,000 trade-in for the defender, while the vendor of B offers us a $4,000 trade-in. The best offer we can get on the open market is $2,000. What do we use as a salvage value when calculating the EUAC of the defender?

Replacement Analysis Pitfalls 1.In estimating the salvage value of the defender, have you taken into account the cost of getting it ready for re-sale? (Cleaning, mending, crating, shipping, etc.) 2.In estimating the purchase price of the challenger, have you considered the cost of installing, de-bugging, re-training and re-routing?

More Replacement Analysis Pitfalls 3. Discount sunk costs! 5.Include all relevant cash-flows – for example, will the challenger produce fewer defectives? 4.Document all assumptions explicitly, so your analysis can be compared with others.

Mr Smith has a small brewing company. In 2006 he bought a shed for $10,000, then in 2007 he bought a brewing machine for $15,000. The shed depreciates in value by 10% per year, while the machine depreciates by 30% per year. Starting in January 2006, Mr Smith paid an annual fire-insurance premium of $2,000. In December 2009, the shed burns down, destroying the machine. Under the terms of the insurance policy, the insurance company will now pay for a new shed and a new machine. Regarding the insurance payments as an investment and the difference between the value of the depreciated property and its new replacements as income, what is Mr Smith’s IRR on his insurance policy?