1 Depreciation equations and rate tables

Transcription

1 The Chinese University of Hong Kong Department of Systems Engineering & Engineering Management SEG 2510 Course Notes 10 for review and discussion (2009/2010) 1 Depreciation equations and rate tables The idea of using rate tables for depreciation is a simple and useful one. Given a depreciation cost I S over N years, how should one construct a rate table with annual rates R(t), for t = 1, 2,..., N, such that the annual depreciation amount is simply given by (I S)R(t) for year t? The rate table can be constructed for different methods, SL, DB, and DBX, and should satisfy N t=1 R(t) = 1 (or 100%). Consider the SL method with N = 5, we have x d = 1/N = 0.2 = 20% and hence the rate table is constructed as 1.1 Rate table for DBX [20%][20%][20%][20%][20%] In the DB case, the depreciation equation D(t) = I α N α N )t 1 gives R(t) = I α I S N α N )t 1 which reduces further to α N α N )t 1 when S = 0. For simplicity, we would assume zero salvage value S = 0 in subsequent discussions; and this assumption has found practical relevance in MACRS. The assumption simplifies the DBX method in the determination of the DB/SL crossover, because the switching time t s is then given by α N = 1 N t s The expression indicates that, for the 200% DB (i.e. α = 2), t s = N/2. As t s = N 1 α ), the crossover will occur late if α is large. From the balance equations, we can relate a summation of R(t) from t = 1 to t, where t t s = k to a balance ratio w.r.t. I 1 t R(t) = α N )t t=1 1

2 where the summation represents the total depreciation percentge over the first t years. The DBX rate table for R(t) can readily be obtained: 1) Using DB to give α N, α N (1 α N ), α N (1 α N )2,..., from t = 1 to k = t s ; 2) Using SL to give a constant rate α N )k+1 1 N k 1, for the remaining t s, i.e., from t = k + 1 to t = N. [Question 1:] Obtain the depreciation rates table for a car priced at $300,000 for a service life of 10 years, using 150% DB with SL switching and zero salvage value. Are the rates affected if DDB+SL is used instead? Will the latter offer a faster depreciation over the first 5 years? For the 150% DB+SL, t s = 10 1/1.5) = 3.33 and k = 4. The rates R(t) = 1.5/10) 4 /6 = = 8.7% for t = 5, 6,...,10. R(1) = 1.5/10 = 15%, R(2) = ) = 12.75%, R(3) = R(2) 0.15) = 10.84%. R(4) = 9.214%. The total depreciation amount over 5 years is 56.5%. Note that 4 t=1 R(t) = 1 (1.5/10))4 = 0.478, which adds up with R(5) = to give For the 200% DB+SL, t s = k = 10/2 = 5. The rates R(t) = 2/10) 5 /5 = 6.55% for t = 6, 7,...,10. R(1) = 2/10 = 20%, R(2) = ) = 16%, R(3) = = 12.8%, R(4) = 10.24%, R(5) = 8.19%. The total depreciation amount over 5 years is = 67.25% (or using 5 t=1 R(t) = 1 (2/10))5 = ). The DBX using 200% DB depreciates 10.75% more than the one using 150% DB over 5 years. [Question 2:] Sometimes it is not necessary to find all the rates, and only certain specific rate information is required. For example, how would you calculate the annual depreciation amount in the 13th year based on the DBX with 175% DB and a 25-year life, for a luxurious yacht worth $15M? You may assume zero salvage cost. To find a specfic rate, we need to first determine t s and check if SL switching has occurred before or after the 13th year. Since t s = 25(1 1/1.75) = and k = 11, SL switchover occurs in the 12th year. The depreciation rate is given by the SL rate α N )k+1 1 N k 1 = 1.75/25)12 /(25 12) = = 3.22%. Therefore, the annual depreciation amount in the 13th year is 15M = 0.483M. 2

3 To confirm on the results, one can alternatively use the BA II. Keying in [2nd][4] to bring up the depreciation worksheet and use [2nd][Enter] until DBX is shown up. Type in 175 for DBX and setting LIF = 25, CST = 15, SAL = 0, Y R = 13 will give DEP = as the depreciation amount. Use CST = 1 to obtain DEP = as the specific rate R(13) = 3.215%. 2 Tax depreciation using MACRS The Modified Accelerated Cost Recovery System (MACRS) 1 is probably the best known depreciation system which puts the DBX and rates tables into practical usage in taxation. Many engineering economy textbooks have in-depth coverage on the many elements of MACRS, including various rules and tables. Here we focus more on the underlying computations involved in the rates tables, with particular emphasis on the required changes in compliance with the specifications and classifications. 2.1 Fractional year considerations Studying the Half-year convention as used in MACRS serves a good purpose to illustrate the meaning of fractional year and its implication on depreciation. Let us consider a general fraction h, where 0 h 1, over a N-year duration; and h = 1/2 means half-year. How would a fractional year affect the rates table of the simple SL method? The SL rate is given by 1/N. Using the Half-year convention implies that the first-year rate is h/n = 0.5/N and the last year rate is (1 h)/n = 0.5/N. The rate table requires N + 1 slots instead of N. For N = 5, the SL rate table becomes The DB case [10%][20%][20%][20%][20%][10%] The SL method for handling fractional year is simple enough. For the DB with a depreciation rate α N, the closed-form depreciation equations can be 1 Pubication 946: How to depreciate property, US Department of the Treasury, Internal Revenue Service, 2009, 120 pages, 3

4 derived from the iterations starting from Bal DB (0) = I: Bal DB (t) = I hα N ) α N )t 1 for t = 1,..., N D(1) = I hα N D(t) = I α hα N N ) α N )t 2 for t = 2,..., N and finally, we should have D(N + 1) = Bal DB (N) (1 h)α N, and Bal DB(N + 1) = Bal DB (N) D(N + 1). The equations look more complicated and unpleasing. For N = 5, h = 1/2, and 200% DB, the rate table becomes [20%][32%][19.2%][11.52%][6.912%][2.0736%] Some of you may already observe that the rates in the DB rate table do not add up to 100% and puzzle at the cause. Recall that this is a known problem in DB for not being able to settle the depreciation cost completely (see explanations in Course Notes 9 Section 4.1). An obvious solution is to change the last year rate to 100 ( ) = % instead of This may be undesirable and a better solution, as we know, is to use the DBX method which provides the necessary SL switching The DBX case Our understanding of DBX reminds us that it is just following the DB before the crossover to SL. How should the required timing t s be adjusted in order to deal with the fractional year? This requires a focus on the straight line between the endpoints at t e and t s, and noting that t e should now be N + h). Hence, the switching time t s is governed by α N = 1 N + 1 h t s and k = t s. The SL rate after crossover, from t = k + 1 to t = N, is R = = Bal DB (k) I(N + 1 h k) 1 hα N + 1 h k N ) α N )k 1 The last year s rate is then calculated as R(N + 1) = h)r. 4

5 Applying the DBX formulations to the previous example with N = 5, h = 1/2, and α = 2, we obtain t s = = 3 = k, R = 0.2) 0.4) 2 /( ) = 11.52%, and R(6) = 0.5) = 5.66%. The DBX rate table is [20%][32%][19.2%][11.52%][11.52%][5.76%] and summing up the rates gives 100% for confirmation. 2.2 The MACRS percentage tables After going through the mathematical derivations, we should be in a better position to understand how the data given in the many different MACRS percentage tables are generated. There are different rules and classifications applying to a very broad range of property and assets. For example, Asset class (Information Systems) has a class life of 6 and the GDS and ADS recovery periods are both 5 years; while Asset class 00.4 (Industrial Steam and Electric Generation and/or Distribution Systems) has a class life of 22 years, a GDS recovery period of 15 years, and ADS recovery is 22 years. Different classes are further classified based on the recovery periods and the applicable depreciation methods (SL, DBX) and conventions (half-year, mid-quarter, mid-month). The percentge tables then list out the specific depreciation rates with respect to the recovery life applicable to a specific property and asset. A typical example is Table A-1 for 3-, 5-, 7-, 10-, 15-, and 20-Year Propery Half-Year Convention, in which the second column (for 5-year recovery period) gives the applicable rates for Information Systems. The data given in that column are quoted for Year 1 to Year 6: 20%, 32%, 19.2%, 11.52%, 11.52%, 5.76%, which match exactly what we obtained for the DBX rate table in the previous section. [Question 1:] According to the MACRS percentage table for a 15-year property (half-year convention), the depreciation rates for Year 1 to Year 16 are listed as percentages in sequential order: 5.00, 9.50, 8.55, 7.70, 6.93, 6.23, 5.90, 5.90, 5.91, 5.90, 5.91, 5.90, 5.91, 5.90, 5.91, 2.95 Discuss the type of depreciation method being used, and verify the given data based on the method. A quick partial solution is probably to search the IRS document (if avail- 5

6 able) and look for the required information. Suppose you don t have the document, then a quick browse on the given data indicates typical features of the DBX method, with SL switching occurred at Year 7. What is the percentage DB? Since R(1) = hα/n is given as 5%, and N = 15, h = 1/2 (half-year convention), we find α = 5% 15/0.5 = 150%. A quick check is to use the BA II if it is available. Specifying DBX = 150, LIF = 15, SAL = 0, CST = 1, and Y R = 1, 2,...,15 will give DEP = 0.10, 0.090,..., The answers are NOT correct! But why? Remember you haven t specified h = 1/2. There is a parameter M01 for the starting month; so for half-year M01 = 7 and now the results look good for DEP = 0.05, 0.095, ,..., Finally with Y R = 16, DEP = Without the BA II, we have to use the depreciation equations previously derived. First, determine the switching time t s which satisfies α N = 1 N + 1 h t s giving 1.5/15 = 1/( t s ), and hence t s = 5.5. Therefore k = 6,, and the SL rates for R(7) to R(15) are all equal to R = 1 hα N + 1 h k N ) α N )k 1 which evaluates to (0.95)(0.9) 5 /( ) = Finally, R(1) = hα/n = /15 = 0.05, and R(2) to R(6) are obtained from R(t) = α hα N N ) α N )t 2 which evaluates to 0.1(0.95)(0.9) t 2 and gives the values 0.095, ,..., [Question 2:] Show how you would apply iterative balance equations to find the MACRS percentage table for a 3-year property (half-year convention) using the 200% DB+SL crossover method. Closed-form equations should not be used. The initial balance is Bal(0) = 1. For the first (half) year, the 200% DB gives a rate 0.5 α/n = 0.5 2/3 = Therefore, R(1) = = 33.33%. The balance after the first (half year) is Bal(1) = Bal(0) R(1) = For the second year, we need to compare the DB rate and SL rate, which are 6

7 R(2) = Bal(1) α/n = /3 = and Bal(1)/(N 0.5) = , respectively. Since the DB rate is greater than the SL rate, there is no switching. The balance after the second year is Bal(2) = Bal(1) R(2) = For the third year, the DB rate and SL rate are R(3) = Bal(2) α/n = /3 = and Bal(2)/(N 1.5) = /1.5 = , respectively. SL switching takes place R(3) = R = 14.81%. For the last (half) year, the SL rate is 0.5 R = 7.405% = R(4). The 3-year (half-year convention) MARCS percentage table is [33.33%][44.44%][14.81%][7.41%] As a cross check, we apply the closed-form equations to obtain t s = = 2 = k, R(3) = ( /3)(1 2/3)/( ) = 4/27 = , and R(2) = (2/3) 0.5 2/3) = 4/9 = Taxable income and depreciation The relationship between incomes and depreciation is not an obvious one, amid many related terms of gross income, net income, taxable income, expenses, and net cash flows. Depreciation has a distinct feature for being a non cash-flow item in accounting books, simply because no real cash is involved. However, depreciation is playing a very important role in affecting cash flows indirectly. MACRS is a good example to illustrate this, because the depreciation methods and percentage tables are affecting the amount of the money (profits, revenues, or gross income) which are taxable. In other words, taxable income implies that depreciation expenses, together with other cash-flow expenses, could be deducted from the gross income. The tax rates then act on the taxable income, resulting in a real tax cash flow. Some simple equations are useful to summarise on incomes: Taxable income = Gross income - Expenses (including depreciation) Taxes = Taxable income tax rates Net income = Taxable income - Taxes Another equation will summarise the net cash flows (it is not NPV!): Gross income - Expenses (excluding depreciation) - Taxes = Net cash flows Simply adding the equations gives Net income = Net cash flows - depreciation expenses 7