cash flow analysis handouts - isye.gatech.edu flow analysis handouts 1. 2. cash flow equivalence ....

CASH FLOW ANALYSIS HANDOUTS

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CASH FLOW EQUIVALENCE A. Future Value Assume per-period interest r is constant. The value of a cash flow C today n periods into the future is called its future value (FV) and is equal to C(1 + r)n. Example 1. C = 1000. r = 10%. FV of 1000 today 1 period hence = 1000(1.1) = 1100. FV of 1000 today 2 periods hence = 1000(1.1)2 = 1100(1.1) = 1210. FV of 1000 today 3 periods hence = 1000(1.1)3 = 1100(1.1)2 = 1220(1.1) = 1331. FV of 1000 today 4 periods hence = 1000(1.1)4 = 1100(1.1)3 = 1210(1.1)2 = 1331(1.1) = 1461.1 Interest rates are often quoted in terms of an annual percentage rate (APR). The APR indicates the amount of simple interest earned in 1 yr, that is, the amount of interest earned without the effect of compounding. Typically, the APR quote is less than the actual amount of interest earned. To compute the actual amount of interest earned in 1 yr, the APR must be converted to an effective annual rate (EAR).

Example 2. A bank advertises savings accounts with an interest rate of “6% APR with monthly compounding.” Here, you earn 0.06%/12 = 0.5% every month. So an APR with monthly compounding is really quoting a monthly interest rate. Because interest compounds each month, you earn $1*(1.005)12 = 1.061678 at the end of 1 yr for every dollar invested. Thus, the effective annual rate EAR = 6.1678%. Key point: You cannot use the APR itself as an interest rate. You must know the number of compound periods. That is:

Interest Rate per Compounding Period = APR/(m periods/yr) Converting an APR to an EAR: 1 + EAR = (1 + APR/m)m.

Example 3. EAR for a 6% APR with different compound periods.

Compounding interval m EARAnnual 1 (1 + 0.06/1)1 – 1 = 6%

Semiannual 2 (1 + 0.06/2)2– 1 = 6.09%Quarterly 4 (1 + 0.06/4)4 – 1 = 6.1364%Monthly 12 (1 + 0.06/12)12 – 1 = 6.1678%

Daily 365 (1 + 0.06/365)365 – 1 = 6.1831%Continuous ∞ exp(0.06) – 1 = 6.1837%

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B. Present Value Assume per-period interest r is constant. The value today of a cash flow C received n periods into the future is called its present value (PV) and is equal to C/(1 + r)n. Example 3. C = 1000. r = 10%. PV today of receiving 1000 1 period from now = 1000/(1.1) = 909.09 PV today of receiving 1000 2 periods from now = 1000/(1.1)2 = 826.45 PV today of receiving 1000 3 periods from now = 1000/(1.1)3 = 751.31. PV today of receiving 1000 4 periods from now = 1000/(1.1)4 = 683.01. Present value of a stream of cash flows (cash flow stream): Consider a stream of cash flows C0 at date 0, C1 at date 1, … ,CN at date N. Often represented as C = (C0, C1, … , CN). Assume per-period interest rate is r. PV[C] = C0 + PV(C1) + PV(C2) + … + PV(CN) = C0 + C1/(1+r) + C2/(1+r)2 + … + CN/(1+r)N . Canonical Example: (-3000, 1100, 1210, 1331), r = 10%. Example 4. You need a car. You just graduated, you need to borrow the money, and your rich aunt will lend you the money. You agree to pay her back within 3 years, and you offer to pay her the rate of interest she would otherwise receive by putting her money in a savings account. She earns 3% EAR on her account. Based on your earnings and expenses, you figure you can pay her $4000 at the end of yr 1 and $8,000 at the end of yrs 2 and 3. How much can you borrow from your aunt? Cash flows you promise to pay your aunt are (0, -4000, -8000, -8000). She will of course receive a cash flow stream of (0, 4000, 8000, 8000). Now the PV of this cash flow stream at 3% is

4000/(1.03) + 8000/(1.03)2 + 8000/(1.03)3 = 18,745.40. This is the amount your aunt should be willing to lend to you. How can your aunt be sure of this equivalent amount? If your aunt left $18,745.60 in her savings account she would have 18,745.40(1.03)3 = $20,483.60 at the end of 3 yrs. Let’s check out how much money your aunt would have at the end of 3 yrs if she loaned you $18,745.40 instead, and then deposited your payments in her account each yr.

1 2 3Cash flow stream 4,000 8,000 8,000

4000(1.03) = 4,120 12,120 12,120(1.03)=12,483.60 20,483.60

Account balance 4,000 12,120 20,483.60 Account balance here is closely related to the concept of Project Balance.

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C. Formulas for PV for Typical Cash Flow Streams Formulas assume a per-period interest rate of r. 1. Constant Annuity a. C0 = 0, Ck = A, k = 1, 2, … (in perpetuity). PV[C] = A/r. PV of receiving 100 per year forever at 8% is 1250. Its value at time 1 (right before receiving the 100) = 1350. b. C0 = 0, Ck = A, k = 1, 2, … N. PV[C] = A/r * [1 – (1+r)-N] PV of receiving 100 per year forever at 8% for 10 yrs is 1250(1 – (1.08)-10) = 671.01. If cash flow earns 8% per year, then amount earned at end of yr 8

= FV(671.01 in 10 yrs at 8%) = 671.01(1.08)10 = 1448.66. 2. Growing Annuity a. C0 = 0, C1 = A, C2 = A(1+g), C3 = A(1+g)2, … , Ck = A(1+g)k-1 , k = 1, 2, … (in perpetuity).

PV[C] = A/(r-g). [Important: g < r !] PV of receiving 100 in the first year, growing at 3% per year thereafter at 8% is

100/(0.08 – 0.03) = 2000.

b. C0 = 0, C1 = A, C2 = A(1+g), C3 = A(1+g)2, … , CN = A(1+g)N-1 . PV[C] = A/(r-g) * (1 – [(1+r)/(1+g)]–N ). [Important: g < r !]

PV of receiving 100 in the first year, growing at 3% per year thereafter for 10 yrs at 8% is 2000(1 – [(1.08)/(1.03)]-10 ) = 755.01.

What happens if the first year cash flow is A(1+g)? D. Equivalence Example 5. C = (0, 100, 110, 100, 110, 100, 110 …). r = 10%. How to compute its PV? E. Continuous Compounding

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F. After-Tax Interest Rate Example 6. Your best taxable investment opportunity has an EAR of 5%. Your best tax-free investment opportunity has an EAR of 3.5%. If your tax rate is 40%, which opportunity provides the higher after-tax interest rate? After-tax rate = (1- 0.40)5% = 3%, which is less than the tax-free investment that pays 3.5%. After-tax rate = (1 – τ)*r. In the US, interest paid is tax deductible for individuals only for home mortgages or home equity loans (up to certain limits), some student loans and loans made to purchase securities. Interest on other forms of consumer debt (e.g. credit cards) is not tax deductible. Interest on debt is tax deductible for corporations, and interest income earned for individuals is taxable as income unless the investment is held in a tax-sheltered retirement account or the investment is from tax-exempt securities (e.g. municipal bonds). Interest from US Treasury securities is exempt from state and local taxes. Interest income earned by a corporation is also taxed at the corporate tax rate. G. Opportunity Cost of Capital When discounting future after-tax cash flows one should use the investor’s opportunity cost of capital (or cost of capital), which is the best available expected return offered in the market on an investment of comparable risk and maturity to the cash flow being discounted.

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CASH FLOW EQUIVALENCE HOMEWORK PROBLEMS 1. You invest $1000 today in a bank. APR is 4%. How much money will you have (to the nearest penny) if this investment is a. compounded annually for 4 years? b. compounded semi-annually for 4 years? c. compounded continuously for 4 years? 2. You invest $1000 today in a bank. APR is 8%. How much money will you have (to the nearest penny) if this investment is a. compounded annually at 8% for 4 years? b. compounded semi-annually for 4 years? c. compounded quarterly for 7.5 years? d. compounded continuously for 4 years? 3. A new bank offers a 5-year CD (certificate of deposit) that pays 4% APR with monthly compounding.

Your current bank offers to match that rate. Your current bank pays interest every 6 months. What is the minimum APR your current bank must offer?

4. Consider the cash flow stream (C0, C1, C2, C3, C4, C5, C6, C7, C8, … , C40, C41) = (100, 110, 100, 110,

100, 110, … , 100, 110). (The pattern (100, 110) repeats itself.) The appropriate per-period rate of interest is 10%. Determine the present value of this cash flow stream.

5. APR is 10% and annual compounding applies. Determine the value (to the nearest dollar) of the

infinite cash flow stream (1, 1100), (2, 1210), (3, 1331), (4, 1100), (5, 1210), (6, 1331), (7, 1100), (8, 1210), (9, 1331), …., etc.

6. The number of periods n for an investment at a per-period interest rate r to double in value must satisfy

(1+r)n = 2. Using ln 2 = 0.69 and the (1st order Taylor series) approximation ln(1+r) ≈ r valid for small r, show that n ≈ 69/i, where i = is the interest rate percentage = 100r. Using the better (2nd order Taylor series) approximation ln(1+r) ≈ r – r2/2, show that for r ≈ 0.0, n ≈ 72/i.

7. Suppose $1 was invested in 1776 at 4% interest compounded yearly. Approximately how much would

that investment be worth today --- $1,000, $10,000, $100,000 or $1,000,000? What if the interest rate were 8%?

8. A lottery advertises that it will pay the winner $1 million. However, the prize money is paid at the rate

of $50,000 per year (first payment is immediate) for a total of 20 years. What is the actual value of winning the prize at a discount rate of 8%?

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CASH FLOW EQUIVALENCE HOMEWORK PROBLEM SOLUTIONS 1. a. (1000)(1.04)4 = 1169.86. b. 1000(1.02)8 = 1171.66. c. 1000e0.16 = 1173.51. 2. a. 1000(1.08)4 = 1360.49. b. 1000(1.04)8 = 1368.57.

c. 1000(1.02)30 = 1811.36. d. 1000e.32 = 1377.13.

3. (1 + 0.04/12)60 = (1 + APR/2)10. APR = 4.033%. 4. The PV with r = 10% of (C0, C1) = (100, 110) at time 0 is, of course, 200.

The cash flow stream is therefore equivalent to (200, 0, 200, 0, 200, … , 200, 0), where the last 200 occurs at time 40. This cash flow stream is equivalent to 200 at each time t = 0, 1, 2, .... , 20, where the period length is 2 years. Since (1.1)2 – 1 = 0.21, the PV equals 200[1 + (1 – (1.21)-20 )/0.21] = 1131.34.

5. PV at time 0 of the cash flow stream (1, 1100), (2, 1210), (3, 1331) at 10% is 3000. The given infinite cash flow stream is equivalent to receiving a cash flow of 3000 at time 0 and thereafter every three years. The appropriate interest rate for three years is 33.1%. Thus, PV = 3000 + 3000/0.331 = 12063.

6. n(r – r2/2) = nr(1 – r/2) ≈ 0.96nr. 7. At 4% money doubles about every 72/4 = 18 years. Over 235 years this equates to 18

doubling periods so that $1 would grow to 213 = 23210 ≈ 8(1,000) = 8,000. When the interest doubles, money doubles twice as frequently, in this case 9 years. So $1 would grow to 226 = (213)2 = 64 million.

8. PV = 50,000[1 + (1 – (1.08)-19 )/0.08] = 530,180.

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NPV, Project Balance, IRR and Payback Period A. NPV

Simple Investment Project after-tax cash flows: (-C0, C1, C2, … , CN), Ck > 0 for all k.

NPV[C] = -C0 + PV[C1, C2, … , CN]. Accept project if NPV[C] > 0.

NPV criterion applies for general after-tax cash flow stream. B. Project Balance Project Balance Schedule {PB(k), k = 0, 1, 2, … , N}: PB(k) = (1+r)PB(k-1) + Ck, PB(0) = C0.

Consider once again Example 4 of Cash Flow Equivalance Handout: You borrow money from your rich aunt who earns 3% EAR on her savings account. You agree to pay her $4000 at the end of yr 1 and $8000 at the end of yrs 2 and 3. We determined that you could borrow $18,745.40. Let’s consider the project balance schedule from your aunt’s perspective (as the one who is providing the loan):

0 1 2 3

Cash Flows -18,745.40 4,000.00 8,000.00 8,000.00 -18,745.40(1.03) = -19,307.76 -15,307.76(1.03) = -15,766.99 -7,766.99(1.03) = 8,000.00

Project Balances -18,745.40 -15,307.76 -7,766.99 0.00

PB(t) = FV at time t of the PV of the cash flows up to and including time t

= (1+r)t * {C0 + C1/(1+r) + … + Ct/(1+r)t}. Project balance at end of project is positive if and only if the NPV is positive. Suppose the loan cash flows remained the same but the EAR = 2%? 4%? C. Internal Rate-of-Return (IRR) Assume a simple investment project. IRR is that interest rate for the PV[C] is zero. Existence? [What about non-simple investment projects?] IRR is also an interest rate for which the project balance at the end of the project is zero. Why? Accept project if IRR > r, where r = cost of capital. Equivalent to NPV criterion. Why? Scale Problem Consider the following example:

Project t = 0 t = 1 NPV@25% IRR A -10 40 22 300% B -25 65 27 160% “B-A” -15 25 5 67%

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. Timing Problem

Consider the following example:

Project t = 0 t = 1 t = 2 t = 3 IRR NPV@0% NPV@10%

NPV@15%

A -10,000 10,000 1,000 1,000 16.04% 2,000 669 109 B -10,000 1,000 1,000 12,000 12.94% 4,000 751 -484 “B-A” 0 -9,000 0 11,000 10.55% 2,000 83 -593

Project Balance Schedules Projects 0 1 2 3 A @ 16.04% -10,000 -1,604 -861 0 A @ 0% -10,000 0 1,000 2,000 A @ 10% -10,000 -1,000 -100 890 A @ 15% -10,000 -1,500 -725 166 B @ 16.04% -10,000 -10,294 -10,626 0 B @ 0% -10,000 -9,000 -8,000 4,000 B @ 10% -10,000 -10,000 -10,000 1,000 B @ 15% -10,000 -10,500 -11,075 -736 “B-A” @ 10.55% 0 -9,000 -9,950 0 “B-A” @ 0% 0 -9,000 -9,000 2,000 “B-A” @ 10% 0 -9,000 -9,900 110 “B-A” @ 15% 0 -9,000 -10,350 -903 Incremental rate-of-return makes sense only for projects that have similar risk and maturity. D. Payback and Discounted Payback Periods Payback period: First period k for which -C0 + C1 + … + Ck > 0. Discounted payback period: First period k for which -C0 + C1/(1+r) + … + Ct/(1+r)k > 0. Also equals the first period k for which PB(k) > 0. Why? Discounted payback period ≥ Payback period. Why?

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CAPITAL BUDGETING I. A firm is considering 7 proposed projects. Available budget is 500. (All numbers in 000’s.) 1 2 3 4 5 6 7 Initial outlay 100 20 150 50 50 150 150 PW 300 50 350 110 100 250 200 Benefit-Cost ratio 3.00 2.50 2.33 2.20 2.00 1.67 1.33 What is the optimal set of projects? Solution: A. Allocate capital using the benefit-cost ranking: Select projects 1-5. PW=910, Cost=370, NPV = 540. B. Use integer programming to find optimal set of projects: The problem may be expressed as MAX 200x1 + 30x2 +200x3 +60x4 +50x5 +100x6 +50x7

subject to: 100x1 + 20x2 + 150x3 + 50x4 + 50x5 + 150x6 + 150x7 ≤ 500, where it is understood that each xi ε {0, 1}. The maximum NPV is 610. It is achieved by selecting projects 1, 3, 4, 5, and 6 at a cost of 500 and a

PW of 1100. II. A county transportation authority is reviewing these transportation alternatives:

Transportation Alternatives * Project Cost (000’s) NPV (000’s) Road between City A and B 1. Concrete, 2 lanes 2,000 4,000 2. Concrete, 4 lanes 3,000 5,000 3. Asphalt, 2 lanes 1,500 3,000 4. Asphalt, 4 lanes 2,200 4,300 Bridge at Broad St 5. Repair existing 500 1,000 6. Add lane 1,500 1,500 7. New structure 2,500 2,500 Traffic control in Town C 8. Traffic lights 100 300 9. Turn lanes 600 1,000 10. Underpass 1,000 2,000 * Adapted from Table 5.2, p. 107 in Investment Science, D. Luenberger, 1998. Total available budget is $5M. At most one project can be selected for each major alternative. What is the optimal set of projects?

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Solution: The problem may be expressed as MAX 4000x1 + 5000x2 +3000x3 +4300x4 +1000x5 +1500x6 +2500x7+ 3008 +1000x9 +2000x10

subject to: 2000x1 + 3000x2 +1500x3 +2200x4 +500x5 +1500x6 +2500x7+ 1008 +600x9 +1000x10 ≤ 5000 x1 + x2 + x3 + x4 ≤ 1 x5 + x6 + x7 ≤ 1 x8+ x9 + x10 ≤ 1 where it is understood that each xi ε {0, 1}. The maximum NPV is … It is achieved by selecting projects … for a cost of … .

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Fixed-Rate Mortgages

1 A Motivating Example

We begin with a motivating example.

Example 1. Smith takes out a 5-yr fixed rate mortgage for 10,000 at a loan interest rate of10% (compounded annually). Smith makes equal annual payments to the bank over the life ofthe loan.

We adopt the following notation:

• N = the number of periods of the loan.

• i = the per-period loan interest rate.

• LB(n) = the outstanding loan balance at time n, n = 0, 1, . . . , N .

• IP (n) = the interest payment made at time n for period n.

• PP (n) = the principal payment made at time n.

• TP (n) = the total payment made at time n.

Table 1 shows the respective cash flow steams for this example, calculated as follows:

Table 1: Cash flows for Example 1.

0 1 2 3 4 5Loan Balance (LB) 10000.00 8362.03 6560.25 4578.32 2398.18 0.00

Total Payment (TP) 2637.97 2637.97 2637.97 2637.97 2637.97Interest Payment (IP) 1000.00 836.20 656.03 457.83 239.82

Principal Payment (PP) 1637.97 1801.77 1981.94 2180.14 2398.15

Total Payment. The bank determines the total payment (TP) so that the value of the totalpayment cash flow steam

(TP (1), TP (2), . . . , TP (N)) = (TP, TP, . . . , TP ) (1)

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at the agreed-upon loan interest rate equals the initial loan balance. Thus,

LB(0) = TP

[1− (1 + i)−N

i

], (2)

TP =iLB(0)

1− (1 + i)−N=

0.10(10000)1− (1.1)−5

= 2637.97. (3)

Interest Payments. IP (n) = iLB(n− 1).

Principal Payments. For this type of loan PP (n) = TP − IP (n).

Observations:

1. The loan balance decreases by the principal payment amount. Bank earns interest on theoutstanding loan balance over the period. The new loan balance equals the old loanbalance plus interest earned minus the cash the bank receives (the total payment madeby Smith). That is,

LB(n) = LB(n− 1) + IP (n)− TP (n) (4)= LB(n− 1) + IP (n)− [IP (n) + PP (n)] (5)= LB(n− 1)− PP (n). (6)

2. Principal payments grow exponentially.

PP (n) = TP − IP (n) (7)= TP − iLB(n− 1) (8)= TP − i[LB(n− 2)− PP (n− 1)] using (6) (9)= [TP − iLB(n− 2)] + iPP (n− 1) (10)= [TP − IP (n− 1)] + iPP (n− 1) using (7) (11)= PP (n− 1) + iPP (n− 1) (12)= (1 + i)PP (n− 1). (13)

3. The outstanding loan balance at time n always equals the present value of the future totalpayment cash flow stream. Here, total payment cash flow stream is the one associatedwhen the loan was originated. For example, the loan balance at time 3 equals

4578.32 =2637.97

1.1+

2637.971.12

.

At time n there are N − n payments remaining. Thus,

LB(n) = TP

[1− (1 + i)−(N−n)

i

](14)

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=[

iLB(0)1− (1 + i)−N

] [1− (1 + i)−(N−n)

i

](15)

= LB(0)

[1− (1 + i)−(N−n)

1− (1 + i)−N

]. (16)

Thus, the ratio of the outstanding loan balance at time n to the initial loan balance isgiven by:

LB(n)LB(0)

=1− (1 + i)−(N−n)

1− (1 + i)−N. (17)

To calculate the outstanding loan balance for this type of loan, you can either (i) firstcalculate the total payment (TP) for the original loan in (3) and then substitute it into (14)or (ii) you can directly use (16). Once you know LB(n) you can calculate the upcomingperiod’s interest and principal payments.

4. The loan balance at the end of the loan always equals 0. Since the loan balance decreaseseach period by the principal payment amount, the sum of the principal payments alwaysequals the initial loan balance. (Can you show this?)

2 Conventional Fixed-Rate Mortgages

Typically, a home mortgage has monthly payments with terms of either 15, 20 or 30 years.Interest is compounded monthly. The equations above still apply but the per-period interestrate i equals the loan interest rate divided by 12. Do not forget this!

Example 2. Consider a 30-year, fixed-rate mortgage for 125,000 at 6.00%. Here i = 0.06/12 =0.005.

• What is the monthly total payment?

TP =(0.005)(125, 000)1− (1.005)−360

= 749.438. (18)

• What is the loan balance after 10 years and 8 months? Here, n = 128 periods and thereare 360 − 128 = 232 periods remaining. Keep in mind you have to convert tomonthly periods! Using (14),

LB(128) = 749.438

[1− (1.005)−232

0.005

]= 102, 763.92. (19)

Alternatively, using (17),

LB(128)LB(0)

=1− (1.005)−232

1− (1.005)−360= 0.822111375,

which gives the same loan balance.

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We now address two practical questions often asked by a mortgagee.

(1) “If I want to pay off my loan in M months, then how much more do I have to payper month?” The present value of the new total payment cash flow stream (each totalpayment now equals TP + A) over the remaining M months must equal the currentoutstanding loan balance. That is,

LB(0) =(TP +A)(1− (1 + i)−M )

i, (20)

from which one can solve for A.

(2) “If I pay an additional A dollars per month, then how many months M will it take topay off my loan?” To answer this question, we can use (20), except here the value of A isknown and we seek the value of M .

Example 3. Consider once again the loan of Example 2, and suppose the remaining durationof the loan is 19 years and 4 months. Recall that TP = 749.438. Since the remaining durationequals 232 months we also know that the outstanding loan balance equals 102, 763.92. If 100 isadded to the monthly payment, then we have

102, 763.92 =849.4380.005

[1− (1.005)−M ], (21)

which implies that M = 186.18 months (or 15.5153 years).

How much must be added to the monthly payment to pay off the loan in 10 years (or 120payments periods)? Let A denote the additional amount. We have that

(M +A) =102, 763.92(0.005)

1− (1.005)−120= 1140.89, (22)

which implies that A = 391.45.

Example 4. Consider a 15-year conventional fixed-rate mortgage for 200,000 at 6.25%. (Thesolution is in brackets.)

• What is the monthly payment? [M = 1714.85.]

• What is the loan balance after 4 years, 3 months? [LB(51) = 160, 792.27.]

• Suppose the remaining duration of the loan is 10 years and 9 months. If we pay 2000each month how quickly will the loan be paid off? [104.44 months or 8.70 years.]

• Suppose the remaining duration of the loan is 10 years, 9 months. How much do we haveto add to our monthly payment to pay off the loan in 5 years? [1412.44.]

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3 Refinancing

The US government allows one to deduct interest payments from their tax obligations. Conse-quently, the after-tax cash flow stream is different than the constant total payment cash flowstream. This difference should be considered when deciding on whether to refinance. First, welook at the motivating example’s after-tax cash flow stream, which is summarized in Table 2.The tax rate τ = 0.40.

Table 2: After-tax cash flows for Example 1 when i = 10%.

0 1 2 3 4 5Loan Balance (LB) 10000.00 8362.03 6560.25 4578.32 2398.18 0.00


Principal Payment (PP) 1637.97 1801.77 1981.94 2180.14 2398.15Tax Shield (TS) 400.00 334.48 262.41 183.13 95.93

After-tax cash flow 10000.00 -2237.97 -2303.49 -2375.56 -2454.84 -2541.99

Remark 1. The effective after-tax loan interest rate equals (1 − τ)i = 0.60(0.10) = 0.06 or6%. What is the present value of the after-tax cash flow stream using the after-tax loan interestrate? Is this always true? Can you prove it? [Hint: Apply the project balance concept.]

The loan of Example 1 could have been originated many years ago; however, there are onlyfive years left now. At the time of origination the prevailing market rate of interest (for loansof this type of risk) was 10%. Let us suppose that interest rates have come down and theprevailing market rate of interest is now 8%. Should Smith refinance?

To answer this question, we first compute the after-tax cash flow stream assuming Smithdid refinance, as shown in Table 3. If Smith were to refinance then the incremental cash flow

Table 3: After-tax cash flows for Example 1 when i = 8%.

0 1 2 3 4 5Loan Balance (LB) 10000.00 8295.44 6454.52 4466.32 2319.07 0.00




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stream is given in Table 4.

Table 4: Incremental after-tax cash flow stream.

0 1 2 3 4 5Incremental After-Tax Cash Flow 0.00 53.41 64.39 77.54 93.20 111.64

What is the value of this cash flow stream? It depends on the choice of the discount rate.An appropriate discount rate is the after-tax loan interest rate equal to (1 - 0.40)(0.08) = 0.048or 4.8%, which represents the cost of funds to Smith. In this case, the value of refinancingequals

342.53 =53.411.048

+64.391.0482

+77.541.0483

+93.201.0484

+111.641.0485

. (23)

If Smith took out a hypothetical loan at 8% in the amount of 342.53, then the repaymentschedule associated with the incremental after-tax cash flows would be as shown in Table 5. Byconstruction, the value of the after-tax cash flows associated with the hypothetical loan is zero.

Table 5: After-tax cash flows for Hypothetical Loan.

0 1 2 3 4 5Loan Balance (LB) 342.53 305.56 255.83 190.57 106.52 0.00




Remark 2. Given a cash flow stream, discount it at the after-tax loan interest rate. Take outa loan in this amount at the pre-tax interest rate. Assume the principal payments are suchthat the after-tax cash flow stream so generated precisely matches the given cash flow stream.This requirement determines the interest and total payment cash flow streams, as shown, forexample, in Table 5. By construction, the present value of the given cash flow stream at theafter-tax loan interest rate is zero. The loan value at the end of the loan will always be zero,and thus the implied repayment plan is valid. (Can you prove this?) (Hint: Apply the projectbalance concept.)

Refinancing entails costs (e.g. origination fees, appraisal, Smith’s time). If the cost toSmith exceeds the 342.53, then he should not finance. What if the cost is less than this value?Actually, it is possible that the best decision is for Smith to wait — for example, suppose Smith

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18

“knew” interest rates were about to fall shortly? This kind of analysis requires a stochasticmodel of interest rates and a game plan for how to make such decisions that involve risk overtime, a subject to be discussed later.

The value of refinancing critically depends on how long the loan will last. For example,if Smith were to sell the asset and pay off his loan sooner than 5 years, then the value ofrefinancing will be less. Suppose, for example, that Smith intends to pay off the loan after 3years. Then the respective after-tax cash flow streams associated with the original and newloans would be as shown in Table 6. At 4.8% the present value of the revised incremental

Table 6: After-tax cash flows when horizon equals three years.

0 1 2 3After-tax cash flow @ 10% 10000.00 -2237.97 -2303.49 -2375.56 + -4578.32 = -6953.88After-tax cash flow @ 8% 10000.00 -2184.56 -2239.10 -2298.02 + -4466.32 = -6764.34

Incremental after-tax cash flow 0 53.42 64.39 189.54

after-tax cash flow stream is 274.27.

4 Loan Valuation

In this section we derive a general formula for the Loan Value, namely, the value of the after-taxcash flows associated with a loan. We require one assumption: interest is always paid on thecurrent outstanding loan balance, i.e., IP (n) = iLB(n− 1). (This certainly holds true for thefixed-rate mortgage type loans examined above.) The implication of this assumption is that atany time the outstanding loan balance equals the sum of the future principal payments, i.e.,

LB(n) =N∑

k=n+1

PP (k). (24)

In what follows, we let τ denote the appropriate tax rate, ia = (1 − τ)i denote the after-taxloan interest rate, and r denote the appropriate per-period discount rate.

General Formula.

LoanV alue = (1 − iar

) ∗ LoanPrincipalV alue, (25)

where the

LoanPrincipalV alue = LB(0) −N∑

n=1

PP (n)(1 + r)n

(26)

7

19

equals the present value of the loan principal cash flows.

In the derivation to follow, the

AfterTaxInterestV alue = (1− τ)N∑

n=1

IP (n)(1 + r)n

(27)

equals the present value of the after-tax interest cash flows.

Derivation. Since the interest payments are deductible

LoanV alue = LB(0) −N∑

n=1

PP (n) + (1− τ)IP (n)(1 + r)n

(28)

= LoanPrincipalV alue − AfterTaxInterestV alue. (29)

In light of (29) it remains to show that

AfterTaxInterestV alue =iar∗ LoanPrincipalV alue,

which we derive in the following sequence of identities:

AfterTaxInterestV alue = (1− τ)N∑

n=1

IP (n)(1 + r)n

(30)

= (1− τ)N∑

n=1

iLB(n− 1)(1 + r)n

(31)

= (1− τ)N∑

n=1

i∑N

k=n PP (k)(1 + r)n

using (24) (32)

= (1− τ)iN∑

k=1

k∑n=1

PP (k)(1 + r)n

(33)

= ia

N∑k=1

[PP (k)

(k∑

n=1

1(1 + r)n

)](34)

= ia

N∑k=1

PP (k)1− (1 + r)−k

r(35)

=iar

{N∑

k=1

PP (k) −N∑

k=1

PP (k)(1 + r)k

}(36)

=iar

{LB(0) −

N∑k=1

PP (k)(1 + r)k

}(37)

=iar∗ LoanPrincipalV alue. (38)

8

20

Observations:

1. When the discount rate r is chosen equal to ia, then the Loan Value Formula (25) im-mediately shows that the LoanV alue = 0. You can verify this for the after-tax cash flowstreams of Examples 1, 2 and 3.

2. The

TaxShieldV alue = τN∑

n=1

IP (n)(1 + r)n

(39)

is the present value of the tax shield cash flow stream. The TaxShieldV alue is animportant component of the Adjusted Present Value (APV) approach to value corporateprojects financed in part by debt. The derivation of the Loan Value formula shows that

TaxShieldV alue = (τ

1− τ) ∗AfterTaxInterestV alue = τ

i

r∗ LoanPrincipalV alue.

(40)

3. An appropriate discount rate for the purpose of corporate valuation is the loan interestrate, i.e., r = i. In this special case,

(1− iai

) = (1− (1− τ)ii

) = τ (41)

and soLoanV alue = τ ∗ LoanPrincipalV alue = TaxShieldV alue. (42)

The second identity follows from (40). That is, when r = i, then the LoanV alue is theTaxShieldV alue. For this special case identity (42) is easily obtained if one recalls thatthe present value of the total payment cash flow stream at the loan interest rate alwaysequals the initial loan balance. (Can you prove this?)

Example 5. What are the LoanV alue and TaxShieldV alue associated with the loan ofExample 1 when the discount rate r equals the loan interest rate of 10%? A direct calculationshows that

LoanV alue = 10000−[

2237.971.1

+2303.49

1.12+

2375.561.13

+2454.84

1.14+

2541.991.15

]= 1021.86.

TaxShieldV alue =400.00

1.1+

334.481.12

+262.411.13

+183.131.14

+95.931.15

= 1021.86.

Of course, we have already shown in (42) that these two values must be the same when r = i.We can also use the Loan Value Formula (25) to calculate either quantity:

LoanV alue = 0.40[10000−

(1637.97

1.1+

1801.771.12

+1981.94

1.13+

2180.141.14

+2398.15

1.15

)]= 0.40(10000− 7445.35) = 1021.86.

9

21

Why use the Loan Value Formula? After all, both calculations of the LoanV alue orTaxShieldV alue use the numbers obtained from the Table. For this type of loan no tableis required when r = i! Here is why. We have shown that the principal payments growexponentially at the loan interest rate. Since the principal payment cash flow stream is beingdiscounted at this same rate, it follows that

7445.35 =1637.97

1.1+

1801.771.12

+1981.94

1.13+

2180.141.14

+2398.15

1.15

=1637.97

1.1+

1637(1.1)1.12

+1637(1.1)2

1.13+

1637(1.1)3

1.14+

1637(1.1)4

1.15

=1

1.1

[1637.97(1.1)

1.1+

1637(1.1)2

1.12+

1637(1.1)3

1.13+

1637(1.1)4

1.14+

1637(1.1)5

1.15

]

=5(1637.97)

1.1.

For this type of loan the general rule for calculating the LoanV alue or TaxShieldV alue whenr = i is this:

LoanV alue = TaxShieldV alue = τ ∗[LB(0) − N ∗ PP (1)

1 + i

](43)

= τ ∗[LB(0) − N ∗ (TP − iLB(0))

1 + i

]. (44)

You only need to calculate the value of TP = iLB(0)/(1− (1 + i)−N ).

Example 6. Consider a conventional 30-yr fixed-rate loan for 100,000 at 6%. What are theLoanV alue and TaxShieldV alue when the discount rate r equals the loan interest rate of 6%?The tax rate is 40%. The calculations are:

TP =0.005(100, 000)1− (1.005)−360

= 599.55

iLB(0) = 0.005(100, 000) = 500.00PP (1) = 599.95− 500.00 = 99.95

LoanV alue = TaxShieldV alue = 0.40[100, 000 − 360(99.95)1.005

] = 26, 678.81.

Example 7. Suppose the loan of Example 6 is paid off at the end of year 10. (This is typicalof a real estate loan.) What are the LoanV alue and TaxShieldV alue associated with thisrepayment schedule when the discount rate r equals the loan interest rate of 6%? To apply thegeneral formula, keep in mind that (i) there will be an additional “balloon” principal paymentat the end of period 120 in the amount of LB(120) whose present value must be accounted for,and (ii) only multiply PP (1) by the number of periods n = 120. The calculations are:

LB(120) = 100, 000 ∗ 1− (1.005)−240

1− (1.005)−360= 83, 685.73.

LoanV alue = TaxShieldV alue = 0.40[100, 000 − 120(99.95)1.005

− 83, 685.731.005120

] = 16, 827.70.

10

22

Example 8. Let’s revisit the refinancing example. We obtained a value of 342.53. Theincremental cash flow stream obtained in Table 4 was discounted at the after-tax loan interestrate of 4.8%. The incremental cash flow stream is the difference between the after-tax cash flowsstreams associated with the loans at 8% and 10%, respectively. So, in effect, the value of 342.53equals the LoanV alue of the 8% loan minus the LoanV alue of the 10% loan both discountedat 4.8%. But the LoanV alue of the 8% loan discounted at 4.8% equals zero, so the value of therefinancing equals minus the LoanV alue of the 10% loan discounted at 4.8%. When applyingthe Loan Value formula to obtain this latter quantity, you can not use the simple expression(44). The reason is that the principal payment cash flow stream grows at 10% but the discountrate equals 4.8%. Here, you must apply the growing annuity formula:

LoanPrincipalV alue = 10, 000−[

1637.971.048

+1801.771.0482

+1981.941.0483

+2180.141.0484

+2398.151.0485

]= 10, 000−

[1637.971.048

+1637(1.1)

1.0482+

1637(1.1)2

1.0483+

1637(1.1)3

1.0484+

1637(1.1)4

1.0485

]

= 10, 000−[1637.97 ∗

(1− (1.048

1.1 )−5

0.048− 0.10

)]= 10, 000− 8629.69 = 1370.31.

Now, we are in position to calculate the LoanV alue as

(1− (1− 0.40)0.100.048

) ∗ (1370.31) = −342.53, (45)

which confirms our earlier calculation. (Keep in mind the refinance value here is minus theLoanV alue.)

Example 9. Suppose your company is considering purchasing a machines that costs 1 million.The manufacturer offers to finance the purchase by lending you the purchase price for 5 yearswith annual interest payments of 5%. The principal of 1 million is paid at the end of year 5.(This is different than the loans we have considered up to now, but the general formula stillapplies.) The local bank will charge you 15% for such a loan. Assume that you will commit torepay the loan repayment schedule. Your tax rate is 40%. What is the value of this loan?

If you take this loan, the cash flows are shown in Table 7: What is the appropriate discountrate? For reasonably safe loans (i.e. very low probability of default), the correct discount rateis your company’s after-tax, unsubsidized borrowing rate, which is 9% in this case. (You canuse the same argument we used to justify the discount rate in the refinance example.) In thiscase, you can verify that the present value of the after-tax cash flow equals 233,379. We canalso apply the Loan Value Formula, as follows:

LoanV alue = (1− 0.030.09

) ∗(

1, 000, 000− 1, 000, 0001.095

)= 233, 379. (46)

11

23

Table 7: After-tax cash flows for Example 9.

0 1 2 3 4 5Loan Balance (LB) 1,000,000 1,000,000 1,000,000 1,000,000 1,000,000 0

Total Payment (TP) 50,000 50,000 50,000 50,000 1,050,000Interest Payment (IP) 50,000 50,000 50,000 50,000 50,000

Principal Payment (PP) 0 0 0 0 1,000,000Tax Shield (TS) 20,000 20,000 20,000 20,000 20,000

After-tax cash flow 1,000,000 -30,000 -30,000 -30,000 -30,000 -1,030,000

In effect, the manufacturer is subsidizing the purchase price by the amount of 233,379. Forpurposes of any other project evaluation, the cost to the company to acquire the machine is766,621.

Remark 3. When the loan repayment schedule is such that interest is paid each period only andthe loan balance is paid at the end of the loan, then the calculation of the LoanPrincipalV alueis elementary: all you have to do is to subtract from the loan amount the present value of theloan amount at the end of the loan!

12

24

FIXED RATE MORTGAGE HOMEWORK PROBLEMS 1. Ms. Jones financed her home purchase with a fixed-rate 20-yr mortgage at 6%. The original loan balance was 400,000.00. With her monthly mortgage just paid her current loan balance is 301,903.98. a. What is Jones’s monthly payment to the bank? b. How many months remain until the loan is paid off? c. Jones would like to pay off her loan sooner. She has decided that she would like to pay off her loan in

10 years, and is willing to add $A per month to her payment. What is the value for A? 2. Smith financed his home purchase with a conventional fixed-rate 30-yr mortgage at 9%. The original loan balance was 200,000.00. With his monthly mortgage just paid his current loan balance is 173,719.16. a. What is Smith’s monthly payment to the bank? b. How many months remain until the loan is paid off? c. Smith would like to pay off his loan sooner. He has decided that he can afford an extra 50 per month.

How many months will it take to pay off his loan? 3. Consider a 15-year fixed-rate mortgage for 200,000 at 6.25%. Provide continuous-time answers: a. What is the monthly payment? b. What is the loan balance after 4 years, 3 months? c. Suppose the remaining duration of the loan is 10 years and 9 months. How quickly will the loan be

paid off if the 2000 is paid each month instead of the original monthly payment? d. Suppose the remaining duration of the loan is 10 years and 9 months. How much must be added to the

original monthly payment to pay off the loan in 5 years? 4. Consider a conventional fixed-rate 30-yr loan for 100,000 at 10%. a. What is the total payment and the total interest paid over the life of the loan? b. In a biweekly program, you pay half of the total monthly payment every two weeks until the loan is

repaid. Assume biweekly compounding. In a biweekly program for this loan, when will it be paid off and what will be the total interest saved over the life of the loan?

5. Consider a conventional fixed-rate 30-yr loan for 500,000 at 12%. Assume a tax rate of 40%. a. What are the LoanValue and TaxShieldValue when the discount rate equals the loan interest rate? b. How do these values change if the discount rate is chosen to be the after-tax loan interest rate? c. Answer parts (a) and (b) if the loan will be paid off after 10 years. d. Suppose this loan can be refinanced now at 9%. What is the value of refinancing? e. Answer part (d) assuming the loan will be paid off after 10 years. 6. Consider a conventional fixed-rate 20-yr loan for 100,000 at 9%. Assume a tax rate of 30%. a. What are the LoanValue and TaxShieldValue when the discount rate equals the loan interest rate? b. How do these values change if the discount rate is chosen to be the after-tax loan interest rate? c. Answer parts (a) and (b) if the loan will be paid off after 5 years. d. Suppose this loan can be refinanced now at 6%. What is the value of refinancing? e. Answer part (d) assuming the loan will be paid off after 5 years. 7. Your company is considering purchasing a machine that costs 1 million. The manufacturer offers to

finance the purchase by lending you the purchase price for 10 years with annual interest payments of 4%. Principal of 1 million is paid at the end of year 10. The local bank will charge you 6% for such a loan. The tax rate is 30%. What is the value of the loan and the net purchase cost of the machine?

25

FIXED-RATE MORTGAGE HOMEWORK PROBLEM SOLUTIONS 1. a. M = (0.005)(400,000)/[1 – (1.005)-240] = 2865.72. b. The ratio LB(t)/LB(0) = 301,903.48/400,000. We know that LB(t)/LB(0) = [1 – (1.005)–(240-t) ]/[1 – (1.005)-240]. Thus, the number of months remaining is (240-t) = 150. Alternatively, 301,903.48 = [2865.72/0.005][1 – (1.005)–n], which gives n = 150. c. We seek the value of A such that [(2865.72 + A)/0.005][1 – (1.005)-120] = 301,903.48. 2. a. M = 200,000(0.09/12)/[1 – (1 + 0.09/12)-360] = 1609.25.

b. 173,719.16/200,000 = [1 – (1.0075)-n]/[1 – (1.0075)-360] and so n = 222. c. Seek n such that 173,719.16 = (1659.25/0.0075) [1 – (1.0075)-n] and so n = 206.

3. a. 1712.16. b. LB(4.25) = 160,833.17.

c. 8.68 years. d. 1409.01. 4. a. M = 877.57. The difference between the total payments and the initial loan balance

equals the total interest payment, and so it equals = 360(877.57) – 100,000 = 215,925. b. 100,000 = (877.57/2)*([ 1 – (1 + 0.10/26)-N]/(0.10/26), which implies N = 545 or 21 yrs.

Total interest payment is now 545(877.57/2) – 100,000 = 139,138, a reduction of 35.6%. 5. a. TP = 5143.06. IP(1) = 5000. PP(1) = 143.06. LoanPrincipalValue = 500,000 – 360(143.06)/1.01 = 449,008. LoanValue = TaxShieldValue = 0.40(449,008) = 179,603. b. Loan Value = LoanPrincipalValue = AfterTaxInterestValue = 0.

LoanPrincipalValue = 500,000 – [143.06*(1 – (1.006/1.01)-360)/(0.006-0.01)] = 386,527. TaxShieldValue = [0.40/(1- 0.40)]*AfterTaxInterestValue = 257,685. c. LB(120) = 467,090. LoanPrincipalValue = 500,000 – 120(143.06)/1.01 – 467,090/(1.01)120 = 341,477. LoanValue = TaxShieldValue = 0.40(374,387) = 136,591.

When r = ia, Loan Value = LoanPrincipalValue = AfterTaxInterestValue = 0. LoanPrincipalValue = 500,000 – [143.06*(1 – (1.006/1.01)-120)/(0.006-0.01)] – 467,090/(1.006)120 = 250,339.

TaxShieldValue = [0.40/(1- 0.40)]*AfterTaxInterestValue = 166,893.

d. LoanPrincipalValue = 500,000 – [143.06*(1 – (1.0045/1.01)-360)/(0.0045-0.01)] = 340,286. Refinance value = -LoanValue = -(1- 0.006/0.0045)*(340,286) = 113,429.

26

e. LoanPrincipalValue = 500,000 – [143.06*(1 – (1.0045/1.01)-120)/(0.0045-0.01)] – 467,090/(1.0045)120 = 203,398. Refinance value = -LoanValue = -(1- 0.006/0.0045)*(203,398) = 67,799.

6. a. TP = 899.73. IP(1) = 750. PP(1) = 149.73. LoanPrincipalValue = 100,000 – 240(149.73)/1.0075 = 64,332. LoanValue = TaxShieldValue = 0.30(64,332) = 19,300.

b. Loan Value = LoanPrincipalValue = AfterTaxInterestValue = 0. LoanPrincipalValue = 100,000 – [149.73*(1 – (1.00525/1.0075)-240)/(0.00525-0.0075)] = 52,742.

TaxShieldValue = [0.30/(1- 0.30)]*AfterTaxInterestValue = 22,604.

c. LB(60) =88,707. LoanPrincipalValue = 100,000 – 60(149.73)/1.0075 – 88,707/(1.0075)60 = 34, 426. LoanValue = TaxShieldValue = 0.30(34,426) = 10,328.

When r = ia, Loan Value = LoanPrincipalValue = AfterTaxInterestValue = 0. LoanPrincipalValue = 100,000 – [149.73*(1 – (1.00525/1.0075)-60)/(0.00525-0.0075)] – 88,707/(1.00525)60 = 25,656.

TaxShieldValue = [0.30/(1- 0.30)]*AfterTaxInterestValue = 10,995. d. LoanPrincipalValue = 100,000 – [149.73*(1 – (1.0035/1.0075)-240)/(0.0035-0.0075)]

=40,182. Refinance value = -LoanValue = -(1- 0.00525/0.0035)*(40,182) =20,091. e. LoanPrincipalValue

= 100,000 – [149.73*(1 – (1.0035/1.0075)-60)/(0.0035-0.0075)] – 88,707/(1.0035)60 = 17,978. Refinance value = -LoanValue = -(1- 0.00525/0.0035)*(17,978) = 8,989.

7. ia = 2.4%. r = 3.6%.

LoanValue = (1 – 0.024/0.036)*[1,000,00 – 1,000,000/(1.036)10] = 99,298.

27

SIMPLE EQUIPMENT SELECTION ANALYSIS I. A company must acquire a piece of equipment. It is considering the following two options:

a. Initial cost = 1,000. Annual cost = 300. Salvage value = 100. 5-year lifetime. b. Initial cost = 1,300. Annual cost = 270. Salvage value = 200. 10-year lifetime. Discount rate = 12%. Ignore taxes and depreciation.

ANALYSIS: I. COMPUTE PRESENT VALUES:

The present value cost of option (a) is:

1000 + 300{ [ 1 – (1.12)-5 ] ÷ 0.12} - 100(1.12)-5 = 2025. The present value cost of option (b) is:

1300 + 270{ [ 1 – (1.12)-10 ] ÷ 0.12} - 200(1.12)-10 = 2761.

2. COMPUTE ANNUAL EQUIVALENTS:

The Annual Equivalent (or Worth) of option (a) is:

2025 { 0.12 ÷ [ 1 – (1.12)-5 ] } = 562,

and its “lifetime cost” is 562 ÷ 0.12 = 4683.

The Annual Equivalent (or Worth) of option (b) is:

2761 { 0.12 ÷ [ 1 – (1.12)-10 ] } = 489,

and its “lifetime cost” is 489 ÷ 0.12 = 4075. 3. USE COMMON CYCLE APPROACH: 10 years = lowest common multiple

Option (a): PV over 10 years = 2025 + 2025/(1.12)5 = 3174. Option (b): PV over 10 years = 2761.

29

A manufacturer requires a chemical finishing process for a product produced under contract for 4 years. Three options are available. Initial cost of process device A = 100,000. It has an annual cost = 60,000 and a salvage value = 40,000 after 4 years. Initial cost of process device B = 150,000. It has an annual cost = 50,000 and a salvage value = 30,000 after 6 years. It is also possible to subcontract at 100,000 per year (option S). Appropriate cost of capital is 10%. Marginal tax rate = 40%. Process devices are classified as 5-year property. Other profitable ongoing operations are sufficient to cover any losses on sale of equipment.

ANALYSIS Process devices A and B are depreciating at assumed constant rates of 15,000 and 20,000 per year, respectively. Using Straight Line (SL) depreciation (with half-year convention):

Process Device A 0 1 2 3 4 Operations cost (60,000) (60,000) (60,000) (60,000) Depreciation (7,500) (15,000) (15,000) (7,500) Total cost (67,500) (75,000) (75,000) (67,500) Tax benefit 27,000 30,000 30,000 27,000 After-tax cost (40,500) (45,000) (45,000) (40,500) Depreciation 7,500 15,000 15,000 7,500 Investment (100,000) 0 0 0 *46,000 After-tax cash flow (100,000) (33,000) (30,000) (30,000) 13,000

Process Device B 0 1 2 3 4 Operations cost (50,000) (50,000) (50,000) (50,000) Depreciation (10,000) (20,000) (20,000) (10,000) Total cost (60,000) (70,000) (70,000) (60,000) Tax benefit 24,000 28,000 28,000 24,000 After-tax cost (36,000) (42,000) (42,000) (36,000) Depreciation 10,000 20,000 20,000 10,000 Investment (150,000) 0 0 0 **78,000 After-tax cash flow (150,000) (26,000) (22,000) (22,000) 52,000

* 46,000 = 40,000 – 0.4(40,000 – 55,000). SV – τ(SV – BV). ** 78,000 = 70,000 – 0.4(70,000 – 90,000). PV @ 10% of after-tax cash flow for A = -168,454. AE for 4 years = -53,142. PV @ 10% of after-tax cash flow for B = -172,830. AE for 4 years = -54,523. PV @ 10% of after-tax cash flow for S = -190,192. AE for 4 years = -60,000. Process device A is the best option.

31

Process devices A and B are depreciating at assumed constant rates of 15,000 and 20,000 per year, respectively. Using MACRS depreciation (with half-year convention):

Process Device A 0 1 2 3 4 Operations cost (60,000) (60,000) (60,000) (60,000) Depreciation (20,000) (32,000) (19,200) (5,760) Total cost (80,000) (92,000) (79,200) (65,760) Tax benefit 32,000 36,800 31,680 26,304 After-tax cost (48,000) (55,200) (47,520) (39,456) Depreciation 20,000 32,000 19,200 5,760 Investment (100,000) 0 0 0 *33,216 After-tax cash flow (100,000) (28,000) (23,200) (28,320) (480)

Process Device B 0 1 2 3 4 Operations cost (50,000) (50,000) (50,000) (50,000) Depreciation (30,000) (48,000) (28,800) (8,640) Total cost (80,000) (98,000) (78,800) (58,640) Tax benefit 32,000 39,200 31,520 23,456 After-tax cost (48,000) (58,800) (47,280) (35,184) Depreciation 30,000 48,000 28,800 8,640 Investment (150,000) 0 0 0 **55,824 After-tax cash flow (150,000) (18,000) (10,800) (18,480) 29,280

* 33,216 = 40,000 – 0.4(40,000 – 23,040). SV – τ(SV – BV). ** 55,824 = 70,000 – 0.4(70,000 – 34,560). PV @ 10% of after-tax cash flow for A = -166,233. AE for 4 years = -52,422. PV @ 10% of after-tax cash flow for B = -169,175. AE for 4 years = -53,370. PV @ 10% of after-tax cash flow for S = -190,192. AE for 4 years = -60,000. Process device A is the best option. SENSITIVITY ANALYSIS: What is the salvage value of B after 4 years that would cause the manufacturer to be indifferent in choosing between it and A? • Let δSV denote the change in salvage value of B for which we would be indifferent between it and A. • The new salvage value = 70,000 + δSV. • The change to after-tax cash flows for B will simply be (1-τ)δSV = 0.60δSV. • The present value of this change @ 10% = 0.60δSV (1.1)-4. • This must equal (-166,233) - (-169,175) = 2,942. • Thus, δSV = 7,179 so that the “break-even” salvage value = 77,179.

32

Depreciation

A. Tax Effect of Depreciation 1. For the current year a company has income of R and cost of C for a net income before tax of R-C for the year. The company’s marginal tax rate is τ. The company’s after-tax cash flow for the year is therefore (1- τ)[R – C].

Example 1. Say R = 3,000 and C = 1,000 with τ = 40%. Then the company’s after-tax cash

flow for the year is (1 – 0.40)[3,000 – 1,000] = 1,200. Note that this may be equivalently expressed as (0.60)[3,000] – (0.60)[1,000] = 1,800 – 600 = 1,200. That is, the company receives a net of 60 cents for each dollar of revenue it takes in but it only costs the company 60 cents for dollar of expense. On the expense side note that

-(1- τ)[C] = -C + τC = -1000 + (0.40)(1000),

and so the net effect of being able to deduct expenses it that the company “receives” a “tax benefit” of τC = 400, i.e., it “wrote a check” for 1,000 during the year and receives back 400 due to the expense write-off. 2. Let’s assume the revenue R is sufficiently high to offset the expenses. In what follows we shall ignore the after-tax cash flow due to revenue, since it is a “constant” as far as the analysis of the after-tax cash flow due to cost.

Now let’s suppose for the current year a company spends I = 1000 on capital investment for equipment. The equipment is classified as 5-year property and straight-line depreciation is used. If the company were able to write-off this capital investment as an expense, then the after-tax cash flow would be –600, exactly as in Example 1.

However, the government’s perspective on this is that the equipment is being used over a 5-

year period and provides economic value over its entire “book” life of 5 years. (The company may choose to use this equipment for more than 5 years.) Consequently, the government will only allow the company to expense each year an “appropriate” amount for the “wear and tear” or depreciation of the equipment. Since straight-line depreciation is being used, the depreciation expense each year is 1000/5 = 200. (Salvage value at end of year 5 is assumed zero and the half-year convention is being ignored here.)

There is NO out-of-pocket depreciation expense each year. That is, the 1000 is spent now;

the 200 is computed only for tax purposes. As explained in #1 above, from the after-tax cash flow perspective, this 200 depreciation “allowance” yields a tax benefit of 0.40(200) = 80 each year for the next 5 years. That is, the after-tax cash flow due to the capital investment for the next 5 years (ignoring the salvage value due to possible sale of equipment) is:

0 1 2 3 4 5 -1,000 80 80 80 80 80

33

Note how the undiscounted sum of the depreciation allowances over the 5-year horizon adds up to 400, the total tax benefit. Again, the total tax benefit of 400 would be added to the –1,000 if this were an ordinary expense (for materials, labor, etc.). Suppose the company were to use a 10% cost of capital for purposes of determining the Net Present Value of this cash flow stream, then the

NPV = -1,000 + {[1 – (1.10)-5]/(0.10)} 80 = -1,000 + 3.791(80) = -696.74,

which translates to a net cost of 69.7% per dollar spent on this capital equipment. Note that if this equipment had been expensed, then the net cost would be 60% per dollar spent. The “loss” of 9.7% is due to the DELAY of receiving the benefit of 80 per year for 5 years, as above, as opposed to receiving the total of 400 right NOW. Which cash flow would a company prefer? For the purpose of after-tax cash flow, the company would prefer to be able to “write-off” as much depreciation as quickly as it can. That is, it would prefer accelerated depreciation. For the purpose of reporting profits, however, the company would prefer to delay the expense! The IRS allows a company to keep “two sets of books,” one for tax purposes and the other for reporting to shareholders. B. Accounting for Depreciation in the Income Statement Example 2. Suppose a company has a net operating income (R – C) for the year of 10,000 and a depreciation expense of 2,000.

According to what we have learned above, the taxes owed to the government this year would be 0.40(10,000 – 2,000) = 3,200. The “Flow In” is therefore 10,000 and the “Flow Out” is 3,200 for a net of 6,800. Here is the income statement approach to this calculation:

Income Statement Approach Flow In – Flow Out Approach Net operating income 10,000 Net operating income 10,000 Depreciation 2,000 Taxes -3,200 Income before tax 8,000 Tax 3,200 Net income 4,800 Adjustments Depreciation 2,000 After tax cash flow 6,800 After tax cash flow 6,800

In general notation, the two approaches to after tax cash flow are, respectively: Income Statement Approach: = (1 - τ)[R – C – Dep] + Dep = (1- τ)[R – C] + τ Dep, Flow In – Flow Out Approach: = [R – C] - τ[R – C – Dep] = (1- τ)[R – C] + τ Dep, which produce the same result, as it should. Note that if the depreciation adjustment “below the line” (i.e. below the Net Income after tax) is NOT made, then the income statement approach would incorrectly report an after tax cash flow of 1,200. The discrepancy would arise because the depreciation allowance is NOT an “out of pocket” expense.

34

of 120 of Publication 946 15:26 - 13-MAY-2010

The type and rule above prints on all proofs including departmental reproduction proofs. MUST be removed before printing.

Chart 1. Use this chart to find the correct percentage table to use for any property other than residential rentaland nonresidential real property. Use Chart 2 for residential rental and nonresidential real property.

MACRSSystem

DepreciationMethod

Appendix AMACRS Percentage Table Guide

General Depreciation System (GDS)Alternative Depreciation System (ADS)

Recovery Period Convention Class

Month orQuarterPlacedin Service Table

GDS 200% GDS/3, 5, 7, 10 (Nonfarm) Half-Year 3, 5, 7, 10 Any A-1

GDS 200% GDS/3, 5, 7, 10 (Nonfarm) Mid-Quarter 3, 5, 7, 10 1st Qtr2nd Qtr3rd Qtr4th Qtr

A-2A-3A-4A-5

GDS 150% GDS/3, 5, 7, 10 Half-Year 3, 5, 7, 10 Any A-14

GDS 150% GDS/3, 5, 7, 10 Mid-Quarter 3, 5, 7, 10 1st Qtr2nd Qtr3rd Qtr4th Qtr

A-15A-16A-17A-18

GDS 150% GDS/15, 20 Half-Year 15 & 20 Any A-1

GDS 150% GDS/15, 20 Mid-Quarter 15 & 20 1st Qtr2nd Qtr3rd Qtr4th Qtr

A-2A-3A-4A-5

GDSADS

SL GDSADS

Half-Year Any Any A-8

GDSADS

SL GDSADS

Mid-Quarter Any 1st Qtr2nd Qtr3rd Qtr4th Qtr

A-9A-10A-11A-12

ADS 150% ADS Half-Year Any Any A-14

ADS 150% ADS Mid-Quarter Any 1st Qtr2nd Qtr3rd Qtr4th Qtr

A-15A-16A-17A-18

Chart 2. Use this chart to find the correct percentage table to use for residential rental and nonresidential realproperty. Use Chart 1 for all other property.

MACRSSystem

DepreciationMethod Recovery Period Convention Class

Month orQuarterPlacedin Service Table

GDS SL GDS/27.5 Mid-Month Residential Rental Any A-6

GDS SLSL

GDS/31.5GDS/39

Mid-Month Nonresidential Real Any A-7A-7a

ADS SL ADS/40 Mid-Month Residential RentalandNonresidential Real

Any A-13

Chart 3. Income Inclusion Amount Ratesfor MACRS Leased Listed Property

Table

Amount A Percentages

Amount B Percentages

A-19

A-20

Page 76 Publication 946 (2009)35



Table A-1. 3-, 5-, 7-, 10-, 15-, and 20-Year PropertyHalf-Year Convention

Year

12345

6789

10

Depreciation rate for recovery period

3-year 5-year 7-year 10-year 15-year 20-year

1112131415

1617181920

21

33.33%44.4514.81

7.41

20.00%32.0019.2011.5211.52

5.76

14.29%24.4917.4912.49

8.93

8.928.934.46

10.00%18.0014.4011.52

9.22

7.376.556.556.566.55

3.28

5.00%9.508.557.706.93

6.235.905.905.915.90

5.915.905.915.905.91

2.95

3.750%7.2196.6776.1775.713

5.2854.8884.5224.4624.461

4.4624.4614.4624.4614.462

4.4614.4624.4614.4624.461

2.231

Table A-2. 3-, 5-, 7-, 10-, 15-, and 20-Year PropertyMid-Quarter ConventionPlaced in Service in First Quarter

Year

12345

6789

10



1112131415

1617181920

21

58.33%27.7812.35

1.54

35.00%26.0015.6011.0111.01

1.38

25.00%21.4315.3110.93

8.75

8.748.751.09

17.50%16.5013.2010.56

8.45

6.766.556.556.566.55

0.82

8.75%9.138.217.396.65

5.995.905.915.905.91

5.905.915.905.915.90

0.74

6.563%7.0006.4825.9965.546

5.1304.7464.4594.4594.459

4.4594.4604.4594.4604.459

4.4604.4594.4604.4594.460

0.565

Publication 946 (2009) Page 7736



Table A-3. 3-, 5-, 7-, 10-, 15-, and 20-Year PropertyMid-Quarter ConventionPlaced in Service in Second Quarter

Year

12345

6789

10



1112131415

1617181920

21

41.67%38.8914.14

5.30

25.00%30.0018.0011.3711.37

4.26

17.85%23.4716.7611.97

8.87

8.878.873.34

12.50%17.5014.0011.20

8.96

7.176.556.556.566.55

2.46

6.25%9.388.447.596.83

6.155.915.905.915.90

5.915.905.915.905.91

2.21

4.688%7.1486.6126.1165.658

5.2334.8414.4784.4634.463

4.4634.4634.4634.4634.462

4.4634.4624.4634.4624.463

1.673

Table A-4. 3-, 5-, 7-, 10-, 15-, and 20-Year PropertyMid-Quarter ConventionPlaced in Service in Third Quarter

Year

12345

6789

10



1112131415

1617181920

21

25.00%50.0016.67

8.33

15.00%34.0020.4012.2411.30

7.06

10.71%25.5118.2213.02

9.30

8.858.865.53

7.50%18.5014.8011.84

9.47

7.586.556.556.566.55

4.10

3.75%9.638.667.807.02

6.315.905.905.915.90

5.915.905.915.905.91

3.69

2.813%7.2896.7426.2375.769

5.3364.9364.5664.4604.460

4.4604.4604.4614.4604.461

4.4604.4614.4604.4614.460

2.788




Table A-5. 3-, 5-, 7-, 10-, 15-, and 20-Year PropertyMid-Quarter ConventionPlaced in Service in Fourth Quarter

Year

12345

6789

10



1112131415

1617181920

21

8.33%61.1120.3710.19

5.00%38.0022.8013.6810.94

9.58

3.57%27.5519.6814.0610.04

8.738.737.64

2.50%19.5015.6012.48

9.98

7.996.556.556.566.55

5.74

1.25%9.888.898.007.20

6.485.905.905.905.91

5.905.915.905.915.90

5.17

0.938%7.4306.8726.3575.880

5.4395.0314.6544.4584.458

4.4584.4584.4584.4584.458

4.4584.4584.4594.4584.459

3.901

Table A-6. Residential Rental PropertyMid-Month ConventionStraight Line—27.5 Years

Year

12–9101112

1314151617

Month property placed in service

7 8 9 10 11 12

1819202122

2324252627

2829

0.152%

654321

0.455%0.758%1.061%1.364%1.667%1.970%2.273%2.576%2.879%3.182%3.485%3.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

1.97

3.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

2.273

3.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

2.576

3.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

2.879

3.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.182

3.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.485

3.6363.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6360.152

3.6363.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6360.455

3.6363.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6360.758

3.6363.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6361.061

3.6363.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6361.364

0.152%3.6363.6363.6373.636

3.6373.6363.6373.6363.637

3.6363.6373.6363.6373.636

3.6373.6363.6373.6363.637

3.6361.667




Appendix B — Table of Class Lives and Recovery Periods

being used and use the recovery pe- improvements. The land improve-The Table of Class Lives and Recov-riod shown in the appropriate column ments have a 13-year class life and aery Periods has two sections. The firstfollowing the description. 7-year recovery period for GDS. If hesection, Specific Depreciable Assets

elects to use ADS, the recovery periodUsed In All Business Activities, Except Property not in either table. If the is 13 years. If Richard only looked atAs Noted, generally lists assets used activity or the property is not included Table B-1, he would select asset classin all business activities. It is shown as in either table, check the end of Table 00.3, Land Improvements, and incor-Table B-1. The second section, Depre- B-2 to find Certain Property for Which rectly use a recovery period of 15ciable Assets Used In The Following Recovery Periods Assigned. This years for GDS or 20 years for ADS.

property generally has a recovery pe-Activities, describes assets used onlyriod of 7 years for GDS or 12 years for Example 2. Sam Plower producesin certain activities. It is shown as Ta-ADS. See Which Property Class Ap- rubber products. During the year, heble B-2.plies Under GDS and Which Recovery made substantial improvements to thePeriod Applies in chapter 4 for the land on which his rubber plant is lo-How To Use the Tables class lives or the recovery periods for cated. He checks Table B-1 and findsGDS and ADS for the following. land improvements under asset classYou will need to look at both Table B-1

and B-2 to find the correct recovery 00.3. He then checks Table B-2 and• Residential rental property andperiod. Generally, if the property is finds his activity, producing rubbernonresidential real property (alsolisted in Table B-1 you use the recov- see Appendix A, Chart 2). products, under asset class 30.1,ery period shown in that table. How- Manufacture of Rubber Products.• Qualified rent-to-own property.ever, if the property is specifically Reading the headings and descrip-listed in Table B-2 under the type of • A motorsport entertainment com- tions under asset class 30.1, Samactivity in which it is used, you use the plex placed in service before finds that it does not include land im-recovery period listed under the activ- January 1, 2010. provements. Therefore, Sam uses theity in that table. Use the tables in the recovery period under asset class• Any retail motor fuels outlet.order shown below to determine the 00.3. The land improvements have arecovery period of your depreciable • Any qualified leasehold improve- 20-year class life and a 15-year recov-property. ment property placed in service ery period for GDS. If he elects to usebefore January 1, 2010. ADS, the recovery period is 20 years.Table B-1. Check Table B-1 for a

• Any qualified restaurant propertydescription of the property. If it is de-Example 3. Pam Martin owns a re-placed in service before Januaryscribed in Table B-1, also check Table

1, 2010. tail clothing store. During the year, sheB-2 to find the activity in which thepurchased a desk and a cash registerproperty is being used. If the activity is • Initial clearing and grading landfor use in her business. She checksdescribed in Table B-2, read the text (if improvements for gas utility

any) under the title to determine if the Table B-1 and finds office furnitureproperty and electric utility trans-property is specifically included in that under asset class 00.11. Cash regis-mission and distribution plants.asset class. If it is, use the recovery ters are not listed in any of the asset

• Any water utility property.period shown in the appropriate col- classes in Table B-1. She then checksumn of Table B-2 following the Table B-2 and finds her activity, retail• Certain electric transmissiondescription of the activity. If the activity store, under asset class 57.0, Distribu-property used in the transmissionis not described in Table B-2 or if the tive Trades and Services, which in-at 69 or more kilovolts of electric-activity is described but the property cludes assets used in wholesale andity for sale and placed in serviceeither is not specifically included in or retail trade. This asset class does notafter April 11, 2005.is specifically excluded from that asset specifically list office furniture or a• Natural gas gathering and distri-class, then use the recovery period cash register. She looks back at Tableshown in the appropriate column fol- bution lines placed in service af- B-1 and uses asset class 00.11 for thelowing the description of the property ter April 11, 2005.

desk. The desk has a 10-year class lifein Table B-1.and a 7-year recovery period for GDS.

Example 1. Richard Green is a pa- If she elects to use ADS, the recoveryTax-exempt use property subject toper manufacturer. During the year, he period is 10 years. For the cash regis-a lease. The recovery period for ADSmade substantial improvements to the ter, she uses asset class 57.0 becausecannot be less than 125 percent of theland on which his paper plant is lo- cash registers are not listed in Tablelease term for any property leasedcated. He checks Table B-1 and findsunder a leasing arrangement to a B-1 but it is an asset used in her retailland improvements under asset classtax-exempt organization, governmen- business. The cash register has a00.3. He then checks Table B-2 andtal unit, or foreign person or entity 9-year class life and a 5-year recoveryfinds his activity, paper manufacturing,(other than a partnership). period for GDS. If she elects to use theunder asset class 26.1, Manufacture ADS method, the recovery period is 9of Pulp and Paper. He uses the recov-Table B-2. If the property is not listed years.ery period under this asset class be-in Table B-1, check Table B-2 to findcause it specifically includes landthe activity in which the property is ■




Table B-1.

Assetclass

00.11

00.12

00.13

00.21

00.22

Description of assets included

Table of Class Lives and Recovery PeriodsRecovery Periods

(in years)Class Life(in years)

GDS(MACRS) ADS

SPECIFIC DEPRECIABLE ASSETS USED IN ALL BUSINESS ACTIVITIES, EXCEPT AS NOTED:

00.2300.241

00.242

00.25

00.2600.2700.28

00.3

00.4

Office Furniture, Fixtures, and Equipment:Includes furniture and fixtures that are not a structural component of a building. Includes suchassets as desks, files, safes, and communications equipment. Does not includecommunications equipment that is included in other classes.

10 7 10

Information Systems:Includes computers and their peripheral equipment used in administering normal businesstransactions and the maintenance of business records, their retrieval and analysis.Information systems are defined as:1) Computers: A computer is a programmable electronically activated device capable ofaccepting information, applying prescribed processes to the information, and supplying theresults of these processes with or without human intervention. It usually consists of a centralprocessing unit containing extensive storage, logic, arithmetic, and control capabilities.Excluded from this category are adding machines, electronic desk calculators, etc., and otherequipment described in class 00.13.2) Peripheral equipment consists of the auxiliary machines which are designed to be placedunder control of the central processing unit. Nonlimiting examples are: Card readers, cardpunches, magnetic tape feeds, high speed printers, optical character readers, tape cassettes,mass storage units, paper tape equipment, keypunches, data entry devices, teleprinters,terminals, tape drives, disc drives, disc files, disc packs, visual image projector tubes, cardsorters, plotters, and collators. Peripheral equipment may be used on-line or off-line.Does not incude equipment that is an integral part of other capital equipment that is includedin other classes of economic activity, i.e., computers used primarily for process or productioncontrol, switching, channeling, and automating distributive trades and services such as pointof sale (POS) computer systems. Also, does not include equipment of a kind used primarily foramusement or entertainment of the user.

6 5 5

Data Handling Equipment; except Computers:Includes only typewriters, calculators, adding and accounting machines, copiers, andduplicating equipment.

6 5 6

Airplanes (airframes and engines), except those used in commercial or contract carryingof passengers or freight, and all helicopters (airframes and engines)

6 5 6

Automobiles, Taxis 3 5 5

Buses 9 5 9

Light General Purpose Trucks:Includes trucks for use over the road (actual weight less than 13,000 pounds) 4 5 5

Heavy General Purpose Trucks:Includes heavy general purpose trucks, concrete ready mix-trucks, and ore trucks, for useover the road (actual unloaded weight 13,000 pounds or more)

6 5 6

Railroad Cars and Locomotives, except those owned by railroad transportationcompanies

15 7 15

Tractor Units for Use Over-The-Road 4 3 4

Trailers and Trailer-Mounted Containers 6 5 6

Vessels, Barges, Tugs, and Similar Water Transportation Equipment, except those usedin marine construction

18 10 18

Land Improvements:Includes improvements directly to or added to land, whether such improvements are section1245 property or section 1250 property, provided such improvements are depreciable.Examples of such assets might include sidewalks, roads, canals, waterways, drainagefacilities, sewers (not including municipal sewers in Class 51), wharves and docks, bridges,fences, landscaping shrubbery, or radio and television transmitting towers. Does not includeland improvements that are explicitly included in any other class, and buildings and structuralcomponents as defined in section 1.48-1(e) of the regulations. Excludes public utility initialclearing and grading land improvements as specified in Rev. Rul. 72-403, 1972-2 C.B. 102.

20 15 20

Industrial Steam and Electric Generation and/or Distribution Systems:Includes assets, whether such assets are section 1245 property or 1250 property, providingsuch assets are depreciable, used in the production and/or distribution of electricity with ratedtotal capacity in excess of 500 Kilowatts and/or assets used in the production and/ordistribution of steam with rated total capacity in excess of 12,500 pounds per hour for use bythe taxpayer in its industrial manufacturing process or plant activity and not ordinarily availablefor sale to others. Does not include buildings and structural components as defined in section1.48-1(e) of the regulations. Assets used to generate and/or distribute electricity or steam ofthe type described above, but of lesser rated capacity, are not included, but are included inthe appropriate manufacturing equipment classes elsewhere specified. Also includes electricgenerating and steam distribution assets, which may utilize steam produced by a wastereduction and resource recovery plant, used by the taxpayer in its industrial manufacturingprocess or plant activity. Steam and chemical recovery boiler systems used for the recoveryand regeneration of chemicals used in manufacturing, with rated capacity in excess of thatdescribed above, with specifically related distribution and return systems are not included butare included in appropriate manufacturing equipment classes elsewhere specified. An exampleof an excluded steam and chemical recovery boiler system is that used in the pulp and papermanufacturing equipment classes elsewhere specified. An example of an excluded steam andchemical recovery boiler system is that used in the pulp and paper manufacturing industry.

22 15 22

EPS Filename: 13081f31 Size - Width = 44 picas Depth = 58 picas


Selecting a Bid Price A municipality needs to dispose of 1 million tons of refuse each year for the next 5 years. It is requesting bids from firms for the business. A bid of R means that the firm contractually agrees to receive $R/ton each year for the next 5 years from the municipality in exchange for disposing of the refuse. Additional information: • A new fleet of trucks to pick up the refuse must be acquired. The cost of the trucks to handle the

proposed volume is $6 million. The projected market value of the trucks is $1 million at the end of 5 years. The trucks are classified as 5-year property and straight line depreciation (with half-year convention) is used.

• A landfill is required for disposal. The municipality’s sole landfill owner is currently charging $5/ton

each year.

• Fuel and direct labor costs are projected at $1/ton per year. An administrative cost of $0.5 million each year to plan and manage operations is needed.

• Firm’s marginal tax rate is 40%. • The cost of unlevered equity capital r0 for this type of project is 10%.

41

I. PROJECT VALUATION WITH NO DEBT 0 1 2 3 4 5Revenues 1000R 1000R 1000R 1000R 1000RCost Landfill cost -5000 -5000 -5000 -5000 -5000 Labor/fuel cost -1000 -1000 -1000 -1000 -1000 Administrative cost -500 -500 -500 -500 -500EBITDA 1000R-6500 1000R-6500 1000R-6500 1000R-6500 1000R-6500 Interest expense 0 0 0 0 0 Depreciation -500 -1000 -1000 -1000 -500Profit before tax 1000R-7000 1000R-7500 1000R-7500 1000R-7500 1000R-7000Taxes -(400R-2800) -(400R-2800) -(400R-2800) -(400R-2800) -(400R-2800) Net Income 600R-4200 600R-4500 600R-4500 600R-4500 600R-4200Adjustments Loan principal cash flow Depreciation 500 1000 1000 1000 500 Investment -6000 *1400Free cash flow -6000 600R -3700 600R-3500 600R-3500 600R-3500 600R-2300*BV = Cost basis – accumulated depreciation = 6000 – 4000 = 2000. 1400 = SV – τ(SV – BV) = 1000 – 0.4(1000-2000). (Assumes losses can be used to offset other income.) NPV(Free cash flow) @ 10% NPV(Free cash flow) @ 10% = -6000 + (600R-3500)[ (1-(1.10)-5)/(0.10)] – 200/1.10 + 1200/(1.10)5

= 2,275R – 18,704 = 0 ⇒ R = $8.22/ton.

0 1 2 3 4 5Revenues 8220 8220 8220 8220 8220Cost Landfill cost -5000 -5000 -5000 -5000 -5000 Labor/fuel cost -1000 -1000 -1000 -1000 -1000 Administrative cost -500 -500 -500 -500 -500EBITDA 1720 1720 1720 1720 1720 Interest expense 0 0 0 0 0 Depreciation -500 -1000 -1000 -1000 -500Profit before tax 1220 720 720 720 1220Taxes -488 -288 -288 -288 -488Net Income 732 432 432 432 732Adjustments Loan principal cash flow Depreciation 500 1000 1000 1000 500 Investment -6000 *1400Free cash flow -6000 1232 1432 1432 1432 2632 NPV(Free cash flow) = -6000+ 1232/1.1 + 1432/(1.1)2 + 1432/(1.1)3 + 1432/(1.1)4 + 2632/(1.1)5 = -8.

42

II. PROJECT VALUE WITH DEBT Assume the firm is able to borrow half of the 6 million from a bank. The loan interest rate is 7% (compounded annually) with principal repaid in five equal annual installments. A. Flow-to-Equity Approach 0 1 2 3 4 5 Revenues 8220 8220 8220 8220 8220 Cost Landfill cost -5000 -5000 -5000 -5000 -5000 Labor/fuel cost -1000 -1000 -1000 -1000 -1000 Administrative cost -500 -500 -500 -500 -500 EBITDA 1720 1720 1720 1720 1720 Interest expense -210 -168 -126 -84 -42 Depreciation -500 -1000 -1000 -1000 -500 Profit before tax 1010 552 594 636 1178 Taxes -404 -221 -238 -254 -471 Net Income 606 331 356 382 707 Adjustments Loan principal cash flow 3000 -600 -600 -600 -600 -600 Depreciation 500 1000 1000 1000 500 Investment -6000 1400 Free equity cash flow -3000 506 731 756 782 2007 a. NPV(Free equity cash flow)@10% = 412. NPV(Free equity cash flow)@10% = NPV(Free cash flow w/o loan)@10% + LoanValue@10% = -8 + 420 = 412. Calculation of LoanValue@10%: 0 1 2 3 4 5 LB 3,000 2,400 1,800 1,200 600 0 IP -210 -168 -126 -84 -42 PP -600 -600 -600 -600 -600 TP -810 -768 -726 -684 -642 tax shield 84 67 50 34 17 Loan cash flow 3,000 -726 -701 -676 -650 -625 LoanValue@10% = NPV(Loan cash flow)@10% = 420 = (1 – 4.2/10)(3,000 – 600[(1 – (1.1)-5) /0.1]). b. Cost of levered equity capital: re = r0 + (1- τ)(r0 – rD)(Debt/Value)

= 0.10 + (1-0.40)(0.10-0.07)(3,000/6,000) = 0.118 ≈ 0.12.

NPV(Free equity cash flow)@12% = 208.

NPV(Free equity cash flow)@12% = NPV(Free cash flow w/o loan)@12% + LoanValue@12% = -336 + 544 = 208. LoanValue@12% = NPV(Loan cash flow)@12% = 544 = (1 – 4.2/12)(3,000 – 600[(1 – (1.12)-5) /0.12]).

43

4

B. WEIGHTED AVERAGE COST OF CAPITAL (WACC) APPROACH Ignore financing but value free cash flow at the weighted average cost of capital, calculated as: rWACC = (3000/6000)(1 – 0.40)(0.07) + (3000/6000)(0.118) = 0.08.

= r0(1 - τ B/(B+S)) = 0.10(1 – 0.4(0.5))!

NPV(Free cash flow w/o loan)@8% = 349. NOTE: This value is lower than 412 (good) but higher than 208. It is higher than 208 because the debt to

value ratio here does not remain constant over time (it gets lower over time). C. ADJUSTED PRESENT VALUE (APV) APPROACH Value free cash flows using cost of unlevered equity capital (r0), and then add to this the TaxShieldValue using the pre-tax cost of debt, i.e.,

Project Value with Debt = Project Value without Debt + TaxShieldValue

TaxShieldValue = 84/1.07 + 67/(1.07)2 + 50/(1.07)3 + 34/(1.07)4 + 17/(1.07)5 = 216 = tax rate * LoanPrincipalValue = 0.40 * (3,000 – 600[(1 – (1.07)-5) /0.07])!

NOTE: With personal taxes a more conservative suggested tax rate

T = 1 – (1 – ΤC)(1- TE)/(1- TD), where TC = corporate tax rate, TE = tax on equity income and TD = tax on ordinary income. For typical values of TC, TE and TD, the value for T is closer to 15 or 20%.

III. GENERAL REMARKS 1. Need to lock-in landfill cost or else the owner can appropriate your economic profit! 2. Formulas used here are derived assuming, among other things, a “perpetual no-growth cash flow”

model. The amount of debt in relation to project value changes over time here. So, in principle, one “should” adjust the re over time to reflect this, which is cumbersome. Ditto for the rWACC. This is why the APV approach here is recommended. For a large firm in which financing is centralized for all projects, the WACC approach is commonly used and is arguably the preferred approach. This is because the WACC approach separates the valuation of the operating cash flows, typically made by operations analysts, from the financing decisions made by financial executives.

3. For the APV approach, strictly speaking, Debt/Value does not equal 0.5 but it i close. (It is possible to

determine the amount of debt so that the loan to value ratio equals the target ratio of 0.50.)

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CASH FLOW ANALYSIS HOMEWORK PROBLEMS Present Value After-Tax Analysis 1. The after-tax project cash flows for the first 5 years are 110,000, 121,000, 133,100, 146,410, 161,051,

respectively. The project cash flows thereafter are assumed to grow at an annual rate of 2% in perpetuity. (So, for example, the after-tax project cash flow in year 6 is 1.02(161,051).) The required investment at time 0 is 1,800,000. The cost of capital is 10%. Should the company invest in this project?

2. The after-tax project cash flows for the first 3 years are 110,000, 121,000, 133,100, respectively. The

project cash flows thereafter are assumed to grow at an annual rate of 7% in perpetuity. (So, for example, the after-tax project cash flow in year 4 is 1.07(133,100).) The required investment at time 0 is 2,200,000. The cost of capital is 12% per year. Should the company invest in the project?

3. A friend of yours just made a non-refundable deposit of the first month’s rent (equal to $2,000) on a 6-

month apartment lease. She learns that there is another apartment she likes equally well and its monthly rate is only $1,800. Should she switch? What if your friend had planned to lease for one year? Assume an interest rate of 12% (compounded monthly).

4. You are considering purchase of a home. The inspector reports that the roof will have to be replaced in

5 years. After consulting several roof contractors, you discover that for this home a new roof will cost $20,000 and will last 20 years. Assuming costs remain constant and that the discount rate is 5%, what is the value of the existing roof?

5. Your best taxable investment opportunity has an EAR of 4%. Your best tax-free investment

opportunity has an EAR of 3%. If your tax rate is 30%, which opportunity provides the higher after-tax interest rate?

6. Your friend recently purchased a new energy-efficient furnace. She received a 7% interest rate (APR,

monthly compounding) from the vendor. This rate was lower than the rate she could have obtained on her home equity loan (8% APR, monthly compounding). Interest on her home equity loan is tax deductible. Her tax rate is 25%. Did your friend finance her purchase with the cheapest loan?

7. You have credit card debt of $25,000 that has an APR (monthly compounding) of 15%. Each month

you pay the minimum amount only. (You are required to pay only the outstanding interest.) You receive an offer in the mail for an otherwise identical credit card with an APR of 12% (monthly compounding). You decide to switch cards, rolling over the outstanding balance on the old card to the new card. How much more can you borrow today on the new card without changing the minimum monthly payment you had been paying?

8. You need a new car and the dealer has offered you a price of $20,000 with two payment options:

(a) pay cash and receive a $2000 rebate, or (b) pay a $5000 down payment and finance the rest with a 0% APR loan over 30 months. You have just enrolled in a graduate program and expect to be in debt for at least the next 2½ years. You plan to use your credit card to pay expenses. The credit card interest rate is 15% APR (monthly compounding). Which payment option is best for you?

45

Project Balance and IRR 1. A company is considering investing in a project whose after-tax cash flows for the next two years are

11200 at the end of the first year and 50176 at the end of the second year. The appropriate cost of capital is 12% (compounded annually). The initial investment at time 0 is 52000.

a. Show the project balance schedule for this project. b. Should the company invest in this project? Explain by applying the project balance concept. c. What can you conclude about the IRR for this investment project? Explain. Do NOT calculate the

IRR. 2. A company is considering investing in a project whose after-tax cash flows for the next two years are

10,000 at the end of the first year and 20,000 at the end of the second year. The appropriate cost of capital is 10% (compounded annually). The initial investment at time 0 is 24,000.

a. Show the project balance schedule for this project. b. Should the company invest in this project? Explain by applying the project balance concept. c. What can you conclude about the IRR for this investment project? Explain. Do NOT calculate the

IRR. 3. Cash flows associated with two mutually exclusive projects are:

0 1 2 3 4 5 Project 1 -100 30 30 30 30 30 Project 2 -150 43 43 43 43 43

Cost of capital is 6%. a. Compute the IRR for each project. b. Which project do you recommend? c. Use incremental IRR and project balance to justify your recommendation. d. What are the payback and discounted payback periods for projects A and B?

46

Equipment Selection 1. A company has two options for a machine it must purchase. The manufacturer’s discount rate is 10%. Ignore taxes and depreciation.

Type I: Initial cost = 250,000. Annual cost = 40,000. 4-yr lifetime. Salvage value = 80,000. Type II: Initial cost = 400,000. Annual cost = 20,000. 7-yr lifetime. Salvage value = 50,000.

Use the annual equivalent method to determine which machine to purchase. 2. A company must purchase a piece of equipment. Two types are being considered: TYPE A: initial cost = 1000, annual cost = 400, salvage value = 300, 5-yr lifetime. TYPE B: initial cost = 2000, annual cost = 200, salvage value = 800, 8-yr lifetime. Discount rate is 10%. Use the annual equivalent method to recommend which equipment type to purchase. Ignore taxes and depreciation. 3. A company must acquire a machine for its operations. Vendor A will lease a capable machine. Vendors B and C sell a capable machine. Each of their machines has a useful life of 5 years. Relevant data are: Initial outlay Annual expense Resale value Vendor A 6,000 8,000 0 Vendor B 30,000 2,000 10,000 Vendor C 35,000 1,600 12,000 Company uses a cost of capital of 10%. a. Use PV analysis to recommend which option is best. b. Use the incremental IRR approach to recommend which option is best.

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Project Cash Flow Analysis 1. A manufacturer must acquire equipment to produce a product for a customer for a period of exactly

two years. Relevant data are: • Revenues are projected to be 600,000 in the first year and 800,000 in the second year. • The manufacturer’s cost of levered equity capital is 20% and its marginal tax rate is 30%. • For tax purposes MACRS depreciation schedule (with half year convention) is used. • The equipment’s market value depreciates at a rate of 30% per year, i.e., its value at the end of the

year is 70% of the value at the beginning of the year. • The equipment will be classified as 5-year property. • The equipment initially costs 350,000, of which 150,000 will be financed by a 2-year loan at 12%

interest, with principal repaid in two equal annual installments. • First year annual operating costs will be 250,000. Each successive year these costs rise by 10%. • Use annual compounding.

Obtain the after-tax cash flows by filling out the following table.

Project cash flow analysis 0 1 2 Revenues Expenses EBITDA Interest Depreciation EBIT Taxes Net Income Adjustments Loan principal cash flow Depreciation Investment Free equity cash flow Net present value

2. A manufacturer must acquire equipment to produce a product for a customer for a period of exactly two years. Relevant data are:

• Revenues are projected to be 500,000 in the first year and 700,000 in the second year. • The manufacturer’s cost of levered equity capital is 18% and its marginal tax rate is 35%. • For tax purposes MACRS depreciation schedule (with half year convention) is used. • The equipment’s market value depreciates at a rate of 40% per year, i.e., its value at the end of the

year is 60% of the value at the beginning of the year. • The equipment will be classified as 5-year property. • The equipment initially costs 250,000, of which 100,000 will be financed by a 2-year loan at 10%

interest, with principal repaid in two equal annual installments. • First year annual operating costs will be 400,000. Each successive year these costs rise by 25%. • Use annual compounding.

Obtain the after-tax cash flows by filling out the following table.

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Project cash flow analysis

0 1 2 Revenues Expenses EBITDA Interest Depreciation Profit before tax Taxes Net Income Adjustments Loan principal cash flow Depreciation Investment Free equity cash flow Net present value

3. A manufacturer must acquire equipment to produce a product for a customer for a period of exactly four years. Relevant data are:

• Revenues are projected to be 300,000 in the first year, and are projected to grow at an annual rate of 10% each year thereafter. First year annual operating costs will be 200,000. Each successive year these costs decline by 5%.

• The equipment initially costs 200,000, of which 100,000 will be financed by a 4-year loan at 10% interest, with principal repaid in four equal annual installments. The equipment’s market value depreciates at a rate of 20% per year. The equipment will be classified as 5-year property. For tax purposes MACRS depreciation schedule (with half year convention) is used.

• The manufacturer’s cost of levered equity capital is 18% and its marginal tax rate is 40%.

Obtain the last project year’s free cash flow by filling out the following table.

Year 4 project cash flow Revenue Cost EBITDA Interest expense Depreciation Profit before tax Income tax Net income Adjustments Loan principal cash flow Depreciation Investment Free equity cash flow

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4. A company is deciding on whether to invest in a new project. Production will require 10 million in initial capital. It is anticipated that 1 million units will be sold each for 5 years at which point the product will be obsolete. Each year’s production will required 10,000 hrs of labor and 100 tons of raw material. The average wage rate is $30/hr and cost of raw material is $100/ton. The sales price is projected to be $3.30 per unit and this price is expected to be maintained in real terms. The company uses a 12% discount rate for projects of this type and its tax rate is 34%. Straight-line depreciation is used and no salvage value is expected. a. What is the NPV over time of this project? Should the company invest in the project? b. Upon further review, you realize that inflation is expected to be 4% per year. This will inflate the

projected unit price, labor and material costs each year (starting after the first year’s projections). How does this change your recommendation in part (a)?

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Capital Budgeting 1. A firm is considering 4 proposed projects. Available budget is 300. (All numbers in 000’s.) 1 2 3 4 Initial outlay 100 125 150 75 NPV 25 30 35 40 What is the optimal set of projects? 2. A firm is considering 5 proposed projects. Available budget is 600. (All numbers in 000’s.) 1 2 3 4 5 Initial outlay 100 300 200 150 150 Present worth 200 500 300 200 250 a. Use the benefit-cost ratio to select among the projects. b. What is the optimal set of projects? 3. A company has identified 7 attractive projects. Each project has a positive NPV but requires an outlay

of funds (i.e. a negative cash flow) in the first two years. Company managers have decided that they can allocate up to $250,000 is each of the first two years to support these projects. Unused funds in the first year can be invested at 10% and used to supplement the second year’s budget. Formulate an integer program to determine which projects should be funded. Which projects should be funded? (All numbers in 000’s.)

1 2 3 4 5 6 7 CF(1) -90 -80 -50 -20 -40 -80 -80 CF(2) -58 -80 -100 -64 -50 -20 -100 NPV 150 200 100 100 120 150 240

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CASH FLOW ANALYSIS HOMEWORK PROBLEM SOLUTIONS

Present Value After-Tax Analysis 1. PV = 400,000 + [161,051/(0.10 – 0.02)]/(1.1)4 = 1,775,000. NPV = -25,000. NO. 2. PV = 194,675 + [133,100/(0.12 – 0.07)]/(1.12)2 = 2,316,805. NPV = 116,805. YES. 3. Consider the incremental cash flow for switching apartments: an outlay now of 1800

followed by a benefit of 200 for each month of rent. Since she plans to stay in the apartment 6 months, and since 1800 > 6(200), it is obvious she should stay in the apartment. (The initial outlay of 2000 is a sunk cost and should not be considered.) Let n denote the total number of months your friend plans to stay in the apartment. The PV of this incremental cash flow stream is -1800 + 200[(1 – (1.01)-(n-1))/0.01] := -1800 + 200X, where X denotes the value of the expression in brackets. Break-even value occurs when X = 9, so n = 11.

4. If you replace the roof today, and every 20 years thereafter, the PV of costs will be 20000 +

20000/s20, where s20 denotes the one-period equivalent interest rate that corresponds to compounding 5% per year for 20 years. Since s20 = (1.05)20 – 1 = 1.653, the PV = 32097. Now suppose you wait 5 years. The PV 5 years in the future is 32097, which has a present value today of 32097/(1.05)5 = 25149. The difference 32097-25149=6948 is the value of the existing roof under these conditions.

5. After-tax rate = (1- 0.30)4% = 2.8%, which is less than the tax-free investment that pays 3%. 6. In terms of EARs, the after-tax cost of the home equity loan is (1- 0.25)(1 + 0.08/12)12 %=

6.22%, and the dealer’s loan is (1 + 0.07/12)12 %= 7.23%. Home equity loan is cheaper. 7. Your monthly interest rate on your current credit card is 15/12 = 1.25%. With an outstanding

loan balance of $25,000, monthly payment is $25,000(0.0125) = $312.50 in perpetuity. Your new interest monthly rate is 1%. The PV of the perpetuity at 1% is $312.50/0.01 = $31,250. So you can borrow an additional $6,250 by switching credit cards.

8. Your discount rate is 1.25% per month.

Option (a): You will pay the dealer $18,000 by financing with your credit card. The cost today of this option is $18,000.

[You pay interest of $18,000(1.25%) = $225 per month for 30 months and pay off the principal balance of $18,000 at the end of 30 months. Your cash flow stream is (0, -225, -225, ... , -225 + -18,000). PV at 1.25% is 18,000. Use project balance.]

Option (b): You will finance the $5,000 by borrowing from your credit card at a cost of $5,000 today, and you will have to finance the monthly payment of $15,000/30 = $500 by borrowing from your credit card, too. At 1% the PV of this cash flow stream is $500(1 – (1.0125)-30 )/0.0125 = $12,444. Total cost of option (b) = $17,444.

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Project Balance and IRR 1. a. PB(0) = -52,000.

PB(1) = -52,000(1.12) + 11,200 = -47,040. PB(2) = -47,040(1.12) + 50,176 = -2,508.80.

b. No. Final project balance is negative, which means the NPV at time 0 is also negative. c. IRR has to be less than 12%, since PB is negative at the end of project.

2. a. PB(0) = -24,000.

PB(1) = -24,000(1.1) + 10,000 = -16,400. PB(2) = -16,400(1.1) + 20,000 = 1,960.

b. Yes. Final project balance is positive, which means the NPV at time 0 is also positive. c. IRR has to be greater than 10%, since PB is positive at the end of project.

3. a. IRRA = 15.24%. IRRB = 13.34%. b. NPVA = -100+4.212(30) = 26.37. NPVB = -150+4.212(43) = 31.13. Select B.

c. Incremental cash flow stream A→B = (-50, 13, 13, 13, 13, 13). IRRA→B = 9.43%. Since this is greater than 6%, select B.

Project Balance Schedule 0 1 2 3 4 5 Project A -100 -76 -50.56 -23.59 4.99 35.29 Project B -150 -116 -79.96 -41.76 -1.26 41.66 A→B -50 -40 -29.4 -18.16 -6.25 6.37

Observe that 35.29 = 26.37(1.06)5, 41.66 = 31.13(1.06)5 and 6.37 = 41.66 – 35.29.

d. Payback period for both projects is 4 years. For a simple investment project, the discounted payback period is the first time the project balance becomes positive. This is 4 years for project A and 5 years for project B.

Equipment Selection 1. PVI = 250,000 + 40,000[1 – (1.1)-4]/(0.1) – 80,000/(1.1)4 = 322,154. AEI = [0.1PVI]/[1 – (1.1)-4] = 101,630. PVII = 400,000 + 20,000[1 – (1.1)-7]/(0.1) – 50,000/(1.1)7 = 471,710. AEII = [0.1PVII]/[1 – (1.1)-7] = 96,892. Recommend Type II machine. 1. PVA = 1000 + (400/0.1)[1 – (1.1)-5] - 300(1.1)-5 = 2330.04.

AEA = 0.1(2330.04)/[1 – (1.1)-5] = 614.66. PVB = 2000 + (200/0.1)[1 – (1.1)-8] - 800(1.1)-8 = 2693.78. AEB = 0.1(2693.78)/[1 – (1.1)-8] = 504.93. Recommend equipment type B.

3. a. PVA = 36,326. PVB = 31,372. PVC = 33,614. Select option B. b. Incremental cash flow stream A→B = (-24000, 6000, 6000, 6000, 6000, 16000). IRRA→B = 16.62% > 10%, so B is preferred to A. Incremental cash flow stream B→C = (-5000, 400, 400, 400, 400, 2400). No need to calculate IRR here because the cost 5000 > sum of benefit = 4000.

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Project Cash Flow Analysis 1.

Project cash flow analysis 0 1 2 Revenues 600,000 800,000Expenses 250,000 275,000EBITDA 350,000 525,000Interest 18,000 9,000Depreciation 70,000 56,000Profit before tax 262,000 460,000Taxes 78,600 138,000Net Income 183,400 322,000Adjustments Loan principal cash flow 150,000 (75,000) (75,000) Depreciation 70,000 56,000 Investment (350,000) *187,250Free equity cash flow (200,000) 178,400 490,250Net present value 289,118 586,942 490,250

*187,250 = 171,500 – 0.3(171,500 – 224,000) 2.

Project cash flow analysis 0 1 2 Revenues 500,000 700,000Expenses 400,000 500,000EBITDA 100,000 200,000Interest 10,000 5,000Depreciation 50,000 40,000Profit before tax 40,000 155,000Taxes 14,000 54,250Net Income 26,000 100,750Adjustments Loan principal cash flow 100,000 (50,000) (50,000) Depreciation 50,000 40,000 Investment (250,000) *114,500Free equity cash flow (150,000) 26,000 205,250Net present value 19,441 199,941 205,250

*114,500 = 90,000 – 0.35(90,000 – 160,000)

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3. Year 4 project cash flow

Revenue 399,300Cost 171,475EBITDA 227,825Interest expense 5,000Depreciation 11,520Profit before tax 211,305Income tax 84,522Net income 126,783Adjustments Loan principal cash flow (25,000) Depreciation 11,520 Investment *67,584Free equity cash flow 180,887

*67,584 = 81,920 – 0.4(81,920 – 46,080) 4. a. Reject project.

Project cash flow (in 000’s) without inflation 0 1 2 3 4 5 Revenue 3300 3300 3300 3300 3300 Cost 310 310 310 310 310 EBITDA 2990 2990 2990 2990 2990 Depreciation 2000 2000 2000 2000 2000 Profit before tax 990 990 990 990 990 Income tax 337 337 337 337 337 Net income 653 653 653 653 653 Adjustments Depreciation 2000 2000 2000 2000 2000 Investment -10,000 0 0 0 0 0 Free cash flow -10,000 2653 2653 2653 2653 2653 NPV -437 10711 9025 7137 5022 2653 b. Accept project.

Project cash flow (in 000’s) with inflation 0 1 2 3 4 5 Revenue 3300 3432 3569 3712 3861 Cost 310 322 335 349 363 EBITDA 2990 3110 3234 3363 3498 Depreciation 2000 2000 2000 2000 2000 Profit before tax 990 1100 1234 1363 1498 Income tax 337 377 420 464 509 Net income 653 653 653 653 653 Adjustments Depreciation 2000 2000 2000 2000 2000 Investment -10,000 0 0 0 0 0 Free cash flow -10,000 2653 2732 2814 2900 2989 NPV 89 11299 9684 7786 5569 2989

56

Capital Budgeting 1. Projects 1, 2 and 4 for an NPV of 95,000. 2. a. The benefit-cost ratios for the 5 projects are 2, 5/3, 3/2, 4/3 and 5/3, respectively. Projects 1,

2 and 5 would be recommended using the approximate method based on ranking by benefit-cost ratios.

b. Here, the optimal set of projects is the same. Note that projects 1, 2 and 3 provide the same aggregate NPV and use the entire budget.

3. The problem may be expressed as MAX 150x1 + 200x2 +100x3 +100x4 +120x5 +150x6 +240x7

subject to: 90x1 + 80x2 + 50x3 + 20x4 + 40x5 + 80x6 + 80x7 + Y ≤ 250 58x1 + 80x2 + 100x3 + 64x4 + 50x5 + 20x6 + 100x7 ≤ 250 + 1.1Y, where it is understood that each xi ε {0, 1} and that Y ≥ 0. The maximum NPV is 610, which can be achieved in two ways: (1) Fund projects 4, 5, 6,

and 7 at a cost of 220 in year 1 and 234 in year 2, or (2) fund projects 1, 4, 5, and 7 at a cost of 230 in year 1 and 272 in year 2. Note that the first plan costs less.

57

A Firm Valuation Model Company X has gross earnings of Yn in year n, and decides to invest a portion u of this amount each year in order to attain earnings growth. Next year’s gross earnings follow the equation

Yn+1 = [1 + g(u)]Yn , and next period’s installed capital follows the equation

Kn+1 = (1- δ)Kn + uYn , The resulting growth rate in gross earnings, g(u), is a function of the firm’s characteristics, and will be taken to be

g(u) = 0.12[1 – e5(δ - u)], where δ denotes the depreciation rate on installed capital. Here, we set δ = 0.10. Note that when u = 0.10 the growth rate in earnings is zero, which reflects the fact that investment in new capital is just sufficient to keep up with the depreciation of installed capital. Note also that the maximum growth rate is 12%. The company needs to decide on its investment policy, namely, the constant u. It seeks to choose the investment policy that will maximize the firm’s value, which will be taken to be the present value of the free cash flows. The appropriate discount rate for this firm is 15%, and its tax rate is 34%. Current earnings Y0 = 10 million, and the initial capital K0 = 19.8 million.

U = 0.15 0 1 2 3 Cash flow from operations 10,000,000 10,265,439 10,537,924 10,817,641 Installed capital 19,800,000 19,320,000 18,927,816 18,615,723 Depreciation 1,980,000 1,932,000 1,892,782 1,861,572 Profit before tax 8,020,000 8,333,439 8,645,142 8,956,069 Taxes 2,726,800 2,833,369 2,939,348 3,045,064 Net income 5,293,200 5,500,070 5,705,794 5,911,005 Adjustments Depreciation 1,980,000 1,932,000 1,892,782 1,861,572 Investment 1,500,000 1,539,816 1,580,689 1,622,646 Free cash flow 5,773,200 5,892,254 6,017,887 6,149,931



59

PRO FORMA CASH FLOW ANALYSIS EXAMPLE

Assumptions

Period 0 1 2 3 4 5 6Price 30 27.67 25.51 23.53 21.7 20 Quantity 200 230 264 303 349 400 Unit Cost 9 8.6 8.1 7.7 7.4 7

Cost of capital 13% first 6 years (continuously compounded)12% after year 6 (continuously compounded)

Terminal value 4% growth after year 6 (continuously compounded)

0 1 2 3 4 5 6 7Sales 6000 6364 6735 7130 7573 8024 CGS -1800 -1978 -2138 -2333 -2583 -2807 Rent -200 -200 -200 -200 -200 -200 SGA -600 -637 -676 -718 -763 -810 EBITDA 3400 3549 3721 3879 4027 4207 Depreciation -3500 -3500 -3500 -3500 -3500 -3500 Profit before tax -100 49 221 379 527 707 Taxes 0 20 88 152 211 283 Net Income -100 29 133 227 316 424 Terminal Value 51012Adjustments Depreciation 3500 3500 3500 3500 3500 3500 Investment -35000 0 0 0 0 0 0 Free Cash Flow (FCF) -35000 3400 3529 3633 3727 3816 3924 51012Present Value 34707 36125 37611 39199 40914 42778 44793 Net Present Value -293 39525 41140 42832 44641 46594 48717 FCF Percentage 0.086 0.0858 0.0848 0.0835 0.0819 0.0805

CGS = Cost of Goods SoldSGA = Selling, General, and AdministrativeEBITDA = Earnings Before Interest, Taxes and Depreciation Allowance

61

CASH FLOW PROJECTIONS Exponential Costing An engineering group has finished a preliminary design of a large refinery. Its capacity is 3M barrels per year, and the estimated cost is $27M. a. Estimate the cost for a facility with capacities of 50% and 200% of the existing design and the

associated cost per barrel. The cost-exponent factor is 0.67. b. Suppose the refinery will be built in 5 years. Construction costs are projected to increase at a rate of

3% per year. Estimate the future cost of the refinery. Solution: For 1.5M cap refinery:

9(1.5/3.0)0.67 = 5.66/barrel today or 5.66(1.03)5(1.5M) = 9.84M facility cost 5 yrs from now. For 4.5M cap refinery:

9(4.5/3.0)0.6 = 11.81/barrel today or 11.81(1.03)5(4.5M) = 61.6M facility cost 5 yrs from now.

Cost-Exponent Factors * Process industrial equipment Material handling equipment General industrial equipment Item Exponent Item Exponent Item Exponent Agitators 0.3-0.5 Bagging machines 0.8 Air compressors Centrifuges 0.7-1.3 Conveyors 0.7 Cranes 0.6 Evaporators 0.5-0.7 Conveyors (roller) 0.9 Electric motors 0.8 Heat exchangers 0.7-0.9 Elevators 0.4 Steam boilers 0.5 Piping 0.7-0.9 Hoppers 0.7-0.9 Building Pumps 0.5-0.9 Single-story 0.8 Tanks (rectangular) 0.5 Two-story 0.7-0.8 *Adapted from Table 3-5, p.80 in Capital Investment Analysis for Engineering and Management 3rd ed, J.R. Canada, W.G. Sullivan, J.A. White and D. Kulonda, 2005.

Facility Cost-Exponent Factors *

Facility Exponent LP gas recovery in refineries 0.70 Polymerization, small plants 0.73 Polymerization, large plants 0.91 Steam generation, large, 200 psi 0.61 Steam generation, large, 1000 psi 0.81 Power generation, 2,000-20,000 kW 0.88 Power generation, oil field, 20-200 kW 0.50 Sulfur from H2S 0.64 Oxygen plant 0.65 Styrene plant 0.65 Ammonia, nitric acid or urea plant 0.98 Chorine plant, electrolytic 0.75 Refineries, small 0.57 Refineries, large 0.67 Hydrogen sulfide removal 0.55 * Adapted from Exhibit 16.10, p. 458 in Engineering Economy, 3rd ed, T. Eschenbach, 2011.

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Learning Curves An aircraft manufacturer has a contract to produce 40 jets. The manufacturer’s learning curve percentage for labor is 0.8. It estimates the first jet will require 8,000 person-days. The cost of a person-day is $250. a. What is the estimated labor cost to produce the 2nd unit? 16th unit? 32nd unit? b. Estimate the total labor cost for the contract. c. Suppose the manufacturer can implement a productivity improvement program that will lower its

learning curve percentage for labor by 10%. Estimate the percentage reduction in the total labor cost. Solution: a. 8000(0.8) = 6400. 8000(0.8)4 = 3277. 8000(0.8)5 = 2621. b. 0.8 = 2–b. b = 0.3219.

Total cost ≈ 250[(8000)(401-b/(1-b))] ≈ 36.0M. (Actual cost = 34.5M.) c. New learning curve percentage = 0.9(0.8) = 0.72 = 2–b. b = 0.4739.

Total cost ≈ 250[(8000)(401-b/(1-b))] ≈ 26.5M. (Actual cost = 23.9M.) ≈ 25% reduction.

Learning-Curve Exponents * Activity Exponent Prototype assembly 0.65 Final assembly – complex 0.70 Final assembly – simple 0.80 Packaging 0.80 Shearing metal plates 0.82 Printed circuit-board fabrication 0.83 Welding shop 0.85 Machine shop 0.95 Grinding, chipping, blast cleaning 0.985 * Adapted from Exhibit 16.15, p. 462 in Engineering Economy, 3rd ed, T. Eschenbach, 2011.

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Growth Curves A manufacturer is considering production of a new product. Marketing estimates the upper limit on annual demand will be 5,000, and that the market price will be $150 for the first year and will drop 3% per year thereafter. It expects demand to reach 10% of its limit within 2 years and that it will take 12 years to reach 90% of its limit. a. Estimate annual demand and revenue over the next 10 years assuming a Pearl curve. b. Estimate annual demand and revenue over the next 10 years assuming a Gompetz curve. Solution: Pearl Curve: ln(L/d(t) - 1) = ln a – bt.

ln 9 = ln a – 2b. ln 1/9 = ln a – 12b. b = 0.4394, a = 21.674.

d(t) = 5,000[1/(1+21.674exp(-0.4394t))].

1 2 3 4 5 6 7 8 9 10 demand 334 500 735 1055 1467 1959 2500 3041 3533 3945 Rev 50,112 72,750 103,783 144,496 194,814 252,386 312,365 368,517 415,340 449,812

Gomptez Curve: ln[ln(L/d(t))] = ln a – bt.

ln[ln 9] = ln a – 2b. ln[ln 1/9] = ln a – 12b. b = 0.3084, a = 4.267.

d(t) = 5,000exp[-4.267*exp(-0.3084t)].

1 2 3 4 5 6 7 8 9 10 Demand 217 500 921 1443 2007 2557 3055 3482 3833 4113 Rev 32,641 72,750 130,021 197,584 266,528 329,401 381,752 422,014 450,613 469,042

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CASH FLOW PROJECTION HOMEWORK PROBLEMS Exponential Costing 1. A steam-generating boiler in a utility plant produces 80,000 lb/hr of saturated steam. Its purchase cost was $400,000. a. Estimate the cost of a boiler that produces 240,000 lb/hr of saturated steam if the cost-exponent factor

for this type of boiler is 0.5. b. Suppose this boiler had been purchased 10 years ago, and the price index for this type of boiler has

increased at an average rate of 8% per year for the past 10 years. What is your new estimate of the cost of this boiler?

2. A plant engineering staff is considering a 150 kW diesel electric set to power a small plant. After

some research, you find out that a 100 kW diesel electric set cost $250,000 5 years ago. The cost-exponent factor is 0.7. The price index for this type of equipment 5 years ago was 134 and is now 161. Estimate the cost of the proposed diesel electric set.

3. An engineering group has finished a preliminary design of a small refinery. Its capacity is 2.5M

barrels per year, and the estimated cost is $20M. Estimate the cost for a facility with capacities of 50% and 200% of the existing design and the associated cost per barrel. The cost-exponent factor is 0.6.

Learning Curves 4. An aircraft manufacturer has a contract to produce 50 jets. The manufacturer’s learning curve

percentage for labor is 0.85. It estimates the first jet will require 12,000 person-days. The cost of a person-day is $300.

a. What is the estimated labor cost to produce the 2nd unit? 16th unit? 32nd unit? b. Estimate the total labor cost for the contract. c. Suppose the manufacturer can implement a productivity improvement program that will lower its

learning curve percentage for labor by 10%. Estimate the percentage reduction in the total labor cost. 5. A company assembles custom-designed products that are labor-intensive. A new product within this

line is to be assembled. The contract specifies a total production of 1000 units. A detailed estimate of the number of labor hours for the first unit is 18. For similar products within this product line there is a 10% reduction on average in the number of labor hours per unit for the 20th unit as compared to the 10th unit.

a. Estimate the number of labor hours to produce the final unit without computing the learning curve

coefficient. b. Estimate the total number of labor hours for this contract. c. How sensitive is your answer to (b) to the assumed percentage reduction?

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Growth Curves 6. A manufacturer is considering production of a new product. Marketing estimates the upper limit on

annual demand will be 100,000, and that the market price will be $80 for the first year and will drop $5 per year thereafter. It expects demand to reach 20% of its limit within 2 years and that it will take 8 years to reach 80% of its limit.

a. Estimate annual demand and revenue over the next 10 years assuming a Pearl curve. b. Estimate annual demand and revenue over the next 10 years assuming a Gompetz curve. 7. A software company introduced a new product two years ago. Ten firms acquired this software in the

first year and 50 firms acquired it last year. The software company estimates there are a maximum of 1000 firms who might purchase this software.

a. Estimate annual demand for the next 10 years assuming a Pearl curve. b. Estimate annual demand or the next 10 years assuming a Gompetz curve.

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CASH FLOW PROJECTION HOMEWORK PROBLEM SOLUTIONS 1. a. Estimated cost = 400,000(240/80)0.5 ≈ 700,000. b. Since 72/7 ≈ 10, price has doubled to 1.4M. 2. [250,000(161/134)]*[(150/100)0.7] ≈ 400,000. 3. For 1.25M cap refinery: 8(1.25/2.5)0.6 = 5.28/barrel.

For 5M cap refinery: 8(5.0/2.5)0.6 = 12.13/barrel.

4. a. 10200, 6264, 5324. b. 0.85 = 2–b. b = 0.2345.

Total cost ≈ 300[(12,000)(501-b/(1-b))] ≈ 94M. (Actual cost = 91.85M.) c. New learning curve percentage = 0.9(0.85) = 0.765 = 2–b. b = 0.3865.

Total cost ≈ 300[(12,000)(501-b/(1-b))] ≈ 65M. (Actual cost = 61.13M.) ≈ 30%. 5. a. Since 1000 ≈ 210, final unit cost = 18(0.9)10 = 6.28. b. 0.9 = 2–b. b = 0.152.

Total labor hrs ≈ 18[10001-b/(1-b)] ≈ 7400. (Actual labor hrs = 7428.) c. Learning curve percentage 0 5 15 b 0 0.074 0.2345 Approximate labor hrs 18,000 11,659 4,655 Actual labor hrs 18,000 11,654 4,643 6. Pearl Curve: ln(L/d(t)-1) = ln a – bt.

ln 4 = ln a – 2b. ln 0.25 = ln a – 8b. b = 0.4621, a = 10.08. d(t) = 100,000[1/(1+10.08exp(-0.462t))].

1 2 3 4 5 6 7 8 9 10 demand 13606 20000 28410 38649 50000 61351 71590 80000 86394 90974 Rev 1088480 1500000 1988700 2512185 3000000 3374305 3579500 3600000 3455760 3184090

Gomptez Curve: ln[ln(L/d(t))] = ln a – bt.

ln[ln 5] = ln a – 2b. ln[ln 1.25] = ln a – 8b. b = 0.3293, a = 3.11. d(t) = 100,000exp[-3.11*exp(-0.3293t)].

1 2 3 4 5 6 7 8 9 10 demand 10677 20000 31416 43474 54921 64977 73332 80000 85169 89093 Rev 854160 1500000 2180220 2825810 3295260 3573735 3666600 3600000 3406760 3118255

7. Pearl Curve: ln(L/D(t)-1) = ln a – bt. [Here, D(t) refers to cumulative demand.]

ln 99 = ln a – b. ln 19 = ln a – 2b. b = 1.8436, a = 625.60. D(t) = 100,000[1/(1+10.08exp(-0.462t))].

1 2 3 4 5 6 7 8 9 10 Demand 10 60 287 718 942 990 998 1000 1000 1000 demand 10 50 227 431 124 48 8 2 0 0 Gomptez Curve: ln[ln(L/D(t))] = ln a – bt. [Here, D(t) refers to cumulative demand.]

ln[ln 100] = ln a – b. ln[ln 20] = ln a – 2b. b = 0.4928, a =7.538. D(t) = 100,000exp[-7.538*exp(-0.4928t)].

1 2 3 4 5 6 7 8 9 10 Demand 10 60 179 350 527 676 787 864 915 947 demand 10 50 119 171 177 149 111 77 51 32

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ACTIVITY-BASED COSTING*

A company manufactures two products. Products A and B have annual sales volume of 25,000 and 40,000 units, respectively. Company has used a traditional (absorption-based) cost accounting system based on direct-labor hours. The costs for the current year are: Product A Product BDirect-material cost $350,000 $160,000Direct-labor cost 125,000 600,000Direct-labor hours 12,500 60,000Total manufacturing indirect costs: $698,250 The company is considering switching to an Activity-Based Costing (ABC) System. It has identified 6 main activities that drive the manufacturing indirect costs. Relevant data are: Activity Costs Cost Driver Product A Product B Total Units Production $280,000 Machine hrs 5,000 9,000 14,000Engineering 18,000 Eng. Change orders 100 50 150Materials handling 225,000 Material moves 50,000 40,000 90,000Receiving 12,000 Batches 600 1,000 1,600Quality control 49,500 Inspections 2,500 2,000 4,500Pack and ship 113,750 Products 25,000 40,000 65,000 I. Traditional Costing Method: Indirect cost rate = (Total indirect costs)/(Total direct-labor hrs)

= $698,250/72,500 = $9.631 per direct-labor hr.

Activity Product A Product BDirect material $350,000 $160,000Direct labor 125,000 600,000Allocated indirect 120,388 577,862Total 595,388 717,250Manufacturing cost/unit 23.82 33.45 * Adapted from Example A.3, p. 574-5 in Engineering Economy, 3rd ed by T. Eschenbach.

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II. ABC Method: Activity Applied activity rate Product A Product BProduction $20/machine hr 100,000 180,000Engineering $120/eng. change order 12,000 6,000Materials handling $2.50 move 125,000 100,000Receiving $7.50/batch 4,500 7,500Quality control $11/inspection 27,500 22,000Pack and ship $1.75/product 43,750 70,000Total indirect 312,750 385,500Direct material 350,000 160,000Direct labor 125,000 600,000Total costs 787,750 1,145,500Manufacturing cost/unit 31.51 28.64 Comparison of two methods: Product A Product BTraditional costing system $23.82 $33.45ABC system 31.51 28.64

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