1

2

A. Future
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)
FV of 1000 today 3 periods hence = 1000(1.1)
FV of 1000 today 4 periods hence = 1000(1.1) 4
2
3
= 1100(1.1) = 1210.
= 1100(1.1)
= 1100(1.1)
2
3
= 1220(1.1) = 1331.
= 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
Annual
Semiannual
Quarterly
Monthly
Daily
Continuous m
1
2
4
12
365
∞
EAR
(1 + 0.06/1) 1 – 1 = 6%
(1 + 0.06/2) 2 – 1 = 6.09%
(1 + 0.06/4) 4
(1 + 0.06/12) 12
– 1 = 6.1364%
(1 + 0.06/365) 365
– 1 = 6.1678%
– 1 = 6.1831% exp(0.06) – 1 = 6.1837%
3

B . Present
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)
PV today of receiving 1000 3 periods from now = 1000/(1.1)
PV today of receiving 1000 4 periods from now = 1000/(1.1)
2
3
4
= 826.45
= 751.31.
= 683.01.
Present value of a stream of cash flows (cash flow stream):
Consider a stream of cash flows C
C = (C
0
, C
0
at date 0, C
1
at date 1, … ,C
N
at date N. Often represented as
1
, … , C
N
). Assume per-period interest rate is r.
PV[C] = C
0
+ PV(C
1
) + PV(C
2
) + … + PV(C
N
) = C
0
+ C
1
/(1+r) + C
2
/(1+r) 2 + … + C
N
/(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 3
Cash flow stream 4,000 8,000 8,000
4000(1.03) = 4,120
12,120 12,120(1.03)=12,483.60
Account balance
20,483.60
4,000 12,120 20,483.60
Account balance here is closely related to the concept of
Project Balance
.
4

C. Formulas for PV for Typical Cash Flow Streams
Formulas assume a per-period interest rate of r. a. C
0
= 0, C k
= 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. C
0
= 0, C k
= 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. a. C
0
= 0, C
1
= A, C
2
= A(1+g), C
3
= A(1+g)
PV[C] = A/(r-g). [Important: g < r !]
2 , … , C k
= A(1+g) k-1 , k = 1, 2, … (in perpetuity).
PV of receiving 100 in the first year, growing at 3% per year thereafter at 8% is
100/(0.08 – 0.03) = 2000. b. C
0
= 0, C
1
= A, C
2
= A(1+g), C
3
= A(1+g)
PV[C] = A/(r-g) * (1 – [(1+r)/(1+g)] –N
2 , … , CN = A(1+g) N-1
). [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?
5

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 taxsheltered 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 .
6

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 (C
0
, C
1
, C
2
, C
3
, C
4
, C
5
, C
6
, C
7
, C
8
, … , C
40
, C
41
) = (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 (1 r, show that n ≈ 69/i, where i = is the interest rate percentage = 100r. Using the better (2 series) approximation ln(1+r) ≈ r – r 2 st order Taylor series) approximation ln(1+r) ≈ r valid for small
/2, show that for r ≈ 0.0, n ≈ 72/i. nd order Taylor
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%?
7

1. a. (1000)(1.04) 4 = 1169.86.
1000(1.02) 8 = 1171.66.
1000e 0.16
= 1173.51.
1000(1.08)
1000(1.04) c. 1000(1.02)
4
8
30
= 1360.49.
= 1368.57.
= 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 (C
0
, C
1
) = (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 – r 2 /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 2
(2 13 ) 2 = 64 million.
13 = 2 3 2 10
≈ 8(1,000) = 8,000. When the interest doubles, money doubles twice as frequently, in this case 9 years. So $1 would grow to 2 26 =
8. PV = 50,000[1 + (1 – (1.08) -19 )/0.08] = 530,180.
8

A. NPV
Simple Investment Project after-tax cash flows
: (-C
0
, C
1
, C
2
, … , C
N
), Ck > 0 for all k.
NPV[C] = -C
0
+ PV[C
1
, C
2
, … , C
N
]. 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) + C k
, PB(0) = C
0
.
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):
Cash Flows -18,745.40 4,000.00
-18,745.40(1.03) = -19,307.76
8,000.00
-15,307.76(1.03) = -15,766.99
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 * {C
0
+ C
1
/(1+r) + … + C t
/(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
65
67%
9

. Timing Problem
Consider the following example:
Project t = 0 t = 1 t = 2 t = 3 IRR NPV@0% NPV@10
%
10,000
B -10,000
“B-A” 0 -9,000 10.55% 2,000 83
NPV@15
%
-593
Project Balance Schedules
Projects 0 1 2 3
A @ 16.04% -10,000 -1,604 -861 0
A @ 0%
A @ 10%
-10,000
-10,000
0
-1,000
1,000
-100
2,000
890
A @ 15%
B @ 16.04%
B @ 0%
B @ 10%
B @ 15%
“B-A” @ 10.55%
“B-A” @ 0%
-10,000
-10,000
-10,000
-10,000
-10,000
0
0
-1,500
-10,294
-9,000
-10,000
-10,500
-9,000
-9,000
-725
-10,626
-8,000
-10,000
-11,075
-9,950
-9,000
“B-A” @ 10%
“B-A” @ 15%
0
0
-9,000
-9,000
-9,900
-10,350
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 -C
0
+ C
1
+ … + C k
> 0.
Discounted payback period
: First period k for which -C
0
+ C
1
/(1+r) + … + C
Also equals the first period k for which PB(k) > 0. Why? t
/(1+r) k > 0.
Discounted payback period ≥ Payback period. Why?
166
0
4,000
1,000
-736
0
2,000
110
-903
10

I. A firm is considering 7 proposed projects. Available budget is 500. (All numbers in 000’s.)
Initial outlay
PW
100 20 150 50 50 150 150
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
200x
1
+ 30x
2
+200x
3
+60x
4
+50x
5
+100x
6
+50x
7 to:
100x
1
+ 20x
2
+ 150x
3
+ 50x
4
+ 50x
5
+ 150x
6
+ 150x
7
≤ 500, where it is understood that each x i
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 *
Road between City A and B
1. Concrete, 2 lanes
2. Concrete, 4 lanes
3. Asphalt, 2 lanes
4. Asphalt, 4 lanes
Bridge at Broad St
5. Repair existing
6. Add lane
2,000
3,000
1,500
2,200
500
1,500
4,000
5,000
3,000
4,300
1,000
1,500
7. New structure
Traffic control in Town C
8. Traffic lights
9. Turn lanes
2,500
100
600
2,500
300
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?
11

Solution:
The problem may be expressed as
MAX 4000x
1
+ 5000x
2
+3000x
3
+4300x
4
+1000x
5
+1500x
6
+2500x
7
+ 300
8
+1000x
9
+2000x
10 subject to:
2000x
1
+ 3000x x
1 x
5 x
8
2
+1500x
3
+ x
2
+ x
6
+ x
9
+ x
3
+ x
7
+ x
≤
+2200x
4
1
4
+ x
10
≤ 1
≤ 1
+500x
5
+1500x where it is understood that each x i
ε {0, 1}.
6
+2500x
7
+ 100
8
+600x
9
+1000x
10
≤ 5000
The maximum NPV is … It is achieved by selecting projects … for a cost of … .
12

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 of
10% (compounded annually). Smith makes equal annual payments to the bank over the life of the 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 .
• P P ( n ) = the principal payment made at time n .
• T P ( 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 5
Loan Balance (LB) 10000.00
8362.03
6560.25
4578.32
2398.18
Total Payment (TP) 2637.97
2637.97
2637.97
2637.97
0.00
2637.97
Interest Payment (IP)
Principal Payment (PP)
1000.00
1637.97
836.20
1801.77
656.03
1981.94
457.83
2180.14
239.82
2398.15
Total Payment.
The bank determines the total payment (TP) so that the value of the total payment cash flow steam
( T P (1) , T P (2) , . . . , T P ( N )) = ( T P, T P, . . . , T P ) (1)
1
13

at the agreed-upon loan interest rate equals the initial loan balance. Thus,
LB (0) = T P
"
1 − (1 + i )
− N #
, i
T P = iLB (0)
1 − (1 + i ) − N
=
0 .
10(10000)
1 − (1 .
1) − 5
= 2637 .
97 .
Interest Payments.
IP ( n ) = iLB ( n − 1).
Principal Payments.
For this type of loan P P ( n ) = T P − IP ( n ).
(2)
(3)
Observations:
1.
The loan balance decreases by the principal payment amount.
Bank earns interest on the outstanding loan balance over the period. The new loan balance equals the old loan balance plus interest earned minus the cash the bank receives (the total payment made by Smith). That is,
LB ( n ) = LB ( n − 1) + IP ( n ) − T P ( n )
= LB ( n − 1) + IP ( n ) − [ IP ( n ) + P P ( n )]
= LB ( n − 1) − P P ( n ) .
(4)
(5)
(6)
2.
Principal payments grow exponentially.
P P ( n ) =
=
=
=
T P
T P
T P
(1 +
−
−
− i )
IP i [
P P
( iLB n
LB
(
(
(
) n n n
−
−
−
1)
2)
1) .
− P P ( n − 1)]
= [ T P − iLB ( n − 2)] + iP P ( n − 1) using (6)
= [ T P − IP ( n − 1)] + iP P ( n − 1) using (7)
= P P ( n − 1) + iP P ( n − 1)
3.
The outstanding loan balance at time n always equals the present value of the future total payment cash flow stream.
Here, total payment cash flow stream is the one associated when the loan was originated. For example, the loan balance at time 3 equals
4578 .
32 =
2637 .
97
1 .
1
+
2637 .
97
.
1 .
1 2
At time n there are N − n payments remaining. Thus,
LB ( n ) = T P
"
1 − (1 + i )
− ( N − n ) # i
(14)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
2
14

= iLB (0)
1 − (1 + i ) − N
"
1 − (1 + i )
− ( N − n ) i
= LB (0)
"
1 − (1 + i )
− ( N − n )
1 − (1 + i ) − N
#
.
#
(15)
(16)
Thus, the ratio of the outstanding loan balance at time n to the initial loan balance is given by:
LB
LB
( n )
(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) first calculate 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 upcoming period’s interest and principal payments.
4.
The loan balance at the end of the loan always equals 0.
Since the loan balance decreases each period by the principal payment amount, the sum of the principal payments always equals the initial loan balance. (Can you show this?)
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 interest rate 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?
T P =
(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 there are 360 − 128 = 232 periods remaining .
Keep in mind you have to convert to monthly 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.
3
15

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 pay per month?” The present value of the new total payment cash flow stream (each total payment now equals T P + A ) over the remaining M months must equal the current outstanding loan balance. That is,
LB (0) =
( T P + A )(1 − (1 + i )
− M
)
, i from which one can solve for A .
(20)
(2) “If I pay an additional A dollars per month, then how many months M will it take to pay off my loan?” To answer this question, we can use (20), except here the value of A is known and we seek the value of M .
Example 3.
Consider once again the loan of Example 2, and suppose the remaining duration of the loan is 19 years and 4 months. Recall that T P = 749 .
438. Since the remaining duration equals 232 months we also know that the outstanding loan balance equals 102 , 763 .
92 .
If 100 is added to the monthly payment, then we have
102 , 763 .
92 =
849 .
438
[1 − (1 .
005)
− M
0 .
005
] , which implies that M = 186 .
18 months (or 15 .
5153 years).
(21)
How much must be added to the monthly payment to pay off the loan in 10 years (or 120 payments periods)? Let A denote the additional amount. We have that
( M + A ) =
102 , 763 .
92(0 .
005)
1 − (1 .
005) − 120
= 1140 .
89 , which implies that A = 391 .
45.
(22)
Example 4.
Consider a 15-year conventional fixed-rate mortgage for 200,000 at 6.25%. (The solution 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 2000 each 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 have to add to our monthly payment to pay off the loan in 5 years? [1412 .
44.]
4
16

The US government allows one to deduct interest payments from their tax obligations. Consequently, the after-tax cash flow stream is different than the constant total payment cash flow stream. This difference should be considered when deciding on whether to refinance. First, we look 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 5
Loan Balance (LB) 10000.00
8362.03
6560.25
4578.32
2398.18
Total Payment (TP) 2637.97
2637.97
2637.97
2637.97
0.00
2637.97
Interest Payment (IP)
Principal Payment (PP)
1000.00
1637.97
836.20
1801.77
656.03
1981.94
457.83
2180.14
239.82
2398.15
Tax 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 or
6%. What is the present value of the after-tax cash flow stream using the after-tax loan interest rate? 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 only five years left now. At the time of origination the prevailing market rate of interest (for loans of this type of risk) was 10%. Let us suppose that interest rates have come down and the prevailing market rate of interest is now 8%. Should Smith refinance?
To answer this question, we first compute the after-tax cash flow stream assuming Smith did 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 5
Loan Balance (LB) 10000.00
8295.44
6454.52
4466.32
2319.07
Total Payment (TP) 2504.56
2504.56
2504.56
2504.56
0.00
2504.56
Interest Payment (IP)
Principal Payment (PP)
800.00
1704.56
663.64
1840.92
516.36
1988.20
357.31
2147.25
185.53
2319.03
Tax Shield (TS) 320.00
265.46
206.54
142.92
74.21
After-tax cash flow 10000.00
-2184.56
-2239.10
-2298.02
-2361.64
-2430.35
5
17

stream is given in Table 4.
Table 4: Incremental after-tax cash flow stream.
0 1 2 3 4 5
Incremental 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) = 0.048
or 4.8%, which represents the cost of funds to Smith. In this case, the value of refinancing equals
342 .
53 =
53 .
41
1 .
048
+
1
64 .
39
.
048 2
+
1
77 .
54
.
048 3
+
1
93 .
20
.
048 4
+
111 .
64
.
1 .
048 5
(23)
If Smith took out a hypothetical loan at 8% in the amount of 342.53, then the repayment schedule associated with the incremental after-tax cash flows would be as shown in Table 5. By construction, 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 5
Loan Balance (LB) 342.53
305.56
255.83
190.57
106.52
Total Payment (TP) 64.37
74.17
85.73
99.30
0.00
115.04
Interest Payment (IP)
Principal Payment (PP)
27.40
36.97
24.44
49.73
20.47
65.26
15.25
84.05
8.52
106.52
Tax Shield (TS) 10.96
9.78
8.19
6.10
3.41
After-tax cash flow 342.53
-53.41
-64.39
-77.54
-93.20
-111.64
Remark 2.
Given a cash flow stream, discount it at the after-tax loan interest rate. Take out a loan in this amount at the pre-tax interest rate. Assume the principal payments are such that 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, for example, in Table 5. By construction, the present value of the given cash flow stream at the after-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 project balance concept.)
Refinancing entails costs (e.g. origination fees, appraisal, Smith’s time). If the cost to
Smith 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
6
18

“knew” interest rates were about to fall shortly? This kind of analysis requires a stochastic model of interest rates and a game plan for how to make such decisions that involve risk over time, 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 of refinancing will be less. Suppose, for example, that Smith intends to pay off the loan after 3 years. Then the respective after-tax cash flow streams associated with the original and new loans 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 3
After-tax cash flow @ 10% 10000.00
-2237.97
-2303.49
-2375.56 + -4578.32 = -6953.88
After-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.
In this section we derive a general formula for the Loan Value , namely, the value of the after-tax cash flows associated with a loan.
We require one assumption: interest is always paid on the current outstanding loan balance, i.e., IP ( n ) = iLB ( n − 1). (This certainly holds true for the fixed-rate mortgage type loans examined above.) The implication of this assumption is that at any time the outstanding loan balance equals the sum of the future principal payments, i.e.,
LB ( n ) =
N
X k = n +1
P P ( k ) .
(24)
In what follows, we let τ denote the appropriate tax rate, i a
= (1 − τ ) i denote the after-tax loan interest rate , and r denote the appropriate per-period discount rate.
General Formula.
where the
LoanV alue = (1 − i a
) ∗ LoanP rincipalV alue, r
LoanP rincipalV alue = LB (0) −
N
X
P P ( n ) n =1
(1 + r ) n
(25)
(26)
7
19

equals the present value of the loan principal cash flows .
In the derivation to follow, the
Af terT axInterestV alue = (1 − τ )
N
X
IP ( n ) n =1
(1 + r ) n equals the present value of the after-tax interest cash flows .
Derivation.
Since the interest payments are deductible
N
X
P P ( n ) + (1 − τ ) IP ( n )
LoanV alue = LB (0) −
(1 + r ) n n =1
= LoanP rincipalV alue − Af terT axInterestV alue.
In light of (29) it remains to show that
Af terT axInterestV alue = i a r
∗ LoanP rincipalV alue, which we derive in the following sequence of identities:
Af terT axInterestV alue = (1 − τ )
N
X n =1
IP ( n )
(1 + r ) n
= (1 − τ )
N
X n =1 iLB ( n − 1)
(1 + r ) n
= (1 − τ )
N
X n =1 i P
N k = n
P P ( k )
(1 + r ) n
= (1 − τ ) i
N
X k
X k =1 n =1
P P ( k )
(1 + r ) n using (24)
= i a
N
X
"
P P ( k ) k
X k =1
1 n =1
(1 + r ) n
!#
= i a
N
X
P P ( k ) k =1
1 − (1 + r r )
− k
=
=
= i a r
( N
X
P P ( k ) − k =1
N
X
P P ( k )
) k =1
(1 + r ) k i a r
(
LB (0) −
N
X
P P ( k ) k =1
(1 + r ) k i a r
∗ LoanP rincipalV alue.
)
8
(27)
(28)
(29)
(33)
(34)
(35)
(30)
(31)
(32)
(36)
(37)
(38)
20

Observations:
1. When the discount rate r is chosen equal to i a
, then the Loan Value Formula (25) immediately shows that the LoanV alue = 0. You can verify this for the after-tax cash flow streams of Examples 1, 2 and 3.
2. The
N
X
IP ( n )
T axShieldV alue = τ n =1
(1 + r ) n
(39) is the present value of the tax shield cash flow stream.
The T axShieldV alue is an important component of the Adjusted Present Value (APV) approach to value corporate projects financed in part by debt. The derivation of the Loan Value formula shows that
τ
T axShieldV alue = (
1 − τ i
) ∗ Af terT axInterestV alue = τ r
∗ LoanP rincipalV alue.
(40)
3. An appropriate discount rate for the purpose of corporate valuation is the loan interest rate, i.e., r = i . In this special case,
(1 − i a
) = (1 − i
(1 − τ ) i
) = τ i
(41) and so
LoanV alue = τ ∗ LoanP rincipalV alue = T axShieldV alue.
(42)
The second identity follows from (40). That is, when r = i , then the LoanV alue is the
T axShieldV alue . For this special case identity (42) is easily obtained if one recalls that the present value of the total payment cash flow stream at the loan interest rate always equals the initial loan balance. (Can you prove this?)
Example 5.
What are the LoanV alue and T axShieldV alue associated with the loan of
Example 1 when the discount rate r equals the loan interest rate of 10%? A direct calculation shows that
LoanV alue = 10000 −
T axShieldV alue =
400
1 .
.
1
00
+
2237 .
97
1
1
334
.
1
.
.
1
48
2
+
2303 .
49
+
1 .
1 2
262 .
41
1 .
1 3
+
+
2375
183
1 .
1
.
1 .
13
4
1
.
3
56
+
+
1 .
1
2454
95 .
93
5
1 .
1
.
4
84
+
= 1021 .
2541
86 .
1 .
1
.
5
99
= 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 .
1 .
1 2
77
= 0 .
40(10000 − 7445 .
35) = 1021 .
86 .
+
1981 .
94
1 .
1 3
+
2180 .
14
1 .
1 4
+
2398 .
15
1 .
1 5
9
21

Why use the Loan Value Formula?
After all, both calculations of the LoanV alue or
T axShieldV alue use the numbers obtained from the Table. For this type of loan no table is required when r = i !
Here is why.
We have shown that the principal payments grow exponentially at the loan interest rate. Since the principal payment cash flow stream is being discounted at this same rate, it follows that
7445 .
35 =
=
=
1637
1637
1
1 .
1
1 .
1
.
.
97
97
+
+
1801
1 .
1637(1
1
1
.
.
2
1
77
2
"
1637 .
97(1 .
1)
.
+
1)
+
1981 .
94
+
1 .
1 3
+
1637(1 .
1)
2
1637(1
1 .
1
.
1)
3
2
2180 .
14
+
1 .
+
1 4
1
+
.
1 4
2398
1 .
1637(1 .
1)
3
+
1
.
5
15
+
1637(1 .
1)
4
1 .
1 5
+
1 .
1 1 .
1 1 .
1 2
1637(1 .
1)
3
1 .
1 3
1637(1 .
1)
4
1 .
1 4
1637(1 .
1)
5 #
1 .
1 5
=
5(1637 .
97)
.
1 .
1
For this type of loan the general rule for calculating the LoanV alue or T axShieldV alue when r = i is this:
N ∗ P P (1)
LoanV alue = T axShieldV alue = τ ∗ LB (0) − (43)
1 + i
N ∗ ( T P − iLB (0))
= τ ∗ LB (0) − .
(44)
1 + i
You only need to calculate the value of T P = iLB (0) / (1 − (1 + i )
− N
).
Example 6.
Consider a conventional 30-yr fixed-rate loan for 100,000 at 6%. What are the
LoanV alue and T axShieldV alue when the discount rate r equals the loan interest rate of 6%?
The tax rate is 40%. The calculations are:
T P =
0 .
005(100 , 000)
1 − (1 .
005) − 360
= 599 .
55 iLB (0) = 0 .
005(100 , 000) = 500 .
00
P P (1) = 599 .
95 − 500 .
00 = 99 .
95
LoanV alue = T axShieldV alue = 0 .
40[100 , 000 −
360(99 .
95)
] = 26 , 678 .
81 .
1 .
005
Example 7.
Suppose the loan of Example 6 is paid off at the end of year 10. (This is typical of a real estate loan.) What are the LoanV alue and T axShieldV alue associated with this repayment schedule when the discount rate r equals the loan interest rate of 6%? To apply the general formula, keep in mind that (i) there will be an additional “balloon” principal payment at the end of period 120 in the amount of LB (120) whose present value must be accounted for, and (ii) only multiply P P (1) by the number of periods n = 120. The calculations are:
LoanV alue =
LB (120) = 100 , 000 ∗
1 − (1 .
005)
− 240
1 − (1 .
005)
− 360
= 83 , 685 .
73 .
T axShieldV alue = 0 .
40[100 , 000 −
120(99 .
95)
1 .
005
−
83 , 685 .
73
] = 16 , 827 .
70 .
1 .
005 120
10
22

Example 8.
Let’s revisit the refinancing example.
We obtained a value of 342.53.
The incremental cash flow stream obtained in Table 4 was discounted at the after-tax loan interest rate of 4.8%. The incremental cash flow stream is the difference between the after-tax cash flows streams associated with the loans at 8% and 10%, respectively. So, in effect, the value of 342.53
equals the LoanV alue of the 8% loan minus the LoanV alue of the 10% loan both discounted at 4.8%. But the LoanV alue of the 8% loan discounted at 4.8% equals zero , so the value of the refinancing equals minus the LoanV alue of the 10% loan discounted at 4.8%. When applying the 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 discount rate equals 4.8%. Here, you must apply the growing annuity formula:
1637 .
97
LoanP rincipalV alue = 10 , 000 −
1 .
048
= 10 , 000 −
"
1637 .
97
1 .
048
+
1801 .
77
1 .
048 2
+
1981 .
94
1 .
048 3
+
+
1637(1 .
1)
1 .
048 2
+
1637(1 .
1)
2
1 .
048 3
2180
+
.
1 .
048
14
4
+
2398 .
15
1 .
048 5
1637(1 .
1)
1 .
048 4
3
+
1637(1 .
1)
4 #
1 .
048 5
= 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 .
10
) ∗ (1370 .
31) = − 342 .
53 ,
0 .
048
(45) which confirms our earlier calculation. (Keep in mind the refinance value here is minus the
LoanV 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 years with 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 still applies.) The local bank will charge you 15% for such a loan. Assume that you will commit to repay 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 discount rate? For reasonably safe loans (i.e. very low probability of default), the correct discount rate is your company’s after-tax, unsubsidized borrowing rate , which is 9% in this case. (You can use the same argument we used to justify the discount rate in the refinance example.) In this case, you can verify that the present value of the after-tax cash flow equals 233,379. We can also apply the Loan Value Formula, as follows:
LoanV alue = (1 −
0 .
03
0 .
09
) ∗ 1 , 000 , 000 −
1 , 000 , 000
1 .
09 5
= 233 , 379 .
(46)
11
23

Table 7: After-tax cash flows for Example 9.
0 1 2 3 4 5
Loan Balance (LB) 1,000,000 1,000,000 1,000,000 1,000,000 1,000,000
Total Payment (TP) 50,000 50,000 50,000 50,000
0
1,050,000
Interest Payment (IP)
Principal Payment (PP)
50,000
0
50,000
0
50,000
0
50,000
0
50,000
1,000,000
Tax Shield (TS)
After-tax cash flow 1,000,000
20,000
-30,000
20,000
-30,000
20,000
-30,000
20,000 20,000
-30,000 -1,030,000
In effect, the manufacturer is subsidizing the purchase price by the amount of 233,379. For purposes of any other project evaluation, the cost to the company to acquire the machine is
766,621.
Remark 3.
When the loan repayment schedule is such that interest is paid each period only and the loan balance is paid at the end of the loan, then the calculation of the LoanP rincipalV alue is elementary: all you have to do is to subtract from the loan amount the present value of the loan amount at the end of the loan!
12
24

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

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.
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 = i a
, 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 = i a
, 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. i a
= 2.4%. r = 3.6%.
LoanValue = (1 – 0.024/0.036)*[1,000,00 – 1,000,000/(1.036)
10
] = 99,298.
27

28

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

30

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 0 1 2 3 4
Depreciation
(60,000) (60,000) (60,000) (60,000)
(7,500) (15,000) (15,000) (7,500)
Depreciation
(67,500) (75,000) (75,000) (67,500)
27,000 30,000 30,000 27,000
(40,500) (45,000) (45,000) (40,500)
7,500 15,000 15,000 7,500
After-tax cash flow (100,000) (33,000) (30,000) (30,000) 13,000
Process 0 1 2 3 4
Depreciation
(50,000) (50,000) (50,000) (50,000)
(10,000) (20,000) (20,000) (10,000)
Depreciation
(60,000) (70,000) (70,000) (60,000)
24,000 28,000 28,000 24,000
(36,000) (42,000) (42,000) (36,000)
10,000 20,000 20,000 10,000
(22,000) 52,000 After-tax cash flow (150,000) (26,000) (22,000)
* 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 0 1 2 3 4
Depreciation
(60,000) (60,000) (60,000) (60,000)
(20,000) (32,000) (19,200) (5,760)
Depreciation
(80,000) (92,000) (79,200) (65,760)
32,000 36,800 31,680 26,304
(48,000) (55,200) (47,520) (39,456)
20,000 32,000 19,200 5,760
After-tax cash flow (100,000) (28,000) (23,200) (28,320) (480)
Process 0 1 2 3 4
Depreciation
(50,000) (50,000) (50,000) (50,000)
(30,000) (48,000) (28,800) (8,640)
Depreciation
(80,000) (98,000) (78,800) (58,640)
32,000 39,200 31,520 23,456
(48,000) (58,800) (47,280) (35,184)
30,000 48,000 28,800 8,640
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 δ
SV
denote the change in salvage value of B for which we would be indifferent between it and A.
• The new salvage value = 70,000 + δ
SV
.
• The change to after-tax cash flows for B will simply be (1τ ) δ
SV
= 0.60
δ
SV
• The present value of this change @ 10% = 0.60
δ
SV
(1.1) -4 .
.
• This must equal (-166,233) - (-169,175) = 2,942.
• Thus, δ
SV
= 7,179 so that the “break-even” salvage value = 77,179.
32

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 +
τ
C = -1000 + (0.40)(1000), and so the net effect of being able to deduct expenses it that the company “receives” a “tax benefit” of
τ
C = 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 5year 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 halfyear 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
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
Net operating income 10,000
Flow In – Flow Out Approach
Net operating income 10,000
Depreciation 2,000
Income before tax 8,000
Tax 3,200
Net income 4,800
Adjustments
Depreciation
After tax cash flow
2,000
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

Page 76 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.
Appendix A
MACRS Percentage Table Guide
General Depreciation System (GDS)
Alternative Depreciation System (ADS)
Chart 1.
Use this chart to find the correct percentage table to use for any property other than residential rental and nonresidential real property. Use Chart 2 for residential rental and nonresidential real property.
MACRS
System
Depreciation
Method Recovery Period Convention Class
Month or
Quarter
Placed in Service Table
GDS
GDS
GDS
GDS
GDS
GDS
GDS
ADS
GDS
ADS
200%
200%
150%
150%
150%
150%
SL
SL
GDS/3, 5, 7, 10 (Nonfarm)
GDS/3, 5, 7, 10 (Nonfarm)
GDS/3, 5, 7, 10
GDS/3, 5, 7, 10
GDS/15, 20
GDS/15, 20
GDS
ADS
GDS
ADS
Half-Year
Mid-Quarter
Half-Year
Mid-Quarter
Half-Year
Mid-Quarter
Half-Year
Mid-Quarter
3, 5, 7, 10
3, 5, 7, 10
3, 5, 7, 10
3, 5, 7, 10
15 & 20
15 & 20
Any
Any
Any
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
Any
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
Any
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
Any
A-1
A-2
A-3
A-4
A-5
A-14
A-15
A-16
A-17
A-18
A-1
A-2
A-3
A-4
A-5
A-8
ADS
ADS
150%
150%
ADS
ADS
Half-Year
Mid-Quarter
Any
Any
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
Any
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
A-9
A-10
A-11
A-12
A-14
A-15
A-16
A-17
A-18
Chart 2.
Use this chart to find the correct percentage table to use for residential rental and nonresidential real property. Use Chart 1 for all other property.
MACRS
System
Depreciation
Method Recovery Period Convention Class
Month or
Quarter
Placed in Service Table
GDS
GDS
ADS
SL
SL
SL
SL
GDS/27.5
GDS/31.5
GDS/39
ADS/40
Mid-Month
Mid-Month
Mid-Month
Residential Rental Any
Nonresidential Real Any
Any
A-6
A-7
A-7a
A-13 Residential Rental and
Nonresidential Real
Chart 3.
Income Inclusion Amount Rates for MACRS Leased Listed Property
Amount A Percentages
Amount B Percentages
Table
A-19
A-20
Page 76 35 Publication 946 (2009)

Page 77 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.
11
12
13
14
15
6
7
8
9
10
Table A-1.
Year
1
2
3
4
5
3-, 5-, 7-, 10-, 15-, and 20-Year Property
Half-Year Convention
Depreciation rate for recovery period
3-year 5-year 7-year 10-year 15-year
33.33%
44.45
14.81
7.41
20.00%
32.00
19.20
11.52
11.52
14.29%
24.49
17.49
12.49
8.93
10.00%
18.00
14.40
11.52
9.22
5.00%
9.50
8.55
7.70
6.93
5.76
8.92
8.93
4.46
7.37
6.55
6.55
6.56
6.55
6.23
5.90
5.90
5.91
5.90
3.28
5.91
5.90
5.91
5.90
5.91
2.95
16
17
18
19
20
21
4.462
4.461
4.462
4.461
4.462
4.461
4.462
4.461
4.462
4.461
2.231
20-year
3.750%
7.219
6.677
6.177
5.713
5.285
4.888
4.522
4.462
4.461
16
17
18
19
20
11
12
13
14
15
6
7
8
9
10
21
Table A-2.
Year
1
2
3
4
5
3-, 5-, 7-, 10-, 15-, and 20-Year Property
Mid-Quarter Convention
Placed in Service in First Quarter
Depreciation rate for recovery period
3-year
58.33%
27.78
12.35
1.54
5-year
35.00%
26.00
15.60
11.01
11.01
7-year
25.00%
21.43
15.31
10.93
8.75
10-year
17.50%
16.50
13.20
10.56
8.45
15-year
8.75%
9.13
8.21
7.39
6.65
1.38
8.74
8.75
1.09
6.76
6.55
6.55
6.56
6.55
0.82
5.99
5.90
5.91
5.90
5.91
5.90
5.91
5.90
5.91
5.90
0.74
20-year
6.563%
7.000
6.482
5.996
5.546
5.130
4.746
4.459
4.459
4.459
4.459
4.460
4.459
4.460
4.459
4.460
4.459
4.460
4.459
4.460
0.565
Publication 946 (2009) 36 Page 77

Page 78 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.
Table A-3.
Year
1
2
3
4
5
3-, 5-, 7-, 10-, 15-, and 20-Year Property
Mid-Quarter Convention
Placed in Service in Second Quarter
Depreciation rate for recovery period
3-year
41.67%
38.89
14.14
5.30
5-year
25.00%
30.00
18.00
11.37
11.37
7-year
17.85%
23.47
16.76
11.97
8.87
10-year
12.50%
17.50
14.00
11.20
8.96
15-year
6.25%
9.38
8.44
7.59
6.83
6
7
8
9
10
4.26
8.87
8.87
3.34
7.17
6.55
6.55
6.56
6.55
6.15
5.91
5.90
5.91
5.90
11
12
13
14
15
2.46
5.91
5.90
5.91
5.90
5.91
2.21
16
17
18
19
20
21
16
17
18
19
20
21
11
12
13
14
15
6
7
8
9
10
Table A-4.
Year
1
2
3
4
5
3-, 5-, 7-, 10-, 15-, and 20-Year Property
Mid-Quarter Convention
Placed in Service in Third Quarter
Depreciation rate for recovery period
3-year
25.00%
50.00
16.67
8.33
5-year
15.00%
34.00
20.40
12.24
11.30
7-year
10.71%
25.51
18.22
13.02
9.30
10-year
7.50%
18.50
14.80
11.84
9.47
15-year
3.75%
9.63
8.66
7.80
7.02
7.06
8.85
8.86
5.53
7.58
6.55
6.55
6.56
6.55
4.10
6.31
5.90
5.90
5.91
5.90
5.91
5.90
5.91
5.90
5.91
3.69
20-year
4.688%
7.148
6.612
6.116
5.658
5.233
4.841
4.478
4.463
4.463
4.463
4.463
4.463
4.463
4.462
4.463
4.462
4.463
4.462
4.463
1.673
20-year
2.813%
7.289
6.742
6.237
5.769
5.336
4.936
4.566
4.460
4.460
4.460
4.460
4.461
4.460
4.461
4.460
4.461
4.460
4.461
4.460
2.788
Page 78 37 Publication 946 (2009)

Page 79 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.
16
17
18
19
20
21
11
12
13
14
15
9
10
6
7
8
Table A-5.
3-, 5-, 7-, 10-, 15-, and 20-Year Property
Mid-Quarter Convention
Placed in Service in Fourth Quarter
Depreciation rate for recovery period
Year
3-year 5-year 7-year 10-year 15-year
1
2
3
4
5
8.33%
61.11
20.37
10.19
5.00%
38.00
22.80
13.68
10.94
3.57%
27.55
19.68
14.06
10.04
2.50%
19.50
15.60
12.48
9.98
1.25%
9.88
8.89
8.00
7.20
9.58
8.73
8.73
7.64
7.99
6.55
6.55
6.56
6.55
5.74
6.48
5.90
5.90
5.90
5.91
5.90
5.91
5.90
5.91
5.90
5.17
20-year
0.938%
7.430
6.872
6.357
5.880
5.439
5.031
4.654
4.458
4.458
4.458
4.458
4.458
4.458
4.458
4.458
4.458
4.459
4.458
4.459
3.901
Table A-6.
Residential Rental Property
Mid-Month Convention
Straight Line—27.5 Years
Year
18
19
20
21
22
13
14
15
16
17
23
24
25
26
27
1
2–9
10
11
12
28
29
2
3.182%
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
2.273
1
3.485%
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
1.97
3
2.879%
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
2.576
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
2.879
4
2.576%
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.182
Month property placed in service
5 6 7 8
2.273%
3.636
3.637
3.636
3.637
1.970%
3.636
3.637
3.636
3.637
1.667%
3.636
3.636
3.637
3.636
1.364%
3.636
3.636
3.637
3.636
3.636
3.637
3.636
3.637
3.636
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.637
3.636
3.637
3.636
3.637
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.485
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
0.152
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
0.455
10
0.758%
3.636
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
1.061
9
1.061%
3.636
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
0.758
12
3.636
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
1.667
11
0.455%
3.636
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
1.364
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
3.637
3.636
Publication 946 (2009) 38 Page 79

Page 104 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.
The Table of Class Lives and Recov-
ery Periods has two sections. The first section, Specific Depreciable Assets
Used In All Business Activities, Except
As Noted, generally lists assets used in all business activities. It is shown as
being used and use the recovery peimprovements. The land improveriod shown in the appropriate column ments have a 13-year class life and a following the description.
Property not in either table. If the activity or the property is not included in either table, check the end of Table
Table B-1. The second section, Depre- B-2 to find Certain Property for Which ciable Assets Used In The Following
Activities, describes assets used only in certain activities. It is shown as Table B-2.
You will need to look at both Table B-1 and B-2 to find the correct recovery period. Generally, if the property is listed in Table B-1 you use the recovery period shown in that table. However, if the property is specifically listed in Table B-2 under the type of activity in which it is used, you use the recovery period listed under the activity in that table. Use the tables in the order shown below to determine the recovery period of your depreciable property.
Table B-1. Check Table B-1 for a description of the property. If it is described in Table B-1, also check Table
B-2 to find the activity in which the property is being used. If the activity is described in Table B-2, read the text (if any) under the title to determine if the property is specifically included in that asset class. If it is, use the recovery period shown in the appropriate column of Table B-2 following the description of the activity. If the activity is not described in Table B-2 or if the activity is described but the property either is not specifically included in or is specifically excluded from that asset class, then use the recovery period shown in the appropriate column following the description of the property in Table B-1.
Tax-exempt use property subject to
a lease. The recovery period for ADS cannot be less than 125 percent of the lease term for any property leased under a leasing arrangement to a tax-exempt organization, governmental unit, or foreign person or entity
(other than a partnership).
Table B-2. If the property is not listed in Table B-1, check Table B-2 to find the activity in which the property is
Recovery Periods Assigned. This property generally has a recovery period of 7 years for GDS or 12 years for Example 2. Sam Plower produces
ADS. See Which Property Class Ap- rubber products. During the year, he
plies Under GDS and Which Recovery
Period Applies in chapter 4 for the class lives or the recovery periods for
GDS and ADS for the following.
•
•
•
•
•
•
•
•
•
•
Residential rental property and nonresidential real property (also see Appendix A, Chart 2).
Qualified rent-to-own property.
A motorsport entertainment complex placed in service before
January 1, 2010.
Any retail motor fuels outlet.
Any qualified leasehold improvement property placed in service before January 1, 2010.
Any qualified restaurant property placed in service before January
1, 2010.
Initial clearing and grading land improvements for gas utility property and electric utility transmission and distribution plants.
Any water utility property.
Certain electric transmission property used in the transmission at 69 or more kilovolts of electricity for sale and placed in service after April 11, 2005.
Natural gas gathering and distrimade substantial improvements to the land on which his paper plant is located. He checks Table B-1 and finds land improvements under asset class
00.3. He then checks Table B-2 and finds his activity, paper manufacturing, under asset class 26.1, Manufacture
of Pulp and Paper. He uses the recovery period under this asset class because it specifically includes land
7-year recovery period for GDS. If he elects to use ADS, the recovery period is 13 years. If Richard only looked at
Table B-1, he would select asset class
00.3, Land Improvements, and incorrectly use a recovery period of 15 years for GDS or 20 years for ADS.
made substantial improvements to the land on which his rubber plant is located. He checks Table B-1 and finds land improvements under asset class
00.3. He then checks Table B-2 and finds his activity, producing rubber products, under asset class 30.1,
Manufacture of Rubber Products.
Reading the headings and descriptions under asset class 30.1, Sam finds that it does not include land improvements. Therefore, Sam uses the recovery period under asset class
00.3. The land improvements have a
20-year class life and a 15-year recovery period for GDS. If he elects to use
ADS, the recovery period is 20 years.
Example 3. Pam Martin owns a retail clothing store. During the year, she purchased a desk and a cash register for use in her business. She checks
Table B-1 and finds office furniture under asset class 00.11. Cash registers are not listed in any of the asset classes in Table B-1. She then checks
Table B-2 and finds her activity, retail store, under asset class 57.0, Distribu-
tive Trades and Services, which includes assets used in wholesale and
retail trade. This asset class does not specifically list office furniture or a cash register. She looks back at Table
B-1 and uses asset class 00.11 for the bution lines placed in service after April 11, 2005.
desk. The desk has a 10-year class life and a 7-year recovery period for GDS.
Example 1. Richard Green is a paIf she elects to use ADS, the recovery per manufacturer. During the year, he period is 10 years. For the cash register, she uses asset class 57.0 because cash registers are not listed in Table
B-1 but it is an years.
asset used in her retail business. The cash register has a
9-year class life and a 5-year recovery period for GDS. If she elects to use the
ADS method, the recovery period is 9
■
Page 104 39 Publication 946 (2009)

Page 105 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.
EPS Filename: 13081f31 Size - Width = 44 picas Depth = 58 picas
Table B-1.
Table of Class Lives and Recovery Periods
Asset class
00.11
Description of assets included
SPECIFIC DEPRECIABLE ASSETS USED IN ALL BUSINESS ACTIVITIES, EXCEPT AS NOTED:
Office Furniture, Fixtures, and Equipment:
Includes furniture and fixtures that are not a structural component of a building. Includes such assets as desks, files, safes, and communications equipment. Does not include communications equipment that is included in other classes.
00.12
00.13
00.21
Information Systems:
Includes computers and their peripheral equipment used in administering normal business transactions 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 of accepting information, applying prescribed processes to the information, and supplying the results of these processes with or without human intervention. It usually consists of a central processing unit containing extensive storage, logic, arithmetic, and control capabilities.
Excluded from this category are adding machines, electronic desk calculators, etc., and other equipment described in class 00.13.
2) Peripheral equipment consists of the auxiliary machines which are designed to be placed under control of the central processing unit. Nonlimiting examples are: Card readers, card punches, 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, card sorters, 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 included in other classes of economic activity, i.e., computers used primarily for process or production control, switching, channeling, and automating distributive trades and services such as point of sale (POS) computer systems. Also, does not include equipment of a kind used primarily for amusement or entertainment of the user.
Data Handling Equipment; except Computers:
Includes only typewriters, calculators, adding and accounting machines, copiers, and duplicating equipment.
Airplanes (airframes and engines), except those used in commercial or contract carrying of passengers or freight, and all helicopters (airframes and engines)
00.22
00.23
00.241
00.242
00.25
00.26
00.27
00.28
Automobiles, Taxis
Buses
Light General Purpose Trucks:
Includes trucks for use over the road (actual weight less than 13,000 pounds)
Heavy General Purpose Trucks:
Includes heavy general purpose trucks, concrete ready mix-trucks, and ore trucks, for use over the road (actual unloaded weight 13,000 pounds or more)
Railroad Cars and Locomotives, except those owned by railroad transportation companies
Tractor Units for Use Over-The-Road
Trailers and Trailer-Mounted Containers
00.3
00.4
Vessels, Barges, Tugs, and Similar Water Transportation Equipment, except those used in marine construction
Land Improvements:
Includes improvements directly to or added to land, whether such improvements are section
1245 property or section 1250 property, provided such improvements are depreciable.
Examples of such assets might include sidewalks, roads, canals, waterways, drainage facilities, sewers (not including municipal sewers in Class 51), wharves and docks, bridges, fences, landscaping shrubbery, or radio and television transmitting towers. Does not include land improvements that are explicitly included in any other class, and buildings and structural components as defined in section 1.48-1(e) of the regulations. Excludes public utility initial clearing and grading land improvements as specified in Rev. Rul. 72-403, 1972-2 C.B. 102.
Industrial Steam and Electric Generation and/or Distribution Systems:
Includes assets, whether such assets are section 1245 property or 1250 property, providing such assets are depreciable, used in the production and/or distribution of electricity with rated total capacity in excess of 500 Kilowatts and/or assets used in the production and/or distribution of steam with rated total capacity in excess of 12,500 pounds per hour for use by the taxpayer in its industrial manufacturing process or plant activity and not ordinarily available for sale to others. Does not include buildings and structural components as defined in section
1.48-1(e) of the regulations. Assets used to generate and/or distribute electricity or steam of the type described above, but of lesser rated capacity, are not included, but are included in the appropriate manufacturing equipment classes elsewhere specified. Also includes electric generating and steam distribution assets, which may utilize steam produced by a waste reduction and resource recovery plant, used by the taxpayer in its industrial manufacturing process or plant activity. Steam and chemical recovery boiler systems used for the recovery and regeneration of chemicals used in manufacturing, with rated capacity in excess of that described above, with specifically related distribution and return systems are not included but are included in appropriate manufacturing equipment classes elsewhere specified. An example of an excluded steam and chemical recovery boiler system is that used in the pulp and paper manufacturing equipment classes elsewhere specified. An example of an excluded steam and chemical recovery boiler system is that used in the pulp and paper manufacturing industry.
Class Life
(in years)
Recovery Periods
(in years)
GDS
(MACRS) ADS
10
6
6
6
3
9
4
6
15
4
6
18
20
22
7
5
5
5
5
5
5
5
7
3
5
10
15
15
10
5
6
6
5
9
5
6
15
4
6
18
20
22
Publication 946 (2009) 40 Page 105

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 r
0
for this type of project is 10%.
41

I. PROJECT VALUATION WITH NO DEBT
Revenues
Cost
Landfill cost
Labor/fuel cost
Administrative cost
EBITDA
Interest expense
Depreciation
Profit before tax
-5000
-1000
-500
-5000
-1000
-500
-5000
-1000
-500
-5000
-1000
-500
-5000
-1000
-500
0
-500
0
-1000
0
-1000
0
-1000
0
-500
1000R-7000 1000R-7500 1000R-7500 1000R-7500 1000R-7000
Taxes
Net Income 600R-4200 600R-4500 600R-4500 600R-4500 600R-4200
Adjustments
Loan principal cash flow
Depreciation
Investment
Free cash flow
-6000
500 1000 1000 1000 500
*1400
-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
= 2,275R – 18,704
)/(0.10)] – 200/1.10 + 1200/(1.10) 5
= 0
⇒ R = $8.22/ton.
Revenues 8220
Cost
8220 8220 8220
Landfill cost
Labor/fuel cost
Administrative cost
EBITDA
Interest expense
Depreciation
Profit before tax
Taxes
-5000
-1000
-500
0
-500
1220
-5000
-1000
-500
0
-1000
720
-5000
-1000
-500
0
-1000
720
-5000
-1000
-500
0
-1000
720
-5000
-1000
-500
0
-500
1220
Adjustments
Loan principal cash flow
Depreciation
Investment
Free cash flow
-6000
-6000
500
1232
1000
1432
1000
1432
1000
1432
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.
500
*1400
2632
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
Revenues
Cost
Landfill cost
Labor/fuel cost
Administrative cost
EBITDA
Interest expense
Depreciation
Profit before tax
Taxes
-5000 -5000 -5000 -5000 -5000
-1000
-500
-1000
-500
-1000
-500
-1000
-500
-1000
-500
-210
-500
1010
-168
-1000
552
-126
-1000
594
-84
-1000
636
-42
-500
1178
Adjustments
Loan principal cash flow
Depreciation
Investment
Free equity cash flow
3000
-6000
-3000
-600
500
506
-600
1000
731
-600
1000
756
-600
1000
782
-600
500
1400
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%
=
Calculation of LoanValue@10%:
0 1 2 3 4 5
LB
IP
PP
TP
3,000 2,400 1,800 1,200 600 0
-210 -168 -126 -84 -42
-600 -600 -600 -600 -600
-810 -768 -726 -684 -642
84 67 50 34 17
LoanValue@10% = NPV(Loan cash flow)@10% = 420 = (1 – 4.2/10)(3,000 – 600[(1 – (1.1) -5 ) /0.1]). of equity capital: r e
= r
0
+ (1-
)(r
0
– r
D
)(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%
=
LoanValue@12% = NPV(Loan cash flow)@12% = 544 = (1 – 4.2/12)(3,000 – 600[(1 – (1.12) -5 ) /0.12]).
43

B. WEIGHTED AVERAGE COST OF CAPITAL (WACC) APPROACH
Ignore financing but value free cash flow at the weighted average cost of capital, calculated as: r
WACC
= (3000/6000)(1 – 0.40)(0.07) + (3000/6000)(0.118) = 0.08.
= r
0
(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 (r using the pre-tax cost of debt, i.e.,
0
), and then add to this the TaxShieldValue
TaxShieldValue = 84/1.07 + 67/(1.07) 2
= 216
+ 50/(1.07) 3 + 34/(1.07) 4 + 17/(1.07) 5
= 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 – Τ
C
)(1- T
E
)/(1- T
D
), where T
C
= corporate tax rate, T
E typical values of T
C
, T
E
and T
D
= tax on equity income and T
D
= tax on ordinary income. For
, 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 r e
over time to reflect this, which is cumbersome. Ditto for the r
WACC
. 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.)
4
44

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 6month 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 aftertax 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
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
10,000
12,000
Vendor B
Vendor C
30,000
35,000
2,000
1,600
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.
47

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
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.
48

Project cash flow analysis
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
49

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)?
50

Capital Budgeting
1. A firm is considering 4 proposed projects. Available budget is 300. (All numbers in 000’s.)
1 2 3 4
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 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
NPV 150 200 100 100 120 150 240
51

52

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/s
20
, where s
20
denotes the one-period equivalent interest rate that corresponds to compounding 5% per year for 20 years. Since s
20
= (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.
53

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. IRR
A
= 15.24%. IRR
B
= 13.34%.
b. NPV
A
= -100+4.212(30) = 26.37. NPV
B
= -150+4.212(43) = 31.13. Select B. c. Incremental cash flow stream A → B = (-50, 13, 13, 13, 13, 13). IRR
A → B this is greater than 6%, select B.
= 9.43%. Since
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. PV
I
= 250,000 + 40,000[1 – (1.1) -4 ]/(0.1) – 80,000/(1.1) 4 = 322,154.
AE
I
PV
II
AE
II
= [0.1PV
I
]/[1 – (1.1) -4 ] = 101,630.
= 400,000 + 20,000[1 – (1.1) -7
= [0.1PV
II
]/[1 – (1.1) -7 ] = 96,892.
Recommend Type II machine.
]/(0.1) – 50,000/(1.1) 7 = 471,710.
1. PV
A
= 1000 + (400/0.1)[1 – (1.1) -5 ] - 300(1.1) -5 = 2330.04.
AE
PV
A
B
= 0.1(2330.04)/[1 – (1.1) -5 ] = 614.66.
= 2000 + (200/0.1)[1 – (1.1)
AE
B
Recommend equipment type B.
-8 ] - 800(1.1) -8 = 2693.78.
= 0.1(2693.78)/[1 – (1.1) -8 ] = 504.93.
3. a. PV
A
= 36,326. PV
B
= 31,372. PV
C
= 33,614. Select option B. b. Incremental cash flow stream A → B = (-24000, 6000, 6000, 6000, 6000, 16000).
IRR
A → B
= 16.62% > 10%, so B is preferred to A. stream C = (-5000, 400, 400, 400, 400, 2400).
No need to calculate IRR here because the cost 5000 > sum of benefit = 4000.
54

Project Cash Flow Analysis
1.
Project cash flow analysis
Profit before tax 262,000 460,000
Net Income 183,400 322,000
Adjustments
Loan principal cash flow
Depreciation
Investment
Free equity cash flow
Net present value
150,000
(350,000)
(200,000)
289,118
(75,000)
70,000
178,400
586,942
(75,000)
56,000
*187,250
490,250
490,250
*187,250 = 171,500 – 0.3(171,500 – 224,000)
2.
Project cash flow analysis
Profit before tax 40,000 155,000
Net Income 26,000 100,750
Adjustments
Loan principal cash flow 100,000 (50,000) (50,000)
Depreciation
Investment
Free equity cash flow
Net present value
(250,000)
(150,000)
19,441
50,000
26,000
199,941
40,000
*114,500
205,250
205,250
*114,500 = 90,000 – 0.35(90,000 – 160,000)
55

3.
Year 4 project cash flow
Revenue 399,300
Cost 171,475
EBITDA 227,825
Interest expense 5,000
Depreciation 11,520
Profit before tax
Income tax
Net income
Adjustments
Loan principal cash flow
Depreciation
Investment
Free equity cash flow
211,305
84,522
126,783
(25,000)
11,520
*67,584
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
Revenue
Cost
EBITDA
Depreciation
3300 3300 3300 3300 3300
310 310 310 310 310
2990 2990 2990 2990 2990
2000 2000 2000 2000 2000
337 337 337 337 337
653 653 653 653 653
Adjustments
Depreciation 2000 2000 2000 2000 2000
NPV -437 10711 9025 7137 5022 2653 b. Accept project.
Project cash flow (in 000’s) with inflation
0
Revenue
Cost
EBITDA
Depreciation
Profit before tax
3300 3432 3569 3712 3861
310 322 335 349 363
2990 3110 3234 3363 3498
2000 2000 2000 2000 2000
990 1100 1234 1363 1498
337 377 420 464 509
653 653 653 653 653
Adjustments
Depreciation 2000 2000 2000 2000 2000
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 benefitcost 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
150x
1
+ 200x
2
+100x
3
+100x
4
+120x
5
+150x
6
+240x
7 to:
90x
1
+ 80x
2
58x
1
+ 80x
2
+ 50x
3
+ 100x
+ 20x
4
3
+ 64x
+ 40x
5
+ 80x
6
+ 80x
4
+ 50x
5
+ 20x
6
7
+ Y ≤ 250
+ 100x
7
≤ 250 + 1.1Y, where it is understood that each x i
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

58

Company X has gross earnings of Y n
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
Y n+1
= [1 + g(u)]Y n
, and next period’s installed capital follows the equation
K n+1
= (1-
)K n
+ uY n
,
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 – e
5(
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 Y
0
= 10 million, and the initial capital K
0
= 19.8 million.
U = 0.15 0 1 2 3
Cash flow from operations
Depreciation
Profit tax
Taxes
Adjustments
Depreciation
Investment
10,000,000 10,265,439 10,537,924 10,817,641
19,800,000 19,320,000 18,927,816 18,615,723
1,980,000 1,932,000 1,892,782 1,861,572
8,020,000 8,333,439 8,645,142 8,956,069
2,726,800 2,833,369 2,939,348 3,045,064
5,293,200 5,500,070 5,705,794 5,911,005
1,980,000 1,932,000 1,892,782 1,861,572
1,500,000 1,539,816 1,580,689 1,622,646
5,773,200 5,892,254 6,017,887 6,149,931
U = 0.30
Cash flow from operations
Depreciation
Profit before tax
Taxes
Adjustments
Depreciation
Investment
0
10,000,000
1
10,758,545
2
11,574,629
3
12,452,617
19,800,000 20,820,000 21,965,564 23,241,397
1,980,000 2,082,000 2,196,556 2,324,140
8,020,000 8,676,545 9,378,073 10,128,477
2,726,800 2,950,025 3,188,545 3,443,682
5,293,200 5,726,520 6,189,528 6,684,795
1,980,000 2,082,000 2,196,556 2,324,140
3,000,000 3,227,564 3,472,389 3,735,785
4,273,200 4,580,956 4,913,695 5,273,150
U = 0.50
Cash flow from operations
Depreciation
Profit before tax
Taxes
Adjustments
Depreciation
Investment
0
10,000,000
1
11,037,598
2
12,182,856
3
13,446,947
19,800,000 22,820,000 26,056,799 29,542,547
1,980,000 2,282,000 2,605,680 2,954,255
8,020,000 8,755,598 9,577,176 10,492,692
2,726,800 2,976,903 3,256,240 3,567,515
5,293,200 5,778,695 6,320,936 6,925,177
1,980,000
5,000,000
2,282,000
5,518,799
2,605,680
6,091,428
2,954,255
6,723,474
2,273,200 2,541,896 2,835,188 3,155,958
59

60

Assumptions
Period
Price
Quantity
Unit Cost
Cost of capital
Terminal value
PRO FORMA CASH FLOW ANALYSIS EXAMPLE
0 1
30
200
9
2
27.67
230
8.6
3
25.51
264
8.1
13% first 6 years (continuously compounded)
12% after year 6 (continuously compounded)
4% growth after year 6 (continuously compounded)
4
23.53
303
7.7
5
21.7
349
7.4
6
20
400
7
Sales
CGS
Rent
SGA
EBITDA
Depreciation
Profit before tax
Taxes
Net Income
Terminal Value
Adjustments
Depreciation
Investment
Free Cash Flow (FCF)
Present Value
Net Present Value
FCF Percentage
0 1
6000
-1800
-200
-600
3400
-3500
2
6364
-1978
-200
-637
3549
-3500
3
6735
-2138
-200
-676
3721
-3500
4
7130
-2333
-200
-718
3879
-3500
5
7573
-2583
-200
-763
4027
-3500
6
8024
-2807
-200
-810
4207
-3500
7
-100
0
49
20
221
88
379
152
527
211
707
283
-100 29 133 227 316 424
51012
-35000
-35000
34707
-293
3500
0
3400
36125
39525
0.086
3500
0
3529
37611
41140
0.0858
3500
0
3633
39199
42832
0.0848
3500
0
3727
40914
44641
0.0835
3500
0
3816
42778
46594
0.0819
3500
0
3924 51012
44793
48717
0.0805
CGS = Cost of Goods Sold
SGA = Selling, General, and Administrative
EBITDA = Earnings Before Interest, Taxes and Depreciation Allowance
61

62

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
Heat exchangers 0.7-0.9 Elevators 0.4 Steam boilers 0.5
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
3 rd 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
Polymerization, large plants
0.73
0.91
Steam generation, large, 200 psi
Steam generation, large, 1000 psi
Power generation, 2,000-20,000 kW
Power generation, oil field, 20-200 kW
0.61
0.81
0.88
0.50
Sulfur from H
2
Oxygen plant
S 0.64
0.65
Styrene plant
Ammonia, nitric acid or urea plant
Chorine plant, electrolytic
Refineries, small
0.65
0.98
0.75
0.57
Refineries, large 0.67
Hydrogen sulfide removal 0.55
* Adapted from Exhibit 16.10, p. 458 in Engineering Economy , 3 rd ed, T. Eschenbach, 2011.
63

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 2 b. Estimate the total labor cost for the contract. nd unit? 16 th unit? 32 nd unit? 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 b. 0.8 = 2 –b . b = 0.3219.
Total cost ≈ 250[(8000)(40 1-b
= 3277. 8000(0.8) 5 = 2621.
/(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)(40 1-b /(1-b))] ≈ 26.5M. (Actual cost = 23.9M.) ≈ 25% reduction.
Learning-Curve Exponents *
Activity Exponent
Prototype assembly 0.65
Final assembly – complex
Final assembly – simple
0.70
0.80
Packaging 0.80
Shearing metal plates 0.82
Printed circuit-board fabrication
Welding shop
0.83
0.85
Machine shop 0.95
Grinding, chipping, blast cleaning 0.985
* Adapted from Exhibit 16.15, p. 462 in Engineering Economy , 3 rd ed, T. Eschenbach, 2011.
64

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
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
Rev 32,641 72,750 130,021 197,584 266,528 329,401 381,752 422,014 450,613 469,042
65

66

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 costexponent 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 2 b. Estimate the total labor cost for the contract. nd unit? 16 th unit? 32 nd unit? 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 20
10 th unit. th unit as compared to the 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?
67

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.
68

1. a. Estimated cost = 400,000(240/80) 0.5
≈ 700,000.
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)(50 1-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)(50 1-b /(1-b))] ≈ 65M. (Actual cost = 61.13M.) ≈ 30%.
1000 2 10 , final unit cost = 18(0.9) 10 = 6.28. b. 0.9 = 2 –b . b = 0.152.
Total labor hrs ≈ 18[1000 1-b
c.
/(1-b)] ≈ 7400. (Actual labor hrs = 7428.)
Learning curve percentage
Approximate labor hrs
Actual labor hrs
18,000
18,000
11,659
11,654
4,655
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
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
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
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
69

70

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 B
Direct-material cost
Direct-labor cost
Direct-labor hours
Total manufacturing indirect costs: $698,250
$350,000
125,000
12,500
$160,000
600,000
60,000
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 hrs 5,000
Materials handling 225,000 Material moves
50 150
40,000 90,000
Quality control
Pack and ship
49,500 Inspections
113,750 Products
50,000
600
2,500
25,000
2,000
40,000
4,500
65,000
Indirect cost rate = (Total indirect costs)/(Total direct-labor hrs)
= $698,250/72,500 = $9.631 per direct-labor hr.
Activity
Direct material
Product A
$350,000
Product B
$160,000
Direct labor
Allocated indirect
125,000
120,388
Total 595,388
600,000
577,862
717,250
Manufacturing cost/unit 23.82
* Adapted from Example A.3, p. 574-5 in Engineering Economy, 3 rd
ed by T. Eschenbach.
33.45
71

II. ABC Method:
Activity Applied activity rate
Engineering
Materials handling
$120/eng. change order
$2.50 move
Receiving $7.50/batch
Quality control $11/inspection
Pack and ship $1.75/product
Total indirect
Direct material
Direct labor
Total costs
Manufacturing cost/unit
Comparison of two methods:
Traditional costing system
ABC system
Product A
$23.82
31.51
Product A
100,000
12,000
125,000
4,500
27,500
43,750
312,750
350,000
125,000
787,750
31.51
Product B
180,000
6,000
100,000
7,500
22,000
70,000
385,500
160,000
600,000
1,145,500
28.64
Product B
$33.45
28.64
72