Problem 12.3:
To obtain the appraisal value of land, use formula
(12.1):
(12.1) V0 = P0/it
with: P0 = $80 per acre per year
it = 0.04762 per year
V0 = 80/0.04762 = $1,680 per acre
The real interest rate it is obtained
from expression (11.3) in Chapter 11:
(11.3) 1 + i = (1 + it) (1 + if)
with: i = 0.10 per year
it = ?
if = 0.05 per year
1 + 0.10 = (1 + it) (1 + 0.05)
1 + it = 1.10/1.05
it = 1.04762 - 1 = 0.04762
The anticipated value j years into the future can
be calculated from a generalization of formula (11.8) in Chapter
11 (see solution to Problem 11.6 at end of Chapter 11):
(11.8) Vj =
= V0 (1 + if)j
Hence, the anticipated values one and two years into
the future are, respectively,
V1 = (1,680) (1 + 0.05) = $1,764 per
acre
V2 = (1,680) (1 + 0.05)2 =
$1,852 per acre
That is, the nominal value of land is anticipated
to increase at the inflation rate.
Problem 12.4:
To calculate the appraisal value of land with nonzero
real growth in expected cash rent, apply formula (12.4):
(12.4) V0 = 
with: P0 = $80 per acre per year
g = 0.01 per year
it = 0.04762 per year
V0 = 
= $2,148 per acre
From equation (12.8):
(12.8) it = 
+
g
= (real current rate of return) + (real rate
of capital gain)
Real current rate of return = 
= 
= 0.03762
per year
Real rate of capital gain = g
= 0.01 per year
In other words, the 4.762% annual real rate of return
on the land investment is due to a 3.762% of annual real current
rate of return and a 1% annual rate of real capital gain.
Problem 12.5:
In this problem, the solution is as in Problem 12.4
but with g = -0.01 per year (instead of g = 0.01 per year):
V0 = 
= $1,375
per acre
Note that the appraisal value of land increases as
the expected growth in real cash rent goes up (compare the V0s
obtained in Problems 12.3, 12.4, and 12.5).
Problem 12.6:
To evaluate the profitability of the investment in
land, use formula (12.10):
(12.10) NPV = -INV + 
+ 
+ ... + 
+ 
with: INV = $2,000 per acre
P1 = P2 = … = PN = A = $120 per acre per year
i = 0.05 per year
t = 0
N = 10 years
VN = $2,000 per acre
TN = (VN - INV) t = (2,000 - 2,000) (0) = $0
NPV = -2,000 + (120) [USPV0.05,10] +
= -2,000 + (120) (7.7217) + 2,000 (0.6139)
= $154 per acre
The investment is acceptable because NPV = $154 >
0.
Problem 12.7:
In the presence of anticipated price changes, the
profitability of the land investment can be calculated by employing
the formulas in Table 12.2:
NPV = -INV + 
+ 
+ ... + 
+ 
= -INV + (1 - t) P0 [USPVr,N]
+ 
where: gP = anticipated rate of growth in nominal cash flows
gl = anticipated rate of growth in nominal land values
r = (i - gp)/(1 + gp)
The nominal interest rate i is obtained from
expression (11.3) in Chapter 11:
(11.3) 1 + i = (1 + it) (1 + if)
In this particular problem, the values to use are:
INV = $2,000 per acre
P0 = $120 per acre per year
it = 0.05 per year
if = 0.06 per year
gp = 0.06 per year
gl = 0.06 per year
t = 0
N = 10 years
NPV = -2,000 + (120) [USPV0.05,10] +
= -2,000 + (120) (7.7217) + 2,000 (0.6139)
= $154 per acre
The investment is acceptable because NPV = $154 > 0.
This example is very special in that inflation does
not affect the NPV (compare the solutions of Problems 12.7 and
12.6). The reasons for this result are that (i) the inflation
rate, the rate of growth in nominal cash flows, and the rate of
growth in nominal land values are assumed to be the same (i.e.,
if = gp = gl), and that (ii)
the tax rate is zero (t = 0). In general, however, inflation
will affect NPVs (see, for example, Problems 12.8 and 12.9 below).
Problem 12.8:
This problem is solved as Problem 12.7 but employing
gl = 0.05 per year (instead of gl = 0.06
per year):
NPV = -2,000 + (120) [USPV0.05,10] + 
= -2,000 + (120) (7.7217) + 2,000 (0.5584)
= $43 per acre
The investment is acceptable because NPV = $43 >
0.
Problem 12.9:
The solution to this problem is obtained as for Problem
12.7, but using t = 0.20 (instead of t = 0):
NPV = -2,000 + (1 - 0.20) (120) [USPV0.05,10]
+ 
= -2,000 + (96) (7.7217) + {(2,000) (1.7908)
- [(2,000) (1.7908) - 2,000] (0.20)} (0.6139) (0.5584)
= -$139 per acre
The investment is not acceptable because NPV = -$139
< 0.
Problem 12.10:
The formula to apply in order to solve this problem
is (12.15):
(12.15) NPV = -INV + 
+ 
+ ... + 
- 
- 
- ... - 
+ 
= -INV + (1 - t) P0 [USPVr,N]
- 
- 
- ... -
+ 
+ 
- 
where: gP = anticipated rate of growth in nominal cash flows
gl = anticipated rate of growth in nominal land values
r = (i - gp)/(1 + gp)
The nominal interest rate i is obtained from
expression (11.3) in Chapter 11:
(11.3) 1 + i = (1 + it) (1 + if)
The values to use for this particular problem are:
INV = $2,000 per acre
INVdp = (2,000) (0.20) = $400 per acre
P0 = $120 per acre per year
it = 0.05 per year
if = 0.06 per year
gP = 0.06 per year
gl = 0.06 per year
t = 0.20
N = 10 years
Then:
(1 - t) P0 [USPVr,N] = (1 -
0.20) (120) [USPV0.05,10] = (96) (7.7217) = 741
= 
= (1,600) (0.6139) = 982
= 
= (400) (0.6139) (0.5584) = 137
In order to determine the principal outstanding (Dn),
and the principal (Prn)
and interest (In) amounts
of each annual loan repayment, the easiest procedure is to calculate
first the amount of each annual loan repayment by means of the
annuity formula (9.5) in Chapter 9:
(9.5) D0 = Aloan [
]
with: D0 = (2,000) (1 - 0.20) = $1,600 per acre
Aloan = ?
iloan = 0.11 per year
Nloan = 30 years
Aloan = D0/[USPV0.11,30]
= 1,600/8.6938 = $184 per acre per year
The principal outstanding at each period (Dn)
is the present value (as of that period) of the future annual
payments, which can be calculated using the annuity formula (9.5)
in Chapter 9:
(9.5) Dn = Aloan [
]
where n denotes the period at which the outstanding
principal is to be calculated. Here:
Period 0: D0 = (184) [USPV0.11,30] = (184) (8.6938) = $1,600
Period 1: D1 = (184) [USPV0.11,29] = (184) (8.6501) = $1,592
Period 2: D2 = (184) [USPV0.11,28] = (184) (8.6016) = $1,583
Period 3: D3 = (184) [USPV0.11,27] = (184) (8.5478) = $1,573
Period 4: D4 = (184) [USPV0.11,26] = (184) (8.4881) = $1,562
Period 5: D5 = (184) [USPV0.11,25] = (184) (8.4217) = $1,550
Period 6: D6 = (184) [USPV0.11,24] = (184) (8.3481) = $1,536
Period 7: D7 = (184) [USPV0.11,23] = (184) (8.2664) = $1,521
Period 8: D8 = (184) [USPV0.11,22] = (184) (8.1757) = $1,504
Period 9: D9 = (184) [USPV0.11,21] = (184) (8.0751) = $1,486
Period 10: D10 = (184) [USPV0.11,20]
= (184) (7.9633) = $1,465
The interest payment at each period (In)
is the product of the loan interest rate (iloan) by
the loan's outstanding principal in the previous period (Dn-1):
In = iloan Dn-1
Period 1: I1 = (0.11) (1,600) = $176
Period 2: I2 = (0.11) (1,592) = $175
Period 3: I3 = (0.11) (1,583) = $174
Period 4: I4 = (0.11) (1,573) = $173
Period 5: I5 = (0.11) (1,562) = $172
Period 6: I6 = (0.11) (1,550) = $170
Period 7: I7 = (0.11) (1,536) = $169
Period 8: I8 = (0.11) (1,521) = $167
Period 9: I9 = (0.11) (1,504) = $165
Period 10: I10 = (0.11) (1,486) = $163
The principal payment at each period (Prn)
is equal to the loan's annual installment (Aloan) minus
the interest payment at that period (In):
Prn = Aloan - In
Period 1: Pr1 = 184 - 176 = $8
Period 2: Pr2 = 184 - 175 = $9
Period 3: Pr3 = 184 - 174 = $10
Period 4: Pr4 = 184 - 173 = $11
Period 5: Pr5 = 184 - 172 = $12
Period 6: Pr6 = 184 - 170 = $14
Period 7: Pr7 = 184 - 169 = $15
Period 8: Pr8 = 184 - 167 = $17
Period 9: Pr9 = 184 - 165 = $19
Period 10: Pr10 = 184 - 163 = $21
Therefore:
= 
= (149) (0.9524) (0.9434) = 134
= 
= (149) (0.9070) (0.8900) = 120
= 
= (149) (0.8638) (0.8396) = 108
= 
= (149) (0.8227) (0.7921) = 97
= 
= (150) (0.7835) (0.7473) = 88
= 
= (150) (0.7462) (0.7050) = 79
= 
= (150) (0.7107) (0.6651) = 71
= 
= (151) (0.6768) (0.6274) = 64
= 
= (151) (0.6446) (0.5919) = 58
= 
= (151) (0.6139) (0.5584) = 52
= 
= (1,465) (0.6139) (0.5584) = 502
Substitution of the preceding values in (12.15) yields:
NPV = - 400 + 741 - 134 - 120 - 108 - 97 - 88 - 79
- 71- 64 - 58 - 52 + 982 + 137 - 502 = $88 per acre
The land investment with borrowing is acceptable because NPV = $88 > 0.
Solving this problem by hand is extremely tedious
and provides no additional insight. Therefore, it is strongly
recommended to use a computer spreadsheet (e.g., Lotus 123, Excel,
Quattro Pro) to find the NPV. Such a spreadsheet might look like
Table 12.1 below.
Table 12.1: Excel spreadsheet to solve for net present
value of land investment.
Problem 12.11:
The land investment in Problem 12.10 is profitable but not feasible because it has negative cash flows for the first seven years of the investment. In other words, the cash returns from land are not sufficient to repay the loan and pay for taxes until year 8. The investor should either have other sources of cash to cover for the cash shortages originated by the land investment in the first seven years. Alternatively, feasibility could be achieved by changing the terms of the loan, for example, reducing the loan interest rate, lengthening the loan maturity, making a balloon payment at year 10, and/or postponing principal repayment.
The enclosed Table 12.2 shows the effect of changing
the interest rate on the loan to 7% (as opposed to 11%), whereas
Table 12.3 depicts the impact of a loan with 40 years to maturity
(as opposed to 30 years). It can be seen that feasibility is
achieved practically by getting a loan interest rate of 7%. Lengthening
the loan maturity to 40 years improves the cash flow situation,
but is not enough to achieve feasibility. It is important to
note that these alternative loan arrangements will generally change
the profitability of the investment.
Table 12.2: Net present value and feasibility of
land investment with loan interest rate of 7%.
Table 12.3: Net present value and feasibility of
land investment with loan maturity of 40 years.
Problem 12.13:
Note that, according to the data in Problem 12.6,
there is neither inflation (if = gp = 0),
nor taxes (t = 0), nor external financing (INV = INVdp).
Hence, the maximum bid price (INV) can be obtained by means of
the simple formula (12.19):
(12.19) INV = 
where from (11.3):
(11.3) 1 + i = (1 + it) (1 + if)
with: A = $120 per acre per year
it = 0.05 per year
if = 0 per year
gl = 0 per year
N = 10 years
INV = 
= 
= $2,400 per acre
Given the data in Problem 12.6, the maximum bid price the investor can afford to pay is $2,400 per acre. This price is consistent with the results from Problem 12.6, where it was found that the land investment with a price of $2,000 per acre was profitable.
To find the maximum bid price for the data given
in Problem 12.7, it is no longer possible to use the simple formula
(12.19). In this case, the solution can be obtained by means
of the formulas in Table 12.2:
NPV = -INV + 
+ 
+ ... + 
+ 
= -INV + (1 - t) P0 [USPVr,N]
+ INV [
]
where from (11.3):
(11.3) 1 + i = (1 + it) (1 + if)
with: NPV = $0 per acre
INV = ?
P0 = $120 per acre per year
it = 0.05 per year
if = 0.06 per year
gp = 0.06 per year
gl = 0.06 per year
t = 0
N = 10 years
r = (i - gp)/(1 + gp)
0 = -INV + (1 - 0) (120) [USPV0.05,10]
+ INV [
]
0 = -INV + (120) (7.7217) + INV [
]
INV - 0.6139 INV = 926.604
INV = 
= $2,400 per acre
Therefore, the maximum bid price the investor can afford to pay is $2,400 per acre. This price is consistent with the results from Problem 12.7, where it was found that the land investment with a price of $2,000 per acre was profitable. Note also that the maximum bid price is the same as for the data in Problem 12.6 because, as discussed in the solution to Problem 12.7, in this particular example inflation does not affect the investment's profitability.
Problem 12.19:
To see how land values should respond to changes
in different variables, use formula (12.4):
(12.4) V0 = 
a. V0 as P0
b. V0 as g
c. V0 as it