, Alternatively, Claude may learn nothing from the experience ontinue
and c his unethical
practices.
1.9 i = [(3,885,000 -3,500,000)/3,500,000]*100% = 11% per year
1.10 (a) Amount paid first four years = 900,000(0.12)
= $108,000
(b) Final payment = 900,000 + 900,000(0.12) = $1,008,000
1.11 i = (1125/12,500)*100 = 9%
i = (6160/56,000)*100 = 11%
i = (7600/95,000)*100 = 8%
The $56,000 investment has the highest rate of return
.
1.12 Interest on loan = 23,800(0.10) = $2,380
Default insurance = 23,800(0.05) = $1190
Set
-up fee = $300
Total amount paid = 2380 + 1190 + 300 = $3870
Effective interest rate = (3870/23,800)*100
= 16.3%
1.13 The market interest rate is usually 3 – 4 % above the expected inflation rate. Therefore,
Market rate is in the range 3 + 8 to 4 + 8 = 11 to 12% per year
1.14 PW = present worth; PV = present value; NPV = net present value; DCF = discounted cash
flow; and CC = capitalized cost
1.15 P = $150,000; F = ?; i = 11%; n = 7
1.16 P = ?; F = $100,000; i = 12%; n = 2
1.17 P = $3.4 million; A = ?; i =0%;
1 n=8
1.18 F= ?; A = $100,000 + $125,000?; i = 15%; n = 3
1.19 End-of-period convention means that all cash flows are assumed to take place at the end of
the interest period in which they occur.
1.20 fuel cost
: outflow; pension plan contributions : outflow; passenger fares: inflow;
maintenance: outflow ; freight revenue: inflo
w; cargo revenue: inflow ; extra bag charges:
Inflow; water and sodas: outflow ; advertising: outflow; landing fees
: outflow; seat
preference fees: nflow.
i
2
,1.21 End-of-period amount for June = 50 + 70 + 12020
+ = $260
End-of-period amount for Dec = 150 + 90 + 40 + 110 = $390
1.22 Month Receipts, $1000 Disbursements, $1000 Net CF, $
1000
Jan 500 300 +200
Feb 800 500 +300
Mar 200 400 -200
Apr 120 400 -280
May 600 500 +100
June 900 600 +300
July 800 300 +500
Aug 700 300 +400
Sept 900 500 +400
Oct 500 400 +100
Nov 400 400 0
Dec 1800 700 +1100
Net Cash flow = $2,920 ($2,920,000)
1.23
1.24
3
, 1.34 F = 6,000,000(1 + 0.09)
(1 + 0.09)(1 + 0.09)
= $7,770,174
1.35 4,600,000 = P(1 + 0.10)(1 + 0.10)
P = $3,801,653
1 + 0.20)n
1.36 86,400 = 50,000(
log (86,400/50,000) = n(log 1.20)
0.23754 = 0.07918n
n = 3 years
1.37 Simple: F = 10,000 + 10,000(3)(0.10)
= $13,000
Compound: 13,000 = 10,000(1 +(1i)+ i) (1 + i)
(1 +3i)= 1.3000
3log(1 + i) = log 1.3
3log1( + i) = 0.1139
log(1 + i) = 0.03798
1 + i = 1.091
=i 9.1%per year
1.38 Minimum attractive rate of return is also referred to as hurdle rate, cutoff rate, benchmark
rate, and minimum acceptable rate of return.
1.39 bonds -debt; stock sales – equity
; retained earnings – equity
; venture capital – debt
; short
term loan – debt
; capital advance from friend – debt
; cash on hand – equity
; credit card–
debt; home equity loan debt
- .
1.40 WACC = 0.30(
8%) + 0.70(13%) = 11.5%
1.41 WACC = 10%(0.09) + 90%
(0.16) = 15.3%
The company should undertake
the inventory, technology, and warehouse projects.
1.42 (a) PV(i%,n,A,F)finds the present value P
(b) FV(i%,n,A,P)finds the future value F
(c) RATE(n,A,P,F) finds the compound interest irate
(d)IRR(first_cell:last_cell)finds the compound interest rate
i
(e) PMT(i%,n,P,F)finds the equal periodi c paymentA
(f) NPER(i%,A,P,F) finds the number of periods n
5
and c his unethical
practices.
1.9 i = [(3,885,000 -3,500,000)/3,500,000]*100% = 11% per year
1.10 (a) Amount paid first four years = 900,000(0.12)
= $108,000
(b) Final payment = 900,000 + 900,000(0.12) = $1,008,000
1.11 i = (1125/12,500)*100 = 9%
i = (6160/56,000)*100 = 11%
i = (7600/95,000)*100 = 8%
The $56,000 investment has the highest rate of return
.
1.12 Interest on loan = 23,800(0.10) = $2,380
Default insurance = 23,800(0.05) = $1190
Set
-up fee = $300
Total amount paid = 2380 + 1190 + 300 = $3870
Effective interest rate = (3870/23,800)*100
= 16.3%
1.13 The market interest rate is usually 3 – 4 % above the expected inflation rate. Therefore,
Market rate is in the range 3 + 8 to 4 + 8 = 11 to 12% per year
1.14 PW = present worth; PV = present value; NPV = net present value; DCF = discounted cash
flow; and CC = capitalized cost
1.15 P = $150,000; F = ?; i = 11%; n = 7
1.16 P = ?; F = $100,000; i = 12%; n = 2
1.17 P = $3.4 million; A = ?; i =0%;
1 n=8
1.18 F= ?; A = $100,000 + $125,000?; i = 15%; n = 3
1.19 End-of-period convention means that all cash flows are assumed to take place at the end of
the interest period in which they occur.
1.20 fuel cost
: outflow; pension plan contributions : outflow; passenger fares: inflow;
maintenance: outflow ; freight revenue: inflo
w; cargo revenue: inflow ; extra bag charges:
Inflow; water and sodas: outflow ; advertising: outflow; landing fees
: outflow; seat
preference fees: nflow.
i
2
,1.21 End-of-period amount for June = 50 + 70 + 12020
+ = $260
End-of-period amount for Dec = 150 + 90 + 40 + 110 = $390
1.22 Month Receipts, $1000 Disbursements, $1000 Net CF, $
1000
Jan 500 300 +200
Feb 800 500 +300
Mar 200 400 -200
Apr 120 400 -280
May 600 500 +100
June 900 600 +300
July 800 300 +500
Aug 700 300 +400
Sept 900 500 +400
Oct 500 400 +100
Nov 400 400 0
Dec 1800 700 +1100
Net Cash flow = $2,920 ($2,920,000)
1.23
1.24
3
, 1.34 F = 6,000,000(1 + 0.09)
(1 + 0.09)(1 + 0.09)
= $7,770,174
1.35 4,600,000 = P(1 + 0.10)(1 + 0.10)
P = $3,801,653
1 + 0.20)n
1.36 86,400 = 50,000(
log (86,400/50,000) = n(log 1.20)
0.23754 = 0.07918n
n = 3 years
1.37 Simple: F = 10,000 + 10,000(3)(0.10)
= $13,000
Compound: 13,000 = 10,000(1 +(1i)+ i) (1 + i)
(1 +3i)= 1.3000
3log(1 + i) = log 1.3
3log1( + i) = 0.1139
log(1 + i) = 0.03798
1 + i = 1.091
=i 9.1%per year
1.38 Minimum attractive rate of return is also referred to as hurdle rate, cutoff rate, benchmark
rate, and minimum acceptable rate of return.
1.39 bonds -debt; stock sales – equity
; retained earnings – equity
; venture capital – debt
; short
term loan – debt
; capital advance from friend – debt
; cash on hand – equity
; credit card–
debt; home equity loan debt
- .
1.40 WACC = 0.30(
8%) + 0.70(13%) = 11.5%
1.41 WACC = 10%(0.09) + 90%
(0.16) = 15.3%
The company should undertake
the inventory, technology, and warehouse projects.
1.42 (a) PV(i%,n,A,F)finds the present value P
(b) FV(i%,n,A,P)finds the future value F
(c) RATE(n,A,P,F) finds the compound interest irate
(d)IRR(first_cell:last_cell)finds the compound interest rate
i
(e) PMT(i%,n,P,F)finds the equal periodi c paymentA
(f) NPER(i%,A,P,F) finds the number of periods n
5
