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Transcript of Chapter 03
Chapter 3: Interest Rate and Economic Equivalence Types of Interest
3.1
• Simple interest:
yearsN
NN
5.1208.01
2)08.01()08.01(000,5$000,10$
==
=++=
• Compound interest: yearsN
N
N
2.102)07.01(
)07.01(000,5$000,10$
==+
+=
3.2
• Simple interest: 400$)5)(000,1)($08.0( === iPNI
• Compound interest:
469$)14693.1(000,1$]1)1[( =−=−+= NiPI
3.3 • Option 1: Compound interest with 8%:
408,4$)4693.1(000,3$)08.01(000,3$ 5 ==+=F
• Option 2: Simple interest with 9% 350,4$)45.1(000,3$)509.01(000,3$ ==×+
• Option 1 is better
3.4
End of Year Principal Repayment
Interest payment
Remaining Balance
0 $10,000 1 $1,671 $900 $8,329 2 $1,821 $750 $6,508 3 $1,985 $586 $4,523 4 $2,164 $407 $2,359 5 $2,359 $212 $0
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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2
Equivalence Concept
3.5 402,9$)7835.0(000,12$)5%,5,/(000,12$ === FPP
3.6 328,23$)1664.1(000,20$)2%,8,/(000,20$ === PFF
Single Payments (Use of F/P or P/F Factors)
3.7 (a) 388,7$)8%,5,/(000,5$ == PFF (b) 208,3$)12%,3,/(250,2$ == PFF (c) 161,65$)31%,7,/(000,8$ == PFF (d) 700,45$)7%,9,/(000,25$ == PFF
3.8 (a) 105,3$)6%,10,/(500,5$ == FPP (b) 338,3$)15%,6,/(000,8$ == FPP (c) 418,20$)5%,8,/(000,30$ == FPP (d) 851,3$)8%,12,/(000,15$ == FPP
3.9 (a) 428,5$)5%,13,/(000,10$ == FPP (b) 763,40$)4%,13,/(000,25$ == PFF
3.10
yearsNN
PPF N
69.9)12.1(log3log
)12.01(3
==
+==
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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3
3.11
• yearsN
NPPF N
96.4)15.1(log2log
)15.01(2
==
+==
• Rule of 72: 72 /15 4.80 years=
Uneven Payment Series
3.12 (a) Single-payment compound amount factors for ),,/( NiPF
n 9% 10% 35 20.4140 28.1024 40 31.4094 45.2593
To find , first, interpolate for )38%,5.9,/( PF 38=n
n 9% 10% 38 27.0112 38.3965
Then, interpolate for %5.9=i
7039.32)38%,5.9,/( =PF As compared to formula determination
4584.31)38%,5.9,/( =PF
(b) Single-payment compound amount factors for ),,/( NiFPn 45 50 0.0313 0.0213
Then, interpolate for 47=n
0273.0)47%,8,/( =FP As compared with the result from formula
0269.0)47%,8,/( =FP
3.13
437,114$08.1
000,28$08.1
000,46$08.1000,43$
08.1000,32$
5432 =+++=P
3.14 231,11$)11%,6,/(000,2$)13%,6,/(800,1$)15%,6,/(500,1$ =++= PFPFPFF
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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4
3.15
$3,000,000 $2,400,000( / ,8%,1)$3,000,000( / ,8%,10)$20,734,618
P PP F
= ++=
F +
A
Or,
$3,000,000 $2,400,000( / ,8%,5)$3,000,000( / ,8%,5)( / ,8%,5)$20,734,618
P PP A P F
= ++=
3.16 484,14$)7%,6,/(000,5$)5%,6,/(000,6$)2%,6,/(500,7$ =++= FPFPFPP
Equal Payment Series
3.17 (a) With deposits made at the end of each year
816,13$)10%,7,/(000,1$ == AFF
(b) With deposits made at the beginning of each year 783,14$)07.1)(10%,7,/(000,1$ == AFF
3.18 (a) 713,16$)5%,7,/(000,3$ == AFF (b) 043,77$)12%,25.8,/(000,4$ == AFF (c) 575,267$)20%,4.9,/(000,5$ == AFF (d) 236,134$)12%,75.10,/(000,6$ == AFF
3.19 (a) 166,1$)13%,6,/(000,22$ == FAA (b) 388,4$)8%,7,/(000,45$ == FAA (c) 5.479$)25%,8,/(000,35$ == FAA
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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5
(d) 361,1$)8%,14,/(000,18$ == FAA
3.20 $30,000 $1,500( / , 7%, )
( / , 7%, ) 2012.94 13
F A NF A N
N years
=== ≈
3.21 $15,000 ( / , 11%, 5)
$2,408.56A F A
A==
3.22 (a) 310,2$)5%,5,/(000,10$ == PAA (b) 70.723,1$)4%,7.9,/(500,5$ == PAA (c) 85.975,2$)3%,5.2,/(500,8$ == PAA (d) 171,3$)20%,5.8,/(000,30$ == PAA
3.23 • Equal annual payment:
5.132,11$)3%,16,/(000,25$ == PAA
• Interest payment for the second year:
End of Year Principal Repayment
Interest payment
Remaining Balance
0 $25,000 1 $7,132.5 $4,000 $17,867.5 2 $8,273.7 $2,858.8 $9,593.8 3 $9,593.8 $1,535 -
3.24 (a) 2.781,6$)12%,8.5,/(800$ == APP (b) 25.403,16$)10%,5.8,/(500,2$ == APP
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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6
(c) 61.665,3$)5%,25.7,/(900$ == APP (d) 3.726,30$)8%,75.8,/(500,5$ == APP
3.2
(a) The capital recovery factor NiP for 5
),,/(An 6% 7% 35 0.0690 0.0772 40 0.0665 0.0750
To find , f nterpolate for
)38%,25.6,/( PA irst, i 38=n n 6% 7% 38 0.0675 0.0759
Then, interpolate for %25.6=i ; 6,/( PF 0696.0)38%,25. =
As compared with the result from fo ula rm0694.0)38%,25.6,/( =PF
(b) The equal payment series present-worth factor (P for )85,,/ iAi 9% 10% 11.1038 9.9970
Then, interpolate for
%25.9=i 8271.10)85%,25.9,/( =AP
As compared with the result from for ula m8049.10)85%,25.9,/( =AP
Linear Gradient Series
3.26
3.27
35.988,50$)5%,8,/)(5,%8,/(000,2$)5%,8,/(000,5$
)5%,8,/(000,2$)5%,8,/(000,5$21
=+=+=
+=
AFGAAFGFAF
FFF
47.889,11$)5%,7,/)(5,%7,/(500$)5%,7,/(000,3$
)5%,7,/(500$)5%,7,/(000,3$
=−=−=
PFGPAFGFAFF
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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3.28
26.991$)7%,9,/)](2%,9,/(50$
)4%,9,/(50$)6%,9,/(50$)7%,9,/(100[$100$
=+
+++=FPAF
AFAFAFP
3.29
3.3
)
3.404,10$)12%,8,/(000,1$000,15$
=−= GAA
0 (a 076,372,21$)7%,12%,10,/(000,000,6$ 1 =−= APP
Note that the oil price increases at the annual rate of 5% while the oil production decreases at the annual rate
(b) of 10%. Therefore, the annual
revenue can be expressed as follows:
)945.0(000,000,6$
)1.01(000,1000)05.01(60$−
−−
=
−+=n
nnnA
ies is equivalent to a decreasing geometric gradient series g = -5.5%.
So,
1)055.01(000,000,6$ −−= n
This revenue ser
1
11
with896,847,23$)7%,12%,5.5,/(000,000,6$ 1 =−= APP
(c) sent worth of the remaining series at the end of period 3 gives
1
Computing the pre 4 5 6 7( , , , )A A A A
652,269,14$)7%,12%,5.5,/(460,063,5=P $ =−AP
3.31
20
120
1
120
1
(1 )
(2,000,000) (1.06) (1.06)
1.06(2,000,000 /1.06) ( )1.06
$396, 226, 415
nn
n
n n
n
n
n
P A i
n
n
−
=
− −
=
=
= +
=
=
=
∑
∑
∑
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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8
1
21
Note: if ,[1 ( 1)
(1 )When 6% and 8%,
1 0.98151$2,000,000 0.9815[1 (21)(0.6881) 20(0.6756)]
1.06 0.0003$334,935,843
N NNn
n
i gx N x Nxnx
xg i
gxi
P
+
=
≠
− + +=
−= =
+= =
+− +⎡ ⎤= ⎢ ⎥⎣ ⎦
=
∑
3.32 (a) The withdrawal series would be
Period Withdrawal 11 $5,000 12 $5,000(1.08) 13 $5,000(1.08)(1.08) 14 $5,000(1.08)(1.08)(1.08) 15 $5,000(1.08)(1.08)(1.08)(1.08)
78.518,22$)5%,9%,8,/(000,5$ 110 == APP
Assuming that each deposit is made at the end of each year, then;
$22,518.78 ( / , 9%, 10) 15.1929$1482.19A F A A
A= ==
(b) 85.491,24$)5%,6%,8,/(000,5$ 110 == APP
$24,491.85 ( / ,6%, 10) 13.1808$1858.15A F A A
A= ==
Various Interest Factor Relationships
3.33 (a) 0058.0)2703.0)(0213.0()17%,8,/)(50%,8,/()67%,8,/( === FPFPFP
0058.0)08.01()67%,8,/( 67 =+=FP
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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9
(b) ),,/(1
),,/(NiFP
iNiPA−
=
0394.0)2%,8,/)(40%,8,/()42%,8,/( == FPFPFP
0833.00394.0108.0)42%,8,/( =
−=PA
0833.01)08.1(
)08.1(08.0)42%,8,/( 42
42
=−
=PA
(c) 4996.1208.0
)35%,8,/)(100%,8,/(1),,/(1),,/( =−
=−
=FPFP
iNiFPNiAP
4996.12)08.1(08.0
1)08.1()135%,8,/( 135
135
=−
=PA
3.34 (a)
N
N
NN
ii
iiii
NiAFiNiPF
)1(11)1(
11)1()1(
1),,/(),,/(
+=
+−+=
+−+
=+
+=
(b)
N
N
N
N
N
N
NN
ii
iii
iiiii
iNiAPNiFP
−
−
+=
+−+
−++
=
+−+
−=+
−=
)1()1(
1)1()1()1(
)1(1)1(1)1(
),,/(1),,/(
(c)
1)1(
1)1(]1)1[(
1)1()1(
1)1()1(
1)1(
),,/(),,/(
−+=
−+−+
−−+
+=−
−++
=−+
−=
N
N
N
N
N
N
N
N
ii
iii
iiii
iii
ii
iNiPANiFA
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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10
(d)
1)1()1(
)1(1
)1()1(1)1(
)1()],,/(1[
),,/(
−++
=
+−
++
=−+
+
−=
N
N
NN
NN
N
iii
iii
ii
iiNiFP
iNiPA
(f) & vid(e) (g) Di e the numerator and denominator by and take the
mit .
Equivalence Calculations
3.35
Ni)1( +li ∞→N
49.740$)10%,12,/)](5%,12,/(50$)7%,12,/(50$)9%,12,/(100[$
=++= FPAFAFAFP
3.36
92.373$)4%,8,/(300$
)3%,8,/(120$)2%,8,/(120$)1%,8,/(200$200$)08.1(
=+
++=+
PFP
FPFPFPP
.37 Selecting the base period at n = 0, we find
.38 Selecting the base period at n =0, we find
75.782$)5%,6,/(75.782$
)5%,6,/(100$)4%,6,/(50$)1%,6,/(50$)5%,6,/(100$200$
2
1
===
=+
3
$100( / ,13%,5) $20( / ,13%,3)( / ,13%, 2) ( / ,13%,5)$351.72 $36.98 (3.5172)
$110.51
P A P A P F A P AA
A
+ =+ =
=
3
82.185$
+++=
XAPXP
FPFPFPAPP
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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11
3.39
.40 Establish economic equivalent at
96.0$)12%,10,/(20$)5%,10,/(20$
=−= APGPP
0 1 2 3 4 5 6 7 8 9 10 11 12
$20$40
$60$80
$20
3 8=n :
73.334,1$80.699,10$0164.8
)7833.1(000,6$)2597.1)(08.2(6366.10)2%,8,/(000,6$)3%,8,/)(2%,8,/()8%,8,/(
===−=−
CCCC
APPFAFCAFC
3.41 The original cash flow series is
.42 Establishing equivalence at n = 8, we find
0 0 6 $9001 $800 7 $9202 $820 8 $3003 $840 9 $3004 $860 10 $300 $5005 $880
n nn A n A
−
3
13.297$)4872.9()1436.2(277.092,4$
)7%,10,/()8%,10,/(2)3%,10,/(200$)8%,10,/(300$
=+=
+=+
CCC
AFCPFCAFAF
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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12
.43 Establishing equivalence at 3 5=n
.44 Computing equivalence at
$200( / ,8%,5) $50( / ,8%,1)( / ,8%,5) ($200 )[( / ,8%,2) ( / ,8%,1)]
$1,119.32 (5.8666) ($200 )(2.2464)$185.09
F A F PX F A X F P F P
X XX
−= − + += − +=
3 5=n
2.623,29$)5%,9,/(000,3$)5%,9,/(000,3$ =+= APAFX
3.45 (2), (4), and (6)
3.46 (2), (4), and (5)
3.47
25.91$2638.1)5%,10,/(
32.115$)5%,10,/)](1%,10,/(50$50[$)5%,10,/(50$50($
2
1
==+=
=+−+=
AAPAAAA
PAFPGAA
3.48 (a)
3.49 (b)
.50 (b)
3
793,5$60.458,2$8137.6935,41$
)]6%,10,/)(6%,10,/(000,1$)12%,10,/()6%,10,/(000,30$000,25$
=+=
+=+
CC
FPAPAPCFP
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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13
Solving for an Unknown Interest Rate of Unknown Interest Periods
3.51
3.52 Establishing equivalence at
52 (1 )log 2 5 log (1 )
14.87%
P P ii
i
= += +=
0=n
1 i )/(500,2$)6,,/(000,2$ APiAP 6,%,25,−=
Solving for I with Excel, we obtain %35.92=i
3.53
+
3.54
5$35,000 $10,000( / , ,5) $10,000(1 )28.47%
F P i ii= ==
$1,000,000 $2,000( / ,6%, )(1 0.06) 1500
0.0631 (1 0.06)
log30 log1.0658.37 years
N
N
F A N
NN
=
+ −=
= +==
Short Case Studies
ST 3. uming that they are paid at the beginning of each year 1 Ass(a)
62.58$)3%,6,/(96.15$96.15$ =+ AP It is better to take the offer because of lower cost to renew.
ST 3 payment series at the end of
20 (or beginning of Year 21) is
(b)
$57.12 $15.96 $15.96( / , ,3)7.96%
P A ii= +=
.2 The equivalent future worth of the prize Year
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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14
F F A=
quivalent future worth of the lottery receipts is
F F P= −
resulting surplus at the end of Year 20 is
pute the equivalent present worth (in 2006) for each option at
F +
P P F P FP F
= + + ++
=
(b) sing either Excel or Cash Flow Analyzer, both plans would be
1
$71,819,490=
$1,952,381( / ,6%,20)
The e
2
$109,516,040=($36,100,000 $1,952,381)( / ,6%,20)
The
2 1 $109,516,040 $71,819,490$37,696,550
F F− = −=
ST 3.3 (a) Com %6=i .
Deferred $2,000,000 $566,000( / ,6%,1) $920,000( / ,6%, 2)$1, 260,000( / ,6%,11)
$8,574, 490
P P F PP F
= + ++
=
Non-Deferred $2,000,000 $900,000( / ,6%,1) $1,000,000( / ,6%, 2)$1,950,000( / ,6%,5)
$7, 431,560 ∴ At %6=i , the deferred plan is a better choice. Ueconomically equivalent at %72.15=i
ST 3 ximum amount to invest in the prevention program is
.4 The ma
467,50$)5%,12,/(000,14$ == APP
sing the
ST 3.5 geometric gradient series present worth factor, we can establish e equivalence between the loan amount $120,000 and the balloon payment
s as
Uthserie
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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15
1) 1$120,000 (A= 1
1
1/ ,10%,9%,5) 4.6721$25,684.38
P A A==
2) Paym series
A
ent
n Payment 1 2
$25,684.38 $28,252.82
3 $31,078.10 4 $34,185.91
ST 3.6 1) Compute the required annual net cash profit to pay off the investment and interest.
A
A
5 $37,604.50
$70,000,000 ( / ,1A P A 0%,5) 3.7908$18,465,759
= ==
) Decide the number of shoes, X
2$18,465,759 ($100)
184,657X
X==
ST 3.7
F P
$1,000( / ,9.4%,5) $500( / ,9.4%,5) $4,583.36P F F A$4,583.36( / ,9.4%,60) $1,005,132
+ ==
ain question is whether or not the U.S. government will be able to vest the social security deposits at 9.4% interest over 60 years.
F
(b) A
>
Stay with the original deferred plan.
The min
ST 3.8 (a)
Contract $3,875,000 $3,125,000( / ,6%,1)$5,525,000( / ,6%,2)$8,875,000( / ,6%,7)$39,547,242
P PP FP F
= ++ ++=
…
Bonus
=
$1,375,000 $1,375,000( /P P= + ,6%,7)$9,050,775 $8,000,000
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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16
ST 3.9
e between two payment options. Selecting n = 0 as the base period, we can calculate the equivalent present worth for each option as follows:
Basically we are establishing an economic equivalenc
Option 1: $140,000Option 2: $32,639( / , %,9)
PP P A i==
Or,
P A i$140,000 $32,639( / , %,9)18.10%i
==
If Mrs. Setchfield can invest her money at a rate higher than 18
.10%, it is
better to go with Option 1. However, it may be difficult for her to find an investment opportunity that provides a return exceeding 18%.
Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be
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