1
Module 4
Interest Formulas
Dr Tareq Albahri 2016
Example P74
What is the compound amount on April 1, 2025 that is equivalent to a principal sum of $2,000 on
April 1, 2017?
1. If interest is 9% compound annually
n = 2025 – 2017 = 8
F/P,9, 8
F/P , i, n
F=P(
) = $2,000 ( 1.993 ) =$3,986.
2. If interest is 9% compounded continuously
F/P, 9, 8
F/P , r, n
F=P[
] = $2,000 [ 2.054 ] = $4,108.
Notice that continuous compounding resulted in larger profit.
r
n
1
2
3
4
5
6
7
8
9
10
9
% INTEREST FACTORS FOR CONTINUOUS COMPOUNDING
Single Payment
Equal Payment Series
Uniform
CompoundCompound- SinkingCapital- Gradientamount
Presentamount
fund
Present-worth recovery
Series
Factor
worth Factor
Factor
Factor
Factor
Factor
Factor
to find F
given P
F/P, i, n
to find P
given F
P/F, i, n
to find F
given A
F/A, i, n
to find A
given F
A/F, i, n
to find P
given A
P/A, i, n
to find A
given P
A/P, i, n
to find A
given G
A/G, i, n
1.0942
1.1972
1.3100
1.4333
1.5683
1.7160
1.8776
2.0544
2.2479
2.4596
0.9139
0.8353
0.7634
0.6977
0.6376
0.5827
0.5326
0.4868
0.4449
0.4066
1.0000
2.0942
3.2914
4.6014
6.0347
7.6030
9.3190
11.1966
13.2510
15.4990
1.0000
0.4775
0.3038
0.2173
0.1657
0.1315
0.1073
0.0893
0.0755
0.0645
0.9139
1.7492
2.5126
3.2103
3.8479
4.4306
4.9632
5.4500
5.8948
6.3014
1.0942
0.5717
0.3980
0.3115
0.2599
0.2257
0.2015
0.1835
0.1696
0.1587
0.0000
0.4775
0.9401
1.3878
1.8206
2.2388
2.6424
3.0316
3.4065
3.7674
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
2
If the principal P, the future amount F, and the number of years n are known, the interest rate i may
be determined by equations or linear interpolation in the interest tables.
Example: calculating i
Calculate the interest rate if P = $300, F= $525, and n = 9
F/P , i, n
F=P(
)
$525 = $300(
(
F/P , i, 9
F/P ,i, 9
)
$525
) = $300 = 1.750
A search of the interest tables for annual compounding interest reveals that
F/P,6,9
F/P,7 ,9
F/P,i,9
( 1.689 ) < ( 1.750 ) < ( 1.838 )
(F/P)
1.689
1.750
1.838
6%
i
7%
By linear proportion,
1.689 − 1.75
6−𝑖
=
1.689 − 1.838 6 − 7
i = 6.41 %
Solution for i using equations,
F = P (1+i)n
$525 = $300 (1+i)9
$525
(1+i)9 = $300 = 1.750
9
i = √1.750 – 1
i = 1.0642 – 1= 0.0642, or 6.42%
Minor difference is due to rounding of numbers using the tables
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
3
i
n
1
2
3
4
5
6
7
8
9
10
i
n
1
2
3
4
5
6
7
8
9
10
6
% INTEREST FACTORS FOR DISCRETE COMPOUNDING
Single Payment
Equal Payment Series
CompoundCompound Sinking
Capitalamount
Present-worth
-amount
-fund
Present-worth recover
Factor
Factor
Factor
Factor
Factor
y Factor
to find
to find
to find F
to find P
to find F
A given to find P given A given
given P
given F
given A
F
A
P
F/P, i, n
P/F, i, n
F/A, i, n
A/F, i, n
P/A, i, n
A/P, i, n
1.0600
1.1236
1.1910
1.2625
1.3382
1.4185
1.5036
1.5938
1.6895
1.7908
0.9434
0.8900
0.8396
0.7921
0.7473
0.7050
0.6651
0.6274
0.5919
0.5584
1.0000
2.0600
3.1836
4.3746
5.6371
6.9753
8.3938
9.8975
11.4913
13.1808
1.0000
0.4854
0.3141
0.2286
0.1774
0.1434
0.1191
0.1010
0.0870
0.0759
0.9434
1.8334
2.6730
3.4651
4.2124
4.9173
5.5824
6.2098
6.8017
7.3601
1.0600
0.5454
0.3741
0.2886
0.2374
0.2034
0.1791
0.1610
0.1470
0.1359
Uniform
Gradient
-Series
Factor
to find A
given G
A/G, i, n
0.0000
0.4854
0.9612
1.4272
1.8836
2.3304
2.7676
3.1952
3.6133
4.0220
7
% INTEREST FACTORS FOR DISCRETE COMPOUNDING
Single Payment
Equal Payment Series
Uniform
Compound
Compound Sinking
Capital- Gradient
-amount
Present-worth
-amount
-fund
Present-worth recover
-Series
Factor
Factor
Factor
Factor
Factor
y Factor
Factor
to find
to find
to find F
to find P
to find F
A given to find P given A given to find A
given P
given F
given A
F
A
P
given G
F/P, i, n
P/F, i, n
F/A, i, n
A/F, i, n
P/A, i, n
A/P, i, n A/G, i, n
1.0700
1.1449
1.2250
1.3108
1.4026
1.5007
1.6058
1.7182
1.8385
1.9672
0.9346
0.8734
0.8163
0.7629
0.7130
0.6663
0.6227
0.5820
0.5439
0.5083
1.0000
2.0700
3.2149
4.4399
5.7507
7.1533
8.6540
10.2598
11.9780
13.8164
1.0000
0.4831
0.3111
0.2252
0.1739
0.1398
0.1156
0.0975
0.0835
0.0724
0.9346
1.8080
2.6243
3.3872
4.1002
4.7665
5.3893
5.9713
6.5152
7.0236
1.0700
0.5531
0.3811
0.2952
0.2439
0.2098
0.1856
0.1675
0.1535
0.1424
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
0.0000
0.4831
0.9549
1.4155
1.8650
2.3032
2.7304
3.1465
3.5517
3.9461
4
If the principal sum P, its future amount F, and the interest rate i are known, the number of years n
may be determined by equations or interpolation in the interest tables.
Example: calculating n
Calculate the number of years required to double the amount of money from P = $400 to F = $800
for an interest rate of 9%.
F=P(
F/P ,i,n
)
$800 = $400 (
(
F/P ,9,n
)=
$800
$400
F/P ,9,n
)
=2
A search of the 9% interest table reveals that
F/P,9, 8
F/P,9, n
F/P, 9, 9
( 1.993 ) < ( 2.000 ) < ( 2.172 )
8 yrs
n
9 yrs
1.993
2.0
2.172
By linear interpolation,
𝑛−8
2 − 1.993
=
9 − 8 2.172 − 1.993
n = 8.04 years
Solving for n using equation
F = P (1 + i) n
$800 = $400 (1 + 0.09) n
800
(1.09)n = 400 = 2.000
n ln (1.09) = ln (2.000)
n = 8.043 years
Again small difference is due rounding of numbers using tables, with equations more accurate.
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
5
i
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
40
45
50
55
60
65
9
% INTEREST FACTORS FOR DISCRETE COMPOUNDING
Single Payment
Equal Payment Series
CompoundCompound- SinkingPresentCapitalamount
Present-worth
amount
fund
worth
recovery
Factor
Factor
Factor
Factor
Factor
Factor
to find F
to find P given
to find F
to find A
to find P
to find A
given P
F
given A
given F
given A
given P
F/P, i, n
P/F, i, n
F/A, i, n
A/F, i, n
P/A, i, n
A/P, i, n
1.0900
1.1881
1.2950
1.4116
1.5386
1.6771
1.8280
1.9926
2.1719
2.3674
2.5804
2.8127
3.0658
3.3417
3.6425
3.9703
4.3276
4.7171
5.1417
5.6044
6.1088
6.6586
7.2579
7.9111
8.6231
9.3992
10.2451
11.1671
12.1722
13.2677
14.4618
15.7633
17.1820
18.7284
20.4140
31.4094
48.3273
74.3575
114.4083
176.0313
270.8460
0.9174
0.8417
0.7722
0.7084
0.6499
0.5963
0.5470
0.5019
0.4604
0.4224
0.3875
0.3555
0.3262
0.2992
0.2745
0.2519
0.2311
0.2120
0.1945
0.1784
0.1637
0.1502
0.1378
0.1264
0.1160
0.1064
0.0976
0.0895
0.0822
0.0754
0.0691
0.0634
0.0582
0.0534
0.0490
0.0318
0.0207
0.0134
0
0
0
1.0000
2.0900
3.2781
4.5731
5.9847
7.5233
9.2004
11.0285
13.0210
15.1929
17.5603
20.1407
22.9534
26.0192
29.3609
33.0034
36.9737
41.3013
46.0185
51.1601
56.7645
62.8733
69.5319
76.7898
84.7009
93.3240
102.7231
112.9682
124.1354
136.3075
149.5752
164.0370
179.8003
196.9823
215.7108
337.8824
525.8587
815.0836
1260.0918
1944.7921
2998.2885
1.0000
0.4785
0.3051
0.2187
0.1671
0.1329
0.1087
0.0907
0.0768
0.0658
0.0569
0.0497
0.0436
0.0384
0.0341
0.0303
0.0270
0.0242
0.0217
0.0195
0.0176
0.0159
0.0144
0.0130
0.0118
0.0107
0.0097
0.0089
0.0081
0.0073
0.0067
0.0061
0.0056
0.0051
0.0046
0.0030
0.0019
0.0012
0.0008
0.0005
0.0003
0.9174
1.7591
2.5313
3.2397
3.8897
4.4859
5.0330
5.5348
5.9952
6.4177
6.8052
7.1607
7.4869
7.7862
8.0607
8.3126
8.5436
8.7556
8.9501
9.1285
9.2922
9.4424
9.5802
9.7066
9.8226
9.9290
10.0266
10.1161
10.1983
10.2737
10.3428
10.4062
10.4644
10.5178
10.5668
10.7574
10.8812
10.9617
11
11
11
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
1.0900
0.5685
0.3951
0.3087
0.2571
0.2229
0.1987
0.1807
0.1668
0.1558
0.1469
0.1397
0.1336
0.1284
0.1241
0.1203
0.1170
0.1142
0.1117
0.1095
0.1076
0.1059
0.1044
0.1030
0.1018
0.1007
0.0997
0.0989
0.0981
0.0973
0.0967
0.0961
0.0956
0.0951
0.0946
0.0930
0.0919
0.0912
0.0908
0.0905
0.0903
Uniform
GradientSeries
Factor
to find A
given G
A/G, i, n
0.0000
0.4785
0.9426
1.3925
1.8282
2.2498
2.6574
3.0512
3.4312
3.7978
4.1510
4.4910
4.8182
5.1326
5.4346
5.7245
6.0024
6.2687
6.5236
6.7674
7.0006
7.2232
7.4357
7.6384
7.8316
8.0156
8.1906
8.3571
8.5154
8.6657
8.8083
8.9436
9.0718
9.1933
9.3083
9.7957
10.1603
10.4295
10.6261
10.7683
10.8702
6
If the compound amount F, the annual payments A, and the number of years n are known, the
interest rate i may be determined by equation or by interpolation in the interest tables.
Example P77: Find i
Calculate the interest rate of F = $441.10, A= $100, and n = 4
Using tables
F=A(
F/A, i, n
)
$441.10 = $100(
(
F/A, i, 4
)=
$441.10
$100
F/A, i, 4
)
= 4.411
This value falls between the 6% and 7% table for n = 4. By linear interpolation,
4.375−4.411
i = 6 + (7˗6) 4.375−4.440 = 6.55%
Using equations
(1 + 𝑖)𝑛 − 1
F=A [
]
𝑖
(1 + 𝑖)4 − 1
$441.1 = $100 [
]
𝑖
Using solver or programmable calculator to find i
i = 0.0655 or 6.55%
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
7
Example P78: find n
Suppose that an equal annual cash flow of $100 each year exists, how long will it take to accumulate
$2,000 if the interest rate is 8% compounded continuously?
F=A[
F/A, r, n
]
$2,000 = $100 [
[
F/A, 8, n
]=
$2,000
$100
F/A, 8, n
]
= 20
Interpolate between 12 and 13 years for 8% continuous compounding table
19.351−20.000
n = 12 + (13 ˗ 12) 19.351−21.963 = 12.25 yrs
Using the interest formula requires solution for n in terms of F, A, and r as follows:
𝑒 𝑟𝑛 −1
F = A [ 𝑒 𝑟 −1 ]
𝐹
𝑒 𝑟𝑛 – 1 = 𝐴 (𝑒 𝑟 – 1)
𝐹
ln(𝑒 𝑟𝑛 ) =ln [𝐴 (𝑒 𝑟 − 1) + 1]
𝐹
rn = ln [ 𝐴 ( 𝑒 𝑟 – 1) + 1]
𝑛=
𝐹
𝐴
𝑙𝑛[ (𝑒 𝑟 −1)+1]
𝑟
For F = $2,000, A = $100, and r = 8%
𝑛=
𝑙𝑛[
$2,000 0.08
(𝑒
−1)+1]
$100
0.08
=
ln[20 (0.08330)+1]
0.08
=
ln(2.666)
0.08
=
0.981
0.08
= 12.26 years.
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
8
F & P continuous compounding
F = P er n
𝐹
ln ( ) = ln(𝑒 𝑟𝑛 ) = 𝑟 𝑛
𝑃
𝑛=
𝐹
ln(𝑃)
𝑟=
𝐹
ln(𝑃 )
𝑟
𝑛
F & A discrete compounding
 (1 + i) n  1
F = A

i


𝐹
( ) 𝑖 = (1 + 𝑖)𝑛 − 1
𝐴
𝑙𝑛 (
𝐹𝑖
+ 1) = 𝑛 𝑙𝑛(1 + 𝑖)
𝐴
𝑛=
𝐹𝑖
𝑙𝑛 ( 𝐴 + 1)
𝑙𝑛(1 + 𝑖)
P & A discrete compounding
 ( 1  i) n  1
P = A
n 
 i (1  i ) 
Pi (1  i ) n
= (1  i) n  1
A
Take (1 i) n as a common factor
 Pi 
(1  i) n 1    1
 A


1
ln (1  i) n  ln 

1  Pi A 
 1 
ln 
1  Pi A 
n 
ln (1  i)
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
9
P & A continuous compounding
1  e  rn 
P = A r

 e 1 
P r
e  1 = 1  e rn
A
P r
1
e  1 = e  rn
A
P r


ln 1 
e  1  = ln e rn  rn
 A







 


P r


ln 1 
e 1 
A

n= 
r
Example: P & A (continuous compounding)
Find the present equivalent amount that would generate $4,000 withdrawals each year for 10 years if
the interest rate earned is 7% compounded continuously, using table or equation:
Solution
P=A[
P=A[
P /A,r, n
P/A,7,10
] = $4,000 [ 6.9428 ] = $27,771
1−𝑒 −(𝑟)(𝑛)
𝑒 (𝑟) −1
1−𝑒 −(0.07)(10)
] = $4,000 [
𝑒 0.07 −1
] = $27,771
Example: Uniform Gradient Series (continuous compounding)
Suppose there is a series of 5 payments beginning at $800 and decreasing by $100 each. If r = 12%
compounded continuously, find the amount A equivalent to this series
Solution
A= F1 + G [
A/G, i, n
A/G,12, 5
] = $800 – $100 [ 1.7615 ] = $624 per yr
Alternatively, we can convert “r” to “i” and use equation or tables for discrete compounding
i = exp(0.12) –1 = 0.1275 (12.75% per year compounded annually)
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
10
Example
Calculate the annual equivalent for the following cash flow over 5 years and 4% interest rate
$100
0
1
2
3
4
5
Solution
P/F, 4, 3
PW = $100 ( 0.889 ) = $88.9
A/P,4, 5
A/P,4, 5
AE = PW ( 0.2246 ) = $88.9 ( 0.2246 ) = $19.97
Another solution:
F/P, 4, 2
FW = $100 ( 1.0816 ) = $108.16
A/F, 4, 5
A/F, 4, 5
AE = FW ( 0.1846 ) = $108.16 ( 0.1846 ) = $19.97
Example P73
An individual who won a 20 million jackpot in a national lottery is offered a choice of $12.5 million
now or $2 million per year for the next 10 years. The individual will use 12% interest rate in his
evaluation. The patterns of receipts are shown in Table 4.1.
Table 4.1. Pattern of receipts for two alternatives
End of year Receipts, Alternative A Receipts, Alternative B
0
$12,500,000
0
1
0
$2,000,000
2
0
$2,000,000
3
0
$2,000,000
4
0
$2,000,000
5
0
$2,000,000
6
0
$2,000,000
7
0
$2,000,000
8
0
$2,000,000
9
0
$2,000,000
10
0
$2,000,000
Total Receipts
$12,500,000
$20,000,000
Cannot compare between money today and money in the future (i = 0)
P/A,12,10
P= $2,000,000 ( 5.6502 ) = $11,300,000
So Alternative A is better than B
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
11
Example P82: Present worth for multiple cash flows
Determine the present worth that is equivalent to the following cash flow for an interest rate of 12%:
$300 end of year 6; $60 end of years 9, 10, 11 and 12; $210 end of year 13; $80 end of years 15, 16,
and 17 (these payments are illustrated in Figure 4.1). Also, find the annual equivalent over 17 years.
Partial present equivalent amounts
End
of
year
0
$313.05
(P1=$151.98) +
(P2=$73.61) + (P3=$48.13) + (P4=$39.33)
1
2
3
4
5
6
$300
P/F,12,6
$300( 0.5066 )
7
8
P/F,12,8
$182.24( 0.4039 )
9
$60
10
$60
11
$60
12
$60
13
$210
P/A,12,4
$60( 3.0374 )
P/F,12,13
$210( 0.2292 )
14
15
$80
16
$80
17
$80
F/A,12,3
$80( 3.374 )
P/F,12,17
$269.92( 0.1457 )
To find the equal annual payment series equivalent to the above series of cash flows
A/P ,i, n
A=P(
A/P,12,17
) = $313.05 ( 0.1405 ) = $43.98
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
12
Example P84: Present worth for multiple cash flows
Find the annual equivalent payments that are equivalent to the cash flow in figure 4.2 for an interest
rate of 10%.
Partial present equivalent amounts
End
of
year
0
$612.05
(P1=$347.10)+
1
$200
2
$200
3
0
60.09
4
$20
60.09
5
$40
60.09
6
$60
7
$80
60.09
8
$100
60.09
9
$120
60.09
10
$140
60.09
(P2=$264.95)
P/A,10,2
$200( 1.7355 )
P/F,10,2
320( 0.8265 )
A/G,10,8
$20( 3.0045 )
60.09
P/A,10,8
$60.09( 5.3349 )
To find the equal annual payment series equivalent to the above series of cash flows
A=P(
A/P ,i, n
A/P,10,10
) = $612.05 ( 0.1628 ) = $99.64
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
13
Example: Different interest rates
The cash flow in Figure 4.4 represents three different interest rates applicable over the 5-year time
span. Calculate (a) the equivalent amount P at the present, (b) future amount at year 5, and (c) the
annual equivalent over 5 years for this set of conditions,
$200
Solution:
(a)
A = 4, B = 1
$100
$100
3
4
$100
0.12 4
𝑖 = (1 + 4 ) − 1= 0.1255
i = 12.55% per year compounded annually
0
1
r1 = 12%
compounded
quarterly
P/F,12.55,1
𝑃1 = $200 ( 0.8885 ) = $177.7
P/F,7,1
P/F,12.55, 2
P/A,10, 2
P/F,7,1
2
i2 = 7%
compounded
annually
𝑃2 = $100 ( 0.9346 ) ( 0.7894 ) = $73.77
5 years
i3 = 10%
compounded
annually
P
P/F,12.55,2
𝑃3 = $100 ( 1.7355 ) ( 0.9346 ) ( 0.7894 ) = $128
P = 𝑃1 + 𝑃2 + 𝑃3 = $380
(b)
F/P,12.55,2
F/P, 7, 1
F/P,10,2
F = $380 ( 1.267 ) ( 1.07 ) ( 1.210 ) = $623
(c)
P/F,12.55,2
P/F,12.55,1
P = 𝐹1 ( 0.8885 )+ 𝐹2 ( 0.7894 )
P/F,7,1
F1
F2
F3
F4
F5
1
2
3
4
5 years
P/F,12.55,2
+ 𝐹3 ( 0.9346 ) ( 0.7894 )
P/F,7,1
P/F,10,1
P/F,12.55,2
+ 𝐹4 ( 0.9091 ) ( 0.9346 ) ( 0.7894 )
P/F,10,2
P/F,7,1
P/F,12.55,2
0
+ 𝐹5 ( 0.8265 ) ( 0.9346 ) ( 0.7894 )
i1 = 12.55%
compounded
annually
= $380
F1 = F2 = F3 = F4 = F5 = A
A = 103 $/yr
i2 = 7%
compounded
annually
i3 = 10%
compounded
annually
P
Note: you can also use equations instead of tables
(
(
P /F, i, n
F/P , i, n
1
) = ( 1+𝑖 )𝑛
) = ( 1 + 𝑖 )𝑛
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
14
Example: Calculating the interest rate (i) for a cash flow
0
1
482
482
2
3
482
4
5
482
482
6
7
-250
-500
Set
-1000
pw(i) = 0
Remember (
P /F, i, n
)
1
(1+𝑖)𝑛
then
Pw(i) = –
Let x =
1000
500
482
482
482
250
482
482
–
+
+
+
–
+
+
=0 →
0
1
2
3
4
5
6
(1+𝑖)
(1+𝑖)
(1+𝑖)
(1+𝑖)
(1+𝑖)
(1+𝑖)
(1+𝑖)
(1+𝑖)7
1
(1+i)
(1)
→ (2)
Pw(i) = –1000 – 500x + 482x2 + 482x3 + 482x4 – 250x5 + 482x6 + 482x7 = 0
Use solver or scientific calculator → x = 0.90909
1
x
–1=
1
0.90909
– 1 = 0.1
or 10%
Graphical solution of Eq (1)
assume i = 0
→ pw (0) = 660
assume i = 0.05 → pw (0.05) = 280
assume i = 0.1
→ pw (0.1) = 0
assume i = 0.15 → pw (0.15) = -212
Draw Equation (1): i versus pw(i)
x-axis intercept 0.1 (10%) is the answer
pw(i), $
i=
then into equation (2)
800
700
600
500
400
300
200
100
0
-100 0
-200
-300
0.05
0.1
i
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
0.15
0.2
15
Example: page 90 – Semiannual payments
If payment of $100 occurs semiannually at the end of each 6-months period for 3 years and the
nominal interest rate is 12% compounded semiannually. Calculate the present worth P.
Solution:
Sine A is on semiannual basis it would be better to use everything on semiannual basis so n would be
6 semesters instead of 3 years and i has to be per semester compounded semiannually.
12%
i = 2 𝑠𝑒𝑚𝑒𝑠𝑡𝑒𝑟𝑠 = 6% per semester compounded semiannually
n = (3 years) (2 semesters per year) = 6 semesters
P=A(
P /A,i, n
P/A,6,6
) = $100 ( 4.9173 ) = $491.73
Example: Page 90 – Monthly payments
Suppose that a person borrows $2,000 and is to repay this amount in 24 equal monthly installments
of $99.80 over the next 2 years. Interest is compounded monthly on the unpaid balance of the loan.
What are (a) the effective interest rate per month and (b) the effective annual rate of interest on the
loan?
Solution:
Sine A is on monthly basis it would be better to use everything on monthly basis so n would be 24
months instead of 2 years
A=P(
A/P ,i, n
)
$99.80 = $2,000 (
(
A/P ,i, 24
)
A/P ,i, 24
)
= 0.0499
A search of the interest tables reveals that the preceding factor value is found for i = 1.5 %,
You can also use the equations
 i( 1  i) n 
A = P

n
 (1  i)  1
 i( 1  i) 24 
$99.80  $2,000
 → i = 0.015
24
 (1  i)  1
so the interest rate is 1.5% per month compounded monthly.
or
ia  1  im   1 = 1  0.01512  1 = 19.56% per year compounded annually
12
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
16
Example: Page 91 – Quarterly payments
Suppose a deposit of $100 is place in a bank account at the end of each year for the next 3 years. The
bank pays interest at the rate of 6% compounded quarterly. How much will be accumulated in this
account at the end of 3 years?
Method 1: make all calculations on annual basis (corresponding to the annual deposits).
r = 6% per year compounded quarterly
The effective annual interest rate from Equation with A = 4, B = 1
𝑖𝑎 = (1 +
F=A(
0.06 4
4
F/A, i, n
) − 1 = 0.0614 (or 6.14% per year compounded annually)
F=?
F/A, 6.14, 3
) = $100 ( 3.188 ) = $318.80
0
1
2
$100
$100
3
years
$100
Method 2: if you make the calculations based on the compounding periods, which are 3 months in
length, then
r = 6% per year compounded quarterly
F=?
6%
i = 4 𝑞𝑢𝑎𝑟𝑡𝑒𝑟𝑠 = 1.5 % per quarter compound quarterly
0 1 2 3 4 5 6 7 8 9 10 11 12 quarter
The amount accumulated in the account is
F/P,1.5, 8
F/P,1.5, 4
F = $100 ( 1.127 ) + $100 ( 1.061 ) + $100 = $318.80
$100
$100
$100
Although we obtained the same answer, the method is more difficult than the first one
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
17
Example: Page 92
Find the future amount at the end of 5 years that would result from end-of-month deposits of $1,000
made throughout the 5 years period. Assume that the interest earned on these deposits is 15%
compounded continuously.
Solution:
Since the deposits are monthly, we make all calculations on monthly basis
n = (12 months per year) ( 5 years) = 60 months
15%
r = 12 𝑝𝑒𝑟𝑖𝑜𝑑𝑠 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 = 1.25% per month compound continuously
Using the continuous compounding interest tables
F=A[
F/A, r, n
F/A,1.25, 60
] = $1000 [ 88.8027 ] = $88,802.70
Using algebraic expression:
𝑒 𝑟𝑛 −1
𝐹 = 𝐴 [ 𝑒 𝑟 −1 ]= $1,000 [
𝑒 (0.0125)(60) −1
𝑒 0.0125 −1
] = $88,650
Almost the same answer, with equation being more accurate
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
18
Example: Page 92
Suppose that $10,000 is placed into an account where the interest rate is 8% compounded
continuously. What is the size of equal annual withdrawals that can be made over the next 5 years so
that the account balance will equal zero after the last withdrawal?
Solution:
Since A is annual, we make all calculations on annual basis
r
A/P,r, n
1 
Continuous compounding A = P  e - rn
]
 = P[
1  e

Using continuous compounding tables
A/P,8, 5
A = $10,000 [ 0.2526 ] = $2,526
We can also use equations for continuous compounding
 er -1 
 e 0.08 - 1 
A = P
=
$10,000
 $2,526

 rn 
 0.08( 5) 
1  e 
1  e

Discrete compounding A = P  i( 1  ni)
A/P,i, n

=
P
(
)

 (1  i)  1
n
𝑖𝑎 = 𝑒 𝑟 – 1 = 𝑒 0.08 – 1 = 0.0833 (or 8.33% per year compound annually)
Using discrete compounding tables
A/P,8.33,5
A = $10,000 ( 0.2526 ) =$2,526
We can also use equations for discrete compounding
 i( 1  i) n 
 0.0833(1  0.0833)5 
A = P
=
$10,000


  $2,526
n
5
 (1  i)  1
 (1  0.0833)  1 
Although there are four solutions, the best way would be to use continuous compounding tables, then
continuous compounding equations. It is not advisable to convert to discrete compounding, as that
will most likely result in an interest rate that are not a whole number requiring interpolation between
tables, and if you use equations discrete compounding then why not use the equations for continuous
compounding and save yourself the extra step of converting from continuous to discrete?
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
19
Example: Page 93
Suppose that the preceding example is modified so that the equal withdrawals are to be made
quarterly over the 5-year time span.
Solution:
With quarterly payments, we make all calculations on quarterly basis
n = (4 quarters per year) (5 years) = 20 quarters.
Using continuous compounding
8%
r = 4 𝑞𝑢𝑎𝑟𝑡𝑒𝑟𝑠 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 = 2% per quarter compounded continuously
Using equations
 er - 1 
 e0.02 - 1 
A = P
=
P
1  e (0.02)( 20)  = $613
rn 
1  e 


Using tables
A/P ,r, n
A = P[
A/P,2, 20
] = $10,000 [ 0.0613 ] = $613
Using discrete compounding
i = 𝑒 𝑟 – 1 = 𝑒 0.02 – 1 = 2.02% per quarter compounded quarterly
Using equations
 i( 1  i) n 
 0.0202(1  0.0202)20
A = P
=
$2,000

 (1  0.0202) 20  1
n
 (1  i)  1


 = $613

Using tables
A/P,i, n
A = P(
A/P,2.02, 20
) = $10,000 ( 0.0613 ) = $613
Using discrete compounding tables requires interpolation between the 2% table and 3% table to
obtain the A/P factor, which is a cumbersome procedure, so it is easier to use continuous
compounding tables or equation and not convert to discrete.
Also notice that if you divide A from the previous example (annual) = $2,526 / 4 = $631.50 which
does not equal the A in this example (quarterly) and vice versa.
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
20
Example: Page 94 – Compounding less frequent than payments.
If the interest rate is 12% per year compounded quarterly, calculate the future worth for the
following cash flow
250
100
100 100
0
4
1
2
5
6
3
11
7
100 100 100
8
9
10
Months
12
100
-$400
Method 1: (better) Solve on monthly basis (corresponding to the monthly cash flows).
Convert 12% per year compounded quarterly to per % per year compounded annually
A=4,B=1
i = (1 +
0.12 4
4
) = 0.125509
then convert from % year compound Annually to % per month compounded monthly
im  1  ia 
1 / 12
 1 = 1  0.1255091 / 12  1 = 0.0099
F = - 400 (1.0099)12 + 100 (1.0099)11 + 100 (1.0099)10 + 100 (1.0099)9 - 100(1.0099)8 - 100(1.0099)7
- 100(1.0099)6 +250 (1.0099)4 – 100 (1.0099)1 = - 450.19 + 111.45 + 110.353 + 109.27 – 108.2
-107.14 - 106.09 + 260 – 100.99 = - 281.496
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
21
300
0
250
2
1
4
Quarters
3
100
300
-$400
Method 2: (not recommended) Solve on quarterly basis.
Convert 12% per year compounded quarterly to per % per quarter compounded quarterly
12%
4
= 3% per quarter compounded quarterly, then convert payments to coincide with end of each
quarterly period because monthly deposits do not earn monthly interest but rather quarterly.
F = - 400 (1.03)4 + 300 (1.03)3 - 300(1.03)2 +250 (1.03)1 – 100 = - 450.2 + 327.82 – 318.27 + 257.5 100 = -283.15
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
22
Treasury Bonds
Either government or financial institutions can issue bonds.
Government (municipal) bonds are tax-free
Corporate bonds incur taxes on dividends (‫)االرباح‬
Terms: Interest Rate,
Issue Date, and
Maturity Date
Face Value
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
23
Example: Page 94: BONDS
Suppose that an individual can purchase (for $900) a $1,000 municipal bond that pays every 6
months 6% tax-free interest semiannually (per year compounded semiannually). If the bond will
mature to its face value in 7 years, what will the equivalent rate of interest be in semester
compounded semiannually and per year compounded annually?
(In financial terms, what is the bond’s “yield to maturity”)?
r = 6% per year compounded semiannually
i = 3% per semester compounded semiannually
A = 0.03 ($1000) = $30 every six months
$1000
The cash flow diagram therefore will be,
$30 $30 $30 $30 $30 $30 $30 $30 $30 $30 $30 $30 $30 $30 $30
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
semesters
$900
P /F, i, 14
P /A,i, 14
PW (i) = - $900 + $30 (
) + $1,000 (
)=0
Solving gives i = 3.94% per semester compounded semiannually
To convert to annual using equation, the effective annual interest rate (or yield to maturity) for this
bond is
ia  1  is   1 = ia  1  0.0394  1 = 0.08035 (or 8.035% per year compounded annually)
2
2
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
24
LOANS
Example: Page 97: Actual interest charged on a loan
Suppose that an individual wishes to purchase a home appliance for $300. The salesperson indicates
that the interest rate will be 20% added on, and the payments can be over one year.
The total amount owed = $300 + 0.20 ($300) = $360.
With payment over 12 months, the monthly payment (A) = $360/12 = $30.
$300
0
1
2
3
4
5
6
7
8
9
10
11
12
months
$30 $30 $30 $30 $30 $30 $30 $30 $30 $30 $30 $30
The actual or effective interest rate for the loan is calculated by finding the value for i that sets the
receipts equal to the disbursements, using the following equation:
P=A(
P /A,i, n
)
$300 = $30 (
(
P /A,i, 12
P /A,i, 12
)
) = 10
By interpolation from the tables of 2 and 3%, im = 2.9% per month compounded monthly
Using algebraic equation
( 1+𝑖)12 −1
$300 = $30 [𝑖 ( 1+𝑖 )12 ) ]
im = 2.9 % per month compounded monthly.
The annual interest rate is
ia = ( 1+0.029 )12 - 1 = 0.409 (or 40.9% per year compounded annually)
The actual interest rate is more than 40% because payments were made every month not at the end of
the year.
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
25
Real Example: Actual interest rate on a bank loan
Note that is example shows how some local banks calculate the interest rate; do not use this in
your HW or Exam.
An individual wants a loan of 10,000 KD from a bank. The banker maintains they will charge a
simple interest rate of 7% annually and the 10,000 will have to be paid in equal monthly instalments
over a period of 5 years. The interest amount will be deducted in advance so the borrower will
actually receive less than 10,000 KD. Calculate (a) the amount of monthly installments over five
year’s period and (b) the actual annual interest rate charged by the bank?
Solution:
The interest rate over 5 years = 0.07 (%/year) 5 (years) = 0.35
Interest amount = 0.35 x 10,000 KD = 3,500 KD
The interest is taken in advance and the actual cash received = 10,000 - 3,500 = 6,500 KD
n = 12 (months) 5 (years) = 60 months
10,000 𝐾𝐷
The monthly installments (A) = 60 𝑚𝑜𝑛𝑡ℎ𝑠 = 166.667 KD/month
P=A(
P /A,i, 60
166.67
166.67
)
1
6,500 KD = 166.667 KD (
P /A,i, 60
2
3
4
5
6
60
)
Cash Flow Diagram
6,500
Using tables
(
P /A,i, 60
Using equations
P /A,i, 60
) = 39.0
6,500 = 166.67 (
1%
i
2%
)
60
6500  1  i   1 



166.67  i 1  i 60 
44.9550
39.0
34.7609
By interpolation
By solver or trial & error:
im = 1.58417 % per month compounded monthly
im = 1.584209 % per month compounded monthly.
From monthly to annual ia  1  im   1
12
ia  1  0.0158417  1= 0.20757 (or 20.757% per year compounded annually)
12
The actual interest rate is more than 20% because the interest is paid in advance, instead of over 5
years.
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
26
Real Example: Loan on a house
Given
A = 397 , P = 48,892 , n = 186 months (approx.. 15.5 years), find i
Solution
P=A(
P /A,i, n
)
48,892 KD = 397 KD (
P /A,i, 186
)
Using equations (tables are out of range)
 1  i 186  1 
48,892

 123.15  
186 
397
 i 1  i 

By solver or trial & error: im = 0.4768 % per month compounded monthly.
From monthly to annual ia  1  im   1
12
ia  1  0.004768  1 = 0.05874 (or 5.874% per year compounded annually) → Legal
12
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
27
Example: Page 98: Remaining balance on a loan
Suppose that $10,000 is borrowed with the understanding that it will be repaid in equal quarterly
payments over five years at an interest rate of 16% per year compounded quarterly.
(a) What is the quarterly payment A?
(b) Immediately after the 13th payment is made, the borrower wishes to pay off the remaining
balance, U13, so that the obligation will terminate. How much is the unpaid balance?
Solution:
Since payments are quarterly then we use quarterly period for all calculations
iq =
16%
4
= 4% per quarter compounded quarterly
$10,000
The quarterly payments will be
A/P,4, 20
A = $10,000 ( 0.0736 ) = $736
The remaining balance at the end of Q13
0
F/P, 4,13
1
2
3
4
13 quarters
F/A, 4,13
U13 = $10,000 ( 1.665 ) - $736 ( 16.627 ) = $4,413
The equivalent
of the original
amount loaned
$736
$736
$736
$736
the equivalent of
the amount paid
U13 = ?
EZ method: is to find the equivalent of the payments remaining at the time the remaining balance is
to be paid, with only 7 payments remaining,
P/A,4, 7
13
14
15
16
17
18
19
20 quarters
U13 = $736 ( 6.0021 ) = $4,418
$736
$736
$736
$736
U13 = ?
The second approach is simpler but not always possible since the future interest is not always known
U13 = (7)($736) = $5,152
‫إذا ما يطيح الفوائد‬
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
28
Example: Page 99: Remaining balance on a loan (when interest rate changes over time)
Suppose that an individual borrows $6,000 to be repaid in equal monthly payments of $100 with the
remaining balance due at the end of 5 years. The interest rate is to be changed each year to conform
to the market rate in effect. Assume that the market interest rate in the first year was 1.5% per month
compounded monthly and that it is 1.0% per month compounded monthly in the second year. What
is the remaining balance on this loan after the 19th payment?
Solution:
$6,000
1.5% per mo.
comp. mo.
0 1
3
6
9
$100
1% per mo.
comp. mo.
12
15
19
i%
(unknown)
24
$100
months
$100
U19 = ?
We cannot use the EZ method since i after year 2 years (24 months) in unknown.
The remaining balance on this loan, U19, is calculated as follows,
Step 1: use F/P factor to move $6000 to month 12 with interest 1.5%, then use F/P again to move the
result from month 12 to 19 (7 months) with interest 1%.
F/P,1.5,12
F/P,1, 7
$6,000 ( 1.196 ) ( 1.072 ) = $7,692.67
Step 2: use F/A to move the first 12 payments (for months 1 to 12) A ($100) to the future at month
12, then use F/P to move the result from month 12 to 19 (7 months) with interest 1%. That takes care
of the first 12 payments.
F/A,1.5,12
$100 ( 13.041 ) = $1,304.10
F/P,1, 7
$1,304.10 ( 1.072 ) = $1,398
We still have 7 more payments (for months 13 to 19). Use F/A to move these 7 remaining payments
to month 19 with interest 1%.
F/A,1, 7
$100 ( 7.214 ) = $721.40
Step 3: Subtract the total of step 2 from step 1 as follows
U19 = $7,692.67 – [$1,398 + $721.40] = $5,573.27
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
29
Example: Working capital
Suppose a $100,000 investment in a 5-year project requires an additional $20,000 in working capital.
(a) $5,000 cash to cover maintenance and labor costs that may or may not materialize.
(b) $8,000 accounts receivable expected to average over the life of the project.
(c) $7,000 inventories to be carried throughout the project’s life.
If all the other income and expenses are expected to provide a net income of $35,000 per year and
the interest rate is 20%, What is the annual equivalent to this cash flow?
Solution:
$20,000
Working capital
recovered at the end
$35,000
0
1
2
3
4
5
$100,000
$20,000
A/P,20, 5
Working capital dispersed
at the beginning
A/F, 20, 5
AE = - $120,000 ( 0.3344 ) + $35,000 + $20,000 ( 0.1344 ) = -$2,440 per year.
If you ignore the working capital, you will obtain a positive AE, which is incorrect.
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
30
Home Work Problems: (due one week from today)
1. How long will it take for a series of equal annual payments of $1,600 each year to accumulate to
$100,000 at an interest rate of 20% compounded continuously?
2. What is the present equivalent value of the following future receipts?
a. $1,000 5 years from now at 14% compounded semiannually. Answer: $508
b. $9,000 10 years from now at 8% compounded quarterly.
c. $25,000 8 years from now at 9% compounded monthly.
d. $105,000 12 years from now at 13% compounded weekly.
3. What single equivalent amount will accumulate from each of the following investments?
a. $1,500 in 5 years at 12% compounded semiannually. Answer: $2,686
b. $5,000 in 9 years at 16% compounded quarterly.
c. $1,500 in 13 years at 6% compounded monthly.
d. $4,500 in 15 years at 13% compounded daily.
4. What series of equal payments are equivalent to the following present amounts?
a. $13,000 in 6 years at 8% compounded quarterly with quarterly payments. Answer: $688
b. $20,000 in 2 years at 15% compounded monthly with monthly payments.
c. $6,000 in 4 years at 10% compounded quarterly with semiannual payments.
d. $10,000 in 50 months at 9% compounded monthly with monthly payments.
e. $1,500 in 5 years at 13% compounded annually with quarterly payments.
5. What series of equal payments are equivalent to the following future amounts?
a. $8,000 in 6 years at 16% compounded semiannually when payments are semiannual. Answer:
$422
b. $4,000 in 12 years at 6% compounded quarterly when payments are annual.
c. $25,000 in 10 years at 12% compounded monthly when payments are quarterly.
d. $15,000 in 5 years at 8% compounded semiannually when payments are annual.
e. $70,000 in 7 years at 20% compounded quarterly when payments are monthly.
6. What is the present value of the following series of prospective payments?
a. $600 a month for 3 years at 13% compounded annually.
b. $5,000 a year for 5 years at 16% compounded semiannually.
c. $2,500 each year for 10 years at 8% compounded quarterly.
d. $14,000 each quarter for 4 years at 12% compounded monthly.
7. What is the accumulated value of each of the following series of payments? $5,000 semiannually
for 10 years at 6% compounded monthly.
8. Find the equal-annual-payment series that would be equivalent to the following increasing series
of payments if the interest rate is 12% (a) compounded annually; (b) compounded continuously.
Answer: b. $1,105
End of year
Payments
1
$600
2
$800
3
$1,000
4
$1,200
5
$1,400
6
$1,600
7
$1,800
9. Find the present value equivalent to the following geometrically decreasing series of payments. A
first-year base of $10,000 decreasing at 3% per year, to year 7 at an interest rate of 9%
compounded continuously. Answer: $45,879
10. If compounding is quarterly, what effective annual interest rate and what nominal interest rate
will make the following values of P = $1,000; F = $3,000; n = 6 years equivalent. Answer: 20.1%
per year; 18.7% compounded quarterly
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University
31
11. If compounding is monthly, find the effective monthly interest rate, the effective annual interest
rate, and the nominal interest rate that will make the following values of F and A equivalent for
the values n shown.
a. F = $5,000; A = $500 annual payments; n = 8 years.
b. F= $16,500; A = $299 monthly payments; n = 4 years
12. A series of payments ($10,000, first year; $9,000, second year; $8,000 third year; $7,000, fourth
year; and $6,000, fifth year) is equivalent to what present amount at 10% interest compounded
annually; compounded continuously? Solve using the gradient factors and then by using only the
single-payment present-worth factors.
13. A petroleum engineer estimates that the present production of 400,000 barrels of oil during this
year from a group of 10 wells will decrease at the rate of 15% per year for years 2 through 10.
Oil is estimated to be worth $25 per barrel if the interest rate is 10% compounded annually, what
is the equivalent present amount of the prospective future receipts from the wells?
14. A city that was planning an addition to its water supply and distribution system contracted to
supply water to a large industrial user for 10 years under the following conditions: The first 5
years of service were to be paid for in advance, and the last 5 years at a rate of $45,000 a year
payable at the beginning of each year. Two years after the system is in operation the city finds
itself in need of funds and desires that the company pay off the entire contract so that the city can
avoid a bond issue.
a. If the city uses 9% interest compounded annually in calculating a fair receipt on the contract,
what amount can it expect? Answer: $147,328
b. If the company uses 20% interest compounded annually, how much is the difference between
what the company would consider a fair value for the contract and what the city considers to
be a fair value? Answer: $53,864
15. A young couple have decided to make advance plans for financing their 3-year-old son's college
education. Money can be deposited at 7% compounded annually. What annual deposit on each
birthday from the 4th to the 17th inclusive must be made to provide $16,000 on each birthday
from the 18th to the 21st inclusive?
16. A bond is offered for sale for $1,120. Its face value is $1,000 and the interest is 11% payable
annually. What yield to maturity will be received if the bond matures 10 years hence? Find the
bond's current yield. Answer: 9.12%; 9.82%
17. To purchase a used automobile, $6,000 is borrowed immediately. The repayment schedule
requires equal monthly payments of $264.72 to be made over the next 24 months. After the last
payment is made, any remaining balance on the loan will be paid in a single lump sum. The
effective annual rate of interest on this loan is 19.56% based on monthly compounding. Find the
nominal interest rate and the lump sum paid at the end of the loan.
Engineering Economy – © 2016 Dr. Tareq Albahri – Kuwait University