MUST
(GE205)
Oct 2023
Contents:
STAGE 1:
STAGE 2:
STAGE 3:
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THE FUNDAMENTALS
Ch. (1): Foundations of Eng. Eco.
Ch. (2): Factors: How Time and Interest Affect Money
Ch. (3): Combining Factors
Ch. (4): Nominal and Effective Interest Rates
(Delayed after Ch5&6)
BASIC ANALYSIS TOOLS
Ch. (5): Present Worth Analysis
Ch. (6): Annual Worth Analysis
MAKING BETTER DECISIONS
Ch. (11): Replacement and Retention Decisions
1
Oct 2023
MUST
(GE205)
Summary of Stage 1: THE FUNDAMENTAL
Ch. (1): Foundations of Eng. Eco. (Diff. between Simple & Compound Interest)
+
Interest Amount (I)
=
Time n =3
Present value
Pv=1000
Interest Rate
i= 10%
Future value
Fv as a simple
= 1300
Fv as a compound = 1331
Ch. (2,3): Time Value of Money and Combining Factors
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Summer 2023
Final Revision
(GE205)-Must
Single Payment (P Or F)
‫دفعة واحدة‬
Uniform Series (A)
‫دفعات دورية متساوية‬
Gradient Series
‫دفعات دورية متزايدة او متناقصة‬
Arithmetic (G) ‫ النمو بمبلغ ثابت‬Geometric (g) ‫النمو بنسبة ثابتة‬
Aw=114
Pw = 1000
0
n = 3 years
i = `10%
1000
Fw = 1331
?
3
i = 10%
Pw = 248
0
1
2
100
100
Fw = 331
?
3
1331
F/P
P/F
P/A
F/A
1.3310
F = 1000 (F/P,10%,3)
0.7513
P = 1331 (P/F,10%,3)
2.4869
P = 100 (P/A,10%,3)
3.3100
F = 100 (F/A,10%,3)
Aw=112
Fw = 529
?
Pw=361
0
1
A
100
i = 10%
2
A+G
110
100
A=100
G=+10
4
3
A+2G A+3G
0
1
i = 10%
2
3
120
130
PT = PA + PG
i=10%
g=8%
Geo-Pw Table
4.3781
3.1699
PT = 100(P/A,10%,4) +10(P/G,10%,4)
AT = AA + AG
Pg = A1(P/A1,10%,8%,4)
Geo-Fw Table
Fg = A1(F/A1,10%,8%,4)
1.3812
AT = 100+10(A/G,10%,4)
1.4641
Fw = 361 (F/P,10%,4)
4.6410
Or = 114 (F/A,10%,4)
4
4
100
Fw= Pw(F/P,i,n) or Aw(F/P,i,n)
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Fw = 518
?
Pw=354
Aw= Pw(A/P,i,n)
Or Aw= Fw(A/F,i,n)
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MUST
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Case (1)
Case (2)
Shifted Period
0
1
Shifted Period
F=??
2
3
4
0
5
100 100 100
1
2
3
4
F=??
5
100 100 100
Pw= 100(P/A,10%,3)(P/F,10%,2)
Pw= 100(F/A,10%,3)
Fw= 100(F/A,10%,3)
Fw= 100(F/A,10%,3)(F/P,10%,2)
Aw= PT(A/P,10%,5) or FT (A/F,10%,5)
Aw= PT (A/P,10%,5) or FT (A/F,10%,5)
Case (3) at i=8%
P0 = 90 (P/A,8%,3) (P/F,8%,1)
+ [ 90(P/A,8%,4) + 10(P/G,80%,4)] (P/F,8%,5)
= $ 449.28
Then: Fw = 449.28 (F/P,8%,10)
Aw= 449.28 (A/P,8%,10)
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Example (1)
Stereo phonics, Inc., plans to borrow $20,000 from a bank for 1 year at 9% interest for new
recording equipment. Compute the simple interest and the total amount due after 1 year.
Solution
▪ I simple = p × n × I
= 20,000 × (1) × (0.09) = $1,800
▪ Fsimple = p + I
= 20,000 + 1,800 = $ 21,800
Simple Interest
n = 1 year
0
1
i = 9%
P = $20,000
I&F = ??
?
Example (2)
Assume an engineering company borrows $100,000 at 10% per year compound interest and
will pay the principal and all the interest after 3 years. Compute the annual interest and total
amount due after 3 years.
Solution
Comp. Interest
▪ The annual interest:
n = 3 years
J=t-1
It = ] p + ∑ IJ [ . (i)
0
J=1
i = 10%
P = $100,000
I 1 = 100,000 (0.10) = $10,000
I 2 = 110,000 (0.10) = $11,000
I 3 = 121,000 (0.10) = $12,100
2
1
3
I&F = ??
?
▪ Total amount due after 3 years
F
compound
= p ( 1 + i )n
= 100,000 ( 1+ 0.1)3 = $133,100
Example (3)
Green Tree Financing lent an engineering company $100,000 to retrofit an environmentally
unfriendly building. The loan is for 3 years at 10% per year simple interest. How much money
will the firm repay at the end of 3 years?
Solution
Simple Interest
n = 3 years
▪ Fsimple = p ( 1 + n . i )
= 100,000 [ 1 + 3 (0.10) ] = $130,000
0
i = 10%
3
F = ??
?
P = $100,000
Example (4) P1.31
If a company sets aside $1,000,000 now into a contingency fund, how much will the company have in
2 years, if it does not use any of the money and the account grows at a rate of 10% per year?
Solution
F
compound
0
n = 2 years
P = $,1000,000
i = 10%
n
=p(1+i)
= 1,000,000 ( 1+ 0.10)2= $1,210,00
2
F = ??
?
Example (5)
An employee at LaserKinetics.com borrows $10,000 on May 1 and must repay a total of
$10,700 exactly 1 year later. Determine the interest amount and the interest rate paid.
Solution
n = 1 year
0
1
▪ Interest paid:
Fsimple = p + I
10,700 = 10,000 + I
P = 10,000
then:
I = 10,700-10,000 = $700
then:
i = p × n × 100 =
▪ Interest rate:
I simple = p × n × i
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I
i = ??%
F =10,700
?
700
× 100 = 7% per year
10,000(1)
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Oct 2023
MUST
(GE205)
Example (6) P1.32
A company has extra funds to invest for future capital expansion. if the selected investment
pays simple interest, what interest rate would be required for the amount to grow from$60,000
to $90,000 in 5 years?
Solution
i=
(90000-60000)
× 100 = 10% per year
60,000(5)
n = 5years
0
P = 60k
i = ??%
5
5
F =90k
?
Example (7)
If Laurel made a $30,000 investment in a friend’s business and received $50,000 5 years later,
Determine the rate of return.
Solution
F = P( 1 + i )n
50,000 = 30,000( 1 + i% )5
then: i = 10.76%
n = 5years
0
P = 30k
i = ??%
Example (8) P1.36
5
5
F =50k
?
If interest is compounded at 20% per year, how long will it take for $50,000 to accumulate to $86,400?
Solution
n
n = ??
F = P( 1 + i )
0
n
5
86,400 = 50,000(1 + 20%) n
i = 20%
P = 50k
F =86.4k
then: n = 3 years
?
Example (9) P1.11 Quiz
Which of the following 1-year investments has the highest rate of return?
(a) $95,000 that yields $7,600 in interest, (b) $12,500 that yields $1,125 in interest, or
(c) $80,000 that yields $6,600 in interest. (d) $56,000 that yields $4,160 in interest
Solution
i = (7600/95000)×100 = 8%
i = (1125/12500)×100 = 9%
i = (6600/80000)×100 = 8.25%
i = (4160/56000)×100 = 7.43%
The $12,500 investment has the highest rate of return.
Example (10) P1.26 Quiz-F.Jan23
At an interest rate of 10% per year, an investment of $100,000 one year ago is equivalent to
how much now?
(a) $110,000
(b) $111,000
(c) $120,000
(d) 115,000
Solution
Now
F
compound
= p ( 1 + i )n
= 100,000 ( 1+ 0.10)1 = $110,000
-1
P = $,100k
n = 1 year
i = 10%
0
F = ??
?
Example (11)
At an interest rate of 10% per year, the equivalent amount of $10,000 one year ago is closest
to:
(a) $8264 (b) $9091
(c) $11,000
(d) $12,000
Solution
Now
F= p ( 1 + i )n
100,000 = p ( 1+ 10%)1
Then: P =$9091
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-1
P = ??
n = 1 year
i = 10%
0
F = 10k
?
6
Oct 2023
MUST
(GE205)
Example (12) Quiz
How long does it take a single deposit of $58,350 in a saving account that pays 8% per year,
to accumulate $100,000?
A
7 years
B
5 years
C
4 years
D
3 years
Solution
n
F = P( 1 + i )
100000 = 58350(1 + 8%) n
then: n = 7 years
0
n = ??
P = 58.35k
i = 8%
n
5
F =100k
?
Example (13) P1.47
The concept that different sums of money at different points in time can be said to be
equal to each other is known as
(a) Evaluation criterion
(b) Equivalence
(c) Cash flow
(d) Intangible factors
Example (14) P1.49
All of the following are examples of cash outflows except:
(a) Asset salvage value
(b) Income taxes
(c) Operating cost of asset
(d ) First cost of asset
Example (15) P1.55
All of the following are examples of equity financing, except:
(a) Mortgage
(b) Money from savings
(c) Cash on hand
(d ) Retained earnings
Example (16) P1.50
In most engineering economy studies; the best alternative is the one that:
(a) Will last the longest time
(b) Is most politically correct
(c) Is easiest to implement
(d ) Has the lowest cost
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Example (17) L2
A person deposits $5000 into an account which pays interest at a rate of 8% per year. The
amount in the account after 10 years is closest to:
(a) $2792
(b) $9000
(c) $10795
(d) $12165
Solution 0
Fw = ??
10
Fw = 5000(F/P,8%,10)
= 5000(2.1589)
= $10794.5
i = 8%
5000
Example (18) L2
A small company wants to make a single deposit now so it will have enough money to
purchase a backhoe costing $50,000 five years from now. If the account will earn interest of
10% per year, the amount that must be deposited now is nearest to:
(a) $10,000
(b) $31,050
(c) $33,250
(d) $319,160
Solution
Pw = 50000(P/F,10%,5)
= 50000(0.6209)
= $31045
Pw =?? 0
5
i = 10%
50000
Example (19)- F.Jan23
The amount of money that Diamond Systems can spend now for improving productivity in lieu
of spending $30,000 three years from now at an interest rate of 12% per year is closest to:
A
$15,700
B
$17,800
C
$19,800
D
$21,350
Solution
Pw = 30,000(P/F,12%,3)
= $21,354
Example (20) P2.10
How much could the company afford to spend now on new equipment in lieu of spending
$200,000 one year from now and $300,000 three years from now, if the company uses an
interest rate of 15% per year?
Solution
0
1
2
3
P0th = 200,000(P/F,15%,1)
+ 300,000(P/F,15%,3)
= $371,17
Pw = ??
200
300
Example (21) P2.05
A family that won a $100,000 prize on America’s Funniest Home Videos decided to put onehalf of the money in a college fund for their child who was responsible for the prize. If the fund
earned interest at 6% per year, how much was in the account 14 years after it was started?
Fw = ??
Solution 0
F = 50,000(F/P,6%,14)
= $113,045
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50
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Oct 2023
MUST
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Example (22) P2.09
If GHD Plastics purchases a new building now for $1.3 million for its corporate headquarters,
what must the building be worth in 10 years? The company expects all expenditures to earn a
rate of return of at least 18% per year.
Fw = ??
Solution 0
18%
10
F = 1.3(F/P,18%,10)
= 6.8m
1.3 m
Example (23)
Metso Automation, which manufactures addressable quarter-turn electric actuators, is planning
to set aside $100,000 now and $150,000 one year from now for possible replacement of the
heating and cooling systems in three of its larger manufacturing plants. If the replacement
won’t be needed for 4 years, how much will the company have in the account, if it earns
interest at a rate of 8% per year?
Solution
Fw = ??
0
F4th = 100,000(F/P,8%,4)
+150,000(F/P,8%,3)
= $325,00
1
2
3
4
100
150
Example (24)
The cost of lighting and maintaining the tallest smokestack in the United States (at a shuttered
ASARCO refinery) is $90,000 per year. At an interest rate of 10% per year, the present worth
of maintaining the smokestack for 10 years is closest to:
A
$1,015,000
B
$894,000
C
$712,000
D
$553,000
Solution
Pw = 90,000(P/A,10%,10)
= $553,014
Example (25) P2.02
How much can Haydon Rheosystems, Inc., afford to spend now on an energy management
system if the software will save the company $21,300 per year for the next 5 years? Use an
interest rate of 10% per year.
Solution
Pw = ???
Pw = 21,300(P/A,10%,5)
0
2
3
1
4
5
= $80,744
3
4
A1-5 = 21300
Example (26) L2 Summer 2
The present worth of $500 per year starting from the end of year 1 through years8, at an
interest rate of 8% per year is closest to: :
A
$2,310
B
$2,600
C
$2,870
D $1,660
$1,520
D $2,055
Pw = 500(P/A,8%,8)
Pw If an interest rate of 12% per year is closest to
A
$2,280
B
$2,480
C
Pw = 500(P/A,12%,8)
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Example (27) L2 Quiz-Summer 22
An engineer want to modify the production system his company would make saving by $5000
per year. At an interest rate of 10% per year, how much could the company afford to spend
now to just break even over a 5-year project period?
(a) $11,170
(b) $13,640
(c) $15,300
(d) $18,950
Solution
Pw = ???
Pw = 5000(P/A,10%,5)
= 5000(3.7908)
= $18,954
Example (28) L2
0
2
1
3
4
3
5
4
A = 5000
An industrial engineer made a modification to a chip manufacturing process that will save her
company $10,000 per year. At an interest rate of 8% per year, will the saving amount to in 7 years?
Solution
0
1
2
F = 10000(F/A,8%,7)
= $89,228
4
3
Fw=??
5
6
7
A = $10,000
Example (29) P2.19
Henry Mueller Supply Co. sells tamperproof, normally open thermostats (i.e., thermostat
closes as temperature rises). Annual cash flows are shown in the table below. Determine the
future worth of the net cash flows at an interest rate of 10% per year.
Solution
F = 110 (F/A,10%,8) = $1,258
Example (30)
An enthusiastic new engineering graduate plans to start a consulting firm by borrowing
$100,000 at 10% per year interest. The loan payment each year to pay off the loan in 7 years
is closest to:
A
$18,745
B
$20,540
C
$22,960
D
$23,450
Solution
A1-7 = 100,000(A/P,10%,7)
= $20,541
Example (31)
An engineer who believed in “save now and play later” wanted to retire in 20 years with $1.5
million. At 10% per year interest, to reach the $1.5 million goal, starting 1 year from now, the
engineer must annually invest:
A
$26,190
B
$28,190
C
$49,350
D
$89,680
Solution
A1-20 = 1,500,000(A/F,10%,20)
= $26,190
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Example (32)
A company that sells high-purity laboratory chemicals is considering investing in new
equipment. If the new equipment will cost $220,000 to purchase and install, how much must
the company save each year for 3 years in order to justify the investment, if the interest rate is
10% per year?
Solution
A =??
P0th = 220
0
A1-3 = 220,000(A/P,10%,3)
= $88,464
w
2
1
3
Example (33) P2.20 Quiz
A company that makes self-clinching fasteners expects to purchase new production-line
equipment in 3 years. If the new units will cost $350,000, how much should the company set
aside each year, if the account earns 10% per year?
(a) $111,500
(b) $90,471
(c) $105,740
(d) $113,203
Solution
Aw =??
F3th = 350
A1-3 = 350,000(A/F,10%,3)
0
2
3
1
= $105,739
Example (34) L2
The present worth of $400 in year 1 and amounts increasing by $30 per year through year 5 at
an interest rate of 12% per year is closest to:
Solution
PT = P A + PG
0
1
2
3
4
5
=400(P/A,12%,5) +30(P/G,12%,5)
400
= 1,634
430
by using previous exercise data, find the annual worth?
AT = A A + AG
= 400 + 30(A/G,12%,5) = 454
by using previous exercise data, find the future worth?
Fw = 1,634 (F/P,12%,5) = 2,880
or 454 (F/A,12%,5) =2,880
460
A=400
G=+30
490
520
Example (35) -Summer 22
Energy costs for a green chemical treatment have been increasing uniformly for 5 years. If the
cost in year 1 was $26,000 and it increased by $2000 per year through year 5, the present
worth of the costs at an interest rate of 10% per year is closest to:
A
$102,900
B
$112,300
C
$122,100
D
$195,800
Solution
PT = PA+PG
Pw =26,000(P/A,10%,5) + 2000(P/G,10%,5)
= $112,284
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Example (36)
A company have operating cost per machine is $22,000 for year 1 and increases by a
constant $1000 per year through year 5, what is the equivalent uniform annual cost per
machine for the 5 years at an interest rate of 8% per year?
A
$23,850
B
$24,650
C
$25,930
D
Over $26,000
Solution
AT = AA+AG
A1-5 = 22,000 + 1000(A/G,8%,5)
= $23,847
Example (37) P2.27
Rolled ball screws are suitable for high-precision applications such as water jet cutting. Their
total manufacturing cost is expected to decrease because of increased productivity, as shown
in the table. Determine the equivalent annual cost at an interest rate of 8% per year:
Year
1
2
3
4
5
6
7
8
Cost, $1000
$200
$195
$190
$185
$180
$175
$170
$165
Solution
AT = AA + AG
A1-8 = 200 - 5(A/G,8%,8) = $184.51
Example (38) Summer 22-Quiz
Income from a precious metals mining operation has been decreasing uniformly for 5 years. If
income in year 1 was $300,000 and it decreased by $30,000 per year through year 4, the
annual worth of the income at 10% per year is closest to:
A
$310,500
B
$258,600
C
$203,900
D
$164,800
Solution
AT =AA +AG
A1-4 =300,000 – 30,000(A/G,10%,4)
= $258,564
Aw If an interest rate of 8% per year is closest to
(a) $220,400
(b) $300,000
(c) $231,700
(d) $150,600
Example (39) -Quiz-Jan23
At i= 10% per year, the annual worth of the cash flows shown below is close to:
Year
0
1
2
3
4
5
6
$
0
800
700
600
500
400
300
(a) $350
(b) $450
(c) $578
Solution
(d) $678
AT = AA+AG
A1-6= 800-100(A/G,10%,6)
=$577.64
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By using previous exercise data, the future worth of the cashflows is closest to:
(a) $3,600
(b) $4,020
(c) $4,300
(d) $4,460
Solution
Pw
Fw = [800(P/A,10%,6)-100(P/G,10%,6)] (F/P,10%,6) = $4,456
Aw
Or = [800-100(A/G,10%,6)] (F/A,10%,6) = $4,456
Example (40) -Quiz-Jan23
For the following cash flows, the future worth of the cash flows at an interest rate of 105 oer year is
closest to:
EOY 0
1
2
3
4
5
6
$
0
300
400
500
600
700
800
(a) $5,700
(b) $5,600
(c) $4,030
(d) $3,300
Aw
Fw = [300+100(A/G,10%,6)] (F/A,10%,6)
Example (41)
In planning for your retirement, you expect to save $5000 in year 1, $6000 in year 2 and
amounts increasing by $1000 each year through year 20. If your investments earn 10% per
year, the amount you will have at the end of year 20 is closest to:
A
$242,568
B
$355,407
C
$597,975
D
$659,125
Solution
Fw = [5000(P/A,10%,20) + 1000(P/G,10%,20)](F/P,10%,20)
= $659,126
Example (42) L2
Find the present worth of $ 1,000 in year 1 and amounts increasing by 7% per year through
year 10. Use an interest rate of 12% per year.
(a) $6,565
(b) $7,335
(c) $5,350
(d) $2,742
Pw
Solution
Pg = 1000 (P/A1,12%,7%,10)
Pw= ??
0
1
2
3
10
4
1000
= $7333
i=12%
g=7%
Example (43)
Determine the Future worth of maintenance at the end of year 8 and that has a cost of $600 at
the end of year 1 and an annual increase of 8% per year for 8 years. Use an interest rate of
10% per year.
Solution
Fw= ??
Pw
Fw = [1000. (P/A1,10%,8%,8) ] (F/P,10%,8)
0
1
2
3
4
8
5
600
= $ 8789
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i=10%
g=8%
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Oct 2023
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(GE205)
Example (44)
If you can save $5000 in year 1, $5150 in year 2, and amounts increase by 3% each year
through year 20, the amount you will have at the end of year 20 at 10% per year interest is
closest to:
A
$60,810
Pw
B
$102,250
C
$351,500
D Over $410,000
Solution
Fw = [5000 (P/A1,10%,3%,20) ](F/P,10%,20) = $351,528
Example (45) -Summer 22-Quiz
A series of cash flows of $50,000 in year 1 and increase by 6% each year through year 8, The
annual worth of the entire cash flows at an interest rate of 10% per year is closed to:
(a) $50,500
(b) $40,451
(c) $60,090
(d) $40,460
Solution
Pw
Aw= ??
Aw = [50000. (P/A1,10%,6%,10) ] (A/P,10%,8)
=60,090
0
1
2
3
8
50000
i=10%
g=6%
If a series of cash flows of $10,000 in year 1 and increase by 10% each year through year 8,
The annual worth of the entire cash flows at an interest rate of 12% per year is closed to:
(a) $14,500
(b) $15,200
(c) $12,800
(d) $13,510
A series of cash flows of $80,000 in year 1 and increase by 6% each year through year 5, The
annual worth of the entire cash flows at an interest rate of 8% per year is closed to:
(a) $69,750
(b) $98,000
(c) $89,400
(d) $78,500
A series of cash flows of $15,000 in year 1 and increase by 6% each year through year 8, The
annual worth of the entire cash flows at an interest rate of 12% per year is closed to:
(a) $17,930
(b) $16,900
(c) $18,500
(d) $16,250
A series of cash flows of $25,000 in year 1 and increase by 8% each year through year 8, The
annual worth of the entire cash flows at an interest rate of 8% per year is closed to:
(a) $31,500
(b) $12,225
(c) $33,500
(d) $34,900
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Oct 2023
Example (46)
For the cash flows below, Determine the amount in year 1, if the annual worth in years 1
through 9 is $601.17 and the interest rate is 10% per year:
Year
1
2
3
4
5
6
7
8
9
Cost,$1000
A
A+30 A+60 A+90 A+120 A+150 A+180 A+210 A+240
Solution
Aw =601.17
AT = AA + A G
9
0
2
8
1
601.17 = A + 30(A/G,10%,9)
601.17 = A + 30(3.3724) then:
A
A = $500
A+30
A+210
A+240
Example (47)-Mid21
Sales for the XYZ Company at the end of year 1 is $ 120,000. The sales are expected to
increase by 25,000 each year till year 10. For an interest rate of 12%, find:
a) What is the amount of year 5 sales?
b) What is the future worth of the entire cash flow series in year 10?
Solution
a. The amount of year 5 sales
Amount 5th = A + 4G
= 120 + 4 (25) = 220 K
b. The future worth of the entire cash flow series in year 10
0th
Fw = [120 K (P/A 12%, 10) + 25 K (P/G 12%, 10)] (F/P 12%, 10) = 3,678.44 K
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Example (48) -Mid21
How long does it take a single deposit of $58,350 in a saving account that pays 8% per year,
to accumulate $100,000?
A
7 years
B
5 years
C
4 years
D
3 years
Solution
n = ??
F = P (F/P,i%,n)
100,000 = 58,350(F/P,8%,n)
i = 8%
P = 58,350
F = 100,000
(F/P,8%,n)=1.7138
Interpolate in the 8% interest table, n = 7 years
Example (49) P2.39
If the value of Jane’s retirement portfolio increased from $170,000 to $813,000 over a 15-year
period, with no deposits made to the account over that period, what annual rate of return did
she make?
Solution
n = 15
F = P (F/P,i%,n)
0
5
5
813,000 = 170,000(F/P,i,15)
i = ??%
P = 170k
F =813k
(F/P,i,15) = 4.782
?
Interpolate in all interest tables, i=11%
Example (50) L2-Quiz-Summer22
A contractor purchased equipment for $60,000 which provided income of $16,000 per year for
10 years. The annual rate of return of the investment was closest to:
Solution
Aw = 16k
A = P (A/P,i%,n)
0
10
5
16000 = 60000(A/P,i,10)
i = ??%
P = 60k
(A/P,i,10) = 0.26667
Interpolate in all interest tables, i=23%
An investment of $5,759 in equipment will save a company $1000 per year for
9 years, the rate of return on this investment is close to:
A
8%
B
12%
C
15%
D
10%
A
8%
B
15%
C
12%
D
10%
A
15%
B
8%
C
12%
D
10%
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Oct 2023
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Example (51)- Summer22
An investment of $75,000 in equipment will reduce the time for machining self-locking
fasteners will save $20,000 per year. At an interest rate of 10% per year, the number of years
required to recover the initial investment is closest to:
A
6 years
B
5 years
C
4 years
D
3 years
Solution
75,000 = 20,000(P/A,10%,n)
(P/A,10%,n) = 3.75
By interpolation, n = 4.9 years
Example (52)-Quiz -Summer22
If $ 52,052 are deposited each year in a saving account that earns interest at a rate of 8% per
year. The number of years required for the account to accumulate $650,000 is closest to:
A
7 years
B
8 years
C
10 years
D
Solution
9 years
Aw = 52052
F = A(F/A,i%,n)
650000 = 52052(F/A,8%,n)
(F/A,8%,n) = 12.4875
0
n = ?? years
2
3
1
Fw = 650000
Interpolate in the 8% interest table, n = 9 years
Example (53)-Quiz
If $ 13,380 are deposited each year in a saving account that earns interest at a rate of 12% per
year. Find the number of years required for the account to accumulate $85,000
A
7 years
B
5 years
C
4 years
D
Solution
3 years
Aw = 13380k
F = A(F/A,i%,n)
85000 = 13380(F/A,12%,n)
(F/A,12%,n) = 6.3527
0
n = ?? years
2
3
1
Fw = 85000
Interpolate in the 12% interest table, n = 5 year
Example (54)
The number of years required for an account to accumulate $650,000 if Ralph deposits
$50,000 each year and the account earns interest at a rate of 6% per year is closest to:
A
13 years
B
12 years
C
11 years
D
10 years
Solution
50,000(F/A,6%,n) = 650,000
(F/A,6%,n) = 13.0000
By interpolation or NPER function, n = 9.9 years
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Example:(55) L3
The present worth of the cash flow shown below at i=10% is:
0
1
2
3
4
6
5
10,000
Solution
PW = 10,000(P/A,10%,5) (P/F,10%,1)
= $34,462
Example:(56) P3.46
For the diagram shown, the respective values of n for the following equation are:
P0 = 100(P/A ,10%, n )( P/F ,10%, n )
A
6 and 1
B
6 and 2
C
7 and 1
D
7 and 2
Example:(57) L3
How much money would be available in year 10 if $8000 is deposited each year in years 3
through 10 at an interest rate of 10% per year?
Solution
Fw=??
0
9 10
Fw = 8000(F/A ,10%,8)
3 4
= 8000(11.4359)
= $91,487
8000
Example:(58) P3.05-F.Jan23
Find the present worth at i=10% per year for the cash flow series shown below.
Solution
Pw = 200(P/A,10%,3) (P/F,10%,1) + 90(P/A,10%,3) (P/F,10%,5) =$ 494.15
To find Fw & Aw
Fw = 200(F/A,10%,3) (F/P,10%,4) + 90(F/A,10%,3) or 494.15(F/P,10%,8) =xx
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Aw1-8 = 494.15 (A/P,10%,8) or xx (A/F,10%,8)
Aw for the first 4 years: A1-4 = 494.15 (A/P,10%,4)
Aw for the last 5 years: A4-8 = xx
(A/F,10%,5)
Example:(59) L3
Find the Present worth in year 0 for the cash flows shown using an interest rate of 10% per year.
Solution
PW = 5000(P/A,10%,8) (P/F,10%,2) + 2000(P/F,10%,8)
= 5000(5.3349) (0.8264) + 2000(0.4665)
=$ 22,977
Example:(60)
The offshore design group at Bechtel just purchased upgraded CAD software for $5000 now
and annual payments of $500 per year for 6 years starting 3 years from now for annual upgrades.
What is the present worth in year 0 of the payments if the interest rate is 8% per year?
Solution
Pw=?? 0
P0= 5000
8
3
+ 500(P/A ,8%,6) (P/F,8%,2)
= $ 6981.60
0.5k
Example:(61) P3.46
5k
The net present worth in year 0 of the following series of incomes and expenses at 8% per
year is closest to:
(a) $14,300
(b) $15,500
(c) $16,100
(d) $16,500
Solution
P = 11,000 + 600(P/A,8%,6) + 700(P/A,8%,5) (P/F,8%,6) = $15,535
Example:(62)
The net cash flow associated with development and sale of a new product is shown. Determine
the present worth at an interest rate of 12% per year. The cash flow is in $1000 units:
Year
1
2
3
4
5
6
7
8
9
Cash flow, $
-120
-100
-40
+50
+50
+80
+80
+80
+80
Solution
Pw = -120(P/F,12%,1)
- 100(P/F,12%,2)
- 40(P/F,12%,3)
+ 50(P/A,12%,2) (P/F,12%,3)
+ 80(P/A,12%,4) (P/F,12%,5)
= $-17,320
For help, Call: 01000-70-60-26
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Example (63) P3.38
Determine the present worth in year 0 of the cash flows shown at an interest rate of 15% per
year.
Year
1
2
3
4
5
6
7
8
9
10
Cash flow,$
90
90
90
85
80
75
70
65
60
55
Solution
Pw = 90(P/A,15%,2)
+ [90(P/A,15%,8) - 5(P/G,15%,8)] (P/F,15%,2)
= $404.49
Example:(64)
The cash flow associated with making self-locking fasteners is shown below. Determine the
net present worth (year 0) at an interest rate of 10% per year:
Year
Income, $1000
Cost, $1000
0
20
8
1
20
8
2
20
8
3
20
8
4
30
12
5
30
12
6
30
12
7
30
12
8
30
12
9
30
25
5
18
6
18
7
18
8
18
9
5
Solution
Year
Net Cash flow
0
12
1
12
2
12
3
12
4
18
Pw = 12 + 12 (P/A,10%,3) + 18 (P/A,10%,5)(P/F,10%,3) + 5(P/F,10%,9)
= $95,228
Example 65-.Jan23
Using an interest rate of 10% per year. Find the equivalent annual worth series in years 1
through 7 for the following cash flow.
(a) $72.35
(b) $74.28
(c) $70.25
(d) $73.30
First P0 = 50(P/A,10%,2) + [ 50(P/A,10%,5) + 20(P/G,10%,5) ] (P/F,10%,2) = $356.82
Then:
Aw1-7 = 356.82 (A/P,10%,7) = $ 73.30
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Using an interest rate of 12% per year.
(a) $74.28
(b) $70.25
(c) $72.35
(d) $73.30
First Pw = 50(P/A,12%,2) + [ 50(P/A,12%,5) + 20(P/G,12%,5) ] (P/F,12%,2) = $330.18
Then:
Aw1-7 = 330.18 (A/P,12%,7) = $ 72.35
Example (66)
Find the present worth at i = 8% per year for the cash flow series shown below.
Solution
PT = 90 (P/A,8%,3) (P/F,8%,1)
+ [ 90(P/A,8%,4) + 10(P/G,80%,4)] (P/F,8%,5)
= $ 449.28
Example (67) P3.35
Find the present worth in year 0 for the cash flows shown. Let i = 10% per year.
Solution
PT = 50 (P/F,10%,1)
+ [ 50(P/A,10%,3) + 20(P/G,10%,3) ] (P/F,10%,1)
+ 170(P/F,10%,5)
+ [ 130(P/A,10%,3) + 20(P/G,10%,3) ] (P/F,10%,5)
= $ 536
= $21,537
Example:(68) P3.24
New actuator element technology enables engineers to simulate complex computer-controlled
movements in any direction. If the technology results in cost savings in the design of
amusement park rides, what is the future worth in year 5 of savings of $70,000 now and
$20,000 per year in years 1 through 3 at an interest rate of 10% per year?
Solution
0
Fw = 70,000(F/P,10%,5)
3
5
+20,000 (F/A,10%,3) (F/P,10%,2)
= $192,83
20
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21
Oct 2023
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Example (69) L3
For the cash flows shown, find the future worth in year 7 at i = 10% per year:
0
1
2
3
4
5
6
500
700
650
600
7
450
550
Solution
Find P0 then use F/P factor for 7years
First
Pw = [700(P/A,10%,6) - 50(P/G,10%,6)] (P/F,10%,1) = XX
Then Fw = XX (F/P,10%,7) = 4,544
Example (70) -Mar21&Jan22&Quiz P3.50-Summer22-Jan23
The equivalent annual worth in years 1 through 8 at an interest rate of 8% For the following
cash flow is closed to:
(a) $307
(b) $303
(c) $305
Solution
(d) $310
Find P0 then use the A/P factor for 8 years.
First P0 = [470(P/A,8%,6) – 50(P/G,8%,6)] (P/F,8%,1)
+ 470(P/F,8%,8)
= $1778.53
Then A1-8 =1778.53 (A/P,8%,8)
= $ 309.48
At an interest rate of 10%:
(a) $310
(b) $303
(c) $307
(d) $305
At an interest rate of 12%:
(a) $310
(b) $303
(c) $307
(d) $305
For help, Call: 01000-70-60-26
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Oct 2023
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Example (71) -Summer21
First P0 =100+100(P/F,12%,4)+100(P/F,12%,8)+200(P/F,12%,10)
+100(P/F,12%,12) +100(P/F,12%,16) +100(P/F,12%,20)
= 320.69
Then: Aw1-20 = 320.69 (A/P,12%,20) = $ 42.93
Example:(72) P3.60 -Jan23
For the cash flows shown, the equivalent annual worth in periods 1 through 5 at an interest
rate of 10% per year is closest to:
(a) $1120
(b) $1240
Solution
(c) $1350
(d)$1490
A1-5 = 1000(A/P,10%,5) + 1000 + 500(A/F,10%,5)
= 1000(0.26380) + 1000 + 500(0.16380)
= $1345.70
Example (73) Past Quiz
For the following cash flow diagram, determine the equivalent annual worth using an interest
rate of 10% per year.
Solution
First P0 = - [500 (P/A,10%,4) - 100(P/G,10%,4)]
+ 100(P/A,10%,2)( P/F,10%,4)]
= xx
Then A1-6 = xx (A/P,10%,7)
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Oct 2023
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Example :(74)
Using an interest rate of 12% per year, what uniform series over the five years (1-5) is
equivalent to the following cash flow:
EOY
Cash Flow
0
0
1
0
2
250
3
250
4
250
5
250
6
250
Solution
First P0th = 250(P/A,12%,5) (P/F,12%,1) = XXX
Then: A1-5 = XXX (A/P,12%,5) = XX
Example (75) -June21
First P0th = [150(P/A,10%,5)-25(P/G,10%,5)] (P/F,10%,1) = 360.98
Then: A1-5 = 360.98 (A/P,10%,5) = $ 92.23
Example:(76) P3.21
The company wants to have $360,000 available before it announces the product. If the
company sets aside $55,000 now and $90,000 in year 2, what uniform annual amount will it
have to put in an account in years 3 through 5 to have the $360,000? Assume the account
earns interest at 8% per year.
Solution
F5th =360000
0
1
2
3
4
5
A3-5 = ??
55,000
90,000
360,000 = 55,000(F/P,8%,5) + 90,000(F/P,8%,3) + A(F/A,8%,3)
360,000 = 55,000(1.4693) + 90,000(1.2597) + A(3.2464)
A3-5 = $51,076 per year
Example:(77) P3.46
For the cash flows shown, the value of X that will make the present worth in year 0 equal to
$5000 at an interest rate of 10% per year is closest to:
Solution
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MUST
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Solution
5000 = 200 +[ 300(P/A,10%,8) + 100(P/G,10%,8)] + x(P/F,10%,9)
5000 = 200 + 300(5.3349) + 100(16.0287) + x(0.4241)
0.4241x = 1596.66
x = 3764.82
Example:(78)
For the cash flow diagram shown, determine the value of W that will render the equivalent
future worth in year 8 equal to $-500 at an interest rate of 10% per year.
Solution
-500= -40(F/A,10%,4)(F/P,10%,4) - W(F/P,10%,3) - 40(F/A,10%,3)
-500 = -40(4.6410)(1.4641) - W(1.3310) - 40(3.3100)
W = $71.98
300 = 200(0.20541) + 200(2.4869) (0.20541) + x(0.6830)(0.20541) + 200(3.3100)(0.10541)
0.14030x = 300 – 213.03 then x = $619.88
Example:(79) P3.31
Use the cash flow diagram to determine the single amount of money Q 4 in year 4 that is
equivalent to all of the cash flows shown. Use i = 10% per year.
Solution
Q4 = 25(F/A,10%,6) + 25(P/F,10%,1) + 50(P/A/10%,3) (P/F,10%,1)
= $328.66
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