@LECTJULIESOLUTIONSSTUVIA
All 19 Chapters Covered
SOLUTIONS
,Table of Contents
1. Foundations of Engineering Economy
2. Factors: How Time and Interest Affect Money
3. Combining Factors and Spreadsheet Functions
4. Nominal and Effective Interest Rates
5. Analysis Using Present Worth and Future Worth Values
6. Annual Worth Analysis
7. Rate of Return Analysis: One Project
8. Rate of Return Analysis: Multiple Alternatives
9. Benefit/Cost Analysis and Public Sector Economics
10. Project Financing and Noneconomic Attributes
11. Replacement and Retention Decisions
12. Independent Projects with Budget Limitation
13. Breakeven and Payback Analysis
14. Effects of Inflation
15. Cost Estimation and Indirect Cost Allocation
16. Depreciation and Depletion Methods
17. After-Tax Economic Analysis
18. Sensitivity Analysis and Staged Decisions
19. Decision Making under Risk
, @LECTJULIESOLUTIONSSTUVIA
Chapter 1
Foundations of Engineering Economy
Basic Concepts
1.1 Financial units for economically best.
1.2 Morale, goodwill, dependability, acceptance, friendship, convenience, aesthetics, etc.
1.3 Measure of worth is a criterion used to select the economically best alternative. Some
measures are present worth, rate of return, payback period, benefit/cost ratio.
1.4 The color I like, best fuel rating, roomiest, safest, most stylish, fastest, etc.
1.5 Sustainability: Intangible; installation cost: tangible; transportation cost:
tangible; simplicity: intangible; taxes: tangible; resale value: tangible;
morale: intangible;
rate of return: tangible; dependability: intangible; inflation: tangible; acceptance by others: intangible;
ethics: intangible.
1.6 Examples are: house purchase; car purchase, credit card (which ones to use); personal loans
(and their rate of interest and repayment schedule); investment decisions of all types; when to
sell a house or car.
Ethics
1.7 This problem can be used as a discussion topic for a team-based exercise in class.
(a) Most obvious are the violations of Canons number 4 and 5. Unfaithfulness to the client and
deceptive acts are clearly present.
(b) The Code for Engineer’s is only partially useful to the owners in determining sound
bases since the contractor is not an engineer. Much of the language of the Code is
oriented toward representation, qualifications, etc., not specific acts of deceit and
fraudulent behavior. Code sections may be somewhat difficult to interpret in
construction of a house.
(c) Probably a better source would be a Code for Contractor’s or consulting with a real
estate attorney.
1.8 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b.
1.9 Example actions are:
Try to talk them out of doing it now, explaining it is stealing
Try to get them to pay for their drinks
, Pay for all the drinks himself
Walk away and not associate with them again
1.10 This is structured to be a discussion question; many responses are acceptable. Responses
can vary from the ethical (stating the truth and accepting the consequences) to unethical
(continuing to deceive himself and the instructor and devise some on-the-spot excuse).
Lessons can be learned from the experience. A few of them are:
Think before he cheats again.
Think about the longer-term consequences of unethical decisions.
Face ethical-dilemma situations honestly and make better decisions in real time.
Alternatively, Claude may learn nothing from the experience and continue his unethical
practices.
Interest Rate and Rate of Return
1.11 Extra amount received = 2865 - 25.80*100 = $285
Rate of return = 285/2580
= 0.110 (11%)
Total invested + fee 2865 + 50 = $2915
Amount required for 11% return = 2915*1.11
= $3235.65
1.12 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,000
(b) Interest = total amount paid – principal
= 1,936,000- 1,600,000
= $336,000
1.13 i = [(5,184,000 – 4,800,000)/4,800,000]*100% = 8% per year
1.14 Interest rate = interest paid/principal
= (312,000/2,600,000)
= 0.12 (12%)
1.15 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
1.16 Interest on loan = 45,800(0.10) = $4,580
Default insurance = $900
Set-up fee = 45,800(0.01) = 458
Total amount paid = 4,580 + 900 + 458 = $5938
, @LECTJULIESOLUTIONSSTUVIA
Effective interest rate = (5,938/45,800)*100 = 12.97%
Terms and Symbols
1.17 P = ?; F = 8*240,000 = $1,920,000; n = 2; i = 0.10
1.18 P = $20,000,000; A = ?; n = 6; i = 0.10
1.19 P = $2,400,000; A = $760,000: n = 5; i = ?
1.20 P = $1,500,000; F = $3,000,000: n = ?; i = 0.20
1.21 F = $250,000; A = ?: n = 3; i = 0.09
Cash Flows
1.22 Well drilling: outflow; maintenance: outflow; water sales: inflow; accounting: outflow;
government grants: inflow; issuance of bonds: inflow; energy cost: outflow; pension plan
contributions: outflow; heavy equipment purchases: outflow; used-equipment sales:
inflow; stormwater fees: inflow; discharge permit revenues: inflow.
1.23 Let Rev = Revenues; Exp = Expenses
Year 1 2 3 4 5 Total
Rev, $1000 521 685 650 804 929
Exp, $1000 610 623 599 815 789
NCF, $1000 -89 62 51 -11 140 153
Exp/Rev, % 117 91 92 101 85
(a) Total NCF = $153,000
(b) Last row of the table shows the answers
1.24 Month Receipts, $1000 Disbursements, $1000 NCF, $1000
Jan 300 500 -200
Feb 950 500 +450
Mar 200 400 -200
Apr 120 400 -280
May 600 500 +100
June 900 600 +300
July 800 300 +500
Aug 900 300 +600
Sept 900 200 +700
Oct 500 400 +100
Nov 400 400 0
Dec 1800 700 +1100
Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
@Seismicisolation
2@Seismicisolation
, +3,170
Net cash flow = $3,170 ($3,170,000)
1.25 End-of-period amount for March: 50 + 70 = $120; Interest = 120*0.03 = $3.60
End-of-period amount for June: 120 + 120 + 20 = $260; Interest = 260*0.03 = $7.80
End-of-period amount for September: 260 + 150 + 90 = $500; Interest = $15.00
End-of-period amount for Dec: 500 + 40 + 110 = $650; Interest = $19.50
1.26
1.27
Equivalence
1.28 (a) i = (5000-4275)/4275 = 0.17 (17%)
(c) Price one year later = 28,000 * 1.04 = $29,120
, @LECTJULIESOLUTIONSSTUVIA
(d) Price one year earlier = 28,000/1.04 = $26,923
(e) Jackson: Interest rate = (2750/20,000)*100
= 13.75%
Henri: Interest rate = (2295/15,000)*100
= 15.30%
(f) 81,000 = 75,000 + 75,000(i)
i = 6,000/75,000
= 0.08 (8%)
1.29 (a) Profit = 8,000,000*0.28
= $2,240,000
(b) Investment = 2,240,000/0.15
= $14,933,333
1.30 P + P(0.10) = 1,600,000
1.1P = 1,600,000
P = $1,454,545
1.31 Equivalent present amount = 1,000,000/(1 + 0.10)
= $909,091
Discount = 790,000 – 909,091
= $119,091
1.32 Total bonus next year = (this year’s bonus + interest) + next year’s bonus
= [4,000 + 4,000(0.10)] + 4,000
= $8,400
1.33 (a) Early-bird: 20,000 – 20,000(0.10) = $18,000
(b) Equivalent future amount = 18,000(1 + 0.06)
= $19,080
Savings = 20,000 - $19,080
= $920
Simple and Compound Interest
1.34 (a) F = P + Pni
1,000,000 = P + P(3)(0.20)
1.60P = 1,000,000
P = $625,000
Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
@Seismicisolation
2@Seismicisolation
, (b) P(1+i)(1+i)(1+i) = 1,000,000
P = 1,000,000/[(1+0.20)(1+0.20)(1+0.20)]
= $578,704
1.35 F = P + Pni
120,000 = P + P(3)(0.07)
1.21P = 120,000
P = $99,173.55
1.36
F = 240,000(1+ 0.12)3
= $337,183
1.37 (a) F = P + Pni
10,000 = 5000 + 5000(n)(0.12)
5000 = 600n
n = 8.33 years
(b) 10,000 = 5000 + 5000(n)(0.20)
n = 5 years
1.38 (a) Total due; compound interest = 150,000(1.05)(1.05)(1.05)
= $173,644
Total due; simple interest = P + Pni
= 150,000 + 150,000(3)(0.055)
= 150,000 + 24,750
= $174,750
Select the 5% compound interest rate
(b) Difference = 174,750 – 173,644
= $1106
1.39 90,000 = 60,000 + 60,000(5)(i)
300,000 i = 30,000
i = 0.10 (10% per year)
1.40 Simple: F = 10,000 + 10,000(3)(0.10)
= $13,000
Compound: 13,000 = 10,000(1 + i) (1 + i) (1 + i)
(1 + i)3 = 1.3000
3log(1 + i) = log 1.3
, @LECTJULIESOLUTIONSSTUVIA
3log (1 + i) = 0.1139
log(1 + i) = 0.03798
1 + i = 1.091
i = 9.1% per year
Spreadsheet function: = RATE(3,,-10000,13000) displays 9.14%
1.41 Follow plan 4, Example 1.16 as a model
A is 9900 – 2000 = $7900
B is 7900(0.10) = $790
C is 7900 + 790 = $8690
D is 8690 – 2000 = $6690
1.42 (a) Simple: F = P + Pni
2,800,000 = 2,000,000 + 2,000,000(4)(i)
i = 10% per year
(b) Compound: F = P(1 + i) (1 + i) (1 + i) (1 + i)
2,800,000 = 2,000,000(1 + i)4
(1 + i)4 = 1.4000
4
log(1 + i) = log1.400
4log(1 + i) = 0.146
log(1 + i) = 0.0365
(1 + i) = 100.0365
(1 + i) = 1.0877
i = 8.77%
(c) Spreadsheet function: = RATE(4,,-2000000,2800000)
MARR and Opportunity Cost
1.43 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.
1.44 WACC = 0.40(10%) + 0.60(16%) = 13.60%
1.45 WACC = 0.05(10%) + 0.95(19%) = 18.55%
The company should undertake the inventory, technology, warehouse, and maintenance
projects.
1.46 Let x = percentage of debt financing; Then, 1- x = percentage of equity financing
Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
@Seismicisolation
2@Seismicisolation
, 0.13 = x(0.28) + (1-x)(0.06)
0.22x = 0.07
x = 31.8%
Recommendation: debt-equity mix should be 31.8% debt and 68.2% equity financing
1.47 Money: The opportunity cost is the loss of the use of the $5000 plus the $100 interest.
Percentage: The 30% estimated return on the IT stock is the opportunity cost in percentage.
Exercises for Spreadsheets
1.48 (a) PV is P; (b) PMT is A; (c) NPER is n; (d) IRR is i; (e) FV is F; (f) RATE is i
1.49 (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 rate i
(d) IRR(first_cell:last_cell) finds the compound interest rate i
(e) PMT(i%,n,P,F) finds the equal periodic payment A
(f) NPER(i%,A,P,F) finds the number of periods n
1.50 (a) (1) F = ?; i = 8%; n = 10; A = $3000; P = $8000
(2) A = ?; i = 12%; n = 20; P = $-16,000; F = 0
(3) P = ?; i = 9%; n = 15; A = $1000; F = $600
(4) n = ?; i = 10%; A = $-290; P = 0; F = $12,000
(5) F = ?; i = 5%; n = 5; A = $500; P = $-2000
(b) (1) negative
(2) positive
(3) negative
(4) positive (years)
(5) can’t determine if 5% per year will cover the 5 withdrawals of $500
1.51 Spreadsheet shows relations only in cell reference format. Cell E10 will indicate $64 more
than cell C10.
All 19 Chapters Covered
SOLUTIONS
,Table of Contents
1. Foundations of Engineering Economy
2. Factors: How Time and Interest Affect Money
3. Combining Factors and Spreadsheet Functions
4. Nominal and Effective Interest Rates
5. Analysis Using Present Worth and Future Worth Values
6. Annual Worth Analysis
7. Rate of Return Analysis: One Project
8. Rate of Return Analysis: Multiple Alternatives
9. Benefit/Cost Analysis and Public Sector Economics
10. Project Financing and Noneconomic Attributes
11. Replacement and Retention Decisions
12. Independent Projects with Budget Limitation
13. Breakeven and Payback Analysis
14. Effects of Inflation
15. Cost Estimation and Indirect Cost Allocation
16. Depreciation and Depletion Methods
17. After-Tax Economic Analysis
18. Sensitivity Analysis and Staged Decisions
19. Decision Making under Risk
, @LECTJULIESOLUTIONSSTUVIA
Chapter 1
Foundations of Engineering Economy
Basic Concepts
1.1 Financial units for economically best.
1.2 Morale, goodwill, dependability, acceptance, friendship, convenience, aesthetics, etc.
1.3 Measure of worth is a criterion used to select the economically best alternative. Some
measures are present worth, rate of return, payback period, benefit/cost ratio.
1.4 The color I like, best fuel rating, roomiest, safest, most stylish, fastest, etc.
1.5 Sustainability: Intangible; installation cost: tangible; transportation cost:
tangible; simplicity: intangible; taxes: tangible; resale value: tangible;
morale: intangible;
rate of return: tangible; dependability: intangible; inflation: tangible; acceptance by others: intangible;
ethics: intangible.
1.6 Examples are: house purchase; car purchase, credit card (which ones to use); personal loans
(and their rate of interest and repayment schedule); investment decisions of all types; when to
sell a house or car.
Ethics
1.7 This problem can be used as a discussion topic for a team-based exercise in class.
(a) Most obvious are the violations of Canons number 4 and 5. Unfaithfulness to the client and
deceptive acts are clearly present.
(b) The Code for Engineer’s is only partially useful to the owners in determining sound
bases since the contractor is not an engineer. Much of the language of the Code is
oriented toward representation, qualifications, etc., not specific acts of deceit and
fraudulent behavior. Code sections may be somewhat difficult to interpret in
construction of a house.
(c) Probably a better source would be a Code for Contractor’s or consulting with a real
estate attorney.
1.8 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b.
1.9 Example actions are:
Try to talk them out of doing it now, explaining it is stealing
Try to get them to pay for their drinks
, Pay for all the drinks himself
Walk away and not associate with them again
1.10 This is structured to be a discussion question; many responses are acceptable. Responses
can vary from the ethical (stating the truth and accepting the consequences) to unethical
(continuing to deceive himself and the instructor and devise some on-the-spot excuse).
Lessons can be learned from the experience. A few of them are:
Think before he cheats again.
Think about the longer-term consequences of unethical decisions.
Face ethical-dilemma situations honestly and make better decisions in real time.
Alternatively, Claude may learn nothing from the experience and continue his unethical
practices.
Interest Rate and Rate of Return
1.11 Extra amount received = 2865 - 25.80*100 = $285
Rate of return = 285/2580
= 0.110 (11%)
Total invested + fee 2865 + 50 = $2915
Amount required for 11% return = 2915*1.11
= $3235.65
1.12 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,000
(b) Interest = total amount paid – principal
= 1,936,000- 1,600,000
= $336,000
1.13 i = [(5,184,000 – 4,800,000)/4,800,000]*100% = 8% per year
1.14 Interest rate = interest paid/principal
= (312,000/2,600,000)
= 0.12 (12%)
1.15 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
1.16 Interest on loan = 45,800(0.10) = $4,580
Default insurance = $900
Set-up fee = 45,800(0.01) = 458
Total amount paid = 4,580 + 900 + 458 = $5938
, @LECTJULIESOLUTIONSSTUVIA
Effective interest rate = (5,938/45,800)*100 = 12.97%
Terms and Symbols
1.17 P = ?; F = 8*240,000 = $1,920,000; n = 2; i = 0.10
1.18 P = $20,000,000; A = ?; n = 6; i = 0.10
1.19 P = $2,400,000; A = $760,000: n = 5; i = ?
1.20 P = $1,500,000; F = $3,000,000: n = ?; i = 0.20
1.21 F = $250,000; A = ?: n = 3; i = 0.09
Cash Flows
1.22 Well drilling: outflow; maintenance: outflow; water sales: inflow; accounting: outflow;
government grants: inflow; issuance of bonds: inflow; energy cost: outflow; pension plan
contributions: outflow; heavy equipment purchases: outflow; used-equipment sales:
inflow; stormwater fees: inflow; discharge permit revenues: inflow.
1.23 Let Rev = Revenues; Exp = Expenses
Year 1 2 3 4 5 Total
Rev, $1000 521 685 650 804 929
Exp, $1000 610 623 599 815 789
NCF, $1000 -89 62 51 -11 140 153
Exp/Rev, % 117 91 92 101 85
(a) Total NCF = $153,000
(b) Last row of the table shows the answers
1.24 Month Receipts, $1000 Disbursements, $1000 NCF, $1000
Jan 300 500 -200
Feb 950 500 +450
Mar 200 400 -200
Apr 120 400 -280
May 600 500 +100
June 900 600 +300
July 800 300 +500
Aug 900 300 +600
Sept 900 200 +700
Oct 500 400 +100
Nov 400 400 0
Dec 1800 700 +1100
Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
@Seismicisolation
2@Seismicisolation
, +3,170
Net cash flow = $3,170 ($3,170,000)
1.25 End-of-period amount for March: 50 + 70 = $120; Interest = 120*0.03 = $3.60
End-of-period amount for June: 120 + 120 + 20 = $260; Interest = 260*0.03 = $7.80
End-of-period amount for September: 260 + 150 + 90 = $500; Interest = $15.00
End-of-period amount for Dec: 500 + 40 + 110 = $650; Interest = $19.50
1.26
1.27
Equivalence
1.28 (a) i = (5000-4275)/4275 = 0.17 (17%)
(c) Price one year later = 28,000 * 1.04 = $29,120
, @LECTJULIESOLUTIONSSTUVIA
(d) Price one year earlier = 28,000/1.04 = $26,923
(e) Jackson: Interest rate = (2750/20,000)*100
= 13.75%
Henri: Interest rate = (2295/15,000)*100
= 15.30%
(f) 81,000 = 75,000 + 75,000(i)
i = 6,000/75,000
= 0.08 (8%)
1.29 (a) Profit = 8,000,000*0.28
= $2,240,000
(b) Investment = 2,240,000/0.15
= $14,933,333
1.30 P + P(0.10) = 1,600,000
1.1P = 1,600,000
P = $1,454,545
1.31 Equivalent present amount = 1,000,000/(1 + 0.10)
= $909,091
Discount = 790,000 – 909,091
= $119,091
1.32 Total bonus next year = (this year’s bonus + interest) + next year’s bonus
= [4,000 + 4,000(0.10)] + 4,000
= $8,400
1.33 (a) Early-bird: 20,000 – 20,000(0.10) = $18,000
(b) Equivalent future amount = 18,000(1 + 0.06)
= $19,080
Savings = 20,000 - $19,080
= $920
Simple and Compound Interest
1.34 (a) F = P + Pni
1,000,000 = P + P(3)(0.20)
1.60P = 1,000,000
P = $625,000
Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
@Seismicisolation
2@Seismicisolation
, (b) P(1+i)(1+i)(1+i) = 1,000,000
P = 1,000,000/[(1+0.20)(1+0.20)(1+0.20)]
= $578,704
1.35 F = P + Pni
120,000 = P + P(3)(0.07)
1.21P = 120,000
P = $99,173.55
1.36
F = 240,000(1+ 0.12)3
= $337,183
1.37 (a) F = P + Pni
10,000 = 5000 + 5000(n)(0.12)
5000 = 600n
n = 8.33 years
(b) 10,000 = 5000 + 5000(n)(0.20)
n = 5 years
1.38 (a) Total due; compound interest = 150,000(1.05)(1.05)(1.05)
= $173,644
Total due; simple interest = P + Pni
= 150,000 + 150,000(3)(0.055)
= 150,000 + 24,750
= $174,750
Select the 5% compound interest rate
(b) Difference = 174,750 – 173,644
= $1106
1.39 90,000 = 60,000 + 60,000(5)(i)
300,000 i = 30,000
i = 0.10 (10% per year)
1.40 Simple: F = 10,000 + 10,000(3)(0.10)
= $13,000
Compound: 13,000 = 10,000(1 + i) (1 + i) (1 + i)
(1 + i)3 = 1.3000
3log(1 + i) = log 1.3
, @LECTJULIESOLUTIONSSTUVIA
3log (1 + i) = 0.1139
log(1 + i) = 0.03798
1 + i = 1.091
i = 9.1% per year
Spreadsheet function: = RATE(3,,-10000,13000) displays 9.14%
1.41 Follow plan 4, Example 1.16 as a model
A is 9900 – 2000 = $7900
B is 7900(0.10) = $790
C is 7900 + 790 = $8690
D is 8690 – 2000 = $6690
1.42 (a) Simple: F = P + Pni
2,800,000 = 2,000,000 + 2,000,000(4)(i)
i = 10% per year
(b) Compound: F = P(1 + i) (1 + i) (1 + i) (1 + i)
2,800,000 = 2,000,000(1 + i)4
(1 + i)4 = 1.4000
4
log(1 + i) = log1.400
4log(1 + i) = 0.146
log(1 + i) = 0.0365
(1 + i) = 100.0365
(1 + i) = 1.0877
i = 8.77%
(c) Spreadsheet function: = RATE(4,,-2000000,2800000)
MARR and Opportunity Cost
1.43 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.
1.44 WACC = 0.40(10%) + 0.60(16%) = 13.60%
1.45 WACC = 0.05(10%) + 0.95(19%) = 18.55%
The company should undertake the inventory, technology, warehouse, and maintenance
projects.
1.46 Let x = percentage of debt financing; Then, 1- x = percentage of equity financing
Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
@Seismicisolation
2@Seismicisolation
, 0.13 = x(0.28) + (1-x)(0.06)
0.22x = 0.07
x = 31.8%
Recommendation: debt-equity mix should be 31.8% debt and 68.2% equity financing
1.47 Money: The opportunity cost is the loss of the use of the $5000 plus the $100 interest.
Percentage: The 30% estimated return on the IT stock is the opportunity cost in percentage.
Exercises for Spreadsheets
1.48 (a) PV is P; (b) PMT is A; (c) NPER is n; (d) IRR is i; (e) FV is F; (f) RATE is i
1.49 (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 rate i
(d) IRR(first_cell:last_cell) finds the compound interest rate i
(e) PMT(i%,n,P,F) finds the equal periodic payment A
(f) NPER(i%,A,P,F) finds the number of periods n
1.50 (a) (1) F = ?; i = 8%; n = 10; A = $3000; P = $8000
(2) A = ?; i = 12%; n = 20; P = $-16,000; F = 0
(3) P = ?; i = 9%; n = 15; A = $1000; F = $600
(4) n = ?; i = 10%; A = $-290; P = 0; F = $12,000
(5) F = ?; i = 5%; n = 5; A = $500; P = $-2000
(b) (1) negative
(2) positive
(3) negative
(4) positive (years)
(5) can’t determine if 5% per year will cover the 5 withdrawals of $500
1.51 Spreadsheet shows relations only in cell reference format. Cell E10 will indicate $64 more
than cell C10.
