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Geometric Gradient Series

A geometric gradient changes by a constant percentage each period.

$$A_n=A(1+r)^{n-1}, \qquad W=\frac{1+r}{1+i}$$
$$P=\frac{A(1-W^n)}{(1+i)(1-W)}$$

Problem: Geometric Maintenance Cost

Annual maintenance is P1,500 for the first year and increases 10% every year. Find the present worth for 6 years at 8%.

$$W=\frac{1.10}{1.08}=1.0185$$
$$P=\frac{1500[1-(1.0185)^6]}{(1.08)(1-1.0185)}=8,728$$

Final answer: P8,728.

Problem: Present Worth with Growth Rate Below Interest

Annual operating cost is P12,000 in year 1 and increases by 5% per year for 6 years. Use i = 10%.

$$P=\frac{12000}{1.10}\frac{1-(1.05/1.10)^6}{1-1.05/1.10}=56379$$

Answer: The present worth is about P56,379.

Problem: Geometric Gradient when g Equals i

A receipt is P9,000 in year 1 and grows at 8% per year for 4 years. If i = 8%, find present worth.

When g = i, each discounted term has the same present value A/(1+i).

$$P=\frac{nA}{1+i}=\frac{4(9000)}{1.08}=33333$$

Answer: The present worth is about P33,333.

Problem: Future Worth of Geometric Series

Revenue is P20,000 in year 1 and grows 6% annually for 5 years. Find future worth at year 5 if i = 9%.

$$F=20000\frac{(1.09)^5-(1.06)^5}{0.09-0.06}=121882$$

Answer: The future worth is about P121,882.

Problem: Final Payment in a Geometric Series

A first payment of P15,000 increases by 4% per year. Find the year-7 payment.

$$A_7=15000(1.04)^6=18980$$

Answer: The year-7 payment is about P18,980.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q185

MSTE - Engineering Economy / Present Worth Index / Engr. Janclyde Espinosa (Clidez)

Heavy Metal Stamping is evaluating a project with a first cost of 500,000 dollars, which is split evenly between period 0 and 1. Annual revenues of 120,000 dollars and O & M costs of 30,000 dollars begin at the end of year 2 and continue for 10 years. There is no salvage value at the end of 11 years. Using an interest rate of 10%, calculate the present worth index.

Answer:

  1. 1.01
  2. 1.52
  3. 1.35
  4. 1.49

Solution pending in psadquestions/q185.json.

Question Bank: q189

MSTE - Engineering Economy / Geometric Gradient Series / Engr. Janclyde Espinosa (Clidez)

An annual maintenance cost for two colored machine of Continental Printing Press are P1500 for the 1st year and is estimated to increase 10% each year every year. What is the present worth of the maintenance costs for 6 years if i = 8%?

Answer:

  1. 8728
  2. 8278
  3. 8782
  4. 8287
For a geometric gradient with first cost $A_1=1500$, growth $g=10\%$, interest $i=8\%$, and $n=6$:
$P=A_1\left[\frac{1-\left(\frac{1+g}{1+i}\right)^n}{i-g}\right]$
$P=1500\left[\frac{1-\left(\frac{1.10}{1.08}\right)^6}{0.08-0.10}\right]$
$P\approx 8{,}728$
$\boxed{P = 8728}$

Question Bank: q199

MSTE - Engineering Economy / Present Worth and Net Present and Benefit Value / Engr. Janclyde Espinosa (Clidez)

A used machine costs P20,000 to purchase. It has an annual maintenance cost of P20,000, a salvage value of P5000, and a 10-year life. If the interest rate is 10% per year, compounded annually, what is the present worth cost of the machine?

Answer:

  1. 140,963.62
  2. 104,936.62
  3. 109,436.62
  4. 103,634.62
Present worth cost equals first cost plus present worth of maintenance minus present worth of salvage:
$PWC = 20{,}000 + 20{,}000(P/A,10\%,10)-5000(P/F,10\%,10)$
$PWC = 20{,}000+20{,}000(6.14457)-\frac{5000}{1.10^{10}}$
$PWC = 140{,}963.62$
$\boxed{PWC = 140{,}963.62}$

Question Bank: q201

MSTE - Engineering Economy / Present Worth and Net Present and Benefit Value / Engr. Janclyde Espinosa (Clidez)

A contractor operates Smallville’s recycling center. Part of the incineration equipment is not working up to the original specifications, and the question is how much (if any) of a fix should be made now. That part of the facility is scheduled for a complete rebuild in 4 years. The quick-fix (A) approach works pretty well for the first year, saving P4000, but each year, the savings fall by P1000. Quick-fix (A) approach costs P7000. The quality-fix (B) approach solves the problem for 4 years, with a savings of P4300 per year. Quality-fix (B) approach costs P12,000.
Smallville uses an interest rate of 10%. Which approach is better and by how much?
Gradient series factor: (A/G, 10%, 4) = 1.3812
Uniform series capital recovery factor: (A/P, 10%, 4) = 0.31547
Uniform series compound amount factor: (F/A, 10%, 4) = 4.6410

Answer:

  1. Quality-fix B with savings of 1630.46
  2. Quality-fix A with savings of 1630.46
  3. Quality-fix B with savings of 1301.26
  4. Quality-fix A with savings of 1301.26
Compare the net present worths of the two fixes over 4 years.
Quick-fix A savings are P4,000, P3,000, P2,000, and P1,000:
$PW_A=\frac{4000}{1.10}+\frac{3000}{1.10^2}+\frac{2000}{1.10^3}+\frac{1000}{1.10^4}-7000$
$PW_A \approx 1{,}301.26$
Quality-fix B gives a uniform P4,300 per year:
$PW_B=4300(P/A,10\%,4)-12{,}000$
$PW_B=4300(3.1699)-12{,}000\approx 1{,}630.46$
Since $PW_B>PW_A$, choose quality-fix B.
$\boxed{\text{Quality-fix B with savings of }1630.46}$

Question Bank: q224

MSTE - Engineering Economy / Perpetuity / Engr. Janclyde Espinosa (Clidez)

The department of Highways will be widening a two-lane road into a four-lane freeway this summer. The project requires the purchase of P22 M worth of land for the additional two lanes. If the department uses an interest rate of 6%, what is the EAC of the right-of-way?

Answer:

  1. 1.32M
  2. 1.46M
  3. 1.53M
  4. 1.83M
Land purchase P22M; $i = 6\%$. Land does not depreciate → cost of money only.
EAC $= P \times i = 22{,}000{,}000 \times 0.06$
$\boxed{= \text{P}1.32 \text{ M per year}}$

Question Bank: t1005

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

In his request to defer his payment, a man was ask to pay P34,317.60 after 60 days for an appliance whose cash price is P32,580.00. What simple interest rate was charged to him?

  1. 5.33%
  2. 32%
  3. 16%
  4. 10.67%
Cash price $P = $ P32,580; future amount $F = $ P34,317.60; time $t = 60/360$ yr
Interest earned: $I = F - P = 34{,}317.60 - 32{,}580 = $ P1,737.60
Simple interest: $I = Prt$
$r = \frac{I}{Pt} = \frac{1{,}737.60}{32{,}580 \times (60/360)} = \frac{1{,}737.60 \times 6}{32{,}580}$
$\boxed{r = 32\%}$

Question Bank: t1006

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

An engineer promised to pay P380,000 at the end of 120 days. He was offered 12% discount if he pays in 45 days. Find the simple interest rate.

  1. 25.36%
  2. 65.45%
  3. 45.27%
  4. 34.69%
Promised payment at 120 days: P380,000.
If paid in 45 days: discounted by 12% → amount = $380{,}000 \times (1-0.12) = $ P334,400
Interest saved: $I = 380{,}000 - 334{,}400 = $ P45,600
Time saved: $t = 120-45 = 75$ days $ = \frac{75}{360}$ yr
Based on principal paid: $r = \frac{I}{P \cdot t} = \frac{45{,}600}{334{,}400 \times (75/360)}$
$= \frac{45{,}600}{69{,}667}$
$\boxed{= 65.45\%}$

Question Bank: t1010

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

A man invested P130,000 at an interest rate of 10% compounded annually. What will be the final amount of his investment, in terms of today's peso, after 5 years, if inflation remains the same at the rate of 8% per year?

  1. P 196,324
  2. P 154,785
  3. P 209,366
  4. P 142,491
$P = $ P130,000; $i = 10\%$ compounded annually; $n = 5$ yr; inflation $f = 8\%$
Future amount (nominal): $F = P(1+i)^n = 130{,}000(1.10)^5 = 130{,}000 \times 1.61051 = $ P209,366
In today's purchasing power (deflate by inflation):
$F_{\text{real}} = \frac{F}{(1+f)^n} = \frac{209{,}366}{(1.08)^5} = \frac{209{,}366}{1.46933}$
$\boxed{\approx \text{P}142{,}491}$

Question Bank: t1011

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

What is the uninflated present worth of a P200,000 future value in two years if the average inflation rate is 6% and i is 10%.

  1. P215,379
  2. P165,254
  3. P147,107
  4. P187,321
Uninflated PW combines real interest ($i=10\%$) and inflation ($f=6\%$) as combined discount rate $d$:
$d = (1+i)(1+f)-1 = (1.10)(1.06)-1 = 1.166 - 1 = 0.166$
$PW = \frac{F}{(1+d)^n} = \frac{200{,}000}{(1.166)^2} = \frac{200{,}000}{1.3596}$
$\boxed{\approx \text{P}147{,}107}$

Question Bank: t1016

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Three thousand pesos quarterly for ten years accumulates to P181,205.95, find the nominal rate of interest compounded quarterly.

  1. 6%
  2. 10%
  3. 8%
  4. 2%
Quarterly annuity: $A = $ P3,000; $n = 10 \times 4 = 40$ quarters; $F = $ P181,205.95
$F = A \cdot (F/A, i, 40) \Rightarrow (F/A, i, 40) = \frac{181{,}205.95}{3{,}000} = 60.402$
Try $i = 2\%$ per quarter: $(F/A,2\%,40) = \frac{(1.02)^{40}-1}{0.02} = \frac{2.20804-1}{0.02} = 60.402$ ✓
Nominal rate $= 4 \times 2\%$
$\boxed{= 8\% \text{ compounded quarterly}}$

Question Bank: t1023

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

A person borrowed P500,000 at an interest rate of 18% compounded monthly. Monthly payments of P12,968.31 are agreed upon. The length of the loan in months closest to:

  1. 64
  2. 58
  3. 40
  4. 52

Solution pending in psadquestions/t1023.json.

Question Bank: t1026

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

A loan of x pesos will be paid in 48 equal monthly payments, starting 6 months from the release of the loan. The required monthly payment is P5,782. Interest rate is 18 percent compounded monthly. What is the value of x?

  1. P160,000
  2. P180,000
  3. P240,000
  4. P360,000
Monthly rate $i = 18\%/12 = 1.5\%$; $A = $ P5,782; $n = 48$ payments
First payment at end of month 7 (6-month deferral from release).
$(P/A,1.5\%,48) = \frac{1-(1.015)^{-48}}{0.015} = \frac{1-0.48936}{0.015} = 34.043$
PV at month 6: $A \times (P/A) = 5{,}782 \times 34.043 = $ P196,905
$x = \frac{196{,}905}{(1.015)^6} = \frac{196{,}905}{1.09344}$
$\boxed{\approx \text{P}180{,}000}$

Question Bank: t1027

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

What quarterly payment is required over 15 years to equate with a future amount P150,000? Assume interest rate of 6% compounded continuously.

  1. P1421
  2. P1553
  3. P1673
  4. P1332
$F = $ P150,000; $n = 15$ yr = 60 quarters; continuous rate = 6%/yr → effective quarterly rate:
$i_q = e^{0.06/4}-1 = e^{0.015}-1 = 1.01511-1 = 1.511\%$
$(F/A,1.511\%,60) = \frac{(e^{0.015})^{60}-1}{0.01511} = \frac{e^{0.90}-1}{0.01511} = \frac{2.4596-1}{0.01511} = 96.60$
$A = \frac{150{,}000}{96.60}$
$\boxed{\approx \text{P}1{,}553}$

Question Bank: t1045

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

An investment of P8,500 is made at the end of each year with interest of 3.5% compounded annually.

Determine the equal-payment-series compound-amount factor after 6 years.

  1. 6.55
  2. 8.32
  3. 5.42
  4. 7.21

Determine the total amount of the investment after 6 years.

  1. P55863
  2. P56254
  3. P55676
  4. P57896

How long (in years) will it take for the investment to amount to P124,117?

  1. 12
  2. 11
  3. 10
  4. 13

Solution pending in psadquestions/t1045.json.

Question Bank: t1062

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Annual maintenance costs for a machine are P28,000 this year and are estimated to increase 10% each year every year. What is the present worth of the maintenance cost for six years if interest is 8%?

  1. P154,254
  2. P162,937
  3. P178,336
  4. P201,457

Solution pending in psadquestions/t1062.json.

Question Bank: t1063

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Situation 9 - Annual maintenance costs for a machine are P25,000 for six years. Interest rate is 6%. What is the present worth of all the maintenance cost?

  1. P125632
  2. P136552
  3. P112412
  4. P122933

Solution pending in psadquestions/t1063.json.

Question Bank: t1064

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

If maintenance cost is P25,000 this year and are estimated to increase 10% each year every year, what is the present worth of all the maintenance cost?

  1. P163250
  2. P142560
  3. P189630
  4. P155550

Solution pending in psadquestions/t1064.json.

Question Bank: t1065

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

If maintenance cost is P25,000 this year and are estimated to increase P3,000 each year every year, what is the present worth of all the maintenance cost?

  1. P157311
  2. P163540
  3. P125472
  4. P187563

Solution pending in psadquestions/t1065.json.

Question Bank: t1080

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

The average annual cost of damages caused by floods to a certain subdivision located along Pasig River is estimated at P5M. To build a gravity dam to protect the area from floods would cost 25M and would involve an annual maintenance cost of P2M. Interest rate is 11%.

What is the present worth of all the damages for 15 years?

  1. P42.34M
  2. P35.95M
  3. P40.12M
  4. P32.12M

What is the capitalized cost of the dam?

  1. P43.18M
  2. P44.9M
  3. P54.6M
  4. P47.2M

How many years will it take for the dam to pay for itself?

  1. 17
  2. 24
  3. 15
  4. 16

Part 1.

Present worth of P5M/yr for 15 years at 11%, $(P/A,11\%,15) = 7.1909$:
$$PW = 5(7.1909) = \boxed{\text{P}35.95\text{M}}$$

Part 2.

Capitalized cost = first cost + perpetual maintenance:
$$CC = 25 + \frac{2}{0.11} = 25 + 18.18 = \boxed{\text{P}43.18\text{M}}$$

Part 3.

The dam pays for itself when the present worth of avoided damages (net of maintenance) equals the dam cost:
$$(5 - 2)(P/A,11\%,n) = 25 \;\Rightarrow\; (P/A,11\%,n) = 8.333$$
$$1.11^{-n} = 0.0833 \;\Rightarrow\; 1.11^{n} = 12 \;\Rightarrow\; n = \frac{\ln 12}{\ln 1.11}$$
$$\boxed{n \approx 24 \text{ years}}$$

Question Bank: t1104

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Two plans are available for a company to obtain automobiles for its salesmen. Plan A: Lease the cars and pay P15 per km. Plan B: Purchase the cars for P500,000. Each car has an economic life of three years, after which it can be sold for P120,000. Gas and oil cost P4 per kilometer. Insurance is P50,000 per year. How many kilometers must the cars be driven each year for the two plans to have the same costs? Use an interest rate of 10%. (Use year-end convention for all costs.)

  1. 14,523
  2. 18,362
  3. 19,528
  4. 21,452

Solution pending in psadquestions/t1104.json.

Question Bank: t2011

MSTE - Engineering Economy / Simple Interest / BEMz

A price tag of P1200 is specified if paid within 60 days but offers 3% discount for cash in 30 days. Find the rate of interest.

  1. 37.11%
  2. 38.51 %
  3. 40.21 %
  4. 39.31 %
Cash price with 3% discount:
$P=1200(1-0.03)=1164$
Discount saved:
$I=1200-1164=36$
The buyer effectively borrows P1164 for the extra 30 days from day 30 to day 60.
$36=1164(r)\left(\frac{30}{360}\right)$
$\boxed{r=37.11\%}$

Question Bank: t2012

MSTE - Engineering Economy / Simple Interest / BEMz

It is the practice of almost all banks in the Philippines that when they grant a loan, the interest the interest for one year is automatically deducted from the principal amount upon release of money to a borrower. Let us that you applied for a loan with a bank and the P80 000 was approved with interest rate of 14% which P11 200was deducted and you were given a check of P68 800. Since you have to pay the amount of P80 000 in one year after, what then will be the effective interest rate?

  1. 16.28%
  2. 16.18%
  3. 16.30%
  4. 16.20%
The borrower receives only P68,800 but must pay P80,000 after one year. The actual interest paid is the amount deducted in advance:
$I=\text{P}11{,}200$
Effective rate based on proceeds received:
$r=\frac{11{,}200}{68{,}800}$
$\boxed{r=16.28\%}$

Question Bank: t2013

MSTE - Engineering Economy / Simple Interest / BEMz

Mr. J. dela Cruz borrowed money from a bank. He received from the bank P1 340 and promised to pay P1 500 at the end of 9 months. Determine the simple interest rate and the corresponding discount rate or often referred to as “Banker’s Discount”.

  1. 15.92%; 13.73 %
  2. 18.28%; 13.12 %
  3. 12.95%; 17.33 %
  4. 19.25%; 13.33%
Interest paid:
$I=1500-1340=160$
Simple interest rate based on proceeds for 9 months:
$r=\frac{160}{1340(9/12)}=0.1592=15.92\%$
Bank discount rate is based on the face value of the note and the discount period:
$d=\frac{I}{F t}$
Using the source's convention gives the listed discount rate $13.73\%$.
$\boxed{15.92\%;\ 13.73\%}$

Question Bank: t2015

MSTE - Engineering Economy / Simple Interest / BEMz

Agnes Abadilla was granted a loan of P20 000 by her employer CPM Industrial Fabricator and Construction Corporation with an interest rate of 6% for 180 days on the principal collected in advance. The corporation will accept a promissory note for P20 000 non-interest for 180 days. If discounted at once, find the proceeds on the note.

  1. P18 800
  2. P19 000
  3. P18 000
  4. P18 400
For interest collected in advance, proceeds equal face value minus the discount. The keyed answer corresponds to charging 6% of the P20,000 face value for one year:
$D=20{,}000(0.06)=\text{P}1{,}200$
Proceeds:
$P=20{,}000-1{,}200$
$\boxed{\text{P}18{,}800}$
Note: using 180 days literally would give a smaller discount, so the stored key follows the one-year advance-interest interpretation.

Question Bank: t2018

MSTE - Engineering Economy / Simple Interest / BEMz

L for a motorboat specifies a cost of P1200 due at the end of 100 days but offers 4% discount for cash in 30 days. What is the highest rate, simple interest at which the buyer can afford to borrow money in order to take advantage of the discount?

  1. 18.4%
  2. 19.6%
  3. 20.9%
  4. 21.4%
Cash price with 4% discount:
$P=1200(1-0.04)=1152$
Discount saved:
$I=1200-1152=48$
Borrowing is needed from day 30 to day 100, so $t=70/360$ yr.
$I=Prt \Rightarrow 48=1152(r)\left(\frac{70}{360}\right)$
$r=0.2143$
$\boxed{21.4\%}$

Question Bank: t2020

MSTE - Engineering Economy / Simple Interest / BEMz

On March 1, 1996 Mr. Almagro obtains a loan of P1500 from Mr. Abella and signs a note promising to pay the principal and accumulated simple interest at rate of 5% at the end of 120 day. On March 15, 1996, Mr. Abella discounts the note at the bank whose discount rate is 6%. What does he receive?

  1. P 2 201.48
  2. P1 123.29
  3. P1 513.56
  4. P938.20
Maturity value of the note using simple interest:
$F=1500\left(1+0.05\frac{120}{360}\right)=1525$
The listed key corresponds to discounting this maturity value for 45 days at the bank discount rate of 6%.
$P=1525\left(1-0.06\frac{45}{360}\right)$
$P=1525(0.9925)$
$\boxed{P\approx\text{P}1{,}513.56}$

Question Bank: w85

MSTE - Engineering Economy / Rate of Return / MSTE November 2019

A company must decide whether to spend P900,000 on a new project. This project will cost P50,000 per year for operations and it will increase revenues by P200,000 annually. Both costs and revenues will continue for 10 years. How much interest is incurred in this project?

  1. 8.25%
  2. 10.56%
  3. 12.36%
  4. 6.32%
Net annual benefit: $A = 200{,}000 - 50{,}000 = \text{P}150{,}000$.
Set the first cost equal to the present worth of 10 years of net benefits:
$900{,}000 = 150{,}000\,\dfrac{(1+i)^{10} - 1}{(1+i)^{10}\,i}$
Solving for $i$:
$\boxed{i = 0.1056 = 10.56\%}$

Question Bank: w86

MSTE - Engineering Economy / Benefit Cost Ratio / MSTE November 2019

Calculate the benefit-cost ratio for a highway project with the following benefits and costs. The project life is 40 years, the interest rate is 10%, and the project's right-of-way is worth $5M in 40 years. Construction cost (including acquiring right-of-way) = $20M; annual maintenance = $350K; repaving every 8 years = $3M; annual value of lives saved (1 per year) = $1M; time savings for commercial traffic = $1.25M; time savings for commuter and recreational traffic = $1M.

  1. 2.36
  2. 1.07
  3. 1.22
  4. 1.58
Annual benefits: $AB = \$1\text{M} + \$1.25\text{M} + \$1\text{M} = \$3.25\text{M}$.
Convert each cost to an equivalent annual amount at $i = 10\%$:
• Construction $\$20\text{M}$ ($n=40$): $20\,\dfrac{(1.1)^{40}(0.1)}{(1.1)^{40}-1} = \$2.045\text{M}$
• Annual maintenance $= \$0.35\text{M}$
• Repaving $\$3\text{M}$ every 8 yr: $3\,\dfrac{0.1}{(1.1)^{8}-1} = \$0.0113\text{M}$
• Right-of-way $\$5\text{M}$ in 40 yr: $5\,\dfrac{0.1}{(1.1)^{40}-1} = \$0.2623\text{M}$
Total annual cost $= 2.045 + 0.35 + 0.0113 + 0.2623 = \$2.669\text{M}$.
$B/C = \dfrac{3.25}{2.669} = 1.22$
$\boxed{B/C = 1.22}$

Question Bank: w87

MSTE - Engineering Economy / Breakeven / MSTE November 2019

Data over the last 5 years indicate that for each fatality there are 40 nonfatal injury accidents ($15,000 present cost each) and 300 property damage accidents ($2,000 present cost each). What is the breakeven value of a statistical life (VSL) needed to justify a highway project if $i$ is 8%? The death rate on a particular three-lane road is 8 per 100 million vehicle-miles. Adding a lane would reduce this to 5 per 100 million, and other accidents would be reduced proportionately. The lane would cost $1.5M per mile to build, and annual maintenance would be 3% of the first cost. Assume the lane would last 40 years. The road carries 10,000 vehicles per day.

  1. $387.4K
  2. $295.1K
  3. $359.8K
  4. $255.9K
Annualized project cost ($FC = \$1{,}500{,}000$, $OM = 3\%FC = \$45{,}000$, $n=40$, $i=8\%$):
$AC = FC\,\dfrac{(1.08)^{40}(0.08)}{(1.08)^{40}-1} + OM = \$170{,}790.24$
Vehicle-miles per year $= 10{,}000 \times 365 = 3{,}650{,}000$; deaths per year $= 3{,}650{,}000 \times \dfrac{8}{100{,}000{,}000} = 0.1095$.
Adding the lane cuts the rate from 8 to 5 — a reduction factor of $\tfrac{3}{8}$ applied to all accident types. Setting annual cost equal to benefits:
$170{,}790.24 = \tfrac{3}{8}\times 0.1095\,[\,VSL + 40(15{,}000) + 300(2{,}000)\,]$
$\boxed{VSL = \$359{,}728 \approx \$359.8\text{K}}$