Solution manual engineering economic analysis 9th edition ch05 present worth analysis

Parts b and c assume earlier payments, hence their PW of Cost is greater... Two assumptions are needed:1 Value of an urn of cherry blossoms plus the cost to have the bank administer the

Chapter 5: Present Worth Analysis5-1

P = {[$400 - $100 (A/G, 8%, 4) + $900 (A/F, 8%, 4)]/0.08 + $1,000}

{P/F, 8%, 5}

= {[$400 - $100 (1.404) + $900 (0.2219)]/0.08 + $1,000} {0.6806}

= $4,588

Alternative Solution: An alternate solution may be appropriate if one assumes that the

$1,000 cash flow is a repeating annuity from time 13 to infinity (rather than indicating the repeating decreasing gradient series cycles)

In this case P is calculated as:

$1,000

Determine the cash flow:

Year Cash Flow

= $220 (0.9434) + $1,320 (0.8900) + $1,980 (0.8396) + $1,540 (0.7921) - $4,400

= -$135.41NPW is negative Do not purchase equipment

For end-of-year disbursements,

PW of wage increases = ($0.40 x 8 hrs x 250 days) (P/A, 8%, 10)

+ ($0.25 x 8 hrs x 250 days) (P/G, 8%, 10)

= $800 (6.710) + $500 (25.977)

= $18,356This $18,356 is the increased justifiable cost of the equipment

Dec 31,1999Jan 1, 2000

NPW

12/31/99

= $122,400(c) PW of Cost = F Σ (n= 1 to 6) [(er – 1)/(rern)]

= $35,500 [((e0.18 – 1)/(0.18e(0.18)(1))) + ((e0.18 –

1)/(0.18e(0.18)(2)) + ]

= $35,000 [ (0.1972/0.2155) + (0.1972/0.2580) +

(0.1972/0.3089) + (0.1972/0.3698) + (0.1972/0.4427) + (0.1972/0.5300)]

= $33,500 (3.6686)

= $122,897(d) Part (a) assumes end-of-year payments Parts (b) and (c) assume earlier payments, hence their PW of Cost is greater

P

$15

0

$300

$450

$600

$750

A = $100,000 (A/P, 3%, 40) = $100,000 (0.0433) = $4,330

P = $4,330 (P/A, 3%, 20) = $4,330 (14.877) = $64,417(b) Service Charge = 0.05 P

Amount of new loan = 1.05 ($64,417) = $67,638

Quarterly payment on new loan = $67,638 (A/P, 2%, 80)

= $67,638 (0.0252)

= $1,704Difference in quarterly payments = $4,330 - $1,704 = $2,626

Since NPW is positive, the process should be automated

PW Receipts = ($550,000) (90) (P/A, 9%, 10) + ($50,000) (90) (P/G, 9%,

10) + ($1,000,000) (90) (P/A, 9%, 70) (P/F, 9%, 10)

= $849,000,000NPW = $849,000,000 - $811,000,000 = $38,000,000

This project meets the 9% minimum rate of return as NPW is positive

(b) Other considerations:

Engineering feasibility

Ability to finance the project

Effect on trade with Brazil

Military/national security considerations

P = the first cost = $980,000

F = the salvage value = $20,000

AB = the annual benefit = $200,000

Remember our convention of the costs being negative and the benefits being positive Also,

remember the P occurs at time = 0.

NPW = - P + AB (P/A, 12%, 13) + F (P/F, 12%, 13)

= -$980,000 + $200,000 (6.424) + $20,000 (0.2292)

= $309,384Therefore, purchase the machine, as NPW is positive

Two assumptions are needed:

1) Value of an urn of cherry blossoms (plus the cost to have the bank administer the trust) – say $50.00 / year

2) A “conservative” interest rate—say 5%

…

Compute an A that is equivalent to $100,000 at the end of 10 years.

$100,000

……

P

The trust fund has three components:

(1) P = $1 million

(2) For n = ∞ P= A/i = $150,000/0.06 = $2.5 million

(3) $100,000 every 4 years: First compute equivalent A Solving one portion of the

perpetual series for A:

A = $100,000 (A/F, 6%, 4) = $100,000 (0.2286)

= $22,860

P = A/i = $22,860/0.06 = $381,000Required money in trust fund

= $1 million + $2.5 million + $381,000 = $3.881 million

By buying the “lifetime” muffler the car owner will avoid paying $50 two years hence Compute how much he is willing to pay now to avoid the future $50 disbursement

P = $50 (P/F, 20%, 2) = $50 (0.6944)= $34.72

Since the lifetime muffler costs an additional $15, it appears to be the desirable alternative

5-48

Compute the PW of Cost for a 25-year analysis period

Note that in both cases the annual maintenance is $100,000 per year after 25 years Thus after 25 years all costs are identical

Single Stage Construction

Three One-Year Subscriptions

The revenues are common; the objective is to minimize cost

(a) Present Worth of Cost for Option 1:

PW of Cost = $200,000 + $15,000 (P/A, 10%, 30)

= $341, 400Present Worth of Cost for Option 2:

PW of Cost = $150,000 + $150,000 (P/F, 10%, 10) + $10,000 (P/A, 10%,

30) + $10,000 (P/A, 10%, 20) (P/F, 10%, 10)

$10,000 (8.514) (0.3855)

= $334,900Select option 2 because it has a smaller Present Worth of Cost

(b) The cost for option 1 will not change The cost for option 2 will now be higher

PW of Cost = $150,000 + $150,000 (P/F, 10%, 5) + $10,000 (P/A, 10%,

30) + $10,000 (P/A, 10%, 25) (P/F, 10%, 5)

= $394,300Therefore, the answer will change to option 1

+ $30,000 (P/F, 9%, 10)

= -$80,000 - $1,000 (6.418) + $12,000 (6.418) + $30,00O (0.4224)

= +$3,270(a) Buy Model B because it has a positive NPW

(b) The NPW of Model A is negative; therefore, it is better to do nothing or look for more alternatives

5-55

Machine A

NPW = - First Cost + Annual Benefit (P/A, 12%, 5) – Maintenance &

Operating Costs (P/A, 12%, 5) + Salvage Value (P/F, 12%, 5)

= -$250,000 + $89,000 (3.605) - $4,000 (3.605) + $15,000 (0.5674)

= $64,936

Machine B

NPW = - First Cost + Annual Benefit (P/A, 12%, 5) – Maintenance &

Operating Costs (P/A, 12%, 5) + Salvage Value (P/F, 12%, 5)

Natural Gas $30,000 $7,500 > Fuel Oil

Fuel Oil $55,000

For fixed output, minimize PW of Cost:

Natural Gas

PW of Cost = $30,000 + $7,500 (P/A, 8%, 20) + PW of Fuel Oil Cost

= $30,000 + $7,500 (9.818) + PW of Fuel Oil Cost

= $103,635 + PW of Fuel Oil Cost

Fuel Oil

PW of Cost = $55,000 + PW of Fuel Oil Cost

Coal

PW of Cost = $180,000 - $15,000 (P/A, 8%, 20) + PW of Fuel Oil Cost

= $180,000 - $15,000 (9.818) + PW of Fuel Oil Cost

= $32,730 + PW of Fuel Oil CostInstall the coal-fired steam boiler

Machine A

NPW4 = -$52,000 + ($38,000 - $15,000)(P/A, 12%, 4) + $13,000(P/F, 12%, 4) = -$52,000 + $69,851 + $8,262

= $93,497

Machine C is the correct choice

5-61

It appears that there are four alternative plans for the ties:

1) Use treated ties initially and as the replacement

PW of Cost = $6 + $4 (P/F, 8%, 10) – $0.50 (P/F, 8%, 15)

= $6 + $4 (0.4632) - $0.50 (0.3152)

= $7.703) Use untreated ties initially Replace with treated ties

PW of Cost = $4.50 + $5.50 (P/F, 8%, 6) - $0.50 (P/F, 8%, 15)

= $4.50 + $5.50 (0.6302) - $0.50 (0.3152)

= $7.814) Use untreated ties initially, then two replacements with untreated ties

= $8.45Choose Alternative 1 to minimize cost.

5-62

This is a situation of Fixed Input Therefore, maximize PW of benefits By inspection, one can see that C, with its greater benefits, is preferred over A and B Similarly, E is preferred over D The problem is reduced to choosing between C and E

5-63

Compute the Present Worth of Benefit for each share

From the 10% interest table: (P/A, 10%, 4) = 3.170

(P/F, 10%, 4) = 0.683

PW of Future Price

PW of Dividends

PW of BenefitWestern House $32 x 0.683 + 1.25 x 3.170 = 21.86 + 3.96 = $25.82

Fine Foods $45 x 0.683 + 4.50 x 3.170 = 30.74 +

14.26

= $45.00Mobile Motors $42 x 0.683 + 0 x 3.170 = 28.69 + 0 = $28.69

shares of the various shares It would buy 83 shares of Spartan Products, but only 42 shares of Western House The criterion, therefore, is to maximize NPW for the amount invested This could be stated as Maximize NPW per $1 invested

Buy Spartan Products

Eight mutually exclusive alternatives:

Plan Initial Cost Net Annual

Benefit x (P/A, 10%, 10) 6.145

PW of Benefit

NPW = PW of Benefit minus Cost

An additional $360 now instead of n annual payments of $15 each Compute n

P = A (P/A, 4%, n)

$360 = $15 (P/A, 4%, n)

(P/A, 4%, n) = $360/$15

= 24

From the 4% interest table, n = 82

Lifetime (patron) membership is not economically sound unless one expects to be active for

82 + 1 = 83 years (But that’s probably not why people buy patron memberships or avoid buying them.)

5-67

Cap CostA = $500,000 + $35,000/0.12 + [$350,000(A/F, 12%, 10)]/0.12

= $500,000 + $35,000/0.12 + [$350,000 (0.0570)]/.12 = $957,920

Cap CostB = $700,000 + $25,000/0.12 + [$450,000 (A/F, 12%, 15)]/0.12

= $700,000 + $25,000/0.12 + [$450,000 (0.0268)]/0.12 = $1,008,830

Type A with its smaller capitalized cost is preferred

5-68

Full Capacity Tunnel

Capitalized Cost = $556,000 + ($40,000 (A/F, 7%, 10))/0.07

Lifetime Membership

Capitalized Cost = $402,000 + [($32,000 (0.0724))/0.07] + [$2,000/0.07]

= $463,700

Second Half-Capacity Tunnel

20 years hence the capitalized cost of the second half-capacity tunnel equals the present capitalized cost of the first half

Capitalized Cost = $463,700 (P/F, 7%, 20)

= $463,700 (0.2584)

= $119,800Capitalized Cost for two half-capacity tunnels = $463,700 + $119,800

= $583,500Build the full capacity tunnel

PW of Cost of 30 years of Itis

NPW = $6,000 (P/A, 10%, 12) + $1,000 (P/G, 10%, 12) - $10,000 – ($10,000

- $1,000) [(P/F, 10%, 2) + (P/F, 10%, 4) + (P/F, 10%, 6) + (P/F, 10%, 8) + (P/F, 10%, 10)]

= -$51.41NPW3 = $100 (P/A, 8%, 10) - $700 + $180 (P/F, 8%, 10)

= +$54.38NPW4 = $0

$2,000

$12,000

$3,000

$5,000

(b) 12% interest

NPW1 = $135 (P/A, 12%, 10) - $500 - $500 (P/F, 12%, 5)

= -$20.95NPW2 = ($100 + $250) (P/A, 12%, 10) - $600 - $350 (P/F, 12%, 5)

= -$153.09NPW3 = $100 (P/A, 12%, 10) - $700 + $180 (P/F, 12%, 10)

= -$77.04NPW4 = $0

Problem 5-78 will repay the largest loan because the payments start at the end of the first month, rather than waiting until the end of the year

Problem 5-80 has the same effective interest rate as 5-77, but the rate on 5-79 is lower