Solution Manual Engineering Economy Sullivan Wicks 17th Edition – Updated 2024

Complete Solution Manual With Answers

Sample Chapter Is Below

Solutions to Chapter 2 Problems

A Note To Instructors: Because of volatile energy prices in today’s world, the instructor is encouraged to vary energy

prices in affected problems (e.g. the price of a gallon of gasoline) plus and minus 50 percent and ask students to

determine whether this range of prices changes the recommendation in the problem. This should make for stimulating in-

class discussion of the results.

2-1 The total mileage driven would have to be specified (assumed) in addition to the variable cost of fuel per

unit (e.g. $ per gallon). Also, the fixed cost of both engine blocks would need to be assumed. The

efficiency of the traditional engine and the composite engine would also need to be specified

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-2 (a) 4 – sunk

(b) 5 – opportunity

(c) 3 – fixed

(d) 2 – variable

(e) 6 – incremental

(f) 1 – recurring

(g) 7 – direct

(h) 8 – nonrecurring

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-3 (a) # cows =

miles/year 000 , 000 ,1 miles/day) (15 days/year) 365 (

= 182.6 or 183 cows

Annual cost = (1,000,000 miles/year)($10 / 60 miles) = $166,667 per year

Annual cost of gasoline = 1,000,000 miles/year

30 miles/gallon × $4/gallon = $133,333 per year

It would cost $33,334 more per year to fuel the fleet of cars with gasoline.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-4 Cost Site A Site B

Rent Hauling = $5,000 (4)(200,000)($1.50) = $1,200,000 = $100,000

(3)(200,000)($1.50) = $900,000

Total $1,205,000 $1,000,000

Note that the revenue of $8.00/yd3 is independent of the site selected. Thus, we can maximize

profit by minimizing total cost. The solid waste site should be located in Site B.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-5 Present cost to company = $1,000,000 + 40 repairs × 24 hrs/repair × $2,500/hr = $3,400,000

With Ajax, the cost to your company = $2,000,000 + 40 repairs × R hrs/repair × $2,500/hr

Ajax is preferred if

2,000,000 + 100,000R ≤ $3,400,000

R ≤ 14 hours/breakdown.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-6 The $97 you spent on a passport is a sunk cost because you cannot get your money back. If you decide

to take a trip out of the U.S. at a later date, the passport’s cost becomes part of the fixed cost of making

the trip (just as the cost of new luggage would be).

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-7 If the value of the re-machining option ($60,000) is reasonably certain, this option should be chosen.

Even if the re-machined parts can be sold for only $45,001, this option is attractive. If management is

highly risk adverse (they can tolerate little or no risk), the second-hand market is the way to proceed to

guarantee $15,000 on the transaction.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-8 The certainty of making $200,000 – $120,000 = $80,000 net income is not particularly good. If your

friend keeps her present job, she is turning away from a risky $80,000 gain. This “opportunity cost” of

$80,000 balanced in favor of a sure $60,000 would indicate your friend is risk averse and does not want

to work hard as an independent consultant to make an extra $20,000 next year.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-9 (a) If you purchase a new car, you are turning away from a risky 20% per year return. If you are a risk

taker, your opportunity cost is 20%, otherwise; it is 6% per year.

(b) When you invest in the high tech company’s common stock, the next best return you’ve given up is

6% per year. This is your opportunity cost in situation (b).

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-10 Let X equal dollars per item delivered, and set total revenue equal to total cost:

$X (15 items/hr) = $42.00/hr + ($0.50/item)(15 items/hr)

X = ($49.50/hr) / (15 items/hr)

X = $3.30 per item

At least $3.30 per item delivered on Sunday will be needed to break even.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-11 (a) Students can use Equation (2-10) to determine D*

75−30

= 225 units/month

0.2

and

(b) Then

p = 75 – 0.1(225) = $52.50 per unit

TR = 225 units × $52.50/unit = $11,812.50

CT = $1,000 + $30/unit × 225 units = $7,750

Maximum profit = TR – CT = $11,812.50 − $7,750 = $4,062.50 per month

Using the quadratic equation to solve for the breakeven points:

45 ± √(−45)2 − 4(0.1)(1,000)

𝐷′

0.2

45 ± 40.3

𝐷′

0.2

′

𝐷1

= 24 (rounded up from 23.5)

′

𝐷2

= 426 (rounded down from 426.5)

Thus, the profitable range of demand is from 24 to 426 units per month.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-12 Re-write the price-demand equation as follows: p = 2,000 – 0.1D. Then,

TR = p  D = 2,000D – 0.1D2

The first derivative of TR with respect to D is

d(TR) / dD = 2,000 – 0.2D

This, set equal to zero, yields the D

that maximizes TR. Thus,

2,000 – 0.2 D

= 0

D

= 10,000 units per month

What is needed to determine maximum monthly profit is the fixed cost per month and the variable cost

per lash adjuster.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-13 p = 150  0.01D CF = $50,000 cv = $40/unit

Profit = 150D  0.01D2

 50,000  40D = 110D  0.01D2

 50,000

d(Profit)/dD = 110  0.02D = 0

D

= 5,500 units per year, which is less than maximum anticipated demand

At D = 5,500 units per year, Profit = $252,500 and p = $150  0.01(5,500) = $95/unit.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-14 (a) (b) D* = (a – Cv) / 2b = (700 – 131.50) / 0.10 = 568.5 / 0.10 = 5,685 103 board-feet

Price = $700 – (0.05)(5,685) = $415.75 per 103 board feet

Profit (per month) = pD – CF – CvD

= $415.75(5,685) – $1,000,000 – $131.50(5,685)

= 2,363,539 – 1,000,000 – 747,578

= $615,961/ month

–bD2 + (a – Cv)D – CF = 0

b= 0.05

(a – Cv) = 568.5

CF = 1,000,000

−0.05D2 + 568.5D – 1,000,000 = 0

D’1,2 = -568.5 ± [ (568.5)2

– 4(-0.05)(-1,000,000)]1/2

D1 D2 2 (-0.05)

= -568.5 ± [ 323,192.25 – 200,000]1/2

-0.10

= (-568.5 + 351) / (-0.10) = 2,175

= (-568.5 – 351) / (-0.10) = 9,195

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.D

+ 1700

2700

 

5000



2-15 (a) Profit = D

 

38 2  

-D

1000

–

5000

= 38D + 2700 –

-1000 – 40D

Profit = -2D – 5000

(Profit)

= -2 + 5000

D2 = 0

5000

or, D2

= 2500 and D*

= 50 units per month

(b) d

2 (Profit)

D2 = 10 000 ,

D3 < 0 for D > 1

Therefore, D* = 50 is a point of maximum profit.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-16 Profit = Total revenue – Total cost

= (15X – 0.2X2) – (12 + 0.3X + 0.27X2)

= 14.7X – 0.47X2 – 12

Profit

0.94X – 14.7 = 0 =

X = 15.64 megawatts

Profit

Note: profit maximizes megawatts 15.64 = X thus,0.94- =

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-17 (a) Using Equation (2-10),

180 − 40

𝐷∗

2(5)

= 14 units per week

(b) Total profit = −5(142) + (180 – 40)(14)

= −980 + 140(14)

= $980/wk

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-18 20,000 tons/yr. (2,000 pounds / ton) = 40,000,000 pounds per year of zinc are produced.

The variable cost per pound is $20,000,000 / 40,000,000 pounds = $0.50 per pound.

(a) Profit/yr = (40,000,000 pounds / year) ($1.00 – $0.50) – $17,000,000

= $20,000,000 – $17,000,000

= $3,000,000 per year

The mine is expected to be profitable.

(b) If only 17,000 tons (= 34,000,000 pounds) are produced, then

Profit/yr = (34,000,000 pounds/year)($1.00 – $0.50) – $17,000,000 = 0

Because Profit =0, 17,000 tons per year is the breakeven point production level for this mine. A loss

would occur for production levels < 17,000 tons/year and a profit for levels > 17,000 tons per year.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-19 (a) (b) (c) BE = $1,500,000 / ($39.95 − $20.00) = 75,188 customers per month

New BE point = $1,500,000 / ($49.95 − $25.00) = 60,120 per month

For 63,000 subscribers per month, profit equals

63,000 ($49.95 − $25.00) − $1,500,000 = $71,850 per month

This improves on the monthly loss experienced in part (a).

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-20 (a)  D = C

p – c

$2,000,000

($90 – $40) / unit = 40,000 units per year

$10,000,000

$9,000,000

$8,000,000

Profit

$6,000,000

Breakeven

Point

$6,000,000

$4,000,000

Loss

$2,000,000

Fixed Cost

D’ = 40,000 units

0 20,000 40,000 60,000 80,000 100,000

Number of Units

(b) Profit (Loss) = Total Revenue – Total Cost

(90% Capacity) = 90,000 ($90) – [$2,000,000 + 90,000 ($40)]

= $2,500,000 per year

(100% Capacity) = [90,000($90) + 10,000($70)] – [$2,000,000 + 100,000($40)]

= $2,800,000 per year

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-21 Annual savings are at least equal to ($60/lb)(600 lb) = $36,000. So the company can spend no more than

$36,000 (conservative) and still be economical. Other factors include ease of maintenance / cleaning,

passenger comfort and aesthetic appeal of the improvements. Yes, this proposal appears to have merit

so it should be supported.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-22 Jerry’s logic is correct if the AC system does not degrade in the next ten years (very unlikely). Because

the leak will probably get worse, two or more refrigerant re-charges per year may soon become

necessary. Jerry’s strategy could be to continue re-charging his AC system until two re-charges are

required in a single year. Then he should consider repairing the evaporator (and possibly other faulty

parts of his system).

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-23 Over 81,000 miles, the gasoline-only car will consume 2,700 gallons of fuel. The flex-fueled car will

use 3,000 gallons of E85. So we have

(3,000 gallons)(X) + $1,000 = (2,700 gallons)($3.89/gal)

and

X = $3.17 per gallon

This is 18.5% less expensive than gasoline. Can our farmers pull it off – maybe with government

subsidies?

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-24 (a) 𝐶𝑦 = 𝑐𝑣𝑋 Eqn. (2-8)

$2,000,000 = 𝑐𝑣(2000 𝑡𝑜𝑛𝑠)

𝑐𝑣 = $1,000/ton = $0.50/ lb.

Profit = [𝑠𝑝 – 𝑐𝑣]X – 𝐶𝐹

= [ $0.80/lb – $0.50/ lb] 4,000,000 lb – $700,000

= $500,000

(b) X’

= 𝐶𝐹 / (𝑠𝑝 – 𝑐𝑣) Eqn. (2-13)

= $700,000 / ($0.80/lb – $0.50/ lb)

= 2,333,333 lbs

𝑐𝐹 = 𝐶𝐹 / X’ = $700,000 / 2,333,333 lbs = $0.30 / lb

X’(𝑠𝑝 – 𝑐𝑣) = 𝑐𝐹 X’ at breakeven point

𝑐𝐹 = (𝑠𝑝 – 𝑐𝑣) = ($0.80/lb – $0.50/ lb) = $0.30/ lb

(c) 𝑐𝑇 = (X𝑐𝑣 + 𝐶𝐹) / X

= [ 4,000,000 lb ($0.50/lb) + $700,000] / 4,000,000 lb

= $0.675/ lb

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-25 (a) Ownership cost = $120 + $0.60X where X = horsepower rating.

Operating cost = $0.055

hp-hour (1

X) × 9000 hp-hour

Total annual cost = 120 + 0.60X + 0.055(9000)

= 120 + 0.6X + 495

dAC

dX = 0.6 – 495X-2 = 0

(b) X = 28.72 horsepower

For X to be a minimum of the function in part (a), show that d2AC

dX2 > 0.

d2AC

dX2 = 990X-3 which is greater than 0 for all positive values of X.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-26

+ C knv 2

$1,500n

= C +

coT

1,500

– 2kv = 0 =

kv =

750 –

750

= v

To find k, we know that

o = $100/mile at v = 12 miles/hr

o 2 2

= kv = k(12) = 100

and

so, v =

= 10.25 miles / hr

k = 100 / 144 = 0.6944

750

0.6944

The ship should be operated at an average velocity of 10.25 mph to minimize the total cost of operation

and perishable cargo.

Note: The second derivative of the cost model with respect to velocity is:

3,000 +1.388n =

The value of the second derivative will be greater than 0 for n > 0 and v > 0. Thus we have found a

minimum cost velocity.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-27 Solve for k: CG / n = kv v in miles/hr

1/18 mils/gal = k (70 miles/hr) ; k = 1 hr – gal / 1,260 mi2 = 0.000794 hr-gal/mi2

CG = (0.000794 hr-gal/mi2)(3000)(v)($3.60/gal) = 8.5752 hr-gal/mi2

CFSS = ($15,000/hr)(1/v)

Find CT

CT = ($8.5752 hr/mi2)(v mi/hr) + ($15,000/hr)(v

-1hr/mi)($/mi)

d CT / dv = 8.5752 −15,000/v

2 = 0

2 = 15,000/8.5752

* = 41.82 mi/hr

Check d2 CT / dv

2 = 30,000v

-3 which is positive for v > 0, therefore we have minimized the total cost

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-28 (293 kWh/106 Btu)($0.15/kWh) = $43.95/106 Btu

R11 R19 R30 R38

A. Investment cost $2,400 $3,600 $5,200 $6,400

B. Annual Heating Load (106 Btu/yr) 74 69.8 67.2 66.2

C. Cost of heat loss/yr $3,252 $3,068 $2,953 $2,909

D. Cost of heat loss over 25 years $81,308 $76,693 $73,836 $72,737

E. Total Life Cycle Cost = A + D $83,708 $80,293 $79,036 $79,137

Select R30 to minimize total life cycle cost.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-29 (a) d

C C C t = 0

  

 

2 R

or, 2 = CI/CRt

and, * = (CI/CRt)1/2; we are only interested in the positive root.

(b) d

C 2C for > 0

 

2 0



3  

Therefore, * results in a minimum life-cycle cost value.

(c) Investment cost versus total repair cost

CR··t



exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-30 180 rpm 1 cycle = (12 jacks/refurb.)(1 +4 refurb.) / (7 jacks/hr) = 60/7 = 8.57 hrs

Cost per cycle = 1 brush —- $90

4 refurb. — 4($30)

Oper. Cost – 8.57 hr. ($70/hr)

= $810/cycle

Cost per jack = $810/60 = $13.50/jack

240 rpm 1 cycle = (8 x 3) / 10 = 2.4 hrs

Cost per cycle = $90 + 2($30) + 2.4hr ($70/hr) = $318/cycle

Cost per jack = $318/24 = $13.25/jack

300 rpm 1 cycle = (6 x 2) / 12 = 1hr

Cost per cycle = $90 + 1($30) + 1hr ($70/hr) = $190/cycle

Cost per jack = $190/12 = $15.83/jack

Select 240 rpm

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-31 (a) (b) With Dynolube you will average (20 mpg)(1.01) = 20.2 miles per gallon (a 1% improvement).

Over 50,000 miles of driving, you will save

miles 50,000





of gallons 75 . 24

gasoline.

mpg 20

mpg 20.2

This will save (24.75 gallons)($4.00 per gallon) = $99.

Yes, the Dynolube is an economically sound choice.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-32 The cost of tires containing compressed air is ($200 / 50,000 miles) = $0.004 per mile. Similarly, the

cost of tires filled with 100% nitrogen is ($220 / 62,500 miles) = $0.00352 per mile. On the face of it,

this appears to be a good deal if the claims are all true (a big assumption). But recall that air is 78%

nitrogen, so this whole thing may be a gimmick to take advantage of a gullible public. At 200,000 miles

of driving, one original set of tires and three replacements would be needed for compressed-air tires.

One original set and two replacements (close enough) would be required for the 100% nitrogen-filled

tires. What other assumptions are being made?

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-33 (a)

Speed Drilling Rate

(ft/min)

Bit life at this Speed

(min)

A 2 10

B 3 6

C 4 3

Given: rock drill with 3 operating speeds

Cost= bit cost + operator cost (+ blasting penalty)

1 cycle = 96 ft. of drilling

Operator cost = $30/hr = $0.50/min

Bit cost = $10.00/each

Speed A – Cycle = 96ft / 2fpm = 48 min

Bit cost- 48 min/ 10 min/bit = 4.8 bits x $10 each = $48.00

Oper. Cost- 48 min x $0.50/min $24.00

$72.00/cycle

$0.75/ft

Cost/ft. = $72.00/cycle / 96 ft/cycle = Speed B – Cycle = 96ft / 3fpm = 32 min

Bit cost- 32 min/ 6 min/bit = 5.33 bits x $10 each = $53.33

Oper. Cost- 32 min x $0.50/min $16.00

$69.33/cycle

$0.72/ft

Cost/ft. = $69.33/cycle / 96 ft/cycle = Speed C – Cycle = 96ft / 4fpm = 24 min

Bit cost- 24 min/ 3 min/bit = 8 bits x $10 each = Oper. Cost- 24 min x $0.50/min $80.00

$12.00

$92.00/cycle

Cost/ft. = $92.00/cycle / 96 ft/cycle = $0.96/ft

Choose Speed B

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-33 (b) Penalty of $60.00/hour for cycle time greater than 30 minutes

$60/hr = $1.00/min

Speed A- Cycle Time = 48 min

Total cost/cycle = $72.00 + (48-30)$1.00 = $90.00

(includes penalty) ($0.9375/ft)

Speed B- Cycle Time = 32 min

Total cost/cycle = $69.33 + (32-30)$1.00 = $71.33

(includes penalty) ($0.743/ft)

Speed C- Cycle Time = 24 min (no penalty

Total cost/cycle = $92.00

($0.958/ft)

Speed B still has the lowest cost per cycle, now the cost per foot is $0.743/ft

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-34 (a) Manufacturing Option A

Labor = (40 x 5)($10) = $2000/wk.

Rental = $20,000

Σ = $22,000

Material = $15/unit

Purchase Option B = $20/unit

$22,000 + $15(x) = $20(x)

$22,000/5 = 𝑥̂

𝑥̂ = 4,400 units/wk = Breakeven amount

(b) $22,000 + $15(3500) [< or > ] $20 (3500)

$74,500 > $70,000

Purchase this item

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-35 Strategy: Select the design which minimizes total cost for 125,000 units/year (Rule 2). Ignore the sunk

costs because they do not affect the analysis of future costs.

(a) Design A

Total cost/125,000 units = (12 hrs/1,000 units)($18.60/hr)(125,000)

+ (5 hrs/1,000 units)($16.90/hr)(125,000)

= $38,463, or $0.3077/unit

Design B

Total cost/125,000 units = (7 hrs/1,000 units)($18.60/hr)(125,000)

+ (7 hrs/1,000 units)($16.90/hr)(125,000)

= $33,175, or $0.2654/unit

(b) Select Design B

Savings of Design B over Design A are:

Annual savings (125,000 units) = $38,463 − $33,175 = $5,288

Or, savings/unit = $0.3077 − $0.2654 = $0.0423/unit.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-36 Profit per day = Revenue per day – Cost per day

= (Production rate)(Production time)($30/part)[1-(% rejected+% tested)/100]

– (Production rate)(Production time)($4/part) – (Production time)($40/hr)

Process 1: Profit per day = (35 parts/hr)(4 hrs/day)($30/part)(1-0.2) –

(35 parts/hr)(4 hrs/day)($4/part) – (4 hrs/day)($40/hr)

= $2640/day

Process 2: Profit per day = (15 parts/hr)(7 hrs/day)($30/part) (1-0.09) –

(15 parts/hr)(7 hrs/day)($4/part) – (7 hrs/day)($40/hr)

= $2155.60/day

Process 1 should be chosen to maximize profit per day.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-37 At 70 mph your car gets 0.8 (30 mpg) = 24 mpg and at 80 mph it gets 0.6(30 mpg) = 18 mpg. The extra

cost of fuel at 80 mph is:

(400 miles/18mpg – 400 miles/24 mpg)($4.00 per gallon) = $22.22

The reduced time to make the trip at 80 mph is about 45 minutes. Is this a good tradeoff in your

opinion? What other factors are involved?

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-38 (a) (b) Operation 1 cycle time = 1 hr + 0.333 hr = 1.333 hr/cycle

Cycles/day = (8hr/day)(1 cycle/ 1.333 hr) = 6 cycle/day

Value added = (2000 parts/cycle)(6 cycles/day)($0.40/part)

= $4,800/day

Cost1 = 8 hr/day ($20/hr) = $160/day

Value – cost = $4,800 – $160 = $4,640/day

Operation 2 cycle time = 2 hr + 0.5 hr = 2.5 hr/cycle

Cycles/day = (8hr/day)(1 cycle/ 2.5 hr) = 3.2 cycle/day

Value added = (3500 parts/cycle)(3.2 cycles/day)($0.40/part)

= $4,480/day

Cost2 = 8 hr/day ($11/hr) = $88/day

Value – cost = $4,480 – $88 = $4,392/day

Select Operation 1 to maximize profit

Output/day for Operation 1 = 12,000 parts and output/.day for Operation 2 = 11,200 parts.

Downtime for Operation 1 = 6 x 20 min = 120 minutes/day and downtime for Operation 2 = 3.2

x 30 = 96 minutes/day. So increased production for Operation 1 is being traded off for increased

tool changing time (downtime), and the balance is favorable for Operation 1 compared to

Operation 2.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-39 Apache: (24 hr/day)(7 days/wk) – 4 = 164 hrs/wk uptime

(90 hits/min) (60 min/hour) = 5,400 hits/hr

(5,400 hits/hr) (164 hrs/wk) = 885,600 hits/wk @ $0.015/hit

= $13,284/wk

Profit/yr. = ($13,284/wk)(52 wk/yr) = $690,768

Windows IIS: (24 hr/day)(7 days/wk) – 0.75 = 167.25 hrs/wk uptime

(5,400 hits/hr) (167.25 hrs/wk) = 903,150 hits/wk @ $0.015/hit

= $13,547.25/wk

Profit/yr. = ($13,547.25/wk)(52 wk/yr) – $5,000 = $699,457

Go with Windows software.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-40 Option A (Purchase):

CT = (10,000 items)($8.50/item) = $85,000

Option B (Manufacture):

Direct Materials = $5.00/item

Direct Labor = $1.50/item

Overhead = $3.00/item

$9.50/item

CT = (10,000 items)($9.50/item) = $95,000

Choose Option A (Purchase Item).

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-41 The two alternatives are “no jig with a skilled machinist” and “use a jig with a lesser skilled machinist.”

No jig: (1.5 min/housing)/(60 min/hr)($25/hr)(4000 housings) = $2,500

With jig: (1 min/2 housings)/(60 min/hr)($15/hr)(4000 housings) + $500 = $1,000

Recommend using a jig to reduce production cost.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-42 Assumptions: You can sell all the metal that is recovered

Method 1: Recovered ore = (0.62)(100,000 tons) = 62,000 tons

Removal cost = (62,000 tons)($23/ton) = $1,426,000

Processing cost = (62,000 tons)($40/ton) = $2,480,000

Recovered metal = (300 lbs/ton)(62,000 tons) = 18,600,000 lbs

Revenues = (18,600,000 lbs)($0.8 / lb) = $14,880,000

Profit = Revenues – Cost = $14,880,000 – ($1,426,000 + $2,480,000)

= $10,974,000

Method 2: Recovered ore = (0.5)(100,000 tons) = 50,000 tons

Removal cost = (50,000 tons)($15/ton) = $750,000

Processing cost = (50,000 tons)($40/ton) = $2,000,000

Recovered metal = (300 lbs/ton)(50,000 tons) = 15,000,000 lbs

Revenues = (15,000,000 lbs)($0.8 / lb) = $12,000,000

Profit = Revenues – Cost = $12,000,000 – ($750,000 + $2,000,000)

= $9,250,000

Select Method 1 (62% recovered) to maximize total profit from the mine.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-43 Profit per ounce (Method A) = $1,750 – $550 / [(0.90 oz. per ton)(0.90)] = $1,750 – $679

= $1,071 per ounce

Profit per ounce (Method B) = $1,750 – $400 / [(0.9 oz. per ton)(0.60) =$1,750 – $741

= $1,009 per ounce

Therefore, by a slim margin we should recommend Method A.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-44 (a) False; (d) False; (g) False; (j) False; (m) True; (p) False; (s) False

(b) False; (e) True; (h) True; (k) True; (n) True; (q) True;

(c) True; (f) True; (i) True; (l) False; (o) True; (r) True;

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-45 (a)



coal lb



Btu 1,750,000

 

Btu 12,000

 

 

Loss



of lbs 486

coal

0.30

(b) 486 pounds of coal produces (486)(1.83) = 889 pounds of CO2 in a year.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-46 (a) Let X = breakeven point in miles

Fuel cost (car dealer option) = ($2.00/gal)(1 gal/20 miles) = $0.10/mile

Motor Pool Cost = Car Dealer Cost

 X$0.10/mi + $0.20/mi + ($30/day) days) (6 = X($0.36/mi)

$0.30X + 180 = $0.36X and X = 3,000 miles

(b) 6 days (100 miles/day) = 600 free miles

If the total driving distance is less than 600 miles, then the breakeven point equation is given by:

($0.36/mi)X = (6 days)($30 /day) + ($0.10/mi)X

X = 692.3 miles > 600 miles

This is outside of the range [0, 600], thus renting from State Tech Motor Pool is best for distances

less than 600 miles.

If driving more than 600 miles, then the breakeven point can be determined using the following

equation:

($0.36/mi)X = (6 days)($30 /day) + ($0.20/mi)(X – 600 mi) + ($0.10/mi)X

X = 1,000 miles The true breakeven point is 1000 miles.

(c) The car dealer was correct in stating that there is a breakeven point at 750 miles. If driving less

than 900 miles, the breakeven point is:

($0.34/mi)X = (6 days)($30 /day) + ($0.10/mi)X

X = 750 miles < 900 miles

However, if driving more than 900 miles, there is another breakeven point.

($0.34/mi)X = (6 days)($30/day) + ($0.28/mi)(X-900 mi) + ($0.10/mi)X

X = 1800 miles > 900 miles

The car dealer is correct, but only if the group travels in the range between 750 miles and 1,800

miles. Since the group is traveling more than 1,800 miles, it is better for them to rent from State

Tech Motor Pool.

This problem is unique in that there are two breakeven points. The following graph shows the two

points.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-46 continued

Car Dealer v. State Tech Motor Pool

Total Cost

$1,000

$900

$800

$700

$600

$500

$400

$300

$200

$100

X2‘ = 1,800 miles

X1‘ = 750 miles

Car Dealer

State Tech

0 500 1000 1500 2000 2500

Trip Mileage

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-47 This problem is location specific. We’ll assume the problem setting is in Tennessee. The eight years

($2,400 / $300) to recover the initial investment in the stove is expensive (i.e. excessive) by traditional

measures. But the annual cost savings could increase due to inflation. Taking pride in being “green” is

one factor that may affect the homeowner’s decision to purchase a corn-burning stove.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-48

A B C D E F G H I J K L M N O P

Fixed cost/

mo. = 73,000$

Demand Start

point (D) = 0

Net Income

Variable

cost/unit = 83$

Demand

Increment = 250

a = 180$

b = 0.02$

$40,000

Monthly

Demand

Price per

Unit

Total

Revenue $20,000

Total Expense Net income

0 180$ –

$ 73,000

$ (73,000)

250 175$ 43,750 $ 93,750

$ (50,000)

500 170$ 85,000 $ 114,500

$ (29,500)

750 165$ 123,750 $ 135,250

$ (11,500)

1000 160$ 160,000 $ 156,000

$ 4,000

Net Income

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

$(20,000)

$(40,000)

1250 155$ 193,750 $ 176,750

$ 17,000

1500 150$ 225,000 $ 197,500

$ 27,500

1750 145$ 253,750 $ $ 35,500

218,250

$(60,000)

2000 140$ 280,000 $ $ 41,000

239,000

2250 135$ 303,750 $ 259,750

$ 44,000

$(80,000)

2500 130$ 325,000 $ 280,500

$ 44,500

2750 125$ 343,750 $ $ 42,500

301,250

$(100,000)

3000 120$ 360,000 $ $ 38,000

322,000

Volume (Demand)

3250 115$ 373,750 $ $ 31,000

342,750

3500 110$ 385,000 $ 363,500

$ 21,500

3750 105$ 393,750 $ 384,250

$ 9,500

4000 100$ 400,000 $ 405,000

$ (5,000)

$600,000

4250 95$ 403,750 $ 425,750

$ (22,000)

4500 90$ 405,000 $ 446,500

$ (41,500)

4750 85$ 403,750 $ 467,250

$ (63,500)

$500,000

5000 80$ 400,000 $ 488,000

$ (88,000)

5250 75$ 393,750 $ 508,750

$ (115,000)

5500 70$ 385,000 $ 529,500

$ (144,500)

$400,000

Summary of impact of changes in cost components on optimum

demand and profitable range of demand.

Cash Flow

$300,000

Total Revenue

Total Expense

Percent Change

$200,000

CF cv D* D1

‘ D2

‘

-10% -10% 2,633 724 4541

$100,000

0% -10% 2,633 824 4443

10% -10% 2,633 928 4339

-10% 0% 2,425 816 4036

0% 0% 2,425 932 3918

10% 0% 2,425 1060 3790

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

-10% 10% 2,218 940 3495

0% 10% 2,218 1092 3343

Volume (Demand)

10% 10% 2,218 1268 3167

$600,000

$500,000

$400,000

$300,000

Cash Flow

Total Revenue

$200,000

Total Expense

Net income

$100,000

$(100,000)

$(200,000)

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

5500

Volume (Demand)

Reducing fixed costs has no impact on the optimum demand value, but does broaden the profitable range

of demand. Reducing variable costs increase the optimum demand value as well as the range of

profitable demand.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-49 New annual heating load = (230 days)(72 °F  46 °F) = 5,980 degree days. Now, 136.7  106 Btu are

lost with no insulation. The following U-factors were used in determining the new heating load for the

various insulation thicknesses.

U-factor Heating Load

R11 0.2940 101.3  106 Btu

R19 0.2773 95.5  106 Btu

R30 0.2670 92  106 Btu

R38 0.2630 90.6  106 Btu

$/kWhr $/106 Btu

Energy Cost $0.086 $25.20

R11 R19 R30 R38

Investment Cost 900 $ 1,350$ 1,950$ 2,400$

Annual Heating Load (106 Btu) 101.3 95.5 92 90.6

Cost of Heat Loss/yr $2,553 $2,406 $2,318 $2,283

Cost of Heat Loss over 25 years $63,814 $60,160 $57,955 $57,073

Total Life Cycle Cost $64,714 $61,510 $59,905 $59,473

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-50 In this problem we observe that “an ounce of prevention is worth a pound of cure.” The ounce of

prevention is the total annual cost of daylight use of headlights, and the pound of cure is postponement

of an auto accident because of continuous use of headlights. Clearly, we desire to postpone an accident

forever for a very small cost.

The key factors in the case study are the cost of an auto accident and the frequency of an auto accident.

By avoiding an accident, a driver “saves” its cost. In postponing an accident for as long as possible, the

“annual cost” of an accident is reduced, which is a good thing. So as the cost of an accident increases,

for example, a driver can afford to spend more money each year to prevent it from happening through

continuous use of headlights. Similarly, as the acceptable frequency of an accident is lowered, the total

annual cost of prevention (daytime use of headlights) can also decrease, perhaps by purchasing less

expensive headlights or driving less mileage each year.

Based on the assumptions given in the case study, the cost of fuel has a modest impact on the cost of

continuous use of headlights. The same can be said for fuel efficiency. If a vehicle gets only 15 miles

to the gallon of fuel, the total annual cost would increase by about 65%. This would then reduce the

acceptable value of an accident to “at least one accident being avoided during the next 16 years.” To

increase this value to a more acceptable level, we would need to reduce the cost of fuel, for instance.

Many other scenarios can be developed.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-51 Suppose my local car dealer tells me that it costs no more than $0.03 per gallon of fuel to drive with my

headlights on all the time. For the case study, this amounts to (500 gallons of fuel per year) x $0.03 per

gallon = $15 per year. So the cost effectiveness of continuous use of headlights is roughly six times

better than for the situation in the case study.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-52 p = 400 – D2

TR = p  D = (400 – D2) D = 400D – D3

TC = $1125 + $100  D

Total Profit / month = TR – TC = 400D – D3

– $1125 – $100D

= – D3 + 300D – 1125

= -3D2 + 300 = 0  D2 = 100  D* = 10 units

= -6D; at D = D*, 2

= – 60

Negative, therefore maximizes profit.

Select (a)

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-53 – D3 + 300D – 1125 = 0 for breakeven

At D = 15 units; -153 + 300(15) – 1125 = 0

Select (b)

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-54 CF = $100,000 + $20,000 = $120,000 per year

CV = $15 + $10 = $25 per unit

p = $40 per unit



= D

$120,000

units/yr 8,000 =

c – p

$25) – ($40

Select (c)

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-55 Profit = pD – (CF + CVD)

At D = 10,000 units/yr,

Profit/yr = (40)(10,000) – [120,000 + (25)(10,000)] = $30,000

Select (e)

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-56 Profit = pD – (CF + CVD)

60,000 = 35D – (120,000 + 25D)

180,000 = 10D; D = 18,000 units/yr

Select (d)

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-57 Annual profit/loss = Revenue – (Fixed costs + Variable costs)

= $300,000  [$200,000 + (0.60)($300,000)]

= $300,000  $380,000

= $80,000

Select (d)

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.2-58 Savings in first year = (7,900,000 chips) (0.01 min/chip) (1 hr/60 min) ($8/hr + 5.50/hr) = $17,775

Select (d)

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

solution manual Engineering Economy Sullivan Wicks Koelling 17th Edition

Solution Manual Engineering Economy Sullivan Wicks 17th Edition – Updated 2024

Complete Solution Manual With Answers

Sample Chapter Is Below

Related

Be the first to review “solution manual Engineering Economy Sullivan Wicks Koelling 17th Edition” Cancel reply

Solution Manual Foundations of Financial Management 18th Edition By Block Hirt Danielsen

Solution Manual Advanced Transport Phenomena Analysis, Modeling, and Computations Ramachandran 1st edition – Updated 2024

Solution Manual Essentials of Accounting for Governmental And Not-for-Profit Organizations 14th edition by Copley

Solution Manual Engineering Design 6th Edition by Dieter Schmidt

Need Help

Explore

Categories

solution manual Engineering Economy Sullivan Wicks Koelling 17th Edition

Share this:

Solution Manual Engineering Economy Sullivan Wicks 17th Edition – Updated 2024

Complete Solution Manual With Answers

Sample Chapter Is Below

Related

Be the first to review “solution manual Engineering Economy Sullivan Wicks Koelling 17th Edition” Cancel reply

Related products

Solution Manual Foundations of Financial Management 18th Edition By Block Hirt Danielsen

Solution Manual Advanced Transport Phenomena Analysis, Modeling, and Computations Ramachandran 1st edition – Updated 2024

Solution Manual Essentials of Accounting for Governmental And Not-for-Profit Organizations 14th edition by Copley

Solution Manual Engineering Design 6th Edition by Dieter Schmidt