Economy EngineeringTEKNIK INDUSTRI

Nama Anggota Kelompok :

Anggina Vanessa ( 1106013624 )

Deo Gratias ( 1106069430 )

Hasna Afifah ( 1106069595 )

Nurman Wibisana ( 1106069563 )

Richard Sihombing ( 1106054473 )

Yonathan Elia Munthe ( 1106054372 )

8.18. A chemical company is considering two processes for isolating DNA

material .The incremental cash flow between the two alternatives, J and S, is

estimated. The company uses a marr of 50 % per year. The rate of return on the

incremental cash flow below is less than 50%, but the company CEO prefers the

more expensive process. The CEO believes she can negotiate the initial cost of

the more expensive process downward. By how much would she have to reduce

the first cost of S, the higher cost alternatives, for it to have an incremental rate

of return of exactly 50 %?

Year Incremental cash flow

0 $-900,000

1 400,000

2 700,000

3 950,000

Answer

1. Calculated incremental cash flowpada in year 0 when NPV=0 with i*=50%

0 = X + $400,000 ( P/F, 50%, 1) + $700,000 ( P/F, 50%, 2) + $950,000 ( P/F, 50%,3 )

0 = X + $400,000 ( P/F, 50%, 1) + $700,000 ( P/F, 50%, 2) + $950,000 ( P/F, 50%,3 )

0 = X + $859,245

X = $-859,245

Reduce the initial cost = -$900,000-(-$859,245)

= -$40,755

8.23.Two roadway designs are under consideration for access to a permanent

suspension bridge. Design 1A will cost $3 million to build and $100,000 per year

to maintain. Design 1B will cost $3.5 million to build and $80,000 per year to

maintain. Use an AW-based rate of return equation to determine which design is

preferred. The marr is 10% per year.

Answer :

1A 1B Incremental

Initial -$ 3 million -$ 3.5 million -$0,5 million

Maintenance /

Year

-$ 100.000 -$ 80.000 $ 20.000

MARR / year 10%

Total -$4.000.000 -$4.300.000 -$300.000

PW A = -$3 million - $100.000 / 10% = -$4.000.000

PW B = -$3.5 million - $80.000 / 10% = -$4.300.000

PW Incremental = -$0.5 million + $20.000 / 10% = -$300.000

The incremental cash flow of the larger investment don’t justify ( -300.000 ) so the cheaper one is Design A

8.28 A metal plating company is considering four different methods for recovering

by product heavy metals from a manufacturing site’s liquid waste. The

investment costs and incomes associated with each method have been

estimated. All methods have a 10 year-life. The marr is 12% per year.(a) if the

methods are independent, because they can be implemented at different plants,

which ones are acceptable? (b)if the methods are mutually exclusive, determine

which one method should be selected, using a ROR evaluation.

Methode First cost,$ Salvage Value Annual Income

A -15,000 +1,000 +4,000

B -18,000 +2,000 +5,000

C -25,000 -500 +6,000

D -35,000 -700 +8,000

Answer :

A) using PW method

Method A B C D0 -15000 -18000 -25000 -350001 4000 5000 6000 80002 4000 5000 6000 80003 4000 5000 6000 80004 4000 5000 6000 80005 4000 5000 6000 80006 4000 5000 6000 80007 4000 5000 6000 80008 4000 5000 6000 80009 4000 5000 6000 8000

10 1000 2000 -500 -700IRR ( i* ) 23% 24% 19% 17%

Method A:

0 = -15000 + 4000(P/A,i*,10) + 1000(p/f, i*,10)

Subtituting MARR (=12%) for i*, the RHS is positive (=7922.8),

i*=23%>MARR therefore select A

Method B :

0 = -18000 + 5000 (P/A, i*,10) + 2000(p/f,i*,10)

Subtituting MARR (=12%) for i*, the RHS is positive (=10895),

i*=24%>MARR therefore select B

Method C :

0 = -25000 + 6000(P/A,I*,10) – 500(P/F,i*,10)

Subtituting MARR(=12%) for i*, the RHS is positive (=8740.2),

i*=19%>MARR therefore select C

Method D :

0 = -35000 + 8000(P/A,i*,10) – 700(P/F,i*,10)

Subtituting MARR(=12%) for i*, the RHS is positive (=9976.2),

i*=17%>MARR therefore select D

Project A, B, C, D selected because their i*>MARR

B)

Year A B C DInitial Cost -15.000 -18.000 -25.000 -35.000salvage Value 1.000 2000 -500 -700Annual Income   4.000 5000 6000 8.000Incremental Comparison A to DN B to A C to B D to B

0 -15.000 -3.000 -7000 -170001 4.000 1000 1000 30002 4.000 1000 1000 30003 4.000 1000 1000 30004 4.000 1000 1000 30005 4.000 1000 1000 30006 4.000 1000 1000 30007 4.000 1000 1000 30008 4.000 1000 1000 30009 4.000 1000 1000 3000

10 1.000 1000 -2500 -2700incremental i* ( ∆i* ) 23% 31% -2% 9%incremental justified ? yes yes no noalternative selected A B B B

A-DN : It is already shown in Athat i* for method A >MARR. Therefore, A is

selected over DN. Alternatively, we can write (using PW method) :

0 = -15000 + 4000(P/A,i*,10) + 1000 (P/F,i*,10)

I*>MARR select A

B-A :

0 = -3000 + 1000(P/A,Di*,10) + 1000(P/F,Di*,10)

It can be shown Di* > 12% RHS is positive therefore select B

C-B :

0 = -7000 + 1000(P/A,Di*,10) -2500(P/F,Di*,10)

With Di* = 12% RHS is negative; Hence Di* < MARR(=12%) therefore select

B again

D-B :

0 = -17000 + 3000(P/A,Di*,10) – 2700(P/F,Di*,10)

Here,again Di* < MARR therefore B is still the selection. Select B

8.30 An independent dirt contractor is trying to determine which size dump truck to

buy. The contractor knows that as the bed size increases, the net income

increases, but he is uncertain whwther the oincremental expenditure required for

the larger trucks is justifie. The cash flow associated with each size truck are

estimated below. The contractor’s MARR is 18% per year, and all trucks are

expected to have a useful life of 8 years. (a) determine which size truck should

be purchased. (b) if two trucks are to be purchased, what should be the size of

the second truck?

Truck Bed Size, Cubic Meters Initial Investment, $

Annual Operating Cost, $

Slavage Value, $

Annual Income, $

8 -10.000 -4.000 2000 6.50010 -14.000 -5.500 2.500 10.00015 -18.000 -7.000 3.000 14.00020 -24.000 -11.000 3.500 20.50025 -33.000 -16.000 6.000 26.500

Answer :

Year 8 10 15 20 25Initial Cost   -10.000 -14.000 -18.000 -24.000 -33.000Annual Operating Cost   -4.000 -5.500 -7.000 -11.000 -16.000salvage Value 2.000 2.500 3.000 3.500 6.000Annual Income   6.500 10.000 14.000 20.500 26.500Annual Profit   2.500 4.500 7.000 9.500 10.500Incremental Comparison 8 to DN 10 to DN 15 to 10 20 to 15 25 to 20 0 -10.000 -14.000 -4.000 -6.000 -9.000

1 1.500 4.500 3.500 2.500 1.000 2 1.500 4.500 3.500 2.500 1.000

3 1.500 4.500 3.500 2.500 1.000 4 1.500 4.500 3.500 2.500 1.000

5 1.500 4.500 3.500 2.500 1.000 6 1.500 4.500 3.500 2.500 1.000

7 1.500 4.500 3.500 2.500 1.000 8 2.000 2500 500 500 2.500incremental i* ( ∆i* ) 5% 27% 86% 37% 1%incremental justified ?   no yes yes yes noalternative selected     10 15 20 20

A ) DN vs 8:0 = -10000(A/P,i,5) + (6500-4000) + 2000(A/F,i,5)

Solve for i by trial and error or excel

I= 5% < MARR

Eliminate 8

DN vs 10:0 = -14000(A/P,i,5) + (10000-5500) + 2500(A/F,i,5)

Solve for i by trial and error or excel

i = 27% > MARR

Eliminate DN

10 vs 15:0 = -4000(A/P,i,5) + (7000 – 4500) + 500(A/F,i,5)

Solve for i by trial and error or excel

I = 86%

Eliminate 10

15 vs 20:0 = -6000(A/P,i,5) + (9500-7000) + 500(A/F,i,5)

Solve for i by trial and error or excel

I = 37% > MARR

Eliminate 15

20 vs 25:0 = -9000(A/P,i,5) + (10.500-9500) + 2500(A/F,i,5)

Solve for i by trial and error or excel

I = 1% < MARR

Eliminate 25

Purchase 20 m3 truck

B) for second truck, purchase truck that was eliminated next to 15 m3

8.31 An engineer at anode metal is considering the projects below, all of which can

be considered to last indefinitely.if the company’s MARR is 13 % per year,

determine which should be selected (a) if they are independent and (b) if they

are mutually exclusive (c)assume the projects are mutually exclusive and that

the incorrect decision rule to select the one project that invests the most

money and has a ROR that exceeds the MARR has been applied. What

project was selected ? why is this an incorrect choice?

First Cost ($) Annual Income ($) Alternative rate of

return (%)

A -$20,000 $4,000 20%

B -$10,000 $2,000 20%

C -$15,000 $2,000 19,3%

D -$70,000 $10,000 14,3%

E -$50,000 $6,000 12%

Answer :

(a) If they are independent we must selected the choice who has ROR >

MARR alternative A, B, C, and D are selected

(b) (b) If they’re mutually exclusive

A B C D E

First Cost ($) -$20,000 -$10,000 -$15,000 -$70,000 -$50,000

Annual Income ($) $4,000 $2,000 $2,000 $10,000 $6,000

Alternative rate of return (%) 20% 20% 19,3% 14,3% 12%

Project E is eliminated due to its ROR which is less than MARR

Now compare B and C, C will the new challenge and B the defender.

∆ i¿fromC to B comparison is determined from

0=−15000− (−10000 )+ 2900∆ i¿

+ 2000∆ i¿

∆ i¿=18%

Since ∆ i¿>MARR so C will be the new defender.

Compare A to C is determined from

0=−20000− (−15000 )+ 4000∆ i¿

−2900∆ i¿

∆ i¿=12%

The ∆ i¿is less than MARR so D eliminated and A remains. So, A is selected.

c) The incorrect decision rule to select the one project that invest the most money and has a ROR that exceeds the MARR has been applied.

A B C D E

First Cost ($) -$20,000

-$10,000

-$15,000

-$70,000

-$50,000

Annual Income ($) $4,000 $2,000 $2,000 $10,000 $6,000

Alternative rate of return (%)

20% 20% 19,3% 14,3% 12%

Find the NPV

Project B

NPV=−10000+ 20000,13

NPV=5384,6

Project C

NPV=−15000+ 29000,13

NPV=7307,69

Project A

NPV=−20000+ 40000,13

NPV=−10769,2

Project D

NPV=−70000+ 100000,13

NPV=−6923,07

9.15 A state highway department is considering the conscruction of a new

highway through a scenic rural area.The road is expected to cost $6million, with

annual upkeep estimated at $20,000 per year. The improved accessibility is

expected to result in additional income from tourists of $350,000 per year. If the

road is expected to have a useful life of 25 years, use the (a) B-C method and

(b)B/C method at an interest rate of 6% per year to determine if the highway

should be constructed .(c) what is the modified B/C value?

Answer :

(a) B-C analysis

AW = C = 469,380 + 20,000 = $ 489,380

Annual benefits = B = $350,000

B-C = 350,000 – 489,380 = - $ 139,380

Since B-C <0, the highway shouldn’t be build.

(b) B/C analysis

AW = C = 469,380 + 20,000 = $ 489,380

Annual benefits = B = $350,000

B/C = 350,000/489,380 = 0.715

Since B/C < 1.0, the highway shouldn’t be build.

(c)For modified B/C ratio,

C = 6,000,000 (0.07823) = $ 469,380

B = 350,000 – 20,000 = $ 330,000

Modified B/C = 330,000/469,380

= 0.703

Since modified B/C < 1.0, the highway shouldn’t be build

9.18 Two routes are under consideration for a new interstate highway segment. The

long route would be 25 kilometers and would have an initial cost of $21 million.

The short transmountain route would span 10 kilometers and would have an

initial cost of $45 million. Maintenance costs are estimated at $40,000 per year

for the long route and $15,000 per year for the short route. Additionally, a major

overhaul and resurfacing will be required every 10 years at a cost of 10% of the

first cost of each route. Regardless of which route is se- lected, the volume of

traffic is expected to be 400,000 vehicles per year. If the vehicle operating

expense is assumed to be $0.35 per kilometer and the value of reduced travel

time for the short route is estimated at $900,000 per year, determjne which route

should be selected, using a conven- tionalBIC analysis. Assume an infirute life

for each road, an interest rate of 6% per year, and that one of the roads wiII be

built.

Answer :

Alternative Comparison

A B

Distance 25 km 10 km

Initial Cost $21 million $45 million

Maintenance

Cost/year

$40.000 $15.000

Overhaul Cost/10

year

10% Initial Cost 10% Initial Cost

Vehicle 400.000 unit 400.000 unit

Operating Vehicle/km $0.35 $0.35

Reduced Travel Time 0 $900.000

Cost

A. FW10 = 21

million

(F/P,6%,10) + 40.000 (F/A,6%,10) + 2.100.000

= 21 million (1.7908) + 40.000 (13.1808) + 2.100.000

= 37.606.800 + 527.232 + 2.100.000

= $40.234.032

B. FW10 = 45 million (F/P,6%,10) + 15.000 (F/A,6%,10) + 4.500.000

= 45 million (1.7908) + 15.000 (13.1808) + 4.500.000

= 80.586.000 + 197.712 + 4.500.000

= $85.283.712

∆C = FWb – FWa

= $85.283.712 - $40.234.032

= $45.049.680

Benefit

A. 0 + 400.000 (0.35)(25) = $3.500.000

B. $900.000 + 400.000 (0.35)(10) = $2.300.000

∆B = A-B = $3.500.000 - $2.300.000 = $1.200.000

B/C = $1.200.000/$45.049.680

= $0.02

Because B/C less than 1, so we choose option A (long road)

A BDistance 25 km 10 kmInitial Cost $21 million $45 millionMaintenance Cost/year $40.000 $15.000Overhaul Cost/10 year 10% Initial Cost 10% Initial CostVehicle 400.000 unit 400.000 unitOperating Vehicle/km $0.35 $0.35Reduced Travel Time 0 $900.000Total Benefit ( B ) 3.500.000 2.300.000Total Cost ( C ) 40.234.032 85.283.712Overall B/C 0,086991033 0,026968807Incremental B ( ∆B ) 1.200.000Incremental cost ( ∆C )   45.049.680∆B/C 0.02

9.20 A privately owned utility is considering two cash rebate programs to achieve

water conservation. Program 1, which is ex- pected to cost an average of $60 per

household, would involve a rebate of 75 % of the purchase and installation costs

of an ultralow-ftush toilet. This program is pro- jected to achieve a 5% reduction

in overall household water use over a 5-year evalua- tion period. This will benefit

the citizenry to the extent of $1 .25 per household per month. Program 2 would

involve grass re- placement with desert landscaping. This is expected to cost

$500 per household, but it will result in reduced water cost at an estimated $8 per

household per month (on average). At a discount rate of 0.5% per month , which

program , if either, should the utility undertake? Use the B/C method.

Answer :

First compare program 1 to do-nothing (DN).

Cost/household/mo = $60(A/P,0.5%,60)

= 60(0.01933)

= $1.16

B/C1 = 1.25/1.16

= 1.08, Eliminate DN

Compare program 2 to program 1.

Δcost = 500(A/P,0.5%,60) – 60(A/P,0.5%,60)

= (500 – 60)(0.01933)

= $8.51

Δbenefits = 8 – 1.25

= $6.75

Incr B/C2 = 6.75/8.51

= 0.79 Eliminate program 2

The utility should undertake program 1.

9.23 The California Forest Service is consider- ing two locations for a new state park.

Lo- cation E would require an investment of $3 million and $50,000 per year in

mainte- nance. Location W would cost $7 million to construct, but the Forest

Service would receive an additional $25,000 per year in park use fees. The

operating cost of loca- tion W will be $65,000 per year. The rev- enue to park

concessionaires will be $500,000 per year at location E and $700,000 per year at

location W. The disbenefits associated with each location are $30,000 per year

for location E and $40,000 per year for location W. Use (a) the B/C method and

(b) the modified B/C method to determine which location , if either, should be

selected, using an in- terest rate of 12% per year. Assume that the park will be

maintained indefinitely.

Answer :

E WFirst Cost $ 3.000.000 $7.000.000Annual / Year $ 50,000 $ 65,000Revenue / Year $ 500,000 $ 700,000Disbenefit $ 30.000 $ 40,000

(a) Location E

AW = C = 3,000,000(0.12) + 50,000

= $410,000

Revenue = B = $500,000 per year

Disbenefits = D = $30,000 per year

Location W

AW = C = 7,000,000 (0.12) + 65,000 - 25,000

= $880,000

Revenue = B = $700,000 per year

Disbenefits = D = $40,000 per year

B/C ratio for location E:

(B – D)/C = (500,000 – 30,000)/410,000

= 1.15

Location E is economically justified. Location W is now incrementally compared

to E.

Δcost of W = 880,000 – 410,000

= $470,000

Δbenefits of W = 700,000 – 500,000

= $200,000

Incr disbenefits of W = 40,000 – 30,000

= $10,000

Incr B/C = (B – D)/C = (200,000 – 10,000)/470,000

= 0.40

Since incr(B – D)/C < 1, W is not justified. Select location E.

(b) Location E

B = 500,000 – 30,000 – 50,000 = $420,000

C = 3,000,000 (0.12) = $360,000

Modified B/C = 420,000/360,000 = 1.17

Location E is justified.

Location W

ΔB = $200,000

ΔD = $10,000

ΔC = (7 million – 3 million)(0.12)

= $480,000

ΔM&O = (65,000 – 25,000) – 50,000

= $-10,000

Note that M&O is now an incremental cost advantage for W.

Modified ΔB/C = 200,000 – 10,000 + 10,000 = 0.42

480,000

W is not justified; select location E.

9.26 Public service company of Texas is considering four size of pipe for a new

pipeline. The costs per kilometer (km) for each size are given in the table.

Assuming that all pipes will last 15 years and the company’s MARR is 8% per

year, which size pipe should be purchased based on a B/C analysis? Installation

cost is considered a part opf the initial cost.

140 160 200 240

Initial

investment,$/km

9,180 10,510 13,180 15,850

Installation cost,

$/km

600 800 800 1,500

Annual cost,

$/km

6,000 5,800 5,200 4,900

Answer :

Size 140 160 200 240

Total first cost 9,180+600 =

9,780

10,510+800 =

11,310

13,180+800 =

14,580

15,850+1,500

= 17,350

140 160 200 240Initial investment,$/km 9,18 10,51 13,18 15,85Installation cost,$/km 600 800 800 1,5Annual cost, $/km 6.000 5.800 5.800 4.900Annual Total Cost ( C ) 1142,6 1321,347 1703,381 2027,001Benefit ( B ) 6.000 5.800 5.200 4.900Overall B/C 5,25119 4,389459 3,052751 2,417365Alternative Compared 160-140 200-160 240-200Incremental Benefits ( ∆B ) 200 600 300Incremental Cost ( ∆C ) 178,7499 382,0341 323,6191∆B/C 1,118882 1,57054 0,927016Incremental Justified yes yes no

Pipeline Selected 160 200 200

160 - 140 mm

DC = (11,310 – 9,780)(A/P,8%,15)

= 1,530(0.11683)

= $178.75

DB = 6,000 – 5,800

= $200

DB/C = 200/178.75

= 1.12 > 1.0 Eliminate 140 mm size.

200 - 160 mm

DC = (14,580 – 11,310)(A/P,8%,15)

= 3270(0.11683)

= $382.03

DB = 5800 – 5200

= $600

DB/C = 600/382.03

= 1.57 > 1.0 Eliminate 160 mm size.

240 - 200 mm

DC = (17,350 – 14,580)(A/P,8%,15)

= 2770(0.11683)

= $323.62

DB = 5200 – 4900

= $300

DB/C = 0.93 < 1.0 Eliminate 240 mm size.

So Select 200mm