Lecture 12

ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Reviewing… Net Present Worth: • Used to select among alternative projects. • Used to compare mutually exclusive alternatives. • If all expenses and revenues are included, select the largest NPW that is greater than zero. • If some or none of the revenues are included, select the largest NPW.

Reviewing… Salvage Value – – the amount of money you can expect to receive by selling an asset when you are done with it. What value does it have when you are done with it? MARR – Minimum Attractive Rate of Return– I expect or need this return in order to be willing to invest my money.

S.V. Atre

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ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Problem 2 Project #1 costs $10,000 and has annual, end of the year revenues of $10,000 over its 5 year life. There is no salvage value. Project #2 costs $20,000 and has annual end of year revenues of $10,000 over its 10 year life. There is no salvage value. Conduct an economic analysis to select the preferred project using a MARR of 15% per year, compounded annually.

Problem 2 GIVEN:

DIAGRAM:

LIFETIME 1ST

ANN. REV. 0

1

2

3

COST

ANN. REV. N ANN. COST SALVAGE MARR

PROJECT 1 PROJECT 2 5 YRS

10 YRS

$10,000

$20,000

$10,000

$10,000

-

-

-

-

15%/YR CPD. ANNUALLY

1st COST NPW1 = - $10,000 + $10,000(P|A,15%,5) NPW2 = - $20,000 + $10,000(P|A,15%,10) = - $10,000 + $10,000(3.3522)

= - $20,000 + $10,000(5.0188)

= - $10,000 + $33,522

= - $20,000 + $50,188

NPW1 = $23,522

NPW2 = $30,188

Worth investing in ???

S.V. Atre

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ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Calculating NPWs… NPW1 = $23,522

NPW2 = $30,188

Why is it wrong to select #2 based on this analysis??? Project life is not the same So… How do we handle comparing projects with unequal durations?

(Net) Present Worth Analysis One possible approaches when project lives are different: Common Multiple Period: Projects are assumed to be repeated until a common multiple point in time is established.

S.V. Atre

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ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Problem 2 Common Multiple Period: Projects are assumed to be repeated until a common multiple point in time is established:

Reinvest $10,000 in project again after Year 5 for 5 years (with similar revenue) NPW1’ = $NPW1 + NPW1(P|F,15%,5) = $23,522 + $23,522(0.4972) = $23,522 + $11,695 NPW1’ = $35,217 NPW2 = $30,188

Pick Project 1 (for similar life time)

Problem 3 A firm is considering the purchase of one of two new machines. The data on each are as below:

Machine

A

B

Service Life

3 years

6 years

Initial Cost

$3,400

$6,500

Annual Net Operating Expense:

$2,000

$1,800

Salvage Value

$100

$500

Use a MARR of 12% compounded annually and the lowest common multiple assumption to determine the alternative to be selected.

S.V. Atre

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ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Problem 3 GIVEN:

DIAGRAM: MACHINE A

MACHINE A MACHINE B

LIFETIME $100 0

1

2

3

4

$100 5

6

$2K CYCLE 2

3 YRS

COST

$3,400

$6,500

None Given

None Given

ANN. COST

$2,000 (NET EXP.)

$1,800 (NET EXP.)

$100

$500

SALVAGE

$3,400

6 YRS

ANN. REV. $2K

CYCLE 1 $3,400

1ST

MARR

12%/YR CPD. ANNUALLY

NPWA1 = - $3,400 - $2,000(P|A,12%,3) + $100(P|F,12%,3) = - $3,400 - $2,000(2.4018) + $100(0.7118) = - $3,400 - $4,804 + $71 NPWA1 = -$8,133

Must be a “have-to-do” decision

Problem 3 0

1

2

$100 3

4

$100 5

6

DIAGRAM: MACHINE A $2K CYCLE 1 $3,400

$2K CYCLE 2 $3,400

NPWA= NPWA1 + NPWA2 = - $8,133 - $8,133(P|F,12%,3) = - $8,133 - $8,133(0.7118) = - $8,133 - $5,789 v NPWA = -$13,922

S.V. Atre

Must be a “have-to-do” decision

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ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Problem 3 DIAGRAM: MACHINE A

NPWB = - $6,500 - $1,800(P|A,12%,6) + $500(P|F,12%,6) = - $6,500 - $1,800(4.1114) + $500(0.5066) = - $6,500 - $7,401 + $353

Pick Machine B if it is a “have-to-do” decision

v NPWB = -$13,548 v NPWA = -$13,922

Or pick neither

(Net) Present Worth Analysis When project lives are different: Common Multiple Period: Projects are assumed to be repeated until a common multiple point in time is established. OR Study Period: Select a study period for both projects and estimate cash flows to conform to the study period.

S.V. Atre

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ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Problem 4 Two alternatives are being considered regarding construction of a new high-voltage transmission line. Alternative I would build the transmission towers and the line at a capacity of 230 kVA, which is expected to be adequate for 15 years. After 15 years the 230 kVA lines would be removed and 560 kVA lines placed on the existing towers. Alternative II would build the transmission towers and the 560 kVA lines immediately. Given below are the pertinent data on the costs of these facilities. Expected Expected Item Present Cost Service Life Salvage Value Trans. Towers $15,000,000 55 years 0 after 30 yrs 230 kVA lines $8,000,000 15 years 10% of 1st cost 560 kVA lines $12,000,000 35 years 10% of 1st cost Salvage values for both transmission lines are 10% of first cost regardless of age at retirement. The cost of 560 kVA lines will inflate at the rate of 10% per year. The MARR is 15%. Use Present Worth analysis to determine which alternative is least expensive for a 35 year study period.

Problem 4 GIVEN:

230 kVA line

560 kVA line

LIFETIME

15(+20) YRS

35 YRS

$15,000,000 (for towers) + $8,000,000 (for line) = $23,000,000

$15,000,000 (for towers) + $12,000,000 (for line) = $27,000,000

ANN. REV.

None given

None given

ADDITIONAL COST

Purchase of new 560 kVA line at year 15

None given

SALVAGE

1) $800,000 from sale of first line after year 15: (10% ) 2) 10% of initial cost for 560 kVA line after year 35

(10% of initial cost) $1,200,000

1ST

COST

MARR

S.V. Atre

15 % APR

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ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Problem 4 For Option 1: Cost of 560 kVA inflates at 10% year (g) 1st cost of 560 kVA line at year 15: = $12,000,000(1+g)N = $12,000,000(1+0.1)15 = $50,126,978

Salvage value of 560 kVA line at year 35: 10% of $50,126,978 = $5,012,698

Problem 4 GIVEN:

230 kVA line

560 kVA line

LIFETIME

15(+20) YRS

35 YRS

$15,000,000 (for towers) + $8,000,000 (for line) = $23,000,000

$15,000,000 (for towers) + $12,000,000 (for line) = $27,000,000

ANN. REV.

None given

None given

ADDITIONAL COST

Purchase of new 560 kVA line at year 15: $50,126,978

None given

SALVAGE

1) $800,000 for 230 kVA line after year 15: (10% ) 2) 5,012,698 for 560 kVA line after year 35 (10%)

$1,200,000 (10% of initial cost)

1ST

COST

MARR

S.V. Atre

15 % APR

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ENGR 390 Lecture 12: Present Worth Analysis - Problems

Winter 2007

Problem 4

= $49,326,978

NPW1 = - $23,000,000 - $49,326,978 (P|F,15%,15) + $5,012698(P|F,15%,35) = - $23,000,000 - $49,326,978(0.1229) + $5,012,698(0.0075) = - $23,000,000 - $6,062,286 + $37,595 NPW1 = -$28,968,691

Problem 4

NPW2 = -$27,000,000 + 1,200,000(P|F,15%,35) = -$27,000,000 + $1,200,000(0.0075) = -$27,000,000 + $9,000 v NPW2 = -$26,991,000

Pick Alt. 2 if it is a “have-to-do” decision

v NPW1 = -$28,968,691

S.V. Atre

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