ENGM 661 Engineering Economics

Replacement AnalysisReplacement Analysis

Replacement / Challenge

Example Car grows older and needs repairsat engine overhaul time should we fix or replace?

Replacement / Challenge

Example Car grows older and needs repairsat engine overhaul time should we fix or replace?

Note: sunk costs are unrecoverable

Example Just put $800 in car, engine needs overhaul, should we repair or replace?The $800 just invested has no bearing number is not part of analysis.

Example: Replacement

Chemical Plant owns filter press purchased 3 yearsago. Operating expense started at $4,000 per year2 years ago and has increased by $1,000 per year.The press could last 5 more years with an estimated salvage of $2,000 at that time. Current market value of the press is $9,000. A new press can be purchased for $36,000 with an estimated life of 10 years. Annual operating costs are 0 in year 1 growing by $1,000 per year.

Cash Flow Approach

End of Operating and End of Operating and SalvageYear, t Maintenance Costs Year, t Maintenance Costs Value, St

-2 -4,000 -2-1 -5,000 -10 -6,000 0 36,0001 -7,000 1 0 30,0002 -8,000 2 -1,000 24,6003 -9,000 3 -2,000 19,8004 -10,000 4 -3,000 15,6005 -11,000 5 -4,000 12,000

6 -5,000 9,0007 -6,000 6,6008 -7,000 4,8009 -8,000 3,60010 -9,000 3,000

Replacement (Cash Flow)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

Replacement (Cash Flow)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

0 1 2 3 4 5

4,000

12,000

Replace

36,000

9,000

1,000

Replacement (Cash Flow)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

NPW = -7,000 (P/A, 15,5) - 1,000 (P/G, 15,

5) + 2,000 (P/F,

15, 5)

= ($28,246)

Replacement (Cash Flow)

0 1 2 3 4 5

4,000

12,000

Replace

36,000

9,000

1,000

NPW = 9,000 - 36,000 -1,000 (P/G,

15,5) + 12,000 (P/F,

15, 5)

= ($26,809)

Replacement (Cash Flow)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

0 1 2 3 4 5

4,000

12,000

Replace

36,000

9,000

1,000

NPWK = ($28,246) NPWR = ($26,809)

Replacement (Cash Flow)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

0 1 2 3 4 5

4,000

12,000

Replace

36,000

9,000

1,000

NPWK = ($28,246) NPWR = ($26,809)

Choose Replace

Replacement (Cash Flow)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

0 1 2 3 4 5

4,000

12,000

Replace

36,000

9,000

1,000

NPWK = ($28,246) NPWR = ($26,809)

Note: NPWR - NPWK = $ 1,437

Replacement (Outsider View)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

9,000

0 1 2 3 4 5

4,000

12,000

Replace

36,000

1,000

Replacement (Outsider View)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

9,000

NPW = - 9,000 -7,000 (P/A, 15,5) - 1,000 (P/G, 15, 5) + 2,000 (P/F, 15,

5)

= ($37,246)

Replacement (Outsider View)

0 1 2 3 4 5

4,000

12,000

Replace

36,000

1,000

NPW = - 36,000 -1,000 (P/G,

15,5) + 12,000 (P/F,

15, 5)

= ($35,809)

Replacement (Outsider View)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

9,000

0 1 2 3 4 5

4,000

12,000

Replace

36,000

1,000

NPWK = ($37,246) NPWR = ($35,809)

Replacement (Outsider View)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

9,000

0 1 2 3 4 5

4,000

12,000

Replace

36,000

1,000

NPWK = ($37,246) NPWR = ($35,809)

Choose Replace

Replacement (Outsider View)

0 1 2 3 4 5

7,000

11,000

2,000

Keep

9,000

0 1 2 3 4 5

4,000

12,000

Replace

36,000

1,000

NPWK = ($37,246) NPWR = ($35,809)

Note: NPWR - NPWK = $ 1,437

With 10 year Horizon

Suppose we now consider a 10 year planninghorizon. We estimate that the old press willstill have a salvage value of $2,000 5 years fromnow but that the new press will only cost$31,000 5 years from now. Further, estimated salvage 5 years hence is $15,000.

Then:

With 10 Year Planning Horizon

0 0 9000 - 36000 = -270001 -7000 02 -8000 -10003 -9000 -20004 -10000 -30005 2000 -31000 -11000= -40000 -40006 0 -50007 -1000 -60008 -2000 -70009 -3000 -800010 15000 - 4000 = 11000 3000 - 9000 = -6000

NPWkeep = ($42,821.85)

NPWrep = ($43,237.92)

Keep old and replace at 5yr. point

Replacement (Cash Flow)

0 1 2 3 4 5 10

7,00011,000

2,000

Keep

31,000

1,000

4,000

15,000

NPW = -7,000(P/A, 15,5) - 1,000(P/G,15,5) -29,000(P/F,15,5) -1,000(P/A,15,5)(P/F,15,5)

+ 12,000(P/F,15,10)= ($42,821)

Replacement (Cash Flow)

0 1 2 3 4 5 . . . 10

4,000

Replace

36,000

9,000

1,000

9,000

3,000

. . . .

NPW = -27,000 - 1,000(P/G,15,10) + 3,000(P/F,15,10)

= ($43,237)

10-Year Horizon

0 1 2 3 4 5 . . . 10

4,000

Replace

36,000

9,000

1,000

9,000

3,000

. . . .

0 1 2 3 4 5 10

7,00011,000

2,000

Keep

31,000

1,000

4,000

15,000

NPWK = (42,821) NPWR = (43,237)

Choose Keep, trade in 5 years

Multiple Alternatives

Suppose Dealer offers a $10,000 trade-in. In addition, we identify 2 new alternatives:

3. New press for $40,000 with salvage after 5 years of $13,000. Trade-in on this machine is $12,000.

4. Lease a press for $7,500 per year during the 5 year horizon. Existing press will be sold on the open market.

Trade - In / Lease Options

TRADE - IN / LEASE OPTIONS

Keep Replace A Replace B LeaseEnd of Net Cash Flows Net Cash Flows Net Cash Flows Net Cash FlowsYear, t A1t A2t A3t A4t

0 0 10000-36000= -26,000 12000-40000= -28,000 9000-7500= 1,5001 -7,000 0 -500 -7,5002 -8,000 -1,000 -1,000 -8,3003 -9,000 -2,000 -1,500 -9,1004 -10,000 -3,000 -2,000 -9,9005 2000-11000= -9,000 12000-4000= 8,000 13000-2500= 10,500 -3,200

NPWK = (28,246) NPWRA = (25,809) NPWRB = (26,100) NPWL = (24,532)

Outsider Viewpoint Approach

Cash Flows for Several Replacement Alternatives - Outsider's Viewpoint Approach

End of Net Cash Flows Net Cash Flows Net Cash Flows Net Cash FlowsYear, t A1t A2t A3t A4t

0 -9,000 -35,000 -37,000 -7,5001 -7,000 0 -500 -7,5002 -8,000 -1,000 -1,000 (-7500-800)= -8,3003 -9,000 -2,000 -1,500 (-7500-1600)= -9,1004 -10,000 -3,000 -2,000 (-7500-2400)= -9,9005 2000-11000= -9000 12000-4000= 8000 13000-2500= 10500 -3,200

NPWK = (37,246) NPWRA = (34,809) NPWRB = (35,100) NPWL = (33,532)

Optimal Replacement

Suppose we have a compressor which costs $2,000 and has annual maintenance costs of $500 increasing by $100 per year. MARR=20%.

Then:

Optimal ReplacementOptimal Replacement

t 4 Yr 5 Yr 6 Yr 7 Yr 8 Yr 9 Yr 10 Yr0 2,000 2,000 2,000 2,000 2,000 2,000 2,0001 500 500 500 500 500 500 5002 600 600 600 600 600 600 6003 700 700 700 700 700 700 7004 800 800 800 800 800 800 8005 900 900 900 900 900 9006 1,000 1,000 1,000 1,000 1,0007 1,100 1,100 1,100 1,1008 1,200 1,200 1,2009 1,300 1,30010 1,400

NPV = 3,624 3,986 4,321 4,628 4,907 5,159 5,385

Optimal ReplacementOptimal Replacement

t 4 Yr 5 Yr 6 Yr 7 Yr 8 Yr 9 Yr 10 Yr0 2,000 2,000 2,000 2,000 2,000 2,000 2,0001 500 500 500 500 500 500 5002 600 600 600 600 600 600 6003 700 700 700 700 700 700 7004 800 800 800 800 800 800 8005 900 900 900 900 900 9006 1,000 1,000 1,000 1,000 1,0007 1,100 1,100 1,100 1,1008 1,200 1,200 1,2009 1,300 1,30010 1,400

NPV = 3,624 3,986 4,321 4,628 4,907 5,159 5,385EUAC = ($1,400) ($1,333) ($1,299) ($1,284) ($1,279) ($1,280) ($1,336)

=NPV(.2,C5:C13)+C4 = PMT(.2,4,C14)

Class Problem

The new president of Angstrom Technologies feels the company must use the newest and finest equipment in its labs. He has recommended that a 2-year-old piece of precision measurement equipment be replaced immediately. Besides, he feels it can be shown that his proposed equipment is economically advantageous at a 15%-per-year return and a planning horizon of 5 years. Perform the replacement analysis for a 5-year period.

Check Out Replacement Excel File

Class Problem

Current ProposedOriginal purchase price $30,000 $40,000Current market value 15,000 ...Estimated useful life, years 5 15Estimated value, 5 years $7,000 $10,000Salvage after 15 years ... 5,000Annual operating cost 5,000 3,000

Solution

7,000

5,000

0 1 2 3 4 5

40,000

15,00010,000

3,000

0 1 2 3 4 5

Keep Replace

EUAW = -5,000 + 7,000(A/F,15,5)

= ($3,962)

EUAW = -25,000(A/P,15,5) -3,000 + 10,000(A/F,15,5)

= ($8,975)