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Replacement Analysis

Replacement analysis compares the economic cost of keeping existing equipment against buying a replacement.

Problem: CE Board May 2019

A standby machine costs $3,000, has economic life 10 years, salvage $600, and annual operating cost $100. Without it, shutdown cost is $350 per year. MARR is 10%. Should it be purchased?

EUAC of shutdown is $350. EUAC of standby machine is $550.59.

Final answer: do not purchase because standby EUAC is greater.

Problem: Replace Existing Machine

A new machine costs $16,000 installed, has 8-year life, $3,000 salvage, and $1,000 annual expenses. Existing machine can be sold for $5,000, has 8 more years, $2,000 salvage, and $1,800 annual operating cost. At MARR 15%, replace?

Present worth cost of new machine is $19,507. Present worth cost of existing machine is $12,423.

Final answer: do not replace the existing machine.

Problem: Comparative Use Value

An old machine can be sold now for $15,000. If kept another year, salvage drops to $13,000 and operating expense is $30,000. A new machine costs $50,000 with 5-year life, $20,000 salvage, and operating costs starting at $18,000 increasing $1,000 per year. MARR is 20%. Should it be replaced?

EUAC of the new machine is $33,671.89. Solving for comparative use value of the old machine gives $13,893.24.

Since comparative use value is less than selling price $15,000, replace the machine.

Problem: Defender Annual Cost

A defender has current market value P150,000, annual operating cost P70,000, and expected salvage P60,000 after 3 years. Use i = 10%.

$$EAC=150000(A/P,10\%,3)-60000(A/F,10\%,3)+70000$$
$$EAC=150000(0.40211)-60000(0.30211)+70000=112190$$

Answer: Defender EAC is about P112,190.

Problem: Defender vs Challenger

A defender has EAC P112,190. A challenger has EAC P105,800 for the same service. Which is economical?

Choose the lower annual cost when service is equivalent.

Answer: Replace with the challenger.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q226

MSTE - Engineering Economy / Replacement Analysis / Engr. Janclyde Espinosa (Clidez)

A plant is considering buying a second-hand machine to use as standby equipment. The machine costs 3000 dollars and has an economic life of 10 years, at which time its salvage value is 600 dollars; expected annual operating cost is 100 dollars. Without a stand-by machine, the plant woud have to shut down an average of seven days a year at a cost of 50 dollars per day. If the MARR is 10% is it expedient to buy the stand-by machine?

Answer:

  1. No, because 550.59>350
  2. Yes, because 550.59>350
  3. Yes, because 350>243.56
  4. No, because 350>243.56
Annual cost of machine: $(A/P,10\%,10) = \frac{0.10(1.10)^{10}}{(1.10)^{10}-1} = 0.16275$
CR = $(3000-600) \times 0.16275 + 600 \times 0.10 = 390.60 + 60 = \$450.60$
Total annual cost = \$450.60 + \$100 = \$550.59
Annual benefit (avoiding shutdown) = 7 days × \$50/day = \$350
Since annual cost (\$550.59) > annual benefit (\$350), it is not expedient to buy.
$\boxed{\text{No — annual cost \$550.59 exceeds benefit \$350}}$

Question Bank: q227

MSTE - Engineering Economy / Replacement Analysis / Engr. Janclyde Espinosa (Clidez)

XYZ Company is considering replacing a machine. The new improved machine will cost P16,000 installed; it will have an estimated service life of 8 years and P3000 salvage value. It is estimated that operating expenses will average P1000 a year. The present machine was purchased for P20,000 four years ago and is estimated to have 8 more years of service life, at the end of which its salvage value will be P2000. Operating costs are P1800 per year. If replaced now, it can presumably be sold for P5000. Using MARR of 15%, determine whether to replace the existing machine.

Answer:

  1. Do not replace for the present worth of the existing machine is 12423
  2. Replace because the present worth of the existing machine is 12423
  3. Do not replace because the present worth of the new machine is 19507
  4. Replace because the present worth of the existing machine is 18980.71
MARR = 15%; $(P/A,15\%,8) = 4.4873$; $(P/F,15\%,8) = 0.32690$
New machine PW: $16{,}000 + 1{,}000(4.4873) - 3{,}000(0.32690) = 16{,}000 + 4{,}487 - 981 = \text{P}19{,}507$
Old machine PW (sell now for P5,000 → defender cost):
$5{,}000 + 1{,}800(4.4873) - 2{,}000(0.32690) = 5{,}000 + 8{,}077 - 654 = \text{P}12{,}423$
PW$_{\text{old}}$ (P12,423) $<$ PW$_{\text{new}}$ (P19,507) → keep old machine.
$\boxed{\text{Do not replace; PW of existing machine} = \text{P}12{,}423}$

Question Bank: q228

MSTE - Engineering Economy / Replacement Analysis / Engr. Janclyde Espinosa (Clidez)

A machine can be sold now for P15,000, if kept for another year, its salvage value will decline to P13,000. The operating expenses for this year are expected to be P30,000. A new machine is available for P50,000, with expected operating expenses of P18,000 for the first year, increasing by P1000 a year because of deterioration. It is believed that after 5 years new technology would make replacement necessary; the new machines salvage value at that time is estimated to be P20,000. The MARR is 20%. Should the new machine be acquired? And how much is the comparative use value of the old machine? (A/G, 20%, 5) = 1.6405

Answer:

  1. The machine should be replaced, comparative value of old machine = 13893.24
  2. The machine should not be replaced, comparative value of old machine = 13893.24
  3. The machine should be replaced, comparative value of old machine = 16739.42
  4. The machine should not be replaced, comparative value of old machine = 16739.42
MARR = 20%; given $(A/G,20\%,5) = 1.6405$; $(A/P,20\%,5) = 0.33438$; $(A/F,20\%,5) = 0.13438$
EAC of new machine:
CR = $50{,}000(A/P,20\%,5) - 20{,}000(A/F,20\%,5) = 50{,}000(0.33438)-20{,}000(0.13438)$
$= 16{,}719 - 2{,}688 = $ P14,031
Gradient OC = $18{,}000 + 1{,}000(1.6405) = $ P19,641
EAC$_{\text{new}} = 14{,}031 + 19{,}641 = $ P33,672
Comparative value $V$ of old machine — set annual cost of keeping one year = EAC$_{\text{new}}$:
$V(1.20) - 13{,}000 + 30{,}000 = 33{,}672$
$1.20V = 33{,}672 - 17{,}000 = 16{,}672$
$V = 13{,}893$
Since current SV (P15,000) $>$ $V$ (P13,893) → replace.
$\boxed{\text{Replace; comparative value of old machine} = \text{P}13{,}893.24}$