EXAMPLE: Evaluating Alternal Three plans are considered for an engineering design problem: Plan A: purchase an equipment at $650,000 with a 10 year lifetime and a $17,000 salva value. Equipment annual operating cost is $50,000 while the annual cost of a control program is $120,000 Plan B: Initial cost is $4 million, annual maintenance cost is $5,000, repairs occurs ev years at $30,000. However, this plan provides a permanent solution Plan C: Initial cost is $6 million and the annual maintenance cost is $3,000 with a life of 50 years, Assume i = 5% We need to compute AW for each plan for one cycle وعی .
EXAMPLE: Evaluating Alternal Three plans are considered for an engineering design problem: Plan A: purchase an equipment at $650,000 with a 10 year lifetime and a $17,000 salva value. Equipment annual operating cost is $50,000 while the annual cost of a control program is $120,000 Plan B: Initial cost is $4 million, annual maintenance cost is $5,000, repairs occurs ev years at $30,000. However, this plan provides a permanent solution Plan C: Initial cost is $6 million and the annual maintenance cost is $3,000 with a life of 50 years, Assume i = 5% We need to compute AW for each plan for one cycle وعی .
Essentials Of Investments
11th Edition
ISBN:9781260013924
Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.
Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.
Chapter1: Investments: Background And Issues
Section: Chapter Questions
Problem 1PS
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hi, could u explain how to solve plan B
thx
Transcribed Image Text:A
EXAMPLE: Evaluating Alternatives by Annual Worth Analysis
Three plans are considered for an engineering design problem:
Plan A: purchase an equipment at $650,000 with a 10 year lifetime and a $17,000 salvage
value. Equipment annual operating cost is $50,000 while the annual cost of a control
program is $120,000
Plan B: Initial cost is $4 million, annual maintenance cost is $5,000, repairs occurs every 5
years at $30,000. However, this plan provides a permanent solution
Plan C: Initial cost is $6 million and the annual maintenance cost is $3,000 with a lifetime
of 50 years, Assume i = 5%
We need to compute AW for each plan for one cycle
Plan A:
.
CR=-650,000(A/P,5%, 10) + 17,000(A/F,5%, 10) = $ -82,824
A₁ = -50,000 and A₂ = -120,000
AW₁= -82,824 + (-50,000) + (-120,000) = $-252,824
Plan B:
CR-4,000,000x5% = $-200,000
A = $-5,000
Repair cost = -30,000(A/F,5%,5) = $-5,429
AWR-200,000+ (-5,000) + (-5,429) = $-210,429
MY
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