arch 449 erna audrey mangaleu toukam 127476 term project

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Page 1: Arch 449 erna audrey mangaleu toukam 127476 term project

LIFE CYCLE COSTING ANALYSISNAME : Erna Audrey mangaleu toukam

STUDENT NUMBER : 127476

COURSE CODE : Arch 449 Economics and Managerial Issues in architecture

TERM PROJECT

SEMESTER: SPRING 2014/2015

INSTRUCTOR: ASSIST PROF. DR. ERCAN HOSKARA

Page 2: Arch 449 erna audrey mangaleu toukam 127476 term project

CASE STUDY 1

• Two cars are under consideration for a newly weeded couple , through the duration of 5 years . Machine A is with initial price $30,000. the operation and maintenance cost are as follows: fuel consumption $400 per annum, Taxes $500 per annum, insurance $450 per annum and service cost S250 per annum, with a salvage value $10,000. Machine B is with initial price $34,000, operation costs : fuel consumption $200 per annum, taxes $400 per annum, insurance $350 per annum, service cost $100 per annum with a salvage value $12,000. which machine should be selected on the basis of their present worth values using an interest rate of 6%.

• Machine A operation and maintenance cost = 400 + 500 + 450 + 250 = $1600

• Machine B operation and maintenance cost = 200 + 400 + 100 + 350 = $1050

Machine A Machine B

First cost, P $30,000 $34,000

Operation and Maintenance cost, A

$1600 $1050

Salvage value, F $10,000 $14,000

Service life (year), n 5 5

Where p = present value, A = annual cost, F = future value, and n = no. of years

Page 3: Arch 449 erna audrey mangaleu toukam 127476 term project

Machine A

Machine B

$30,000

$1600 $1600 $1600 $1600 $1600

$10,000

PA = 30000 + 1600(P/A, 6%, 5) – 10,000(P/F, 6%, 5) = $29,266.84

$34,000

$1050 $1050 $1050 $1050 $1050

$14,000

PB = 34000 + 1050(P/A, 6%, 5) – 14,000(P/F, 6%, 5) = $27,960.82

Machine B should be selected since PB < PA.

1 2 3 4 5

1 2 3 4 5

Page 4: Arch 449 erna audrey mangaleu toukam 127476 term project

CASE STUDY 2• This coming semester I’m recently trying

to decide between two different accommodations in school and I have the choice to choose two dormitories. Sabanci has an initial down payment of $1500 and a service life of 14 months and is 2min from the University and its maintenance costs 100$ . on the other hand Akdeniz has an initial down payment of $1000 and a service of 7 months and is a 10 min walk from the University and its maintenance cost is more than that of sabanci . If their interest rate is 2%, which dormitory should be selected, assuming both have no salvage value?

Where p = present value, A = annual cost, F = future value, and n = no. of years

Page 5: Arch 449 erna audrey mangaleu toukam 127476 term project

CASE STUDY 2

Where p = present value, A = annual cost, F = future value, and n = no. of years

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

$1500 $1500

• P(AKDENIZ ) = $1500+$1500 (P/F, 2%, 14) + 100 (P/A, 2%, 28) =

• $1500 + $1500 (0.7579) + 100 (21.2813) =

• 1500+ 1136.85+ 2128.13 = $4764.98

Page 6: Arch 449 erna audrey mangaleu toukam 127476 term project

1500+ 1136.85+ 2128.13 = $4764.98

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

$1000 $1000 $1000 $1000

P(SABANCI) = $ 1000+$1000 (P/F, 2%, 7) + $1000 (P/F, 2%,14) + $1000 (P/F,

2%, 21) = $1000 + $1000 (0.8706) + 1000 (0.7579) + 1000 (0.6598) =

1000 + 870.6 + 757.9 + 659.8 = $3288.3

Dorm B should be selected because it is a 10 min walk from school against an

hour walk expense and it costs lesser and P(SABANCI ) < P(AKDENIZ ).