Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
First, we need to sort out the data given properly, so that we can answer to the best.
Data Given:
For Chamber D103:
Installed Cost = -400,000 USD
Annual Operating Cost Per Year = -4000 USD
Salvage Value 10% of P = 40,000 USD
Life = 3 years
Similarly,
For Chamber 490G:
For Chamber D103:
Installed Cost = -250,000 USD
Annual Operating Cost Per Year = -3000 USD
Salvage Value 10% of P = 25,000 USD
Life = 2 years
a.
For Chamber D103
AW Chamber D103 = -400,000 x (A/P, 10%, 3) + 40000 x (A/F,10%, 3) - 4000
So,
(A/P, 10%, 3) = 0.40211 (from the compound interest table )
(A/F,10%, 3) = 0.30211
AW Chamber D103 = -400,000 x 0.40211 + 40000 x 0.30211 - 4000
AW Chamber D103 = -152,760 USD
For Chamber 490G:
AW Chamber 490G = -250,000 x (A/P,10%,2) + 25000 x (A/F,10%, 2) - 3000
So,
(A/P,10%,2) = 0.5762
(A/F,10%, 2) = 0.4762
AW Chamber 490G = -250,000 x 0.5762 + 25000 x 0.4762 - 3000
AW Chamber 490G = -135,143 USD
So, after evaluating both the chambers using the AW method, more economical is the chamber 490G.
b.
Now, we need to change the values to check whether the chamber selection can be changed or not:
so, New values for Chamber D103
P = -300,000 and Salvage value of 30,000
For Chamber D103
AW Chamber D103 = -300,000 x (A/P, 10%, 3) + 30000 x (A/F,10%, 3) - 4000
So,
(A/P, 10%, 3) = 0.40211 (from the compound interest table )
(A/F,10%, 3) = 0.30211
AW Chamber D103 = -300,000 x 0.40211 + 30,000 x 0.30211 - 4000
AW Chamber D103 = -115,570 USD
New Values for Chamber D103
P = -500,000 and the Salvage Value = 50,000
AW Chamber D103 = -300,000 x (A/P, 10%, 3) + 30000 x (A/F,10%, 3) - 4000
So,
(A/P, 10%, 3) = 0.40211 (from the compound interest table )
(A/F,10%, 3) = 0.30211
AW Chamber D103 = -500,000 x 0.40211 + 50,000 x 0.30211 - 4000
AW Chamber D103 = -189,950 USD
Hence,
At P = 300,000 will definitely change the selection to D103 Chamber.
= 
= $1,492.11
)= 
= $75.13