Question

Benjamin is an engineer with the Lego Group in Bellund, Denmark, manufacturers of Lego toy construction blocks. He is responsible for the economic analysis of a new production method of special-purpose Lego parts. Method 1 will have an initial cost of $400,000, an annual operating cost of $140,000, and a life of three years. Method 2 will have an initial cost of $600,000, an operating cost of $100,000 per year, and a six-year life. Assume 10% salvage values for both methods. If Lego Industries uses a MARR of 15% per year, which method should it select on the basis of a present worth analysis?

Answer 1

Answer:

Given that person B is an engineer and he is responsible for the economic analysis of new production. The analysis period is 6 yrs. The useful life of method-1 is 3 years and method 2 is 6 years. There will be replacement of 3 years in method 1 and no replacement in method 2.

Calculate the present worth for method 1:

Below is the formula used for the calculation of Present worth:

$PW_{1} = P + A (P / A,i,n) + F_{1} (P / F,i,n) + F_{2} (P / F,i,n)$

Here P is -$400,000, A is -$140,000, F1 is -8360,000 and F2 is $40,000 (10% of 400,000)

PW = -400,000 - 140,000 (P/A,15%,6) -360,000(P/F,15%,3) + 40,000(P/F,15%,6)

PW = -400,000 - 140,000 (3.7845) -360,000 (0.6575) + 40,000 (0.4353)

PW = -400,000 - 529,830 - 236,700 + 17,292

PW = -$1,149,238 .

Calculate the present worth for method 2:

PW= P+ AV/ .4,40+ (P I F,i,n)

Here Pis -$600,000, A is -$100,000, F is -$60,000, interest rate is 15 % and the time-period is 6 years.

PW1 = -600,000 - 100,000(P/A,15%,6) + 60,000(P/F,15%,6)

PW1 = -600,000 - 100,000(3.7845) + 60,000(0.4323)

PW1 = -600,000 - 378,450 + 25,938

PW1 = -$952,512