Question

engineering economics

number 5

of return is decision is to our estimate of the initial cost of A. how much higher than 4000 can the initial cost be and still have A the preferred alternative? (Hint: find the present worth of both alternative to start with) (15 pts) 4200 b. 4200 c.4300 d. 4400 5. On Microsoft Excel Spread sheets, to find P: (10 pts) a. PMT(, n, P. F, Type) b. PV(, n, A, F. Type) c. FVi, n, A, P. Type) d. NPVIL, n, P, F, A, 0)

Answer 1

1. The present worth of both the alternatives can be determined by the following equation

$\large PW = - I + A(P/A, i\%,n)$

In case of alternative A

Initial cost, I = KD 4,000

Annual benefit, A = KD 639

MARR, i= 6%

$\large PW_A = - 4,000 + 639(P/A, 6\%,20)$

$\large PW_A = - 4,000 + 639*11.470$

$\large PW_A = - 4,000 + 7,329.33$

$\large PW_A = KD 3,329.33$

Similarly for alternative B we can calculate PW

$\large PW_B = - 5,000 + 700(P/A, 6\%,20)$

$\large PW_B = - 5,000 + 700*11.470$

$\large PW_B = - 5,000 + 8,029$

$\large PW_B = KD 3,029$

Since, PW of A is greater than B. So, select alternative A.

Now, determine the difference between the two alternatives

PW_A - PW_B = KD 3,329.33 - 3,029 = KD 300.33

So, up to a initial cost of KD 4,300 alternatives A will be preferable over B.

Option C. KD 4,300

5. In case of determining present value we have to use the following excel syntax

=-PV(6%,20,639)

Or, "=-PV(i%,n,A)"

If you wish to determine the Net Present value that is Present Worth then

"=-PV(6%,20,639)+(-4,000)"

Or, "=-PV(i%,n,A)+I"

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