Question

1(a) A new machine is to be bought. MARR is 10%. The following two models are under consideration. Model First cost Economic life Yearly net savings Salvage value MI $100,000 5 years $50,000 $18,000 M2 150,000 5 years 60,000 24,000 Using the present worth and equivalent annual methods, which model should be bought? 1(b) If you discover a third alternative M3, as shown in the table below, which is the best? MARR is 10%. Salvage value Model First cost Economic Yearly net life savings MI $100,000 5 years $50,000 M2 150,000 5 years 60,000 M3 200,000 3 years 75,000 Use the applicable method NOTE: Since economic life is not the same, use EAW method. $18,000 24,000 100,000

Answer 1

1(a)

Case of M1

Present worth of machine 1=PW1=-100000+50000*(P/A,0.10,5)+18000*(P/F,0.10,5)

Let us calculate the interest factors

$(P/A,i,n)=\frac{1-\frac{1}{(1+i)^{n}}}{i}$​​​​​​

1- (1+0.10) (P/A, 0.10,5) = - - 3.790787 0.10

(P/F,0.10,5)=1/(1+0.10)^5=0.620921

Present worth of machine 1=PW1=-100000+50000*3.790787+18000*0.620921=$100715.93

EAW1=PW1/(P/A,0.10,5)=100715.93/3.790787=$26568.61

Case of M2

Present worth of machine 2=PW2=-150000+60000*(P/A,0.10,5)+24000*(P/F,0.10,5)

Present worth of machine 2=PW2=-150000+60000*3.790787+24000*0.620921=$92349.32

EAW2=PW2/(P/A,0.10,5)=92349.32/3.790787=$24361.52

We can observe that

PW is higher in case of machine 1, it should be selected on the basis of present worth analysis.

EAW is higher in case of machine 2, it should be selected on the basis of EAW analysis.

b)

We should go for EAW method as project lives are different.

EAW of machine 3=EAW3=-200000*(A/P,0.10,3)+75000+100000*(A/F,0.10,3)

Let us calculate the interest factors

(A/P,1,n) = 1 -

$(A/P,0.10,3)=\frac{0.10}{1-\frac{1}{(1+0.10)^{3}}}=0.402115$

$(A/F,i,n)=\frac{i}{(1+i)^{n}-1}$

$(A/F,0.10,3)=\frac{0.10}{(1+0.10)^{3}-1}=0.302115$

EAW of machine 3=EAW3=-200000*0.402115+75000+100000*0.302115=$24788.50

We still observe that EAW is highest in case of machine 1. Machine 1 should be selected.