Economic analysis: replace machine now or later

[gfd43tg]

For scenario 2, then the sum would be

$$ \sum_{n=1}^{\infty} \frac {350}{1.1^{n}} + \sum_{n=1}^{\infty} \frac {-22,000}{1.1^{10n-1}} + \sum_{n=1}^{\infty} \frac {7500}{1.1^{10n-1}} $$

 

[Chestermiller]

For scenario 2, then the sum would be

$$ \sum_{n=1}^{\infty} \frac {350}{1.1^{n}} + \sum_{n=1}^{\infty} \frac {-22,000}{1.1^{10n-1}} + \sum_{n=1}^{\infty} \frac {7500}{1.1^{10n-1}} $$

For scenario 2, then the sum would be

$$ \sum_{n=1}^{\infty} \frac {350}{1.1^{n}} + \sum_{n=1}^{\infty} \frac {-22,000}{1.1^{10n-1}} + \sum_{n=1}^{\infty} \frac {7500}{1.1^{10n-1}} $$

My understanding is:
$$ \sum_{n=1}^{\infty} \frac {350}{1.1^{n}} + \sum_{n=0}^{\infty} \frac {-22,000}{1.1^{10n}} +7500 $$

 

[gfd43tg]

I don't understand how you are picking your exponents for the interest rates in the sums. Why 10n, 10n-5??

 

[Chestermiller]

What does that last term mean?

It means that you replace the old machine after 5 years, and then replace each new machine every subsequent 10 years.

 

[Chestermiller]

I don't understand how you are picking your exponents for the interest rates in the sums. Why 10n, 10n-5??

In scenario 2, the 10n (starting with n = 0), means that you replace the machine immediately, and then every subsequent 10 years.

In scenarion 1, the 10n-5 (starting with n = 1) means that you replace the machine after 5 years, and then every subsequent 10 years.

 

[gfd43tg]

Thanks, that is much more clearer now. Now I need to figure out how to actually calculate an infinite series

 

[Chestermiller]

Thanks, that is much more clearer now. Now I need to figure out how to actually calculate an infinite series

a/(1-r)

 

[gfd43tg]

I found scenario 1 NPV to be -$21,953.7, and for scenario 2 it is -$24,804, so it is better to keep using the machine until it has finished its operating life.