Problem 3 A company needs a new grinder Compute the present

Problem 3: A company needs a new grinder. Compute the present worth for the mutually exclusive alternatives and identify which option is better. Assume an interest rate of 6% and that the grinder will be replaced with an identical model at the end of its life.

Option 1: Initial cost of $4500, annual costs of $300, salvage value of $500, life of 5 years

Option B: Initial cost of $5500, annual costs of $400, no salvage value, life of 10 years

Solution

Year

Option 1

Option 2

Discount Factor

Discounted Option 1

Discounted Option 2

0

-4500

-5500

1

-4500

-5500

1

-300

-400

0.9434

-283.02

-377.36

2

-300

-400

0.89

-267

-356

3

-300

-400

0.8396

-251.88

-335.84

4

-300

-400

0.7921

-237.63

-316.84

5

200

-400

0.7473

149.46

-298.92

6

-400

0.705

0

-282

7

-400

0.6651

0

-266.04

8

-400

0.6274

0

-250.96

9

-400

0.5919

0

-236.76

10

-400

0.5584

0

-223.36

PW

-5390.07

-8444.08

Discount factor = 1/(1+0.06)^n, n = number of years

Discounted Option 1 = Option1 values * Discount factor

Discounted Option 2 = Option2 values * Discount factor

Looking at the present worth value, option 1 is a better option as its present value is less and the company can save money

Year	Option 1	Option 2	Discount Factor	Discounted Option 1	Discounted Option 2
0	-4500	-5500	1	-4500	-5500
1	-300	-400	0.9434	-283.02	-377.36
2	-300	-400	0.89	-267	-356
3	-300	-400	0.8396	-251.88	-335.84
4	-300	-400	0.7921	-237.63	-316.84
5	200	-400	0.7473	149.46	-298.92
6		-400	0.705	0	-282
7		-400	0.6651	0	-266.04
8		-400	0.6274	0	-250.96
9		-400	0.5919	0	-236.76
10		-400	0.5584	0	-223.36
			PW	-5390.07	-8444.08