Question

INTERNAL RATE OF RETURN/PAYBACK METHOD

PART 1: Johnson Drill Bits, Inc. has just purchased a new piece of equipment for $66,350. The machine saves the company $13,500 each year of its 6-year life. There is no salvage value. What is the approximate internal rate of return for this equipment?

PART 2: Catalina Company has just purchased a new piece of equipment for $85,000 that will provide an annual cost savings of $18,000. There is no salvage value. What is the payback period?

Here are some present value tables you can use for this problem:

PRESENT VALUE TABLES

Present Value of $1

Periods	4%	6%	8%	10%	12%	14%
4	.855	.792	.735	.683	.636	.592
5	.822	.747	.681	.621	.567	.519
6	.790	.705	.630	.564	.507	.456
7	.760	.665	.583	.513	.452	.400
8	.731	.627	.540	.467	.404	.351
9	.703	.592	.500	.424	.361	.308
10	.676	.558	.463	.386	.322	.270
Present	Value	of	Annuity
Periods	4%	6%	8%	10%	12%	14%
4	3.630	3.465	3.312	3.170	3.037	2.914
5	4.452	4.212	3.993	3.791	3.605	3.433
6	5.242	4.917	4.623	4.355	4.111	3.889
7	6.002	5.582	5.206	4.868	4.564	4.288
8	6.733	6.210	5.747	5.335	4.968	4.639
9	7.435	6.802	6.247	5.759	5.328	4.946
10	8.111	7.360	6.710	6.145	5.650	5.216

Answer 1

Part 1

Let us first try to get an approximate

Annuity factor = Initial investment / Annual inflow = 66,350 / 13,500 = 4.915

In 6th year, the nearest to this is 4.917 which is at 6%.

So the answer to this is, IRR = 6.02%. Here is the proof

Year	Cash flow	PV Factor	Present Value
0	$      (66,350)	1	$       (66,350)
1	$        13,500	0.94325	$         12,734
2	$        13,500	0.88972	$         12,011
3	$        13,500	0.83923	$         11,330
4	$        13,500	0.79161	$         10,687
5	$        13,500	0.74669	$         10,080
6	$        13,500	0.70431	$           9,508
			$                 (0)

Part 2

Payback period = Initial investment / Annual cost saving = 85,000 / 18,000 = 4.72 years