Depreciation, Interest, and Inflation
Assume that the Copper River Valves’ manufacturing equipment evaluated in Example 15.14 will be financed with a 5-year, 11% loan that has equal annual payments. What is the EAC of this equipment? Check this answer by computing the PW of the loan separately and then combining that PW with the PW from Example 15.14. That equipment had a first cost of $250K and annual operating costs of $45K—both in year-0 dollars.

Question

Accepted Answer

The first step is to calculate the annual payment.

Loan payment [latex]= 250K(A/P,11\%,5) = 250K · .2706 = \$67.65K[/latex]

This value is subject to a differential inflation rate of −7%, like the MACRS tax deduction. Note that if inflation and the tax deduction on the interest were not decreasing the true cost of the payments, a firm that uses an interest rate of 9% would not borrow money at 11%.

XHIBIT 15.14 Interest, depreciation, and inflation at Copper River Valves

I	H	G	F	E	D	C	B	A
					first cost for capital equipment			$250,000	1
						MACRS class		7	2
							N	10	3
							loan rate	11%	4
							loan term	5	5
							loan payment	$67,643	6
							operations cash flow	-$45,000	7
							f = CPI	7%	8
							tax rate	35%	9
						real interest rate		9%	10
									11
ATCF in Year-0 $s	ATCF in Nominal $s	Tax	Taxable Income in Nominal $s	Operations Cost in Nominal $s	Principal Portion of Loan	Interest Portion of Loan	MACRS Deduction Nominal $	Year	12
								0	13
-71,790	-76,815	-38,978	-111,364	-48,150	40,143	27,500	35,714	1	14
-62,558	-71,623	-47,540	-135,829	-51,521	44,558	23,084	61,224	2	15
-66,777	-81,805	-40,965	-117,042	-55,127	49,460	18,183	43,732	3	16
-69,111	-90,591	-36,038	-102,965	-58,986	54,900	12,742	31,237	4	17
-70,238	-98,512	-32,246	-92,130	-63,115	60,939	6,703	22,312	5	18
-24,046	-36,087	-31,446	-89,845	-67,533			22,312	6	19
-24,387	-39,160	-33,100	-94,572	-72,260			22,312	7	20
-26,977	-46,352	-30,966	-88,474	-77,318			11,156	8	21
-29,250	-53,775	-28,956	-82,731	-82,731				9	22
-29,250	-57,539	-30,983	-88,522	-88,522				10	23
PW= -$331,731									24

Exhibit 15.14 calculates columns B through H in nominal dollars, then converts them to constant-value dollars in column I. Columns C and D are calculated as they were in Exhibit 15.13. For each year in column F, the taxable income equals the cash flow for operating costs minus the MACRS deduction and minus the interest portion of the loan payment. For each year in column G, the after-tax cash flow (ATCF) equals the cash flows for operating costs minus the loan payment (principal and interest) minus taxes.
From Exhibit 15.14, the PW has changed to −$331,731 with the loan from −$385,166 in Example 15.14. This $53,435 difference occurs because the cost of the loan is less than the firm’s interest rate, which is 9% after taxes. In fact, as detailed in Exhibit 15.15, the loan’s true cost at .14% is about 0%. As stated in Example 15.15, the loan’s cost can be approximated as (pretax rate)(1 − the tax rate) minus f, the inflation rate. This equals:

After-tax interest rate ≈ .09(1 − .35) − 7% = .15% ≈ 0%

Exhibit 15.15 computes the PW of the loan using nominal dollars for the interest portion of the loan payment and for the resulting tax savings. These are converted in column F to year-0 dollars by dividing the nominal values by [latex](1 + f )^{t}[/latex] . The loan’s PW equals $53,436 (the same value as above).

Question 15.16: Depreciation, Interest, and Inflation Assume that the Copper...

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