EAM.8:5.1 - Constructing a four-year policy
Consider a separate constructed asset E supplying the same required service under two technically qualified policies. All amounts are constant-price €million; the comparison uses a 3% real rate, year-end cash and sufficient resources for either policy. Tax and financing differences are held equal. These assumptions do not supply rates or technical support for a real asset.
Policy X retains E: initial work costs 0.40 and supports its duty through year three; a 0.90 overhaul at the end of year three supplies the further support needed for year four. Policy Y replaces E initially for 1.50 and has qualified support throughout the four years. The following account includes recurring operation and routine maintenance separately from the overhaul.
| Consequence and date | Retain and overhaul X | Replace Y |
|---|---|---|
| Initial work at year zero | 0.40 | 1.50 |
| Recurring cost at each year end, years 1–4 | 0.30 | 0.15 |
| Further work at the end of year three | 0.90 | 0 |
| Value of continued use at the end of year four | 0.20 | 0.50 |
The teaching qualifications include permitted eight-hour work windows and temporary service sufficient to preserve the requirement during initial work and X’s later overhaul. The corresponding costs are included in the initial and overhaul amounts. That makes the policies service-comparable; omitting the later window or temporary provision would leave X conditional. No other material consequence distinguishes the eligible policies in this example.
At 3%, the four-year annuity factor is 3.717098, the year-three discount factor 0.915142 and the year-four factor 0.888487. X’s present cost is 0.40 + 0.30 × 3.717098 + 0.90 × 0.915142 − 0.20 × 0.888487 = 2.161060. Y’s is 1.50 + 0.15 × 3.717098 − 0.50 × 0.888487 = 1.613321. Replacement has the lower comparable cost. Omitting the overhaul would report X as 1.337432 and reverse the recommendation while removing the work needed to support year four.
Now move the end of the explicit cash table to year two, keeping both complete policies, the required service, the rate and all later amounts unchanged. Express the remaining year-three and year-four cash, including the stated year-four continuing value, as an equivalent net value at year two:
- X:
V_X(2) = −(0.30 + 0.90)/1.03 + (0.20 − 0.30)/1.03² ≈ −1.259308. - Y:
V_Y(2) = −0.15/1.03 + (0.50 − 0.15)/1.03² ≈ 0.184278.
These values represent the same remaining account on the cost-comparison basis. X’s negative value means its later support payments exceed its discounted year-four continuing value. The common required service remains the same and its equal benefits are outside both cost accounts. The calculation does not estimate a sale price.
Let a₂ = 1/1.03 + 1/1.03² and d₂ = 1/1.03². Using unrounded values, the shortened explicit accounts give X: 0.40 + 0.30 × a₂ − V_X(2) × d₂ = 2.161060; Y: 1.50 + 0.15 × a₂ − V_Y(2) × d₂ = 1.613321. Replacement remains cheaper. Moving the table boundary has moved later consequences into the ending value; it has removed neither their cost nor their service obligations. A different supplied appraisal needs reconciliation of its changed assumptions before it can support a different recommendation.
Actual withdrawal at year two is another case. If the authority ends the service and X incurs a net exit cost of 0.10 while Y yields net sale proceeds of 0.60, replace the continuing values with those consequences. Present costs are then 1.068300 and 1.221463. X is cheaper in this different case because the service requirement and exit consequences have changed. There is no year-three service obligation in this case. If the service remains required, withdrawal alone is incomplete: a qualified replacement service and its consequences must join the policy.