Q:

Resistor Power Rating and headroom for AC circuit

Background
I'm going to put together a resistive load, to load a 100W source with an \$8\Omega\$ or \$16\Omega\$ resistor.  I'm finding that when to use peak, peak-to-peak or RMS values of voltage, current or power is a subject still shrouded in mystery for those new to the subject or not working in the industry.
Understanding which power to use (RMS, Peak or Peak-to-peak)
When selecting a power rating for the resistor

do you assume that the power rating on the component website is for a DC supply
should you calculate the power rating of a component from the RMS power, Peak Power or Peak-to-peak power?

Example:
Without taking into account the extra headroom needed (let's run this resistor at maximum permissible power), 
$$100W_{RMS}=200W_{Peak}=400W_{Peak-to-Peak}$$
 therefore should you select a 400W resistor? 
Headroom
What is the convention when it comes to selecting headroom on the component's power rating.  Is, 1.2 - 1.5 a reasonable factor to multiply by.  Let's for the sake of safety put a fuse in line with the resistor to protect it.
My intuition
My intuition tells me to place a fuse in series and either:

work with the Peak power value in mind because the AC signals will only ever reach a maximum of the peak value, not the peak-to-peak value, and to select headroom in the region of a factor of 1.25 - 1.50.  From the example, this would mean selecting a resistor with power rating of \$200W_{Peak} \cdot 1.2 = 250W\$ or \$200W_{Peak} \cdot 1.50 = 300W\$.

or

work with the peak-to-peak power value which according to the logic in the previous bullet point will surpass the peak power value by a factor of 2 and not chose to use additional headroom. From the example, this would mean selecting a resistor with power rating of \$400W_{Peak-to-peak} = 400W\$

Of these two estimations, the second seems to be sensible due to the apparent large headroom, and therefore component lifespan, but perhaps at an extra financial cost.
Simplification of finding a component
To make it easier to find components within a restricted budged (in the tens of pounds rather than hundreds of pounds) I intend to connect a number of resistors in parallel to distribute the current and therefore power/ component heating. i.e. \$5 \times 40\Omega\$ resistors in parallel \$=8\Omega\$ and \$5 \times 80\Omega\$ resistors in parallel \$=16\Omega\$.
What say you? Thanks in advance
Daniel

A:

For determining the power a voltage will dissipate when applied to a resistance, use the RMS voltage.
One way to look at the RMS voltage is that's the equivalent AC to the DC voltage that would dump the same power into the same resistance.
For example, 5 V across a 2 Ω resistor will dump (5 V)2/(2 Ω) = 12.5 W into the resistor.  That is true whether the 5 V is DC, or 5 V AC RMS.  Actually, the voltage in the equation is always RMS.  For DC, the RMS is the DC voltage.

