Basic question about multiplex PCR
I haven't done this exact experiment. I am just deducing from the known facts about PCR. 

The product of PF and PR2 will serve as a template for both 1kb product (P1) and 2kb product (P2). 

Lets assume that after 2nd cycle there is 1 copy each of P1 and P2 (Delay of 1 cycle to make a smaller length product. See here). 

Lets assume that the primer binding affinity is equal (and for just preliminaries consider that only one primer can bind a template — such as in overlapping binding sites). So half of P2 will bind to PR2 and half to PR1.

Number of copies at nth cycle is:

P2(n) = 1.5 × P2(n-1)

P1(n) = 2 × P1(n-1) + 0.5 × P2(n-2)

However this is not really the scenario; the primers are not competing for the template.
Which means that both primers can bind to the substrate simultaneously. 

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DNA polymerases (except Klenow) will remove any DNA strand bound to the template, ahead of it. In that case P1 will not form. But when PR2 runs out, the unused PR1 can produce P1. 

Now there are additional factors, such as the distance between PR1 and PR2 binding sites. If the distance between the primer binding sites is less than the smaller product then the abovementioned case occurs. If the inter-primer distance is larger than smaller product then both products will form (and just like the competing primers case, P2 will serve as a template for both products). In this case the copy numbers will be:

P2(n) = 2 × P2(n-1)

P1(n) = 2 × P1(n-1) + 1 × P2(n-2)

This means that P1 will dominate. 

Other assumptions:


No limiting concentrations of dNTP, primers etc
PCR efficiencies are equal for both primers and is equal to 2 (in non-competing cases)
Primer annealing temperatures are same i.e binding affinities are same.