Table 2.

Rates of pedigree and overall true genomic coancestry across generations (t) when applying different management strategies (RAN, PED and OVE) in populations of two different sizes (N) and using two mutation rates (μ) to create the base population

N	t	Rate of pedigree coancestry (%)						Rate of genomic coancestry (%)
		μ = 2.5 × 10−3			μ = 2.5 × 10−5			μ = 2.5 × 10−3			μ = 2.5 × 10−5
		RAN	PED	OVE	RAN	PED	OVE	RAN	PED	OVE	RAN	PED	OVE
20	1	2.46	1.28	2.47	2.45	1.28	2.69	2.47	1.32	0.17	2.57	1.31	0.20
	2	2.40	1.30	1.79	2.39	1.30	1.97	2.47	1.25	1.32	2.45	1.31	1.01
	3	2.44	1.30	1.73	2.43	1.30	1.89	2.34	1.24	1.29	2.30	1.32	1.08
	4	2.52	1.30	1.70	2.52	1.30	1.89	2.55	1.30	1.40	2.48	1.32	1.05
	5	2.46	1.30	1.75	2.45	1.30	1.88	2.40	1.35	1.50	2.48	1.30	1.10
	10	2.39	1.30	1.81	2.39	1.30	1.85	2.36	1.28	1.47	2.42	1.35	1.07
100	1	0.50	0.25	0.74	0.52	0.25	1.05	0.50	0.26	−0.16	0.55	0.22	−0.40
	2	0.51	0.25	0.50	0.48	0.25	0.69	0.50	0.25	0.23	0.57	0.25	−0.16
	3	0.49	0.25	0.49	0.46	0.25	0.65	0.50	0.25	0.28	0.44	0.19	−0.15
	4	0.50	0.25	0.48	0.52	0.25	0.63	0.51	0.26	0.31	0.60	0.27	−0.05
	5	0.50	0.25	0.47	0.51	0.25	0.64	0.50	0.26	0.34	0.60	0.25	−0.07
	10	0.50	0.25	0.46	0.50	0.25	0.58	0.51	0.26	0.37	0.47	0.19	−0.03

Average linkage disequilibrium between consecutive SNPs at t = 0 was 0.28 and 0.13 for N = 20 and 100, respectively when μ = 2.5 × 10⁻³, and 0.40 and 0.21 for N = 20 and 100, respectively when μ = 2.5 × 10⁻⁵

RAN contributions are assigned at random, PED contributions are optimised to minimise f _p, OVE contributions are optimised to minimise f _{m_ove}