Theoretical Reproductive Performance in a 120 Cow dairy Herd With 40% Estrus Detection Efficiency
AI service | 1 | 2 | 3 | 4 | 5 | 6 |
Days postpartum 1 | 55 - 76 | 77 - 98 | 99 - 120 | 121 - 142 | 143 - 164 | > 164 |
Total open cows | 120 | 91 | 69 | 52 | 39 | 30 |
Normal cows (P = 0.9)2 | 108 | 82 | 62 | 47 | 35 | 27 |
Detected-mated (P = 0.40)3 | 43 | 33 | 25 | 19 | 14 | 11 |
Calving, this AI (P = 0.675)4 | 29 | 22 | 17 | 13 | 10 | 7 |
Cumulative pregnant cows | 29 | 51 | 68 | 81 | 90 | 97 |
Days Open 5 | 1910 | 1931 | 1830 | 1663 | 1470 | 1269 |
Cumulative Days Open | 1910 | 3841 | 5671 | 7335 | 8805 | 10073 |
1 Cows mated after a voluntary waiting
period of 54 days.
2 Assume 10% of cows are
abnormal due to reproductive
pathology/acyclicity.
3 Assume that 50% of the
normal, cycling cows are detected and mated at each successive
estrus.
4 Cows that successfully conceive,
complete gestation and calve normally from each insemination. The probability
is a combination of 90% chance of conception and 75% chance of successful
gestation/parturition. (P = 0.9 x 0.75 =
0.675).
5 Number of cows calving x median days
postpartum for this insemination.
Mean Days Open (pregnant cows) = 10073/97 = 103.8
But 23 cows were still open after six inseminations and should be culled
for reproductive failure. If these animals remained in the herd for a full
year, this would add 23 x 365 = 8395 days open to the total.
Mean
Days Open (all cows) = 10073 + 8395/120 = 153.9
Services per Pregnancy (97 cows) = 2.76