Theoretical Reproductive Performance in a 120 Cow Dairy Herd With 50% Estrus Detection
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 | 84 | 58 | 41 | 28 | 20 |
Normal cows (P = 0.9)^{2} | 108 | 75 | 52 | 36 | 25 | 18 |
Detected-mated (P = 0.50)^{3} | 54 | 38 | 26 | 18 | 13 | 9 |
Calving, this AI (P = 0.675)^{4} | 36 | 25 | 18 | 12 | 9 | 6 |
Cumulative pregnant cows | 36 | 62 | 79 | 92 | 100 | 106 |
Days Open ^{5} | 2387 | 2221 | 1935 | 1618 | 1315 | 1044 |
Cumulative Days Open | 2387 | 4608 | 6543 | 8161 | 9476 | 10519 |
^{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) = 10519/106 = 99.2
But six 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 14 x 365 = 5110 days open to the total.
Mean Days Open (all cows) = 10519 + 5110/120 = 130.2
Services per Pregnancy (106 cows) = 2.54