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Total Mixed Ration (TMR) for Dairy Cows Dr. István HULLÁR associate professor (2010)

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Presentation on theme: "Total Mixed Ration (TMR) for Dairy Cows Dr. István HULLÁR associate professor (2010)"— Presentation transcript:

1 Total Mixed Ration (TMR) for Dairy Cows Dr. István HULLÁR associate professor (2010)

2 TMR - Mixture of forages, roughages, grains, extr. meals, supplements, etc., - fed according to the total daily requirements of production groups.

3 Control of the TMR TASK Check the following TMR whether it is sufficient for the following dairy cow. STEPS of Solution 1. Data Sheet - TMR, - data regarding the dairy cow; 2. Nutrient Requirements 3. Nutrient Content of the TMR 4. Evaluation (requirements vs. TMR), Corrections.

4 Control of the TMR 1. Data Sheet - TMR meadow hay (average): 5.0 kg, corn silage (predough stage):15.5 kg sunflower meal extr. (average): 2.6 kg, CCM (Corn Cob Mix): 8.4 kg; - Data regarding the dairy cow W: 580 kg, milk production: 25 kg/day, milk protein: 3.2%, milk fat: 3.5%, 2nd lactation.

5 Calculation of the Requirements 2. Nutrient Requirements 2.1. DMI: 2-3% of the W (580 kg) = 11.6-17.4 kg/day, 2.2. CF: 18-23% of the DM content of TMR.

6 Calculation of the Requirements 2. Nutrient Requirements 2.3. Maintenance 2.3.1. Energy reqirement NE l = 0.3514*W 0.75 = 0.3514*580 0.75 = 0.3514*118.19 = 41.53 MJ/day; 2.3.2. Protein requirement MP = 3.41*W 0.75 = 3.41*118.19 = 403 g/day; 2.3.3. Ca requirement: 44*0.001*W = 25.52 g/day; 2.3.4. P requirement: 34*0.001*W = 19.72 g/day.

7 Calculation of the Requirements 2. Nutrient Requirements 2.3. Maintenance 2.3.5. Correction 1st lactation: +20% of the total maintenance req.; 2nd lactation: +10% of the total maintenance req. (see later); 2.4. Milk Production 2.4.1. Energy reqirement NE l = 1.471 + 0.4032*milk fat% = 1.471 + 0.4032*3.5 = 2.88 MJ/kg milk; 25 kg milk: 25*2.88 = 72.06 MJ/day.

8 Calculation of the Requirements 2. Nutrient Requirements 2.4. Milk Production 2.4.2. Protein requirement MP = milk protein (g)/0.65 = 32/0.65 = 49 g/kg milk; 25 kg milk: 25*49 = 1225 g/day. 2.4.3. Ca requirement: 2.8 g/kg milk; 25 kg milk = 25*2.8 = 70.00 g/day; 2.4.4. P requirement: 1.7 g/kg milk; 25 kg milk: 25*1.7 = 42.50 g/day.

9 2. Requirements Sum of the Requirements DMCFNE l MPCaP (kg)(g)(MJ) (g)(g) (g) Maintenance - -41.53 40325.5219.72 +10% - - 4.15 40 2.55 1.97 25 kg milk - -72.06122570.0042.50 TOTAL 11.6-17.4 18-23% of DM 117.74 166898.0764.19

10 3. Nutrient Content of the TMR DMCFNE l MPEMPNCaP Feeds(kg)(g)(MJ) (g) (g) (g)(g) 5.0 kg hay 4.371372 21.84 415 37120.9812.24 15.5 kg silage 5.411087 34.73 395 29817.8513.52 2.6 kg sunfl. 2.33 417 14.79 333 608 7.2217.47 8.4 kg CCM 5.28 312 41.13 491 306 3.1713.20 TOTAL 17.393188112.46 1634158349.2256.43 Which MP value is valid? 

11 4. Evaluation, Corrections RequirementsTMREvaluation DM, kg11.60-17.4017.39 O.K. CF, g 18-23% of the DM 3.188/17.39*100 = 18.33% O.K. Other Nutrients The tolerable difference between the requirement and the TMR: the rquirement of 2 kg milk (  ).

12 4. Evaluation, Corrections NE l (MJ) Requirement:117.74 TMR:112.46 Difference:- 5.28 Tolerable level:2.88*2 =  5.76 Evaluation:O.K. MP (g) Requirement:1668 TMR:1583 Difference:- 85 Tolerable level:49*2 =  98 Evaluation:O.K.

13 4. Evaluation, Corrections Ca (g) Requirement:98.07 TMR:49.22 Difference: - 48.85 Tolerable level:2.8*2 =  5.6 Evaluation:insufficient P (g) Requirement:64.19 TMR:56.43 Difference:- 7.76 Tolerable level:1.7*2 =  3.4 Evaluation: insufficient.

14 4. Evaluation, Corrections Corrections Types of Ca and P supplements NameCa (g)P (g) Limestone380 - AP-17210170 AP-18190180 Phylafor 40125 AP-IV 17170

15 4. Evaluation, Corrections Corrections 1. Correction of the P deficiency by AP-IV: 1000 g AP-IV contains170 g P x g AP-IV contains 7.76 g P x = 7.76/170*1000 = 45.65 g AP-IV 2. Calculation of the amount of Ca given by AP-IV: 1000 g AP-IV contains17 g Ca 45.65 g AP-IV contains x g Ca x = 45.65/1000*17 = 0.78 g Ca.

16 4. Evaluation, Corrections Corrections 3. Calculation of the Ca deficiency remained: 48.85 – 0.78 = 48.07 g; 4. Correction of the Ca deficiency by limestone: 1000 g limestone contains380 g Ca x g limestone contains 48.07 g Ca x = 48.07/380*1000 = 126.50 g limestone.

17 4. Evaluation, Corrections Final Evaluation - DM, CF, NE l, and MP content of the TMR are O.K.; - but 45.65 g AP-IV and 126.50 g limestone supplementations are required.


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