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Repetition of the Calculations Dr. István HULLÁR associate professor

Repetition of the Calculations Dr. István HULLÁR associate professor. Calculation 1 Calculate the organic matter (OM) and the N-free E xtract (NFE) content of the following wheat sample!. Calculation 1. organic matter (OM) , N-free E xtract (NFE) DM: 900 g/kg, CP: 120 g/kg,

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Repetition of the Calculations Dr. István HULLÁR associate professor

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  1. Repetition of the CalculationsDr. István HULLÁR associate professor Calculation 1 Calculate the organic matter (OM) and the N-free Extract (NFE) content of the following wheat sample!

  2. Calculation 1 organic matter (OM), N-free Extract (NFE) DM: 900 g/kg, CP: 120 g/kg, CF: 30 g/kg, Ash: 25 g/kg, EE: 20 g/kg, NEm: 9.6 MJ/kg.

  3. Solution 1. OM: DM – Ash = 900-25 = 875 g/kg; 2. NFE: OM – CP – CF – EE = 875 – 120 – 30 – 20 = 705 g/kg.

  4. Calculation 2 Calculate the protein:energy (P:E) ratio of the following turkey diet by using the necessary data! CP:29.00%,DCP:25.00%, MPE: 12.00%,MPN: 14.00%, DE:15.42 MJ/kg,ME:12.13 MJ/kg, NEm:9.00 MJ/kg, NEl: 8.50 MJ/kg, NEg: 7.00 MJ/kg.

  5. Solution P:E ratio: protein (g/kg)/energy (MJ/kg). But what type of protein and energy are used? Type of protein and energy used in the given species. Turkey: CP/ME CP = 29%, ME = 12.13 MJ/kg; CP/ME = 290 (g/kg)/12.13 (MJ/kg) = 23.91 g/MJ.

  6. Calculation 3 Calculate the CP digestibility of corn according to the necessary data of an experiment carried out by pigs! Feed intake:6000 g, CP content of the feed: 9%, Faeces output: 2500 g, CP content of faeces: 4%, CP content of urine: 14%.

  7. Apparent Digestibility: nutrient intake  nutrient output  = nutrient intake (6000*0.09)  (2500*0.04)  = (6000*0.09) 540 100 440  = = 0.8148 540540 0.8148*100 = 81.48%. Feed intake:6000 g, CP content of the feed: 9%, Faeces output: 2500 g, CP content of faeces: 4%, Solution

  8. Calculation 4 Calculate the TDN content of the corn silage by using the necessary data!

  9. Calculation 4 DM: 32.90%,Corn Silage Nutrients onDM basis (%) CP: 9.20EE: 4.20 CF: 20.10NFE: 61.20 Nutrients’ Digestibility (%) CP: 55.00 EE:80.00 CF: 64.00 NFE: 74.00.

  10. Solution TDN (g/kg DM): DCP + 2.25*DEE + DCF + DNFE Calculation of the TDN Nutrients g/kg DM Dig. factor TDN CP 92 *0.55 - 50.6 EE 42 *0.80 *2.25 75.6 CF 201 *0.64 - 128.6 NFE 612 *0.74 - 452.9 707.7 g/kg DM

  11. Calculation 5 UFP How many grams of urea can be added to the daily ration of a beef cattle when 25 kg of corn silage is fed as an only feed? DM: 32.90%, TDN: 707.7 g/kg DM (if it is not given, the basic data are required for the calculation of TDN); dg: 69% (on DM basis).

  12. Solution UFP (g/kg DM): (1.044*TDN) – dg (g/kg DM) 2.8 (1.044*707.7) – 690  = 17.44 2.8 g urea/1 kg DM of corn silage; DM content of 25 kg of corn silage = 25*0.329 = 8.225 kg; Amount of urea can be added to 25 kg of corn silage: 8.225*17.44 = 143.5 g.

  13. Calculation 6 Check the following ration based on its energy content only whether it is sufficient for a limusine bull having 350 kg of body weight and 1300 g/day body weight gain. (Do not calculate with the CF, CP, Ca and P content of the ration!) Ration: 11 kg corn silage (predough stage), 15 kg sugar beet silage, 1.5 kg corn 0.6 kg sunflower extr., average; Requirements (you will get them) NEm: 31.30 MJ/day; NEg: 19.70 MJ/day.

  14. Solution Ration Feeds DM, kg NEm, MJ NEg, MJ 11 kg corn silage 3.84 26.03 16.28 15 kg sugar beet 2.16 16.76 11.04 1,5 kg corn 1.37 12.53 8.63 0,6 kg sunflower 0.54 3.52 2.17 Total 7.91 58.84 38.12 Requirements: 31.3019.70

  15. Solution Evaluation Tolerable level of Energy:  10% of the requirement; a) NEm 7.91 kg DM contains 58.84 MJ x kg DM contains 31.30 MJ (requirement) x = 31.30/58.84*7.91 = 4.21 kg DM contains the energy level for maintenance; DM remained for gain: 7.91-4.21 = 3.7 kg.

  16. Solution b) NEg7.91 kg DM contains 38.12 MJ 3.70 kg DM contains x MJ x = 3.70/7.91*38.12 = 17.83 MJ (requirement: 19.70 MJ); Difference:19.70-17.83 = 1.87 MJ; How many % is the difference of the requirement? 1.87/19.70*100 = 9.49% (the tolerable level is 10%, O.K.); the ration can cover the energy requirement of the bull.

  17. Calculation 7 1. How many kg of milking concentrate is required for producing 1 kg of milk when the milk fat is 3.7%, the milk protein is 3.4%? Do not calculate with the Ca and P content of the concentrate! The components of the milking concentrate are as follows. 0.4 kg sunflower meal, extr. (average), 0.6 kg corn.

  18. Solution Milking concentrate Feeds DM, kg NEl, MJ MPE, g MPN, g 0.4 kg sunflower 0.358 2.27 51 93 0.6 kg corn 0.547 4.66 61 39 1.0 kg 0.905 6.93 112 132 Which MP value is valid? .

  19. Solution 1. How many kg of milking concentrate is required for producing 1 kg of milk when the milk fat is 3.7%, the milk protein is 3.4%? 1. Calculation on energy basis NEl requirement of 1 kg milk: 1.471 + 0.4032*milk fat (%) = 1.471 + 0.4032*3.7 = 2.96 MJ;, 1 kg milking concentrate contains 6.93 MJ NEl x kg milking concentrate contains 2.96 MJ NEl x = 2.96/6.93 = 0.427 kg milking concentrate is required for producing 1 kg milk.

  20. Solution 1. How many kg of milking concentrate is required for producing 1 kg of milk when the milk fat is 3.7%, the milk protein is 3.4%? 2. Calculation on protein basis MP requirement of 1 kg milk: Milk protein (g/kg)/0.65 = 34/0.65 = 52 g; 1 kg milking concentrate contains 112 g MP x kg milking concentrate contains 52 g MP x = 52/112 = 0.464 kg milking concentrate is required for producing 1 kg milk. On energy basis: 0.427 kg  on protein basis: 0.464 kg??? MP is the limiting factor; Answer 1: 0.464 kg of milking concentrate is required for producing 1 kg of milk.

  21. Solution 2.Supplement the milking concentrate with 0.5% of premix and 0.5% of salt! Which component of the milking concentrate (sunflower or corn?) must be reduced by 1% owning to the supplementation? Answer 2: MP is the limiting factor, supplementation is made in account of the energy rich compound (corn). 3. As the final solution, write down the new receipt of the milking concentrate! Answer 3: 40% sunflower meal 59% corn 0.5% premix 0.5% salt 100%.

  22. Calculation 8 Check the following TMR whether it can cower the Ca and P requirement of the following dairy cow! (Do not calculate with the other nutrients!); In case of Ca or P deficiency please correct the TMR by using the adequate mineral supplements.

  23. Calculation 8 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, 2nd lactation.

  24. Solution 2. Nutrient Requirements Maintenance Ca requirement: 44*0.001*W = 25.52 g/day; P requirement: 34*0.001*W = 19.72 g/day; Correction 2nd lactation: +10% of the total maintenance req. (see later).

  25. Solution 2. Nutrient Requirements Milk Production Ca requirement: 2.8 g/kg milk; 25 kg milk = 25*2.8 = 70.00 g/day; P requirement: 1.7 g/kg milk; 25 kg milk: 25*1.7 = 42.50 g/day.

  26. Solution Sum of the Requirements Ca P (g) (g) Maintenance 25.52 19.72 +10% 2.55 1.97 25 kg milk 70.00 42.50 TOTAL 98.07 64.19

  27. Solution DM, Ca, and PContent of the TMR DM Ca P Feeds (kg) (g) (g) 5.0 kg hay 4.37 20.98 12.24 15.5 kg silage 5.41 17.85 13.52 2.6 kg sunfl. 2.33 7.22 17.47 8.4 kg CCM 5.28 3.17 13.20 TOTAL 17.39 49.22 56.43

  28. Solution 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.

  29. Solution Corrections Types of Ca and P supplements Name Ca (g) P (g) Limestone 380 - AP-17 210 170 AP-18 190 180 Phylafor 40 125 AP-IV 17 170

  30. Solution Corrections 1. Correction of the P deficiency by AP-IV: 1000 g AP-IV contains 170 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 contains 17 g Ca 45.65 g AP-IV contains x g Ca x = 45.65/1000*17 = 0.78 g Ca.

  31. Solution 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 contains 380 g Ca x g limestone contains 48.07 g Ca x = 48.07/380*1000 = 126.50 g limestone.

  32. Solution Final Evaluation - 45.65 g AP-IV and 126.50 g limestone supplementations are required.

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