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

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.

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

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,NE l : 8.50 MJ/kg, NEg: 7.00 MJ/kg.

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 = MJ/kg; CP/ME = 290 (g/kg)/12.13 (MJ/kg) = g/MJ.

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%.

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

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

Calculation 4 DM: 32.90%,Corn Silage Nutrients on DM basis (%) CP: 9.20EE: 4.20 CF: 20.10NFE: Nutrients’ Digestibility (%) CP: EE: CF: NFE:

Solution TDN (g/kg DM): DCP *DEE + DCF + DNFE Calculation of the TDN Nutrientsg/kg DMDig.factorTDN CP 92* EE 42*0.80* CF201* NFE612* g/kg DM

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: g/kg DM (if it is not given, the basic data are required for the calculation of TDN); dg: 69% (on DM basis).

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

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: MJ/day; NEg: MJ/day.

Solution Ration FeedsDM, kgNEm, MJNEg, MJ 11 kg corn silage kg sugar beet ,5 kg corn ,6 kg sunflower Total Requirements:

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

Solution b) NEg7.91 kg DM contains MJ 3.70 kg DM contains x MJ x = 3.70/7.91*38.12 = MJ (requirement: MJ); Difference: = 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.

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.

Solution Milking concentrate Feeds DM, kgNE l, MJ MPE, gMPN, g 0.4 kg sunflower kg corn kg Which MP value is valid? ..

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 NE l requirement of 1 kg milk: *milk fat (%) = *3.7 = 2.96 MJ;, 1 kg milking concentrate contains 6.93 MJ NE l x kg milking concentrate contains 2.96 MJ NE l x = 2.96/6.93 = kg milking concentrate is required for producing 1 kg milk.

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 = kg milking concentrate is required for producing 1 kg milk. On energy basis: kg  on protein basis: kg??? MP is the limiting factor; Answer 1: kg of milking concentrate is required for producing 1 kg of milk.

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%.

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.

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.

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

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

Solution Sum of the Requirements Ca P (g) (g) Maintenance % kg milk TOTAL

Solution DM, Ca, and P Content of the TMR DMCaP Feeds(kg)(g)(g) 5.0 kg hay kg silage kg sunfl kg CCM TOTAL

Solution Ca (g) Requirement:98.07 TMR:49.22 Difference: Tolerable level:2.8*2 =  5.6 Evaluation:insufficient P (g) Requirement:64.19 TMR:56.43 Difference: Tolerable level:1.7*2 =  3.4 Evaluation: insufficient.

Solution Corrections Types of Ca and P supplements NameCa (g)P (g) Limestone380 - AP AP Phylafor AP-IV 17170

Solution 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 = g AP-IV 2. Calculation of the amount of Ca given by AP-IV: 1000 g AP-IV contains17 g Ca g AP-IV contains x g Ca x = 45.65/1000*17 = 0.78 g Ca.

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

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