Ration Balancing, by Hand or by Computer

Dale M. Forsyth, Department of Animal Sciences, Purdue University

Statement of Purpose: I have received numerous requests for a reprint of a paper I wrote entitled Computer Programming of Beef Cattle Diets, a chapter in the book BEEF CATTLE FEEDING AND NUTRITION, 2nd Ed., by T.W. Perry and M.J. Cecava, Academic Press. That article is copyrighted and I cannot distribute it. There seems to be interest in ration balancing by students, producers, and others, at various levels of complication, so I am writing this and making it widely available in order to share what I know about the subject. The principles apply equally well to rations for any species.

My descriptions will be broken into separate parts: Hand calculation methods, simple computer methods, more advanced computer methods.


In all of the following methods, corn is used as an example grain; SBM is used as an example protein source; HAY is used as an example forage, as is CS (corn silage); Dical is used as an example Phosphorus source and Lime stands for limestone, a calcium source.

The EASIEST and most practical way to balance this type of ration by hand is to balance for 1 nutrient, protein or lysine (whichever is desired or more appropriate), leaving slack space for vitamins and minerals to be balanced for later. Additional feeds beyond Corn and Soybean Meal can be added in FIXED AMOUNTS. ENERGY is not specifically balanced for: the ration contains all concentrate feeds and has as much energy as possible without the addition of FAT. Fat can be added on a practical, thumb rule basis: 1 or 2% for dust control purposes and about 6% maximum for added energy without making the diet greasy and unable to feed down in feeders. (If a specific energy level is desired, though, the simultaneous equation method can be used).

One nutrient. With (or without) one or more fixed ingredients. With or without added slack space (if no slack space or fixed ingredients, just make them equal to zero). In this method there can be just 2 variables (feeds of unknown amounts).

Let's let X=Corn, Y=SBM, FF=fixed feeds, Slack=free space for vitamins and minerals.

On % basis:

Total[100] = Corn + SBM + Fixed_feeds + Slack

CP = (Corn)(CPcorn) + (SBM)(CPsbm) + (FF)(CPff)

solving: Total - FF - Slack - SBM = Corn

example: on % basis, total = 100. SBM = 100 - ff - slack - corn.

if ff = 10 and slack = 3 then SBM = 87-corn

Assume you want 16% CP and 10% whey, corn contains 8% CP, SBM contains 44% CP, and whey contains 12% CP.

substitute into 2nd equation and solve.

16 = X (0.08) + (87 - X) (0.44) + 10 (0.12) + 3 (0)

16 = 0.08X + 38.28 - 0.44X + 1.2 + 0

16 - 38.28 - 1.2 = 0.08X - 0.44X

-23.48 = -.36X and X = 65.22 Corn and (87-X) = 21.78 SBM

Description in words:

When balancing a ration on a % basis, all the feeds add up to the total, 100% (or 100 pounds). Two feeds (only) can vary, here corn and soybean meal. Everything that isn't specified, and isn't corn, is soybean meal. If we call corn=X, feed 10% whey and leave 3% slack for vitamins and minerals, then the amount of SBM is 100 - 10 - 3 - X, or 87-X. We multiply the % crude protein (CP) we want by 100, the amount of corn we want (X) by the % CP in corn, the % SBM we want (87-X) by the % CP in SBM, the amount of whey (10) (or other fixed feed) by the % protein in whey (or other fixed feed) and the amount of slack (3) by 0 (there is no CP in the slack). It is easier to do than to write about. Just multiply it out and solve for X, then solve of 87-X. You need 65.22% corn, 10% whey and 21.78% SBM. You still have 3% slack for vitamins and minerals, which we'll deal with presently.

On pounds basis:

Set up and solve the same way, except let the TOTAL equal the number of pounds of dry matter intake (or as-fed intake).

Total = Corn + SBM + Fixed_feeds + Slack

CP[lbs desired] = (Corn)(CPcorn) + (SBM)(CPSBM) + (FF)(CPff)

solving: Total - FF - slack - SBM = Corn

Example: Say intake = 25 lbs; slack = 0.5 lb; ff = 2 lbs

Corn = 25 - 0.5 - 2 - SBM = 22.5 - SBM

Substitute (22.5 - SBM) for corn in second equation and solve.

if 14% CP is desired, then:

(0.14)(25)=(22.5-SBM)(CPcorn) + (SBM)(CPSBM) + (2)(CPff)

Solving for ingredients in the Slack Space


Dicalcium phosphate (and most phosphorus sources) contain both phosphorus AND calcium. Limestone, our source of extra calcium, contains no phosphorus. So balancing for phosphorus (P) first, avoids conflicts.

EXAMPLE: If 0.5% (0.5 lbs in 100 lbs) of Phosphorus is desired, then:

From above: corn=65.22%, Whey=10%, SBM=21.78%; say P levels are: corn = 0.29%, Whey = 0.76%, SBM = 0.60%, Dical = 18.5%

%Pdesired (100) = (%Corn)(%Pin_corn) + (%SBM)(%Pin_sbm) + (%Whey)(%Pin_whey) + %DICAL(%Pin_dical)

0.50 = (65.22)(.0029) + (21.78)(.0060) + (10)(.0076) + (X)(.185)

X = 0.56 lbs of Dical in 100 lbs of feed, or 0.56% Dical.


Limestone (38% Calcium) is our source of calcium. Ca in: Corn = 0.02%, Whey = 0.97%, SBM = 0.25%, Dical = 22%

If 0.6% (0.6 lbs in 100 lbs) of Calcium is desired, then, using the same example:

0.60 = (65.22)(.0002) + (21.78)(.0025) + (10)(.0097) + (.56)(.22) + (X)(.38)

X = 0.82 lbs of Limestone in 100 lbs of feed, or 0.82% Limestone

Other Ingredients

Add salt at a set, practical level 0.25% to 0.50%; If adding a selenium premix, add it as specified; add Trace Mineral Premixes and Vitamin Premixes as formulated and Recommended by the manufacturer; Add any drug at the manufacturers specifications. Add something as filler, to make the whole ration equal 100%; using extra corn for filler is acceptable and common and changes the total ration composition insignificantly.

Power of the Method

The methods just described in the previous section can be employed in a wide variety of situations. For example, if balancing a beef ration for finishing steers, specifying as a fixed feed 15% forage as hay or corn silage, and balancing using corn and SBM exactly as shown works fine. The compositions used in this case should be DRY MATTER composition and the results will be in drymatter, requiring conversion to AS-FED amounts at the end.

If balacing for a Dairy ration, and you want a 40:60 concentrate to forage ratio, set forage as a fixed feed at 60 and balance as shown.

More exact and more complicated methods can be used, however.

Solving for MORE than 1 NUTRIENT

Solving for extra nutrients means adding extra equations, and solving them simultaneously

Let's let X=corn, Y=SBM, Z=hay

eg: 100 = X + Y + Z

CP = X(CPx) +Y(CPy) + Z(CPz)

TDN = (X)(TDNx) + (Y)(TDNy) + (Z)(TDNz)

Solve this equation set to find the amounts of X, Y and Z.

NOTE: You must have exactly 3 feeds to solve for 3 equations. If you add another nutrient, you must add another equation. You cannot have more feeds in unknown amounts. (You could deal with fixed amounts of feeds and slack space, however).


stay tuned for more hand balancing methods to come, though

A METHOD EXISTS FOR DOING THIS (the previous simultaneous equation method) EASILY, BY MATRIX METHODS, especially if you have a computer spreadsheet program!

Solving Simultaneous Equations by Matrix Methods

Using matrix algebra gives us a very powerful method of solving simultaneous equations with the aid of a computer. QUATTRO or LOTUS 123 or EXCEL, for example, will solve matrix problems and some programming languages (such as certain BASIC's) do also.

Let the requirements (also called restrictions) be represented by R. R is a matrix of n rows and 1 column, where n=the number of requirements.

Example: let the R matrix be:


Let the feedstuffs analysis values be placed in a matrix represented by C. C is a matrix of n rows and n columns, where n=the number of nutrients which also must = the number of requirements.

Example: let the C matrix be:

1 1 1

Let the feedstuff amounts, the answers, be represented by B.

Example: the B matrix be:


Then, in matrix notation, R = B C

Therefore: B = Cinverse R

In other words, to get the amount of each feed to feed, just multiply the inverse of C (the composition table) by R (the requirements). The computer programs easily invert the matrix and multiply matrices.

Precaution. This method, just as all simultaneous equations methods, gives exact solutions mathematically, without regard to practicality. Negative numbers are possible, even though they make no sense nutritionally.



It is not a complicated additional step from setting up simultaneous equations to set up the equations for least-cost solutions. Solving the equation set is harder, we'll let a computer program do that.

Conceptually, the equation set could be:

X + Y + Z = 100

X(CPx) +Y(CPy) + Z(CPz) > CP

(X)(TDNx) + (Y)(TDNy) + (Z)(TDNz) > TDN

and we must have a decision function (PRICE) for each feed.

X($X) Y($Y) Z($Z)

Another difference is that now we could have a great number of feeds, not just exactly the same number as 'non-trivial equations' that we have. And the solution would use just the feeds that make up the least-cost solution, in the correct amounts.

In all the equations above, we could have had fixed amounts of feeds (FF) and added that term as appropriate, and the nutrient content of FF, to each line in the sets of equations.


When exact solutions for 2 nutrients are not required, but the requirement for 2 nutrients must be met or exceeded, you can use the following method which requires some judgement and decision on your part.

Start by estimating the feed intake of the animal. It is important that you balance a ration that is realistic in terms of expected feed intake.

Choose a fixed amount of forage to feed. You might choose one-half or three-forths of the expected dry matter intake, for example.

Calculate the contribution toward the requirements provided by the forage and figure how much Protein and Energy you still need.

Using an average ENERGY value for the concentrate you will feed, calculate the pounds of concentrate you will have to feed to meet the energy requirement. (ie, divide the TDN reqd by the avg TDN in the concentrate mix).

Finally, balance the concentrate (between corn and SBM, for example), to provide the additional protein (if any) is still needed.

EXAMPLE: A 1000 lb beef cow requiring about 20 lbs dry matter needs 10.5 lbs TDN and 1.6 lbs CP.

I choose to feed 1.25% of the cow's body weight as forage (expected DM intake = 1.5 to 2.25). The forage I'll use is Timothy Hay (9% CP, 56% TDN).

1000 lb X .0125 = 12.5 lbs timothy hay.

Rqmt still needed

12.5 X .09 = 1.125 lb CP 1.6 - 1.125 = .475 CP

12.5 X .56 = 7 lb TDN 10.5 - 7 = 3.5 TDN

Let's assume our concentrate mix will average 78% TDN. (We can't know exactly before-hand; estimate).

3.5 divided by .78 = 4.48 lbs concentrate mix.

What percent protein should our concentrate mix be?

.475 lb CP divided by 4.48 lbs of feed = 10.6 % CP.

(4.48)(.106)=(x)(.08)+(4.48-x)(.44) x=corn, 4.48-x=SBM

.475=.08x - 1.9172 - .44x 0.08=% CP in Corn,

-1.496 = -.36x 0.44=%CP in SBM

4.16 = x = corn; 4.48-x= 0.32 lb = SBM


To calculate the amount of dry matter from the concentration of dry matter in a feed, take:

(Feed Amount as-fed, lbs) times (% DM) = Lbs of dry matter.

Example: you feed 25 lbs and it is 90% DM. How many lbs of dry matter did you feed? 25 X .9 = 22.5

To calculate the amount of as-fed (read: WET) feed from the dry matter amount and the %DM in the as-fed feed:


(X lbs of as-fed feed) (%DM) = dry matter, lbs

Lbs DM
---------- = Lbs of AS-FED FEED


Example: You feed 20 lbs of corn silage dry matter and corn silage has 30% DM. How much feed do you weigh out to feed?

--- = 66.7 lbs. THAT's a LOT, isn't it.
.30 Remember corn silage is 2/3 water.

(well, really 70% in this example)

posted to the web 3-28-1997

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