
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.


METHODS OF HAND RATION CALCULATION 
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
= 87corn
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 (87X) = 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 87X. 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 (87X) 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 87X.
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 asfed 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)=(22.5SBM)(CPcorn) + (SBM)(CPSBM)
+ (2)(CPff)
Solving for ingredients in the Slack
Space 
ALWAYS SOLVE FOR PHOSPHORUS FIRST! 
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. 
NEXT SOLVE FOR CALCIUM
Limestone (38% Calcium) is our course
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 ASFED
amounts at the end 
If balacing for a Dairy ration,
and you want a 60:40 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). 
COMPUTER METHODS
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:
TOTAL
CP
TDN
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
CP1 CP2 CP3
TDN1 TDN2 TDN3
Let the feedstuff amounts, the answers,
be represented by B.
Example: the B matrix be:
Feed1
Feed2
Feed3
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. 
Solving with INEQUALITIES
LINEAR PROGRAMMING LEAST COST SOLUTIONS
It is not a complicated additional step from setting
up simultaneous equations to set up the equations
for leastcost 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 'nontrivial equations' that we
have. And the solution would use just the feeds that
make up the leastcost 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. 
OTHER PRACTICAL
HAND METHODS OF RATION BALANCING
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 onehalf or threeforths
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 beforehand;
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)+(4.48x)(.44)
x=corn, 4.48x=SBM
.475=.08x  1.9172  .44x 0.08=%
CP in Corn,
1.496 = .36x 0.44=%CP in SBM
4.16 = x = corn; 4.48x= 0.32 lb
= SBM 
HOW TO CALCULATE DRY MATTER AND
ASFED
To calculate the amount of dry matter from the concentration
of dry matter in a feed, take:
(Feed Amount asfed, 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 asfed
(read: WET) feed from the dry matter amount and the
%DM in the asfed feed:
Think:
(X lbs of asfed feed) (%DM) = dry
matter, lbs
Lbs DM
 = Lbs of ASFED FEED
%DM
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?
20
 = 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 3281997



