One of my favorite phrases is: “We are drowning in information while starving for wisdom.” I am sure everyone has felt this at some point in their life.

Dairy farms are great at recording data and collecting information; however, analysis of this information can be challenging. When evaluating data, these three numbers should be reviewed to not lose sight of the big picture.

## Pounds of solids

I am sure you have read or heard someone talking about pounds of solids in an article or a presentation before. This is a calculation shows how many pounds of fat and protein the average cow is producing. When I ask producers where their components are at, they immediately tell me their percentage of fat and protein. But is this really the best answer? Yes, a 4% fat and a 3.1% protein are good, but this answer tells a different story at 95 pounds of milk compared to 85 pounds of milk.

To calculate the pounds of fat and protein, simply add the two percentages (fat and protein) together and multiply the sum by your milk production. For this example, take 4% fat plus 3.1% protein = 7.1%. Multiply 7.1% by 85 pounds to get 6.035 pounds of solids. 7.1% multiplied by 95 pounds is 6.745 pounds. This is a massive difference both in production of the cow and the financial bottom line that cow is generating.

Why is this important? Producers get paid based on pounds of solids and pounds of protein – not the percentages of each. I often challenge producers to think about making decisions based on how many pounds of solids are shipped down their driveways instead of average milk production.

## Pounds per stall

I look at this number when I am trying to help a producer understand their total farm efficiency and a way to increase a farm’s bottom line. Let’s use a four-pen barn with 40 stalls in each pen. This would give us room for 160 cows at 100% stocking rate and 200 cows at 125% stocking rate. Since very few producers are at 100% stocking rate, let’s use the 200-cow example.

These cows are averaging 85 pounds of milk per day. Two hundred cows multiplied by 85 pounds of milk, divided by 160 stalls equals 106.25 pounds of milk per stall.

We have worked with many of our producers to look at making their herds more efficient. We have done this by increasing their milking numbers by 10%. For this example, they would milk 20 more cows, or five more cows per pen. If the herd’s milk production stays the same while milking more cows, then for milk per stall you would take 220 cows multiplied by 85 pounds of milk and then divide by 160 stalls to get 116.88 pounds of milk. This means average milk production can go down to 77.25 pounds of milk and you will be shipping the same amount of milk as a 200-cow dairy. We have experienced a very minimal drop in milk, if any at all, when we have implemented this thought process. Yes, management must increase and there are costs associated with feeding the cows, but your fixed daily costs do not go up – and thus your return over feed cost is very profitable with these extra 20 cows.

Another factor that plays into this number is: What percentage of your herd is first-lactation versus mature cows? A herd with a high percentage of cows in their first lactation (over 40% of the herd) is using stalls in their barn for cows producing 10 to 15 pounds of milk less than mature cows. Therefore, their milk per stall ratio is less, resulting in the entire barn being less efficient.

## Component efficiency

Component efficiency means, “How efficiently are your cows producing their components?” In other words, how many pounds of dry matter does it take to produce the pounds of fat and protein in the milk? Let’s use the same example for a cow producing 85 pounds of milk with a 4% fat and 3.1% protein, or 6.035 pounds of solids. If the average cow eats 57 pounds of dry matter, her component efficiency can be calculated by taking 6.035 pounds of solids and dividing by 57 pounds of dry matter, which equals an efficiency rate of 10.58%. Compared to a cow that may produce 87 pounds of milk with a 4.2% fat and 3.2% protein, or 6.438 pounds of solids, while eating 61 pounds of dry matter or an efficiency rate of 10.55%. On paper, you see the higher milk and higher components and think that herd is doing better but, in reality, because of the higher intake the 85-pound herd is more efficient.

Now this doesn’t mean the herd that is more efficient is more profitable. Many factors play into profitability, like component pay price, percentage of the herd that is milking, feed costs, etc. This just means there may need to be a discussion regarding ration efficiency and what forages and byproducts are being (or could be) fed to make the cows more efficient.

These three numbers are some of the many numbers we can calculate and track to help better understand your herd’s efficiency and profitability. Our goal is always to have healthy cows first and do what we can to help our producers increase their bottom line for their business.

• ### Tim Kinches

• Technical Services Specialist
• Form-A-Feed
• Email Tim Kinches