Extending baker's percentages to preferment recipes

Question

I'm trying to solve a simple problem I created for myself. I'm no mathematician, so any help is greatly appreciated.

Background

In baking and "baker's math", the amount of each ingredient is often expressed as a percentage of the total flour amount. For instance, it's common to use 2.5% salt with pizza and bread doughs. If our recipe calls for 1000g of flour, we would have to add $1000 \cdot 0.025 = 25g$ salt to the dough. Similarly, hydration is a measurement describing how wet the dough is, and is expressed as $h = \frac{w}{f}$, where $w$ is the total amount of water in grams, and $f$ is the total amount of flour in the dough.

Imagine we want to scale a bread recipe to have a total weight, denoted $W$, of 1000g, with a hydration of 65%, 2.5% salt (denoted as variable $a$), and 0.5% dry active yeast (denoted as variable $b$). The total weight of the dough can be calculated by

$$ W = f \cdot (1 + h + a + b) $$

Isolating for $f$, we get

$$ f = \frac{W}{1 + h + a + b} $$

Inserting our input values, we can calculate the total amount of flour

$$ f = \frac{1000g}{1 + 0.65 + 0.025 + 0.005} = 595.24 g $$

When $f$ is known, we can calculate the other ingredients to get the absolute values needed for the recipe.

Extending to preferments

When baking at scale, bakers often use preferments such as poolish or biga. Preferments are a type of dough that contains the rising agents and ferment at a smaller scale before being mixed with the remaining ingredients, typically the day after. Preferments usually have a different hydration and ingredients than the "main dough" (poolish is usually 100% hydrated, whereas biga is 45% hydrated). Scaling these recipes become a bit harder, since the ingredients are now split into two groups, but are still expressed as a function of the total amount of flour across the two groups.

Input variables

Say we know/specify the following variables and use them as inputs:

• $W$ (target total weight in grams)
• $h$ (target total hydration)
• $h_p$ (preferment hydration)
• $r$ (the ratio/percent of the entire dough that consists of preferment. $0 \leq r \leq 1$)
• $a$ (ingredient a, as a percent of total flour. Is only added to preferment)
• $b$ (ingredient b, as a percent of total flour. Is only added to main dough)

Objective

Find closed formulas to calculate the absolute values for all ingredients.

My calculations so far

The total weight of the preferment ($W_p$) and the main dough ($W_m$) can be calculated as follows:

$$ W_p = W \cdot r \\ W_m = W \cdot (1 - r) $$

I believe the total hydration can be expressed as follows

$$ h = r \cdot h_p + (1 -r) \cdot h_m $$

where $h_m$ is the hydration of the main dough, and is the only unknown variable in this formula. Isolating it, we get

$$ h_m = \frac{h - r \cdot h_p}{1-r} $$

The total amount of flour in the recipe must be the sum of flour in the preferment ($f_p$) and in the main dough ($f_m$):

$$ f = f_p + f_m $$

Like in the background section, the total amount of flour can be calculated by

$$ f = \frac{W}{1 + h + a + b} $$

from which we can calculate the total amount of water ($w$), amount of ingredient $a$ ($a_{abs}$), and amount of ingredient $b$ ($b_{abs}$):

$$ w = f \cdot h \\ a_{abs} = f \cdot a \\ b_{abs} = f \cdot b $$

From this, I believe we can calculate the amount of flour and water in the preferment and main dough, respectively:

$$ f_p = \frac{W_p - a_{abs}}{1 + h_p} \\ f_m = \frac{W_m - b_{abs}}{1 + h_m} \\ w_p = f_p \cdot h_p \\ w_m = f_m \cdot h_m $$

However, when I plug in real values, there's an array of issues. Here are the input values:

• $W = 1000 g$
• $h = 0.65$
• $h_p = 1.00$
• $r = 0.2$
• $a = 0.2$
• $b = 0.8$

I input these values into the equations:

$$ W_p = W \cdot r = 1000 \cdot 0.2 = 200 $$ $$ W_m = W \cdot (1-r) = 1000 \cdot (1-0.2) = 800 $$ $$ h_m = \frac{h - r \cdot h_p}{1-r} = \frac{0.65 - 0.2 \cdot 1}{1-0.2} = 0.56 $$ $$ f = \frac{W}{1 + h + a + b} = \frac{1000}{1 + 0.65 + 0.2 + 0.8} = 377.36 $$ $$ w = f \cdot h = 377.36 \cdot 0.65 = 245.28 $$ $$ a_{abs} = f \cdot a = 377.36 \cdot 0.2 = 75.47 $$ $$ b_{abs} = f \cdot b = 377.36 \cdot 0.8 = 301.89 $$ $$ f_p = \frac{W_p - a_{abs}}{1 + h_p} = \frac{200 - 75.47}{1 + 1} = 62.26 $$ $$ f_m = \frac{W_m - b_{abs}}{1 + h_m} = \frac{800 - 301.89}{1 + 0.56} = 318.79 $$ $$ w_p = f_p \cdot h_p = 62.26 \cdot 1 = 62.26 $$ $$ w_m = f_m \cdot h_m = 318.79 \cdot 0.56 = 179.32 $$

The following issues arise:

• The actual hydration (0.63) differs from the target hydration (0.65). I.e. $(w_p + w_m)/(f_p + f_m) = 0.63 \neq 0.65 = h$
• The sum of flour in preferment ($f_p$) and in main dough ($f_m$) (total 381.06 g) does not equal $f$ (377.36 g)
• Similarly, the sum of water in preferment ($w_p$) and in main dough ($w_m$) (total 241.58g) does not equal $w$ (245.28g)

What am I missing here? Is it possible to calculate these values given the input variables?

Answer 1

You say that A and B scale with the total amount of flour, which would mean

$$a = \frac{\text{total amount of salt}}{\text{total amount of flour}} = \frac{a_{\rm abs}}{f_m+fp}$$

and in turn, the weight of each dough component would be

$$\begin{cases} W_m = f_m + w_m + b_{\rm abs} = \left(1+h_m+b\right)f_m + bf_p \\ W_p = f_p + w_p + a_{\rm abs} = \left(1+h_p+a\right)f_p + af_m\end{cases}$$

Now with $(W,r,a,b)=(1000,0.2,0.2,0.8)$ and $\left(W_m,W_p\right)=((1-r)W,rW)$, we have

$$\begin{cases} 800 = \left(1.8 + h_m\right) f_m + 0.8 f_p \\ 200 = \left(1.2 + h_p\right) f_p + 0.2 f_m \end{cases} \tag{$\star$}$$

We want a total hydration of $h=0.65$ and we know the preferment's hydration is $h_p=1.00$, so $w_p=f_p$ and

$$h = \frac{w_m + w_p}{f_m + f_p} = 0.65 \implies w_m = 0.65 f_m - 0.35 f_p$$

Substitute this into $(\star)$ and you will get the expected total flour and water:

$$\begin{cases} 800 = 2.45 f_m + 0.45 f_p \\ 200 = 2.2 f_p + 0.2 f_m \end{cases} \implies \left(f_m,f_p\right) \approx (315.09, 62.26) \\ \implies \begin{cases} \boxed{f_m+f_p \approx 337.36} \\ w_m \approx 183.02 \\ \boxed{w_m+w_p \approx 245.28} \end{cases}$$

We also see that $\frac{f_p}{f_m+f_p} = 0.165 \neq 0.2 = r = \frac{W_p}W$, as suspected.