# Extending baker's percentages to preferment recipes

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?

• Perhaps I'm misunderstanding but it seems you've made mistakes plugging values into the formulae. For example, for $f = W/(1 + h + a + b)$, I get $f = 1000/(1 + 0.65 + 0.2 + 0.8) = 1000/2.65 \approx 377$, not $\approx 425.53$. Commented Jul 16, 2023 at 15:54
• Looking at the total hydration equation: $h=0.65$, $r=0.2$, and $h_p=1$ gives $h_m=0.5625$, and sure enough,$$0.8\times0.5625+0.2\times1=0.65\quad.$$How exactly are you getting your numbers? Commented Jul 16, 2023 at 22:33
• @Peter Sorry, I put in a typo in my worksheet. I have edited my post to reflect the numbers I use. The issues are still the same, though. Commented Jul 17, 2023 at 11:08
• @user170231 Sure, because that is how I define $h$. However. when I plug in the actual amounts of water and flour, the hydration is different from $h$. I have edited my post to explain. Commented Jul 17, 2023 at 11:10
• If I'm following correctly, you're essentially claiming that the ratio $r$ is the same for {flour in preferment} to {total flour} as {preferment weight} to {total dough weight} which I suspect is not correct:$$h = \frac{w_m + w_p}{f_m + f_p} = \frac{h_mf_m+h_pf_p}{f_m+f_p} \\ \implies h_m = \frac{h\left(f_m+f_p\right)-h_p f_p}{f_m} = \frac{h-h_p \frac{f_p}{f_m+f_p}}{\frac{f_m}{f_m+f_p}} \stackrel?= \frac{h-rh_p}{1-r} \\ \implies r=\frac{f_p}{f_m+f_p} = \frac{W_p}W$$ Commented Jul 17, 2023 at 14:03

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.

• +1 but please don't report those spuriously precise decimal values. Commented Jul 17, 2023 at 17:36