**Example 1** (**initial and formation phase using a C**:**N ratio of 20**:**1**) **in a tilapia culture tank that receives 4 kg of feed** (**30**% **of crude protein**) **per day.**

### *Calculation 1* (*C*:*N content in the feed*)

C: 4 kg of feed × 0.9 (90% dry matter) × 0.7 (30% of fish assimilation or 70% of waste that remains in water)/2 (carbon content of the feed is ~50% based on dry matter) = 1260 g of C

N: 4 kg of feed × 0.9 (90% dry matter) × 0.7 (30% of fish assimilation or 70% of waste that remains in water) × 0.3 (30% crude protein content of feed)/6.25 (constant) = 121 g of N. **The results indicated a** ~**10**:**1 C**:**N ratio of feed**.

*Calculation 2* (*adjusting the C*:*N ratio*)

If I want a C:N ratio of 20:1, 121 g of N in feed × 20 = I need 2420 g of C. But I already have 1260 g of C (calculated in feed). So 2420 g–1260 g of C = I really need 1160 g of C.

If the molasses has 50% of carbon content (based on dry matter), 1 kg of molasses represents 500 g of carbon. So, 1160 g of carbon requirement will represent **2320 g** (**or 2**.**3 kg**) **of molasses** (**applied daily until biofloc maturation**).

**Example 2** (**maintenance phase and C**:**N ratio of 6**:**1**) **in a** *L*. *vannamei* **culture tank** (**30 m**<sup>3</sup> ) **that indicates 2**.**0 mg L**<sup>−</sup>**<sup>1</sup> TAN values**.

*Calculation 1* (*TAN in water*)

For 2.0 mg L−1 of TAN in a 30 m3 tank = 0.002 g × 30,000 L = 60 g of TAN

*Calculation 2* (*adjusting the C*:*N ratio*)

If I want a C:N ratio of 6:1, 60 g of TAN in water × 6 = I need 360 g of C. If my molasses has 50% of carbon content (based on dry matter), 1 kg of molasses represents 500 g of carbon. So, 360 g of carbon requirement will represent 720 g (or 0.72 kg) of molasses (one application and checked after 2–3 days).
