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Fertilizer is a major crop production input. Many different products are available at many different prices, so it can be difficult to accurately compare the available nutrient sources. That is, to compare “apples-to-apples” and “oranges-to-oranges”.
This Crop File provides a set of calculation steps to help with cost comparisons for individual nutrient units. The nutrient price is important, but not the only factor to consider. Considerations, like application cost, available equipment, ability to combine field operations, agronomic requirements, and others should weigh into the final purchase decision.
Calculating the cost per pound of nutrient (i.e., per unit) for a single nutrient fertilizer is rather simple:
¶ Step #1: Required Information |
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| Fertilizer | Analysis | Cost Per Ton |
| Anhydrous ammonia | 82-0-0 | $1250 |
| UAN solution | 28-0-0 | $550 |
| Urea | 46-0-0 | $650 |
¶ Step #2 |
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| Fertilizer | lb N/ton | = % N x 2000 lb/ton |
| Anhydrous ammonia | 1650 | = 85% x 2000 |
| UAN solution | 560 | = 28% x 2000 |
| Urea | 920 | = 46% x 2000 |
¶ Step #3 |
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| Fertilizer | $/lb N | = $/ton ÷ lb N/ton |
| Anhydrous ammonia | $0.76 | = $ 1250 ÷ 1640 |
| UAN solution | $0.98 | = $ 550 ÷ 560 |
| Urea | $0.71 | = $ 650 ÷ 920 |
Conclusion: In this example, unit cost of nitrogen was similar for urea and anhydrous ammonia, with urea slightly lower priced. UAN had the highest unit nitrogen cost.
Calculating the nutrient unit price per pound for multiple analysis fertilizer products is more difficult. Only one nutrient in the fertilizer is typically of primary or priority interest. Calculating only for the primary nutrient ignores the value of the other nutrients.
Calculating the primary nutrient requires setting standard costs for the other nutrient components of the mixed fertilizer. These standard values are based on the costs calculated from a single nutrient source.
The standard values of the single nutrient(s) are subtracted from the cost of the mixed fertilizer. The remainder of the cost represents the value of the primary nutrient.
Two fertilizer products each contain both nitrogen and phosphate. Ignoring the value of the nitrogen may skew the cost per unit of phosphate in an individual fertilizer. The value of the nitrogen is subtracted from the fertilizer cost to yield the cost of the phosphate nutrient.
Urea was used in this example as a standard nitrogen source, but another nitrogen fertilizer source may be preferred to calculate standard values.
¶ Step #1: Required Information |
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| Fertilizer | Analysis | Cost Per Ton |
| Diammonium phosphate (DAP) | 18-46-0 | $950 |
| Ammonium polyphosphate (APP) | 10-34-0 | $850 |
| Urea (standard nutrient) | 46-0-0 | $650 |
¶ Step #2 |
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| Nitrogen, N | lb N/ton | = % N x 2000 lb/ton |
| DAP (18-46-0) | 360 | = 18% N x 2000 |
| APP (10-34-0) | 200 | = 10% N x 2000 |
| Urea (46-0-0) | 920 | = 46% N x 2000 |
| Phosphate, P | lb P/ton | = % P x 2000 lb/ton |
| DAP (18-46-0) | 920 | = 46% P2O5 x 2000 |
| APP (10-34-0) | 680 | = 34% P2O5 x 2000 |
¶ Step #3 |
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| Fertilizer | $/lb N | = $/ton ÷ lb N/ton |
| Urea | $0.71 | = $ 650 ÷ 920 lb N/ton |
¶ Step #4 |
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| Fertilizer | $/lb N | = lb N/ton x $/lb N |
| DAP | $256 | = 360 x $0.71 |
| APP | $142 | = 2000 x $0.71 |
¶ Step #5 |
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| Fertilizer | $ value of P2O5/ton | = $/ton fertilizer - $ value of N/ton |
| DAP | $694 | = $ 950 - $ 256 |
| APP | $708 | = $ 850 - $ 142 |
¶ Step #6 |
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| Fertilizer | $/lb P2O5 | = $ value of P2O5/ton ÷ lb P2O5/ton |
| DAP | $0.75 | = $ 694 ÷ 920 |
| APP | $1.04 | = $ 708 ÷ 680 |
Conclusion: In this example, the phosphate cost of ammonium polyphosphate (10-34-0) was substantially higher than diammonium phosphate (18-46-0).
Comparing the nutrient cost of multiple nutrient fertilizers is challenging, but can be more difficult if they are priced using different units of volume or weight.
The comparison process is still the same, but the costs must be compared using equivalent units. This requires additional calculations to convert “cost per gallon” to “cost per ton”. The calculations are the same from Step #2 to Step #6 (in preceding section B).
Ammonium polyphosphate (APP) was priced on “per ton” basis. The other is a clear liquid fertilizer (CLS) that was priced on a “per gallon” basis. Both products contain nitrogen, but the CLS also contains potash.
UAN solution and muriate of potash were used in this example as standard nutrient sources, but other single nutrient fertilizer products may be preferred.
¶ Step #1: Required Information |
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| Fertilizer | Analysis | Cost | Density, lb/gal |
| Clear liquid starter (CLS) | 9-18-9 | $ 4.75/gallon | 11.05 |
| Ammonium polyphosphate (APP) | 10-34-0 | $ 850/ton | 11.65 |
| UAN solution | 28-0-0 | $ 550/ton | 10.76 |
| Muriate of potash, dry (MOP) | 0-0-60 | $ 700/ton | --- |
¶ Step #1a: |
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| Fertilizer | gal/ton | = 2000 lb/ton ÷ lb/gal | |
| CLS | 181.0 | = 2000 ÷ 11.05 | |
| APP | 171.7 | = 2000 ÷ 11.65 | |
| UAN | 187.4 | = 2000 ÷ 10.67 | |
¶ Step #1b: |
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| Fertilizer | $/ton | = 2000 lb/ton ÷ lb/gal | |
| CLS | $860 | = 181.0 x $4.75 | |
| APP | $850 | --- | |
| UAN | $550 | --- | |
¶ Step #2: |
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| Nitrogen, N | lb N/ton | = % nutrient x 2000 lb/ton | |
| CLS (9-18-9) | 180 | = 9% N x 2000 | |
| APP (10-34-0) | 200 | = 10% N x 2000 | |
| UAN (28-0-0) | 560 | = 28% N x 2000 | |
| Phosphate, P2O5 | lb P2O5/ton | = % nutrient x 2000 lb/ton | |
| CLS (9-18-9) | 360 | = 18% P2O5 x 2000 | |
| APP (10-34-0) | 680 | = 34% P2O5 x 2000 | |
| Potash, K2O | lb K2O/ton | = % nutrient x 2000 lb/ton | |
| CLS (9-18-9) | 180 | = 9% K2O x 2000 | |
| MOP (0-0-60) | 200 | = 60% K2O x 2000 | |
¶ Step #3 |
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| Fertilizer | $/lb nutient | = $/ton ÷ lb nutrient/ton |
| UAN | $ 0.98/lb N | = $ 550 ÷ 560 |
| MOP | $ 0.39/lb K2O | = $ 700 ÷ 1200 |
¶ Step #4: |
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| Clear liquid starter | $ value /ton | = lb/ton X $/lb nutrient | |
| N | $176 | = 180 x $ 0.71 | |
| K2O | $70 | = 180 x $ 0.39 | |
| Total | $246 | = $ 176 + $ 70 | |
| Ammonium polyphosphate | lb P2O5/ton | = % nutrient x 2000 lb/ton | |
| N | $142 | = 200 x $ 0.71 | |
¶ Step #5 |
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| Fertilizer | $ of P2O5/ton | = $/ton fertilizer - $ of (N + K2O)/ton |
| CLS | $614 | = $ 860 - $ 246 |
| APP | $708 | = $ 850 - $ 142 |
¶ Step #6 |
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| Fertilizer | $/lb P2O5 | $ value of P2O5/ton ÷ lb P2O5/ton |
| CLS | $1.71 | = $ 614 ÷ 360 |
| APP | $1.04 | = $ 708 ÷ 680 |
Conclusion: In this example, the clear liquid starter (9-18-9) was substantially more expensive as a phosphate source than was ammonium polyphosphate (10-34-0).
It may be useful at times to calculate “price per gallon” from “price per ton” information.
A clear liquid fertilizer (CLS) that was priced on a “per gallon” basis. Ammonium polyphosphate (APP) and UAN solution were priced on “per ton” basis
¶ Step #1: Required Information |
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| Fertilizer | Analysis | Cost | Density, lb/gal |
| Clear liquid starter (CLS) | 9-18-9 | $ 4.75/gallon | 11.05 |
| Ammonium polyphosphate (APP) | 10-34-0 | $ 850/ton | 11.65 |
| UAN solution | 28-0-0 | $ 550/ton | 10.76 |
¶ Step #2 |
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| Fertilizer | gal/ton | = 2000 lb/ton ÷ lb/gal |
| CLS | 181.0 | = 2000 ÷ 11.05 |
| APP | 171.7 | = 2000 ÷ 11.65 |
| UAN | 187.4 | = 2000 ÷ 10.67 |
¶ Step #3 |
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| Fertilizer | $/gal | = $/ton ÷ gal/ton |
| CLS | $4.75 | --- |
| APP | $4.95 | = $ 850 ÷ 171.7 |
| UAN | $2.93 | = $ 550 ÷ 187.4 |
Conclusions: