Lithium Argentina AG (LAR) Asset Utilization Ratio (2018 - 2025)
Lithium Argentina AG's (LAR) Asset Utilization Ratio stood at -0.03 in Q4 2025.
Analysis
Lithium Argentina AG (LAR) Asset Utilization Ratio (2018 - 2025) Analysis & Trends
For FY2025, Lithium Argentina AG's Asset Utilization Ratio came in at -0.01.
- Across earlier years, Asset Utilization Ratio came in at -0.01 in FY2024, 0.05 in FY2022 and 0.00 in FY2021 (+107.6%).
- The Q4 2025 figure is the highest quarterly Asset Utilization Ratio since Q4 2018.
Peer Set
Peer Comparison
| # | Company | Market Cap | Enterprise Value | Gross Profit (Qtr) | Asset Util. (Qtr) |
|---|---|---|---|---|---|
| 1 | Rio Tinto | 181.90 Bn | 148.03 Bn | - | - |
| 2 | Southern Copper | 169.19 Bn | 148.06 Bn | 2.90 Bn | 0.69 |
| 3 | Newmont | 121.55 Bn | 89.73 Bn | 4.03 Bn | 0.45 |
| 4 | Ternium | 108.92 Bn | 73.97 Bn | 941.02 Mn | 0.66 |
| 5 | Freeport-Mcmoran | 100.46 Bn | 96.62 Bn | 2.19 Bn | 0.44 |
| 6 | Agnico Eagle Mines | 91.62 Bn | 91.62 Bn | 2.43 Bn | - |
| 7 | Barrick Mining | 68.12 Bn | 52.86 Bn | 2.90 Bn | 0.39 |
| 8 | Nucor | 53.04 Bn | 43.58 Bn | 2.03 Bn | 0.99 |
| 9 | ArcelorMittal | 50.74 Bn | 31.77 Bn | - | 0.64 |
| 10 | Lithium Argentina AG | 891.61 Mn | 502.25 Mn | - | - |
API Access
Lithium Argentina AG Asset Utilization Ratio API
Pull this series into your own models, spreadsheets and apps with the Business Quant
Historical Metrics API. The request below matches the chart above — change the
frequency, period or values and it follows. Swap YOUR_API_KEY for your own key.
https://data.businessquant.com/historic?slug=asset-utilization-ratio&ticker=LAR&period=max&api_key=YOUR_API_KEY
import requests
url = "https://data.businessquant.com/historic"
params = {"slug": "asset-utilization-ratio", "ticker": "LAR", "period": "max", "api_key": "YOUR_API_KEY"}
data = requests.get(url, params=params).json()
const res = await fetch("https://data.businessquant.com/historic?slug=asset-utilization-ratio&ticker=LAR&period=max&api_key=YOUR_API_KEY");
const data = await res.json();