Medline (MDLN) Accumulated Expenses (2024 - 2026)
Medline (MDLN) recorded Accumulated Expenses of $1.64 billion in Q2 2026, up 7.3% from the prior quarter.
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Analysis
Medline (MDLN) Accumulated Expenses (2024 - 2026) Analysis & Trends
At the end of FY2025, Medline reported Accumulated Expenses of $1.45 billion, down 2.7% from FY2024.
- Across earlier years, Accumulated Expenses came in at $1.49 billion in FY2024.
Peer Set
Peer Comparison
| # | Company | Market Cap | Enterprise Value | Gross Profit (Qtr) |
|---|---|---|---|---|
| 1 | Thermo Fisher Scientific | 249.71 Bn | 228.86 Bn | 4.88 Bn |
| 2 | Abbott Laboratories | 171.00 Bn | 142.08 Bn | 7.27 Bn |
| 3 | Danaher | 155.79 Bn | 139.61 Bn | 3.61 Bn |
| 4 | Intuitive Surgical | 143.91 Bn | 123.46 Bn | 1.96 Bn |
| 5 | Medtronic | 110.42 Bn | 76.29 Bn | 6.34 Bn |
| 6 | Stryker | 105.71 Bn | 91.82 Bn | 4.50 Bn |
| 7 | Boston Scientific | 63.26 Bn | 58.27 Bn | 3.85 Bn |
| 8 | Edwards Lifesciences | 49.80 Bn | 33.91 Bn | 1.35 Bn |
| 9 | Becton Dickinson | 48.78 Bn | 45.93 Bn | 2.32 Bn |
| 10 | Medline | 45.62 Bn | 46.07 Bn | 2.22 Bn |
Historic Data
Download Data
Historic Data
| Date | Value |
|---|---|
| Jun 27, 2026 | 1.64 Bn |
| Mar 28, 2026 | 1.53 Bn |
| Dec 31, 2025 | 1.45 Bn |
| Dec 31, 2024 | 1.49 Bn |
API Access
Medline Accumulated Expenses 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=accumulated-expenses&ticker=MDLN&period=max&api_key=YOUR_API_KEYimport requests
url = "https://data.businessquant.com/historic"
params = {"slug": "accumulated-expenses", "ticker": "MDLN", "period": "max", "api_key": "YOUR_API_KEY"}
data = requests.get(url, params=params).json()const res = await fetch("https://data.businessquant.com/historic?slug=accumulated-expenses&ticker=MDLN&period=max&api_key=YOUR_API_KEY");
const data = await res.json();