Where are the most affordable US cities to retire?

The aging population is more active than ever and communities for retirees are thriving. But where are the most affordable cities in the US for retirement?

An analysis by 55places.com has considered several factors including median home prices, cost of living, state tax laws, local health care options, availability of 55+ communities, and public transit; to compile a ranking of the 20 most affordable US cities for retirement.

The company’s research found that 77% of active adults need their next home to be affordable.

"But a home within a certain price point is only half of the equation,” said Danny Goodman, COO and co-founder of 55places.com. “The surrounding area—whether or not an active adult community is of interest—plays an integral role in quality of life, and nearby attractions and available establishments can greatly enhance that. These destinations provide budget-friendly options without demanding a sacrifice for comfort."

The most affordable places to retire in 2019:

1. Sioux Falls, South Dakota
2. Ocala, Florida
3. Lakeland, Florida
4. Jacksonville, Florida
5. Daytona Beach, Florida
6. Gainesville, Florida
7. New Castle County, Delaware
8. South Bend, Indiana
9. Tampa, Florida
10. Waco, Texas
11. Birmingham, Alabama
12. Memphis, Tennessee
13. San Antonio, Texas
14. Ft. Myers-Cape Coral, Florida
15. Phoenix, Arizona
16. Melbourne, Florida
17. Grand Rapids, Michigan
18. Orlando, Florida
19. Wichita Falls, Texas
20. Indianapolis, Indiana