# Edmond Ionwave #RFP27-010

City of Edmond seeks aggregate and specific stop-loss coverage for self-funded health plans

**Posted:** October 26, 2026
**Respond By:** November 16, 2026
**Identifier:** RFP27-010
**NAICS:** 524130, 524114

The City of Edmond is seeking carrier proposals for stop-loss protection for its self-funded employee medical and prescription plans.
- **Buyer:** City of Edmond, Oklahoma; no sub-agency or office is identified.
- **Requested coverage:** Aggregate and specific stop-loss insurance for medical and prescription drug claims, using paid-contract terms.
  - Provide Employee Only and Family rates.
  - Quote specific deductible options of $210,000, $220,000, and $230,000.
  - Aggregate coverage must include a 125% corridor and a $1,000,000 annual maximum.
- **Plan population:** Approximately 874 active participants and 34 enrolled retirees.
- **Named vendors in plan history:** Aetna (medical administration), ServeYouRx (pharmacy benefit management), Berkshire Hathaway (current stop-loss carrier), and Symetra and Voya (prior stop-loss carriers). These are not identified as awardees.
- **Notable requirements:** Quotes must come from medical or stop-loss carriers, not through third-party administrators. Pricing must be net of commissions and include all costs; proposals should state any enrollment requirements. The City may award one or more coverage components.

### Description

The City of Edmond is requesting proposals for aggregate and specific stop loss coverage for its self-funded medical plans. Coverage is to begin January 1, 2027, and continue through December 31, 2027, with annual renewal options subject to performance and mutual agreement. Proposals should include two-tier Employee Only and Family rates, pricing for the specified specific-deductible options and aggregate coverage, and required submission forms. Responses must be submitted electronically by November 16, 2026, at 2:00 PM CT.

[View original listing](https://edmondok.ionwave.net/PublicDetail.aspx?bidID=4962&SourceType=1)