Banglalink has secured Ookla®’s recognition as the fastest mobile network in the country for the fourth consecutive time. On the recently published Ookla® Speedtest Award™ report for Q3-Q4 2021, Banglalink remains in the top spot with a Speed Score® of 20.04, said a press release. Its median download was 13.77 Mbps, while the median upload speed was 7.54 Mbps. Ookla gives the most accurate and comprehensive view of worldwide network performance, quality, and accessibility based on millions of results obtained through the Ookla Speedtest app each day.
Banglalink has managed to raise its performance bar significantly in recent times providing super-fast 4G internet and high-quality digital services. A total of 9.4 MHz new spectrum purchased last year contributed to winning this award by enhancing Banglalink’s network quality further. With an objective to gain double-digit growth, Banglalink is set to install around 3,000 Base Transceiver Stations (BTS) across the country this year.
Upanga Dutta, Chief Commercial Officer, Banglalink said, “It is a matter of great pleasure to win such a prestigious award four times in a row. This achievement is indicative of our firm commitment to providing the best-in-class digital services to customers with unwavering consistency. We want to keep this momentum going and expand our network for enabling more customers to benefit from our fastest mobile internet.”
Doug Suttles, CEO, Ookla said, “Speedtest Awards are given to the best-performing mobile operators for their superior network speed and coverage. According to the consumer-initiated tests taken with Speedtest, Banglalink’s overall network performance was the best in Bangladesh in Q3-Q4 2021. We are happy to present the award for Fastest Mobile Network to Banglalink once again.”
Banglalink will remain committed to providing quality network coverage and digital services to customers.
To download Bangladesh’s Ookla Speedtest Award report, please visit: https://www.speedtest.net/awards/bangladesh/2021?award_type=carrier&time_period=q3-q4
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