How to calculate effective throughput in nRF52

Question

Hi, 
 I measured the throughput using the ATT_MTU Throughput Example.
I referred to Testing in the following URL.
 infocenter.nordicsemi.com/.../ble_sdk_app_att_mtu.html 
 It is simple to measure the throughput and it is very wonderful. 
 I want to know not only measured values but also calculated values . However, I don’t know how to calculate the throughput. 
 Could you tell me the calculation method(Calculating formula) of effective throughput under the following conditions? 
 ATT_MTU size 247 bytes 
Connection interval 7.5 ms 
Data length extension (DLE) OFF 
PHY data rate 2 Ms/s 
 Thanks

Ole Bauck · Accepted Answer

Since you have turned off DLE, the PDU (packet size) will be 27 bytes, of which 23 bytes will be ATT data. 
 With 247 bytes ATT MTU you will have to send 247/23 ~= 11 packets per ATT MTU . 
 The ATT MTU has 3 byte header (1 byte OPCODE_LENGTH and 2 byte HANDLE_LENGTH), so the effective data payload is 244 bytes per ATT MTU. So effective data per packet will be 244/11 ~= 22 bytes per packet . 
 How many packets you get per connection event is a bit tricky to figure out. With 2Mbps PHY, a 27 byte packet (PDU) takes about 160us to send, an empty packet takes about 40us. Between each RX or TX packet there is a 150us interframing space. One RX/TX pair where you send 27 bytes and receives an empty packet will take about 160us + 150us + 40us + 150us = 500us . You would then guess that you could send 15 packets per connection event when the connection interval is 7.5ms, but there have to be a little bit of space between the last packet sent on one connection event, and the first packet sent the next connection event. I did a quick check and you will get 13 packets sent every connection event with your configuration. 
 Adding all these pieces together the total throughput will be: 
 22 bytes pr packet * 8 bits pr byte * 13 packets pr conn event / 7.5ms conn int = 305kbps

How to calculate effective throughput in nRF52

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