Let the queue length be 1 and the courier's journey be k, then

The courier's return trip means that the first half of the team is k- 1 and the second half is 2-k.

The total journey of the team is 1, and the total journey of the messenger is 2k- 1.

The demand is a = (2k-1)1= 2k-1.

therefore

k:(k- 1)=(k- 1):(2-k)

k？ -2k+ 1=2k-k？

2k？ -4k+ 1=0

k=(2+√2)/2(k> 1)

2k- 1= 1+√2, the value is here.