Suppose pakcet number P is sent at t=0 by the sender. Suppose P is lost then remaining 7 packets i.e P+1 to P+7 will be sent at t=100, 200, ..., 700, since the window size is 8.
Sender stops sending at t=800, since window size is 8 and ACK for P has not arrived. ACK for P+1, P+2 and P+3 will be duplicate ACKs and will arrive at t=900, 1000 and 1100. P is re-transmitted at t=1100. ACK arrives at 1900.
Time lost = 1900-800 = 1100.