Go-Back-N ARQ Protocol Throughput Question
Window size Ws is 5. 
the time to transmit one frame is: tf=3000/2  Mbps=1.5ms
As the frame may be received in error with probability 0.7, each frame need to be transmitted 10/3 times on average. Then the time needed for successfully transmitting one frame is 

tf+7/3*Wstf=

38/3*tf=19ms. 

The effective rate is (3000-100)/19ms=152.6kbps.
ii) 
1001 in binary is 9 in decimal. Frame 8 must have be received by the receiver and the ACK has been sent back to the sender. Therefore, frame 3 must not be included in sender’s window. As frame 9 hasn’t been received yet, 9 must be greater than or equal to Slast. Therefore, Slast can be any of these: 4 to 9 in decimal. Orin binary, 0100, 0101, 0110, 0111, 1000, 1001.