Go-Back-N ARQ Protocol Throughput Question I am an undergraduate studying Data Coms and Networking and this question about go-back-N ARQ popped up that I am struggling to get my head around and cant find anything similar online. Any help at all would be greatly appreciated.

The only thing I have at all is that throughput could be defined by (Packet bits + overhead)/N. Is this at all on the right path?

QUESTION
Consider the Go-Back-N ARQ protocol with N=5 and 4 bits used for sequence numbering. The user data frame is 3000 bits long where 100 bits are overhead for header and error check, etc. The frame transmission rate is 2Mbps. Assume that the medium does not reorder the frames. The outstanding frames will be retransmitted when the sending window becomes empty. The probability that a data frame would be received without any errors is 0.3.

i)  What is the effective data transmission rate of this protocol?

ii) Suppose that at time t, the receiver is waiting for an information frame with binary sequence number 1001.  What is the possible sequence number Slast in binary inside the sender’s window at time t? Justify your answer.
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.