We learned that the stop-and-wait protocol allows the sender to have only one out- standing frame on the link at a time, and we need just 1 bit for the sequence number. The main shortcoming is that it may be far below the link's capacity. A more reason- able way to determine the number of outstanding frames is to use the delaybandwidth product, which can be understood as the amount of data the channel can hold. Now, suppose that we are designing a sliding window protocol for a 2-Mbs point-to- point link to the moon, which has a one-way latency of 1.2 seconds. Assuming that each frame carries 2KB of data, what is the minimum number of bits you need for the sequence number in the following two cases: SWS=RWS, RWS =1