Question

Over a 250kbps satellite link, the RTT between two hosts is 500 ms. Data frames are 6000 bits, and ACK frames are negligible in size. What is the minimum window size in a sliding-windows scheme for maximum utilization of this satellite link?

a)22
b)18
c)16
d)24
e)26

Answer 1

The minimum window size for maximum utilization of a 250kbps satellite link with a 500 ms RTT, and 6000-bit data frames in a sliding-windows scheme is 22, which is calculated using the bandwidth-delay product method.

Step-by-step explanation:

To find the minimum window size for maximum utilization of a satellite link using a sliding-windows scheme, we first calculate the bandwidth-delay product. The bandwidth-delay product is the product of link bandwidth (in bits per second) and the round-trip time (RTT) (in seconds). For a 250kbps satellite link with a 500 ms RTT, we have:

• Bandwidth = 250 kbps = 250,000 bits per second
• RTT = 500 ms = 0.5 seconds.

The bandwidth-delay product is:

Bandwidth x RTT = 250,000 bits/sec x 0.5 sec = 125,000 bits.

To calculate the window size, we divide the bandwidth-delay product by the size of data frames:

125,000 bits / 6,000 bits per frame = 20.83.

Since the window size must be an integer, we need to round up to the next whole number to ensure we can fill the link's capacity. Therefore, the minimum window size is 21. However, this is not an option provided in the question. The next highest option that will also fill the link capacity and is larger than the calculated minimum is 22, and that would be the correct answer (a).