So I am looking at some Operating Systems exercises and we have
A swapping system eliminates holes by compaction. Assuming a random distribution of many holes and many data segments and a time to read or write a 32-bit memory word of 10 nsec, about how long does it take to compact 128 MB? For simplicity, assume that word 0 is part of a hole and that the highest word in memory contains valid data.
The solution is :
128 x 2^20 / 4 = 2^25memory address
compaction time = (read + write) * # of memory access = 2 x 10 x 10 ^ -9 x 2^25 = 671 ms
Problem is I dont understand the
128 x 2^20 / 4 = 2^25 memory address part ? How did we get 2^20 in this case? I gues the 4 is 4 bytes =32 bits so I kinda understand that.
2 x 10 x 10 ^ -9 x 2^25 what is 2 here?