Consider a memory system having address spaced at a distance of m, T = Bank cycle time and n number of banks, then the average data access time per word access in synchronous organisation is

Important Point - In Synchronous Access Organisation, the address lines are common at all the banks.

We know that -

• Distance at which address is spaced - 'm'
• Bank Cycle Time - 'T'
• Number of Banks - 'n'

Explanation - 

• When the distance at which address is spaced is very less than the number of banks, that is, m << n, then the average data access time per word access is - $m\cdot \frac{\mathrm{T}}{\mathrm{n}}$
• When the distance at which address is spaced is much more than the number of banks, that is, m >> n, then the average data access time per word access is - T
• Combining above two points, we get the average data access time per word access in synchronous organisation as - t = $\{\begin{array}{l}\mathrm{m}\cdot \frac{\mathrm{T}}{\mathrm{n}}\text{ for }\mathrm{m}<<\mathrm{n}\\ \mathrm{T}\text{ for }\mathrm{m}>>\mathrm{n}\end{array}$