Final answer:
To solve the equation P=(1+102/128)K+32 in a datapath, one needs to consider the precision and range of the values to decide the number of bits. A 32-bit datapath is typically suitable for single precision calculations, but fewer bits may suffice for smaller ranges and lesser precision.
Step-by-step explanation:
The question involves designing a datapath to solve the equation P=(1+102/128)K+32, which is a computation that could be executed in a digital system.
To determine the number of bits needed for this equation, we need to consider the precision and range of the values. The constant 102/128, when simplified, equals 0.796875, which can be rounded to a fixed-point binary representation that fits within a certain number of bits depending on the required precision.
The input K and the result P will determine the size of the datapath. Typically, a 32-bit datapath could be sufficient for most applications requiring single precision floating-point calculations. However, if K is known to be within a smaller range and less precision is acceptable, fewer bits may be used.