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** Micro-Instruction Format**

If a micro program supports 46μ operations with a parallelism of 2,how many and what size of field exists in micro operation field?

Given size of micro-opn field is 9bits.

**Answer:** Total 9 bis divided in 4 and 5 bit.

Can anyone explain how this division is being done.

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We cannot use horizontal micro programming here as its degree of parallelism is * one or more than one* i.e. it cannot guarantee 2 control signals to be active at a time.

But we can use vertical micro programming with two groups. Given that micro-operation field size is 9 bits, division can be done in one among the following ways :

(1,8), (2,7), (3,6), (4,5)

Also it is known that 46 micro operations are supported, so we need to select the most efficient group considering very small wastage of space

With **(1,8)**, number of control signals possible = $2^{1} + 2^{8} = 2 + 256 = 258$ (which is very large than 46)

Similarly with **(2,7)** we have = $2^{2} + 2^{7} = 4 + 128 = 132$ $> 46$

With **(3,6) **we have = $2^{3} + 2^{6} = 8 + 64 = 72$ control signals

And with** (4,5)** we have = $2^{4} + 2^{5} = 16 + 32 = 48$ control signals which is very close to 46.

So we need two vertical micro programming fields with 4 and 5 bits each to support 46 micro operations efficiently.

However the question would have been more apt if it was "with a parallelism of **atmost** 2" as vertical micro programming works on the principle of * none or one*.

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