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((9+3+5)*9)/8===> 1088Bytes

@Hemanth_13

9+3+5=17

17*9=153

153/8=1.9

how are u getting 1088????? pls explain

@Ashish Goyal it should be * 512 not 9

In vertical  microprogrammed memory we have each entry like

 CONDITIONAL BITS NEXT MEMORY ADDRESS CONTROL SIGNALS

now here we have 8 conditional flags so we need LOG(8)= 3 bits here ...now for next memory address ..aas there are 512 instruction and each need 1 micro instruction ..in control memeory we will have 512 entry correspond to each Instruction so to go to any instruction we need 9 bits ..and for CONTROL signals we use decoder here .as it is vertical micro-programmed control unit .

So for control signals we need log(31)= 5 bits ..

so one word is of 3+9+5=17 bits and there are 512 entries ..so control memory size = 512*17/8 BYTES = 1088 Bytes

1 comment

In control memory

1 word ==> 1 Microinstruction

If Control Unit supports 512 Microinstructions

==> 512 words in Control Memory ..To point each word we need 9 bits ..

But "It is given that vertical microprogrammed supports 512 INSTRUCTIONS"

It should be MIcroInstruction right ??