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Standard VI
Mathematics
Substitution
A byte addres...
Question
A byte addressable memory computer has a memory capacity of
2
m
K bytes and can perform
2
n
operations. An instruction involving 3 operands and one operator, needs a maximum of
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Solution
Memory size
=
2
m
K
,
B
y
t
e
s
=
2
m
.2
10
B
y
t
e
s
=
2
(
m
+
10
)
B
y
t
e
s
Address size = ( m + 10) bits
Number of operations
=
2
n
opcode bits = n-bits
Instruction format:
opcode
add 1
add 2
add 3
n (m+10) (m+10) (m+10)
(3m+n+30) bits
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