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3.
ROS
4.
one
5.
serially
6.
parallel
7.
60
8.
12
9.
8
10.
42
ANSW'ERS -- MICRO PROGRAM
1.
Micro-program
2.
ROS word
3.
control fields
4.
function control, storage control,
branching and address.
5.
8
6.
fUnction control
7.
W,
X
8.
X6, X7
9.
eK
10.
storage
ANSWERS -- ROS WORD FORMAT
1.
8
2.
E
3.
arithmetic
".
lower
s.
binary, decimal
6.
E binary or decimal mode must be
specified.
7.
is set into
8.
V
9.
one microsecond
10.
Many
ANSWERS -- BLOCK EXAMPLE
1.
The data in the R-register after
completing block 04B6 the second
time is 0001 0111.
This is arrived at by exe-
cuting the blocks in the fol-
lowing order.
ADDRESS 04AA:
The data in the
R-register is returned to core
(WRITE.)
The data in the
D-
is presented to both the A-
register and B-register inputs
to ALU.
There is no carry
insert because position 3 of
the S-register is zero.
There
is no carryout as a result of
the addition to set 53 (D+D+C
DC) •
A source
0001
0001
B source
0001
0001
D-register 0010
0010
Position 7 of the S-register
is set to zero (0->S7).
An
unconditional 1, 0 branch is
taken to address 4AE.
ADDRESS 04AE:
The data in the
R-register is DECIMAL added to
itself (RtR+C->RC).
Decimal
mode is specified by the E
line.
53
is still zero,
therefore no carry is
inserted.
Because the data in
the R-register is zero, the
resultant addition provides no
carry out to set 53.
Position
6
of the S-register is set to
zero (0->56), Branch 0,0 to
address 04B4.
ADDRESS 04B4:
The mnemonic
STORE, does nothing for us at
this time because the previous
cycle was not a read.
The
data in the D-register is
again added to itself.
Still,
there are no carries.
A source
0010
0010
B source
0010
0010
D-register 0100
0100
You can see that every time
the data in the D-register is
added to itself, the data
shifts one position to the
left.
6-23
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