By M. MORRIS MANO
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13 Insert a No-op instruction between the two instructions in the example of Problem 9-12 (above). 15 Add R2 to R3 Branch to 104 Increment R1 Store R1 1 I I Use example of Problem 9-14. 16 (a) There are 40 product terms in each inner product, 402 = 1,600 inner products must be evaluated, one for each element of the product matrix. 17 8 + 60 + 4 = 72 clock cycles for each inner product. There are 602 = 3600 Inner products. Product matrix takes 3600 x 72 = 259,200 clock cycles to evaluate. 18 memory array 1 use addresses: 0, 4, 8, 12, …, 1020.
16. 24 X2 places PC onto the bus. 4 A line of code such as: LDA I is interpreted by the assembler (Fig. 2) as a two symbol field with I as the symbolic address. A line of code such as: LDA I I is interpreted as a three symbol field. The first I is an address symbol and the second I as the Indirect bit. Answer: Yes, it can be used for this assembler. 5 The assembler will not detect an ORG or END if the line has a label; according to the flow chart of Fig. 1. Such a label has no meaning and constitutes an error.
The CALL behaves as a JMP since there is no return from FETCH. Cannot add and subtract at the same time. The RET will be executed independent of S. The MAP is executed irrespective of Z or 60. 21 See Fig. 2 (b) for control word example. (a) 16 registers need 4 bits; ALU need 5 bits, and the shifter need 3 bits, to encode all operations. 23 (a) (b) See Fig. 24 P is used to determine the polarity of the selected status bit. When P = 0, T = G because G ⊕ O = G When P = 1, T = G’, because G ⊕ I = G' Where G is the value of the selected bit in MU × 2.
solution manual computer system architecture by M. MORRIS MANO
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