A computer which issues instructions in order, has only 2 registers and 3 opcodes ADD, SUB and MOV. Consider 2 different implementations of the following basic block:Case 1 Case 2t1 = a + b; t2 = c + d;t2 = c + d; t3 = e – t2;t3 = e – t2; t1 = a + b;t4 = t1 – t2; t4 = t1 – t2; Assume that all operands are initially in memory. Final value of computation also has to reside in memory. Which one is better in terms of memory and by how many MOV instructions?

A computer which issues instructions in order, has only 2 registers an

1.	A computer which issues instructions in order, has only 2 registers and 3 opcodes ADD, SUB and MOV. Consider 2 different implementations of the following basic block:Case 1 Case 2t1 = a + b; t2 = c + d;t2 = c + d; t3 = e – t2;t3 = e – t2; t1 = a + b;t4 = t1 – t2; t4 = t1 – t2; Assume that all operands are initially in memory. Final value of computation also has to reside in memory. Which one is better in terms of memory and by how many MOV instructions?
A.	Case 2, 2
B.	Case 2, 3
C.	Case 1, 2
D.	Case 1, 3
Answer» E.

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A computer which issues instructions in order, has only 2 registers and 3 opcodes ADD, SUB and MOV. Consider 2 different implementations of the following basic block:Case 1 Case 2t1 = a + b; t2 = c + d;t2 = c + d; t3 = e – t2;t3 = e – t2; t1 = a + b;t4 = t1 – t2; t4 = t1 – t2; Assume that all operands are initially in memory. Final value of computation also has to reside in memory. Which one is better in terms of memory and by how many MOV instructions?
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