The PEP8 processor IS(Instruction Set) has both Unary (one byte) and Trinary(thr

Question

The PEP8 processor IS(Instruction Set) has both Unary (one byte) and Trinary(three byte) instructions.

Access to main memory ( to load/store a byte) takes 100 ns.

For a given program, consisting of one hundred instructions, 70% of the executed instructions are Trinary.

Of these( the Trinary):

a) 95% require a two byte operand and the rest, one byte.

b) 35% are memory STORE operations, with 20% storing a one byte operand(STBYTE).

Consider(ONLY) the three stages in the Instruction Cycle that access memory, FI/FO/WO.

            What is the Total Memory Access Time (TMAT) in ns. (nanoseconds), that is spent in the

             execution of this program.

             SHOW and EXPLAIN EXACTLY HOW you calculate/arrive at the intermediate and final values.

Explanation / Answer

• Absolute • Register indirect • Based (Base + Offset) • Indexed • Memory indirect • Auto increment/decrement Consider the following high-level language code segments: (a) uint8_t a[100]; // a is allocated in memory for (i = 0; i < 100; i ++) { a[i] = 10; } (b) int a[100]; // a is allocated in memory for (i = 0; i < 100; i ++) { a[i] = 10; } (c) int *p; // p is allocated in memory *p = 100;

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