The PEP8 processor ISInstruction Set has both Unary one byte

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.

Solution

• 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;