ECE 2030 G Computer Engineering Spring 2002
4 problems, 5 pages Exam Three 18 April 2002
1
Problem 1 (3 parts, 25 points) Memory Systems
Imagine using a 4 Mbit DRAM organized as 1 million addresses of 4-bit words to build three
memory systems. The following three parts consider memory systems built using this chip.
Part A (10 points) Consider a memory system organized as 16 million addresses of 8-bit words.
number of chips needed in one bank 8-bit word/bank = 2chips/bank
4-bit word/chip
number of banks for memory system 16M/1M = 16 banks
memory decoder required (n to m)4-to-16 decoder
number of DRAM chips required 2chips/bank * 16 banks = 32 chips
capacity (in Mbits) 32*4Mbit = 128 Mbit
Part B (10 points) Consider a memory system organized as 1 million addresses of 32-bit words.
number of chips needed in one bank 32-bit word/bank = 8 chips/bank
4-bit word/chip
number of banks for memory system 1M/1M = 1 bank
memory decoder required (n to m)no decoder is necessary
number of DRAM chips required 8 chips/bank * 1 bank = 8 chips
capacity (in Mbits) 8*4Mbit = 32 Mbit
Part C (5 points) Consider a memory systems with 4 chips per bank and 2 banks.
number of addresses 1M addresses/bank*2banks = 2M addresses
size of a word 4 chips/word * 4 bits/chip = 16 bits/word
capacity (in Mbits) 4chips/bank*2banks*4Mbits/chip = 32Mbits
Problem 2 (1 part, 15 points) Instruction Formats
Suppose a datapath has three operand busses (two source, one destination), 70 instruction types,
and 128 registers where each register is 64 bits wide. Immediate operands can be in the range of
±16K. Determine the minimum number of bits needed for the following fields of instruction
formats for this data path.
bits needed to specify an opcode 64 < 70 < 128, so 7 bits needed
bits needed to specify a register operand log2(128) = 7 bits
bits needed to specify an immediate operand ±16K=±214, so 15 bits needed
bits needed to specify an R-type instruction 7 + 7 + 7 + 7 = 28 bits
bits needed to specify an I-type instruction 7 + 7 + 7 + 15 = 36 bits
