Download Exam Solutions for ECE 2030 Computer Engineering Fall 2000 - Problem 1 to 4 and more Exams Computer Science in PDF only on Docsity!
4 problems, 3 pages Exam Three Solutions 29 November 2000
Problem 1 (2 parts, 20 points) Memory Systems
Imagine using a 256 Mbit DRAM organized as 32 million addresses of eight bit words to build three memory systems. The following two parts consider memory systems built using this chip.
Part A (10 points) Consider a 256 Mbyte memory system organized as 64 million addresses of 32 bit words.
number of chips needed in one bank (^) 32/8 = 4
number of banks for memory system (^) 64M/32M = 2
memory decoder required ( n to m ) (^) 1 to 2
number of DRAM chips required (^) 4 x 2 = 8
Part B (10 points) Consider an 8 Gbyte memory system organized as 1 billion addresses of 64 bit words.
number of chips needed in one bank (^) 64/8 = 8
number of banks for memory system (^) 1B/32M = 32
memory decoder required ( n to m ) (^) 5 to 32
number of DRAM chips required (^) 8 x 32 = 256
Problem 2 (1 part, 20 points) Instruction Formats
An ancient Egyptian hieroglyphic tablet has been found which may be the earliest record of an instruction set architecture (perhaps alien?). The example shown below has all fields shown with zeros. Do your best to interpret its meaning and answer the questions below.
000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 opcode dest. reg. source 1 reg. immediate value
What is the maximum number of opcodes? (^) 12 bits = 4096
What is the number of registers? (^) 9 bits = 512
What is the range of the signed immediate value? (^) 18 bits = ±128K
What is the instruction word size? (^) 48 bits
How many are bits needed to specify a branch if equal instruction in this format?
48 bits
4 problems, 3 pages Exam Three Solutions 29 November 2000
Problem 3 (6 parts, 35 points) Microcode Reverse Engineering
The microcode fragment below runs on the datapath discussed in class (see exam handout). For each part below, answer the question asked about the microcode execution. Don’t care values (X) have been converted to zeros. For maximum credit, be as specific and concise as you can (e.g., list shift types, amounts, and directions, logical functions, memory addresses and operations, etc.)
cycle X Y Z rwe im en im va au en -a/s lu en lf su en st ld en st en r/-w msel 1 0 0 1 1 1 1234 0 0 1 C 0 0 0 0 0 0 2 1 0 1 1 1 FFF0 0 0 0 0 1 0 0 0 0 0 3 1 0 1 1 1 5678 0 0 1 E 0 0 0 0 0 0 4 1 0 2 1 0 0 0 0 0 0 0 0 1 0 1 1 5 1 0 1 1 1 0010 0 0 0 0 1 2 0 0 0 0 6 1 0 3 1 0 0 0 0 0 0 0 0 1 0 1 1 7 2 3 4 1 0 0 1 1 0 0 0 0 0 0 0 0 8 1 4 0 0 0 0 0 0 0 0 0 0 0 1 0 1
Part A (5 points) What value is stored in register 1 following cycle 2 (in hexadecimal)?
Part B (5 points) What memory location is accessed during cycle 4 (in hexadecimal)?
Part C (5 points) What memory location is accessed during cycle 6 (in hexadecimal)?
Part D (5 points) Describes the operation that occurs during cycle 7 (be specific).
R4 <- R2 - R
Part E (5 points) Describes the operation that occurs during cycle 8 (be specific).
(56781234) <- R
Part F (10 points) Describe three differences between the Branch If Equal (BEQ) and Jump (J) instructions described in class.
- BEQ is a conditional branch; J is an unconditional branch
- BEQ is a relative branch; J is an absolute branch
- BEQ has a 16 branch offset; J has a 26 bit branch value
