CSC-205 Computer Organization
Dr. Chuck Cartledge
10 June 2015
1. Convert the following numbers from binary to decimal, assuming unsigned
A. 10110 => 22
B. 10 => 2
C. 10101 => 21
D. 10000 => 16
E. 1111 => 15
0% (0 out of 14 correct)
The questions marked with
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1. A(n) _-controlled loop is one that terminates when something happens
inside the loop body to signal that the loop should be exited.
The suggested answer is
1. THe series of steps isn't an algorithm because it does not give the step-by-step procedure of events.
Go to school
Go to sleep
Repeat from first step
Below gives more details and procedure:
a. Wake up
b. lay the bed
c. Take a s
If the statement
alpha = Max(a, Max(b, c);
appears in the body of the main function, then this statement is an example of a
recursive function call.
The following definition of N factorial is a valid re
Demonstration Outline- Tylers demonstration
General Purpose; to demonstrate
Specific Purpose; to demonstrate how to play golf
Central idea; Golf is a simple and complex game and anyone can play but not everyone can be really
great at it.
Demonstration Speech Outline.
General purpose; To demonstrate
Specific purpose; To demonstrate how to paint with watercolors to create a coral reef landscape.
Central Idea; Watercolor paintings are fairly easy to do and many people can do it without much
Steve Jobs commencement speech.
General purpose; To introduce/motivate
Specific purpose; to introduce myself and inspire/motivate the graduates
Central idea; Now that college is over for these graduates, they will start a new beginning the most
CSC 201 Assignment 6
December 9 2016
1. A void function begins with void, a void function does not return a value, and a return statement
does not appear in the body of a void function.
3. After execution of the last statement in th
September 16, 2016
13. = should be =
14. Very good
15. Very goodExcellent
16. Put the inner if statement in braces.
18. no, as long as there is not a dangling else you can nest as many if statements as y
CSC 215 11/22/16
a) 256 MB (totl number of sectors * number of bytes per sector: 524,288 Sectors* 512 Bytes per sector =
268,435,456 Bytes or 256 MB)
b) Access time = Latency + seek time = 6ms + 5ms = 11ms
a) 640 MB
Steven Harter CSC 215
VALUE LOADED TO AC
The theoretical speedup that can be achieved with the
pipelined system over the non-pipelined system is 4.8077 or
rounded up, 5.
CSC 215 Assignment 10
November 15, 2016
4. A. the CPU needs to be 50% faster to achieve the overall system speed increase of 25%
B. The disk needs to have a speed increase of 100% to achieve the system speedup of 25%
5. at least 18 kHz
CSC 215 Assignment 12
November 29, 2016
Could be implemented in CISC: Allowing few addressing modes, highly pipelined, small number of simple
instructions, three register operands, complexity in compiler, fixed-length instructions and har
Number of address bits 8M = 223 so 23 bits
Number of bits to select a RAM chip? 1 chip = 512K = 2 10 bits * 29 bits = 219 so 19 bits.
Number of chips or modules? 8M * 16 / 512K * 8 = 8 chips * 2 banks =
3. a) there are 227 bits of main memory (232 / 25)
b) the offset is 5 bits, the block is 3 bits, the tag is 24 bits.
c) 0x13A4498A maps to block 100 or 4
5. a) 18 blocks of main memory
b) tag is 18 bits, offset i
Assignment 9 CSC 215
November 1st 2016
a) block = 11 bits. offset = 5 bits. Tag = 8 bits.
b) 5 bits for the offset field, 19 for the tag.
c) There are 211/22 = 29 sets in cache, so 9 bits are needed for the set field. We still need 5 bit
8. If you enter 4 for guess in the pep/8 program for Figure 40 then the stop statement will execute
because it is not one of the cases and the program won't have a command so it will exit the program.
j: .equate 0
1) Push: add an element onto the stack
Pop: remove an element from the stack
2) Stack.top-1 is the real top element since 0 is the placeholder of the first. Stack.top does not give
the top top element
13 5 12
a) 2^5 (2^8(from the 256 bytes)/2^3(from the 8 bytes) = 32
Block 0 of cache would hold memory blocks of address 80 with a tag value of 100, while
Block 1 hold the memory block of cache 2B with tag value of 0
3) The CPU will check for an interrupt via the fetch-decode-execute cycle. If found, it would use an
interrupts handling routine. Only then does the program continue from there.
a. 23 bits
b. 22 bits
a. 20 bits
b. 32 bit address with 6 in the word field, 9 in the block field, and 17 in the tag field
c. 13A4498A = 00010011101001000 100100110 001010, which implies Block 294
b. 24 bit address with 18 in tag field an
Runtime binding resolves symbols at load time, so that external declarations and linking pass
through, while compile time bindings are checked at compile time and are more likely to find
35) Yes it is
13) Line 3; The operator is inaccurate as the function should be bool
14) Line 3; The const at the end is redundant and unnecessary
15) Line 3 and 11; the return type needs to be discover. No scope resolution is needed for line
2. Did it
3. Did it
4. It changes the uppercase letters into lowercase letters (basically adds numbers to the
hexadecimal or ASCII up since it adds)