CMP 334 Exam 1 (Fall 2017) 1) Fill in the first table with binarydecimal hexadec
ID: 3590508 • Letter: C
Question
CMP 334 Exam 1 (Fall 2017) 1) Fill in the first table with binarydecimal hexadecimal 1191181 equivalent value in each row and values in each column in the indicated base 136 2) Use a truth table to prove or disprove: -(a b)3(b 3) Design a combinational circuit for the function described infomally below a) draw a "black box" with labeled inputs and output(s) for the circuit, b) make the description formal by supplying the appropriate truth table C) derive a Boolean formula (in Disjunctive Normal Form) for the circuit, and d) draw the circuit for this formula (do not try to minimize the formula). Informal description: the input should be a binary number between 0 and 15 and the output should be true if that number is equal to 2 modulo 5 (that is 2,7, or 12) 4) For each (X. Y) pair of 8-bit integers in the second table a) Convert X and Y to binary b) Add X and Y in binary c) Which condition flags would be set by the addition in part b? d) Compute the two's complement of Y e) Use two's complement arithmetic to compute X-Y 8x59 8x2F 8x9F ex61 5) Execution of program P on computer X entails 6 billion fixed-point instructions and 3 billion floating-point instructions (you do not need to be concemed about what these instructions do yet). The CPI for fixed-point instructions on X is 4 and that for floating-point instructions is 12. The clock rate for X is 5 billion Hertz a) What is the average CPI for the instructions of P? b) How long does it take P to execute? c) What would be the overall speed up of an improvement reducing the fixed-point CPI to 3 d) What would be the overall speed up of an improvement reducing the loating-point CPI to 8 CISC, and Stack machine assembly instructions for the following high-level code x=(x-y) (x + y)Explanation / Answer
1)
binary decimal hexadecimal
0b01101101 109 6D
0b10001011 139 8B
0b11010101 213 D5
2)
a b c ~(a.b) + c a' + ~(b.c')
F F F T T
F F T T T
F T F T T
F T T T T
T F F T T
T F T T T
T T F F F
T T T T T
From the truth table, LHS = RHS
3)
Let Input = b3 b2 b1 b0
Output is true when the number is 2 modulo 5. Hence,
Output = b3'b2'b1b0' + b3'b2b1b0 + b3b2b1'b0'
4)
X Y
0x59 0x2F
1011001 0101111
X+Y: 010001000
2's complement of Y: 11010001
X-Y : 0101010
b)
X Y
0x9F 0x61
010011111 01100001
X+Y: 100000000
2's complement of Y: 10011111
X-Y : 0111110
c)
X Y
0x84 0xE5
010000100 011100101
X+Y: 0101101001
2's complement of Y: 100011011
X-Y : 10011111