# (Solved Homework): 1. Assume that you are considering enhancing a machine by adding a special feature to it. The percentage of time (t…

1. Assume that you are considering enhancing a machine by adding a special feature to it. The percentage of time (the original run time before the feature is added) that the machine can be run using this new feature is denote as p%, and, during this portion of time, the machine with this new feature runs f% faster than the old one without it.

(a) Determine the overall speedup for each of the following:

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i. f = 100, p = 100

ii. f = 100, p = 50

iii. f = 400, p = 50

iv. f = 400, p = 20

(b) Draw a graph that plots together each of the two speedup functions for f = 100 and f = 400 in terms of p (y-axis being the speedup and x-axis being p).

2. Redo problem 1(a) if p is defined as percentage of time that the machine runs using the special feature but instead measured against the new overall time when the special feature is used.

3. Provide a design of the sequential circuit for the vending machine controller – one product sells for \$0.35, and the machine takes only quarters (input Q) and dimes (input D) with an output signal I for product release and a signal C for change release (if there is change to return). No other signals are used. Show the complete state transition table.

4. Determine the 8-bit representation of each of the following numbers under two’s complement format, and provide an evaluation of the number using the weight system to verify the result.: (a) 103 (b) -97 5. (35 pts) In a hypothetical 5-bit floating point format, 3 leading bits are used for exponent (E, in 2’s complement format) and 2 remaining bits for mantissa (M), and the value is evaluated as 1.M × 2 E, answer the following questions:

(a) Find the decimal value of the expression 01101.

(b) Find the decimal value of the expression 10010.

(c) Determine the range of this format; i.e. ? ≤ |x| ≤?.

(d) Find the expression (representation) for decimal value 7.

(e) Find the expression (representation) for decimal value 0.375.

(f) How many distinct numbers are represented by this format?

(g) On a number axis, mark all the positive numbers represented by this format.

6. In problem 5, if the value evaluation formula is instead defined as each of the following, determine the corresponding range and the number of distinct numbers that can be represented?

(a) M.0 × 2 E

(b) 0.M × 2 E

7. Determine the IEEE-754 standard format representation for each of the following numbers:

(a) −101.4

(b) 2.5 × 10−3

(c) 1.001 × 2 −142 and present your answer in 8-character HEX format. Note: ignore the effect of rounding in this problem.

8. In adding two numbers a and b, both in the IEEE-754 standard format, determine the result (in HEX) of each of the operations.

(a) a = 3f200000, b = be600000

(b) a = 3f200000, b = ff800000

(c) a = 01100000, b = 80e00000

## Expert Answer

7. (a) :

Single precision ( 32 bits )

—————————-

Bit 31 | Bits 30 – 23 | Bits 22 – 0

Sign Bit | Exponent Field | Significand

——– ————– ———–

1 | 10000101 | 1 .10010101100110011001101

0: + | Decimal value of exponent field and exponent | Decimal value of the significand

1: – 133 – 127 = 6 | 1.5843750

Hexadecimal: C2CACCCD

Decimal: -101.40000

7. (b) :

Single precision ( 32 bits )

—————————-

Bit 31 | Bits 30 – 23 | Bits 22 – 0

Sign Bit | Exponent Field | Significand

——– ————– ———–

0 | 01110110 | 1 .01000111101011100001010

0: + | Decimal value of exponent field and exponent | Decimal value of the significand

1: – 118 – 127 = -9 | 1.2800000

Hexadecimal: 3B23D70A

Decimal: 2.4999999e-3

7. (c) :

Single precision ( 32 bits )

—————————-

Input : 1.001 × 2^-142 = 1.7954557e-43

Bit 31 | Bits 30 – 23 | Bits 22 – 0

Sign Bit | Exponent Field | Significand

——– ————– ———–

0 | 00000000 | 0.0000000000000010000000

0: + | Decimal value of exponent field and exponent | Decimal value of the significand

1: – 0 – 127 = -127 | 0.0000305

Hexadecimal: 00000080

Decimal: 1.7936620e-43

8. (a) :

a = 3f200000, b = be600000

a + b = FD800000

8. (b) :

a = 3f200000, b = ff800000

a + b = 13EA00000

8. (c) :

a = 01100000, b = 80e00000

a + b = 81F00000

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