Exercises 1.1-1.30

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Exercise 1.1 Three level of abstraction

a) Biological abstraction at the cell level:

Your answer:

Solution:

Biologists study cells at many levels. The cells are built from organelles such as the mitochondria, ribosomes, and chloroplasts. Organelles are built of macromolecules such as proteins, lipids, nucleic acids, and carbohydrates. These biochemical macromolecules are built simpler molecules such as carbon chains and amino acids. When studying at one of these levels of abstraction, biologists are usually interested in the levels above and below: what the structures at that level are used to build, and how the structures themselves are built.

b) Abstractions of chemistry about material:

Your answer:

Solution:

The fundamental building blocks of chemistry are electrons, protons, and neutrons (physicists are interested in how the protons and neutrons are built). These blocks combine to form atoms. Atoms combine to form molecules. For example, when chemists study molecules, they can abstract away the lower levels of detail so that they can describe the general properties of a molecule such as benzene without having to calculate the motion of the individual electrons in the molecule.

Exercise 1.2 Using Hierarchy, Modularity, Regularity

a) By automobile designers

Your answer:

Solution:

Automobile designers use hierarchy to construct a car from major assemblies such as the engine, body, and suspension. The assemblies are constructed from subassemblies; for example, the engine contains cylinders, fuel injectors, the ignition system, and the drive shaft. Modularity allows components to be swapped without redesigning the rest of the car; for example, the seats can be cloth, leather, or leather with a built in heater depending on the model of the vehicle, so long as they all mount to the body in the same place. Regularity involves the use of interchangeable parts and the sharing of parts between different vehicles; a 65R14 tire can be used on many different cars.

b) By businessman in daily deals

Your answer:

Solution:

Businesses use hierarchy in their organization chart. An employee reports to a manager, who reports to a general manager who reports to a vice president who reports to the president. Modularity includes well-defined interfaces between divisions. The salesperson who spills a coke in his laptop calls a single number for technical support and does not need to know the detailed organization of the information systems department. Regularity includes the use of standard procedures. Accountants follow a well-defined set of rules to calculate profit and loss so that the finances of each division can be combined to determine the finances of the company and so that the finances of the company can be reported to investors who can make a straightforward comparison with other companies.

Exercise 1.3 Ben Bitdiddle is building a house. Save him money and time with three y's principals

Your answer:

Solution:

Hierarchy:

• Design: how many bedrooms, bathrooms, kitchens and other rooms he would like.
• Jump level up: house layout and dimensions of the house.
• Top level hierarchy - material he would like to use, what kind of roof etc.
• Jump level down: layout of each room with doors, windows etc.

Regularity:

• framing of the house
• scale framing to each room
• choose doors and windows (sizes, material)

Modularity:

• the dimensions of the windows are the same

And he could save by buying some items (like windows) in bulk.

Exercise 1.4 Voltage: in 0-5V range. Measure accuracy: ±50mV. How many bits it convey?

Your answer:

Solution:

±50mV is 100mV intervals. 50 intervals. log₂50=5.65 bits

caution

5.65 bits - is the amount of information. Don't be confused with how many bits of a binary number is necessary to include this value.

Exercise 1.5 Old clock without a minute hand

a) You can read the hour each 15 minutes. How many bits you can convey?

b) If include am and pm, how many bits added?

Your answer:

Solution:

a) 60min/15min=4 12hours*4=48 log₂40=5.58 bits

b) 1 bit

Bit

Two discrete values - are 1 bit. See Bit

Exercise 1.6 4000 years ago Babylonian developed sexagesimal numbers (base 60)

a) How many bits they conveyed with one digit?

b) 4000₁₀ in sexagesimal?

Your answer:

Solution:

a) log₂60 = ~6

b) 1 6 40₆₀

Exercise 1.7 How many different numbers can be represent with 16 bits?

Your answer:

Solution:

2¹⁶ = 65536

Exercise 1.8 What is the largest unsigned 32-bit binary number?

Your answer:

Solution:

22³²-1 = 4 294 967 295

Exercise 1.9 What is the largest 16-bit binary number that can be represented with

a) unsigned numbers?

b) two's complement numbers?

c) sign/magnitude numbers?

Your answer: a) b) c)

Solution:

a) 2¹⁶-1 = 65535

b) 2¹⁵-1 = 32767

c) 2¹⁵-1 = 32767

Exercise 1.10 What is the largest 32-bit binary number that can be represented with

a) unsigned numbers?

b) two's complement numbers?

c) sign/magnitude numbers?

Your answer: a) b) c)

Solution:

a) 2³²-1 = 4 294 967 295

b) 2³¹-1 = 2 147 483 647

c) 2³¹-1 = 2 147 483 647

Exercise 1.11 What is the smallest (most negative) 16-bit binary number that can be represented with

a) unsigned numbers?

b) two's complement numbers?

c) sign/magnitude numbers?

Your answer: a) b) c)

Solution:

a) 0

b) -2¹⁵ = -32768

c) -(2¹⁵-1) = -32767

Exercise 1.12 What is the smallest (most negative) 32-bit binary number that can represented with

a) unsigned numbers?

b) two's complement numbers?

c) sign/magnitude numbers?

Your answer: a) b) c)

Solution:

a) 0

b) -2³¹ = -2 147 483 648

c) -(2³¹-1) = -2 147 483 647

Exercise 1.13 Convert unsigned numbers to decimal. Show your work

a) 1010₂

b) 110110₂

c) 11110000₂

d) 0001100010100111₂

Your answer: a) 1010= b) 110110= c) 11110000= d) 0001100010100111=

Solution:

a) 1010₂ = 1x2³ + 0x2² + 1x2¹ + 0x2⁰ = 8 + 0 + 2 + 0 = 10₁₀

b) 110110₂ = 1x2⁵ + 1x2⁴ + 0x2³ + 1x2² + 1x2¹ + 0x2⁰ = 32 + 16 + 0 + 4 + 2 + 0 = 54₁₀

c) 11110000₂ = 1x2⁷ + 1x2⁶ + 1x2⁵ + 1x2⁴ + 0x2³ + 0x2² + 0x2¹ + 0x2⁰ = 128 + 64 + 32 + 16 + 0 + 0 + 0 + 0= 240₁₀

d) 0001100010100111₂ = 0 + 0 + 0 + 1x2¹² + 1x2¹¹ + 0x2¹⁰ + 0x2⁹ + 0x2⁸ + 1x2⁷ + 0x2⁶ + 1x2⁵ 0x2⁴ + 0x2³ + 1x2² + 1x2¹ + 1x2⁰ = 4096 + 2048 + 128 + 32 + 4 + 2 + 1 = 6311₁₀

Exercise 1.14 Convert the following unsigned binary numbers to decimal. Show your work

a) 1110₂

b) 100100₂

c) 11010111₂

d) 011101010100100₂

Your answer: a) 1110= b) 100100= c) 11010111= d) 011101010100100=

Solution:

a) 1110₂ = 1x2³ + 1x2² + 1x2¹ + 0x2⁰ = 8 + 4 + 2 + 0 = 14₁₀

b) 100100₂ = 1x2⁵ + 0x2⁴ + 0x2³ + 1x2² + 0x2¹ + 0x2⁰ = 32 + 0 + 0 + 4 + 0 + 0 = 36₁₀

c) 11010111₂ =1x2⁷ + 1x2⁶ + 0x2⁵ + 1x2⁴ + 0x2³ + 1x2² + 1x2¹ + 1x2⁰ = 128 + 64 + 0 + 16 + 0 + 4 + 2 + 1 = 215₁₀

d) 011101010100100₂ = 1x2¹³ + 1x2¹² + 1x2¹¹ 0x2¹⁰ 1x2⁹ + 0x2⁸ + 1x2⁷ + 0x2⁶ + 1x2⁵ + 0x2⁴ + 0x2³ + 1x2² + 0x2¹ + 0x2¹ = 8192 + 4096 + 2048 + 0 + 512 + 0 + 128 + 32 + 4 = 15012₁₀

Exercise 1.15 Repeat Exercise 1.13, but convert in hexadecimal

a) 1010₂

b) 110110₂

c) 11110000₂

d) 0001100010100111₂

Your answer: a) 1010= b) 110110= c) 11110000= d) 0001100010100111=

Solution:

a) 1010₂ = A₁₆

b) 11 0110₂ = 36₁₆

c) 1111 0000₂ = F0₁₆

d) 0001 1000 1010 0111₂ = 18A7₁₆

Convert Binary to Hexadecimal

Break a binary number into groups of 4 digits and convert each group separately. After that, just concatenate all the resulting hexadecimal digits.

Exercise 1.16 Repeat exercise 1.14, but convert to hexadecimal

a) 1110₂

b) 100100₂

c) 11010111₂

d) 011101010100100₂

Your answer: a) 1110= b) 100100= c) 11010111= d) 011101010100100=

Solution:

a) 1110₂ = E₁₆

b) 10 0100₂ = 24₁₆

c) 1101 0111₂ = D7₁₆

d) 011 1010 1010 0100₂ = 3AA4₁₆

Exercise 1.17 Convert the following hexadecimal numbers to decimal. Show your work

a) A5₁₆

b) 3B₁₆

c) FFFF₁₆

d) D0000000₁₆

Your answer: a) A5= b) 3B= c) FFFF= d) D0000000=

Solution:

a) A5₁₆ = 10x16¹ + 5x16⁰ = 160 + 5 = 165₁₀

b) 3B₁₆ = 3x16¹ + 11x16⁰ = 48 + 11 = 59₁₀

c) FFFF₁₆ = 15x16³ + 15x16² + 15x16¹ + 15x16⁰ = 61440 + 3840 + 240 + 15 = 65535₁₀ (2¹⁶-1)

d) D000 0000₁₆ = 13x16⁷ = 3 489 660 928₁₀

Exercise 1.18 Convert the following hexadecimal numbers to decimal. Show your work

a) 4E₁₆

b) 7C₁₆

c) ED3A₁₆

d) 403FB001₁₆

Your answer: a) 4E= b) 7C= c) ED3A= d) 403FB001=

Solution:

a) 4E₁₆ = 4x16¹ + 14x16⁰ = 64 + 14 = 78₁₀

b) 7C₁₆ = 7x16¹ + 12x16⁰ = 112 + 12 = 124₁₀

c) ED3A₁₆ = 14x16³ + 13x16² + 3x16¹ + 10x16⁰ = 57344 + 3328 + 48 + 10 = 60730₁₀

d) 403FB001₁₆ = 4x16⁷ + 0x16⁶ + 3x16⁵ + 15x16⁴ + 11x16³ + 0x16² + 0x16¹ + 1x16⁰ = 1 073 741 824 + 0 + 3 145 728 + 983 040 + 45056 + 0 + 0 + 1 = 1 077 915 649₁₀

Exercise 1.19 Repeat exercise 1.17, but convert to unsigned binary

a) A5₁₆

b) 3B₁₆

c) FFFF₁₆

d) D0000000₁₆

Your answer: a) A5= b) 3B= c) FFFF= d) D0000000=

Solution:

a) A5₁₆ = 1010 0101₂

b) 3B₁₆ = 0011 1011₂

c) FFFF₁₆ = 1111 1111 1111 1111₂

d) D0000000₁₆ = 1101 0000 0000 0000 0000 0000 0000 0000₂

Exercise 1.20 Repeat exercise 1.18, but convert to unsigned binary

a) 4E₁₆

b) 7C₁₆

c) ED3A₁₆

d) 403FB001₁₆

Your answer: a) 4E= b) 7C= c) ED3A= d) 403FB001=

Solution:

a) 4E₁₆ = 100 1110₂

b) 7C₁₆ = 111 1100₂

c) ED3A₁₆ = 1110 1101 0011 1010₂

d) 403FB001₁₆ = 100 0000 0011 1111 1011 0000 0000 0001₂

Exercise 1.21 Convert the following two's complement binary numbers to decimal

a) 1010₂

b) 110110₂

c) 01110000₂

d) 10011111₂

Your answer: a) 1010= b) 110110= c) 01110000= d) 10011111=

Solution:

a) 1010₂ = -8 +2 = -6₁₀

tip

This is another way to convert negative two's complement binary numbers to decimal numbers. The most significant bit is taken with a minus sign, and the remaining bits are taken with a plus sign:

1010 <- binary digits
8421 <- position weight
====
-1x8 + 0x4 + 1x2 + 0x0 
-8 + 2 = -6

or you can find the magnitude (modulus) of a negative number by taking the two's complement:

Magnitude

1010
0101 <- invert
+  1
====
0110 = 6 (thus, -6)

b) 110110₂ = -32+16+4+2=-10₁₀ or

Magnitude

110110
001001
+    1
======
001010 = 10 (thus, -10)

c) 01110000₂ = 64 + 32 + 16 = 112₁₀

d) 10011111₂ = -128+16+8+4+2+1=-97₁₀ or

Magnitude

10011111
01100000
+      1
========
01100001 = 97 (thus, -97)

Exercise 1.22 Convert the following two's complement binary numbers to decimal

a) 1110₂

b) 100011₂

c) 01001110₂

d) 10110101₂

Your answer: a) 1110= b) 100011= c) 01001110= d) 10110101=

Solution:

a) 1110₂ = -8+4+2 = -2₁₀ or

Magnitude

1110
0001
+  1
====
0010 = 2 (thus, -2)

b) 100011₂ = -32+2+1 = -29₁₀ or

Magnitude

100011
011100
+    1
======
011101 = 29 (thus, -29)

c) 01001110₂ = 64+8+4+2 = 78₁₀

d) 10110101₂ = -128+32+16+4+1 = -75₁₀ or

Magnitude

10110101
01001010
+      1
========
01001011 = 75 (thus, -75)

Exercise 1.23 Repeat exercise 1.21, assuming the binary numbers are in sigh/magnitude form rather than two's complement representation

a) 1010₂

b) 110110₂

c) 01110000₂

d) 10011111₂

Your answer: a) 1010= b) 110110= c) 01110000= d) 10011111=

Solution:

a) 1010₂ = -2₁₀

b) 110110₂ = -22₁₀

c) 01110000₂ = 112₁₀

d) 10011111₂ = -31₁₀

Exercise 1.24 Repeat exercise 1.22, assuming the binary numbers are in sign/magnitude form rather than two's complement representation

a) 1110₂

b) 100011₂

c) 01001110₂

d) 10110101₂

Your answer: a) 1110= b) 100011= c) 01001110= d) 10110101=

Solution:

a) 1110₂ = -6₁₀

b) 100011₂ = -3₁₀

c) 01001110₂ = 78₁₀

d) 10110101₂ = -53₁₀

Exercise 1.25 Convert the following decimal numbers to unsigned binary numbers

a) 42₁₀

b) 63₁₀

c) 229₁₀

d) 845₁₀

Your answer: a) 42= b) 63= c) 229= d) 845=

Solution:

a) 42₁₀ = 1x32(2⁵) + 0x16(2⁴) + 1x8(2³) + 0x4(2²) + 1x2(2¹) + 0x1(2⁰) = 101010₂

b) 63₁₀ = 1x32(2⁵) + 1x16(2⁴) + 1x8(2³) + 1x4(2²) + 1x2(2¹) + 1x1(2⁰) = 111111₂

c) 229₁₀ = 1x128(2⁷) + 1x64(2⁶) + 1x32(2⁵) + 0x16(2⁴) + 0x8(2³) + 1x4(2²) + 0x2(2¹) + 1x1(2⁰) = 1110 0101₂

d) 845₁₀ = 1x512(2⁹) + 1x256(2⁸) + 0x128(2⁷) + 1x64(2⁶) + 0x32(2⁵) + 0x16(2⁴) + 1x8(2³) + 1x4(2²) + 0x2(2¹) + 1x1(2⁰) = 11 0100 1101₂

Exercise 1.26 Convert the following decimal numbers to unsigned binary numbers

a) 14₁₀

b) 52₁₀

c) 339₁₀

d) 711₁₀

Your answer: a) 14= b) 52= c) 339= d) 711=

Solution:

a) 14₁₀ = 1x8(2³) + 1x4(2²) + 1x2(2¹) + 0x1(2⁰) = 1110₂

b) 52₁₀ = 1x32(2⁵) + 1x16(2⁴) + 0x8(2³) + 1x4(2²) + 0x2(2¹) + 0x1(2⁰) = 110100₂

c) 339₁₀ = 1x256(2⁸) + 0x128(2⁷) + 1x64(2⁶) + 0x32(2⁵) + 1x16(2⁴) + 0x8(2³) + 0x4(2²) + 1x2(2¹) + 1x1(2⁰) = 1 0101 0011₂

d) 711₁₀ = 1x512(2⁹) + 0x256(2⁸) + 1x128(2⁷) + 1x64(2⁶) + 0x32(2⁵) + 0x16(2⁴) + 0x8(2³) + 1x4(2²) + 1x2(2¹) + 1x1(2⁰) = 10 1100 0111₂

Exercise 1.27 Repeat exercise 1.25, but convert to hexadecimal

a) 42₁₀

b) 63₁₀

c) 229₁₀

d) 845₁₀

Your answer: a) 42= b) 63= c) 229= d) 845=

Solution:

a) 42₁₀ = 2x16(16¹) + 10x16(16⁰) = 2A₁₆

b) 63₁₀ = 3x16(16¹) + 15x16(16⁰) = 3F₁₆

c) 229₁₀ = 14x16(16¹) + 5x16(16⁰) = E5₁₆

d) 845₁₀ = 3x16(16²) + 4x16(16¹) + 13x16(16⁰) = 34D₁₆

Exercise 1.28 Repeat exercise 1.26, but convert to hexadecimal

a) 14₁₀

b) 52₁₀

c) 339₁₀

d) 711₁₀

Your answer: a) 14= b) 52= c) 339= d) 711=

Solution:

info

You can take the binary results from exercise 1.26 and convert them to hexadecimal more easily

a) 14₁₀ = 14x16(16⁰) = E₁₆

b) 52₁₀ = 3x16(16¹) + 4x16(16⁰) = 34₁₆

c) 339₁₀ = 1x16(16²) + 5x16(16¹) + 3x16(16⁰) = 153₁₆

d) 711₁₀ = 2x16(16²) + 12x16(16¹) + 7x16(16⁰) = 2C7₁₆

Exercise 1.29 Convert the following decimal numbers to 8-bit two's complement numbers or indicate that the decimal number would overflow the range

a) 42₁₀

b) −63₁₀

c) 124₁₀

d) −128₁₀

e) 133₁₀

Your answer: a) 42= b) −63= c) 124= d) −128= e) 133=

Solution:

a) 42₁₀ = 1x32(2⁵) + 0x16(2⁴) + 1x8(2³) + 0x4(2²) + 1x2(2¹) + 0x1(2⁰) = 0010 1010₂

b) −63₁₀ = 1x32(2⁵) + 1x16(2⁴) + 1x8(2³) + 1x4(2²) + 1x2(2¹) + 1x1(2⁰) = 00111111 = 11000000 + 1 = 1100 0001₂

c) 124₁₀ = 1x64(2⁶) + 1x32(2⁵) + 1x16(2⁴) + 1x8(2³) + 1x4(2²) + 0x2(2¹) + 0x1(2⁰) = 0111 1100₂

d) −128₁₀ = 1000 0000₂

e) 133₁₀ = overflow

Exercise 1.30 Convert the following decimal numbers to 8-bit two's complement numbers or indicate that the decimal number would overflow the range

a) 24₁₀

b) −59₁₀

c) 128₁₀

d) −150₁₀

e) 127₁₀

Your answer: a) 24= b) −59= c) 128= d) −150= e) 127=

Solution:

a) 24₁₀ = 0001 1000₂

b) −59₁₀ = 0011 1011 = 1100 0101₂

c) 128₁₀ = overflow

d) −150₁₀ = overflow

e) 127₁₀ = 0111 1111₂

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