Process model formulation and solution, 3E4 Computer software tutorial - Tutorial 3

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1 Process model formulation and solution, 3E4 Computer software tutorial - Tutorial 3 Kevin Dunn, dunnkg@mcmaster.ca October 2 Tutorial solutions: Elliot Cameron. Tutorial objectives Understand computer representation of decimal numbers and chemical engineering data. Question [] Convert from binary to decimal: (.) 2 Convert from decimal to binary: (32.625) Convert from decimal to binary: (.2). (.) 2 Therefore 2. (32.625) x2 }. x2 x2 2 {{ x2 3 x2 4 x2 5 } = (.) 2 = (.65625) Start to the left of the decimal point (i.e. Representing 32 in binary) log 2 (32) = 5 This helps us determine how many binary digits are required Rounding 5 up to 5 (obviously) Therefore 2 5 = 32 (32 32 = ) x2 5 + x2 4 + x2 3 + x2 2 + x2 + x2 = () 2 = (32) Moving to the right of the decimal place : 2 =.5 ( =.25) 2 2 =.25 (.25 <.2) : therefore we move to the next digit 2 3 =.25 ( =.) x2 5 +x2 4 +x2 3 +x2 2 +x2 +x2 +x2 +x2 2 +x2 3 = (.) 2 = (32.625)

2 3. (.2) Start to the left of the decimal point (i.e. Representing in binary) () = () 2 Moving to the right of the decimal place 2 =.5 (.2 <.5) : therefore we move to the next digit 2 2 =.25 (.2 <.25) : therefore we move to the next digit 2 3 =.25 (.2.25 =.75) 2 4 =.625 ( =.25) 2 5 =.325 (.25 <.325) : therefore we move to the next digit 2 6 =.5625 (.25 <.5625) : therefore we move to the next digit 2 7 =.7825 ( =.46875) 2 8 = ( =.7825) 2 9 = (.7825 <.95325) :move to the next digit 2 = (.7825 < ) : move to the next digit 2 = ( = ) 2 2 = ( = )... x2 + x2 + x2 2 + x2 3 + x2 4 + x2 5 + x2 6 + x2 7 + x2 8 + x2 9 + x2 + x2 + x2 2 (.) 2 = (.2) What this questions aims to show you is, just as in the decimal system, there are numbers that cannot be represented finitely in binary (think /3 in decimal notation). Question 2 []. Write, in binary form, the representation for the most negative floating point number that can be stored in an 8-bit word, using bit for the sign, 3 bits for the signed exponent and the remainder for the significand (in normalized form). 2. What is the decimal equivalent of this number? 3. What is the machine number next to this one (both in binary and in decimal please)? 4. Calculate the maximal interval-to-value ratio around these two values. Does it agree with the limit for theoretical machine precision? 5. As which machine number would be stored if the machine used (a) chopping, or (b) rounding?. We start by recalling that normalized scientific notation assumes that the significand starts with an implied leading digit of (i.e. it falls between the following bounds): b m < 2

3 Where m = significand b = base e = exponent Therefore the most negative 8-bit binary number with bit for the sign, 3 bits for the signed exponent, and 4 bits for the normalized significand is: sign exponent ( x2 + x2 2 + x2 3 + x2 4) x2 (x2 +x2 ) =.9375x2 3 significand 2. sign exponent =.9375x2 3 =.9375x8 = 7.5 significand 3. The next closest machine number is the next physically representable number in the 8-bit floating point system. Since no more space exists above the current significand we must go down in magnitude. Therefore: sign exponent =.875x2 3 = 7. significand 4. We start by estimating the theoretical machine precision: t = 4 : number of significant digits in the significand β = 2 : base of number system ɛ mach = β t = 2 4 =.25 Next we test the maximal interval-to-ratio value on either side of the values from part (a)/(b) and (c) x x = ( 7.) ( 7.5) ( 7.) = ɛ mach x = ( 7.) ( 7.5) ( 7.5) = ɛ mach It is easy to see above that the maximal interval-to-value ratios agree with the upper ɛ mach limit 5. As observed in parts (a) - (c) the two closest machine numbers to are -7. and Therefore the effect of chopping (rounding towards ) and rounding (rounding to the closest avaiable machine number) would result in the same floating point value: -7.. Question 3 [.5]. Increasingly we are seeing cameras being used in chemical processes to monitor and control the process, especially systems that deal with foods and solid products. How much space, in kilobytes, is required to store a digital photo with 64 rows and 48 columns of pixels and 3 layers (red, green and blue) using uint8 integer representation? 2. Computer systems are used to archive data from each electronic measurement, such as temperature, pressure, flow measurements, etc. Each measurement is called a tag. At your plant, you wish to store 6,525 tags, recorded once per second and stored in double precision. How much space would be required on the company s server, in terabytes, to store a single copy of year of data? What difference does it make to store the data in single precision? 3

4 3. How many data points can you store in double precision in 5 megabytes of RAM? (For example, MATLAB on a 32-bit Windows Vista machine cannot create arrays greater than 428 megabytes.) Use the fact that: 24 bytes = kilobyte 24 kilobytes = megabyte 24 megabytes = gigabyte 24 gigabyte = terabyte = 2 4 bytes. Recall that the uint8 integer representation refers to an unsigned integer with no exponential term. Therefore it refers to an integer that can store a value from As such this question is simply asking us how much memory would be required to store a 64x48 pixel image using the 256/256/256 RGB colour notation. Now, if each pixel in each of the RGB matrices is represented by a uint8 then each value will logically take up 8 bits of memory. If each of the three colour matrices contains 64x48 values and we assume a standard 8 bit byte then the total space required to store the photo would be: Space = 3 64 }{{ 48 } 8 }{{ bits } = bits byte kilobyte = 9 kilobytes 8 bit 24 byte rgb pixels uint8 2. Recall that a standard floating point double precision number takes up 64 bits (8 bytes) of memory. We start this problem by calculating the number of values to be stored in one year. Values = 6, 525 values s 6 s 6 min 24 hr 365 day = 52, 32, 4, values min hr day year year Therefore the memory requirement is: Memory = 52, 32, 4, values year 8 bytes value kilobyte 24 byte megabyte 24 kilobyte gigabyte 24 megabyte terabyte = 3.79 terabytes 24 gigabyte A floating point single precision value takes up exactly half as much space as a floating point double precision number (i.e. 32 bits = 4 bytes). Therefore storing the same tags in single precision would take up.9 T B. 3. If we take the absolute value of RAM available (i.e. 5 MB) and the standard definition of a floating point double precision number (i.e. 64 bits = 8 bytes), then, 5 MB 24 KB MB 24 B kb =, 572, 864, bytes Therefore the number of double precision values that could be stored would be. Number =, 572, 864, bytes value 8 bytes = 96, 68, values If we use the maximum MATLAB array size then., 428 MB 24 KB MB 24 bytes kb =, 497, 366, 528 bytes Therefore the number of double precision values that could be stored in MATLAB would be. Number =, 497, 366, 528 bytes value 8 bytes = 87, 7, 86 values 4

5 Question 4 [] Consider the following system of linear algebraic equations: 2x 2 + 4x 3 = x 3 + 4x 3 = 3x + 5x 3 =. Use Gauss elimination (forward elimination and backward substitution) to solve these equations for (x,, x 3 ). 2. Validate your solution in either Python or MATLAB.. We are asked to solve this system of equations using Gauss Elimination (without partial pivoting). 2x 2 +4x 3 = x 3 +4x 3 = x = 3x +5x 3 = 3 5 x 3 Forward elimination Divide row by element (,) x x 3 = Subtract row from row 2 and 3 times row from row 3 to eliminate all elements in column below the diagonal element 2 x 2 2 = 2 x 3 Divide row 2 by element (2,2) 2 2 x x 3 =.5 Subtract 2 times row 2 from row 3 to eliminate all elements in column 2 below the diagonal element 2 x =.5 x 3 Backwards substitution 5

6 Subtract - times row 3 from row 2 and 2 times row 3 from row to eliminate all elements in column 3 above the diagonal element x = 2.5 x 3 Subtract - times row 2 from row to eliminate all elements in column 2 above the diagonal element x =.5.5 x 3 2. Checking this solution in MATLAB: EDU>> A = [2,-2,4;,-3,4; 3,-,5]; EDU>> b = [;-;]; EDU>> x = A\b x = Checking this solution in Python: In []: A = np.array([[2,-2,4],[,-3,4],[3,-,5]]) In [2]: b = np.array([[],[-],[]]) In [3]: x = np.linalg.solve(a,b) In [4]: print(x) [[.5] [-.5] [-. ]] 6