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# CS123 Quiz 2 - PowerPoint PPT Presentation

CS123 Quiz 2. Question 5. HVAC system When t <=74, the fan rate is 2360; When t >= 86, the fan rate is 3870; (a) What is the formula when 74 < t < 86? Multiple ways to solve this problem The easiest way is to use solve function to solve linear equations of 2 unknowns

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### CS123 Quiz 2

• HVAC system

• When t <=74, the fan rate is 2360;

• When t >= 86, the fan rate is 3870;

• (a) What is the formula when 74 < t < 86?

• Multiple ways to solve this problem

• The easiest way is to use solve function to solve linear equations of 2 unknowns

• solve({74*a+b=2360, 86*a+b=3870}, [a,b]);

• (b) What is the fan rate when the t is 75?

• Since 75 is between 74 and 86, we can define a function or procedure based on the formula from (a)

• a and b are known at this point

• use evalf() to approximate the result to a float point number

• To define a procedure

• fanRate:=proc(t) return evalf(a*t+b); end;

• Or, to define a function

• fanRate:=(t)->evalf(a*t+b);

• (c) Given a decimal point number, we need to round it up to the nearest multiple of then

• For example: 2380.8 => 2390

• In order to do so, we need to divide it by 10

• For example: 2380.8/10 => 238.08

• Then take its ceiling, to round it up(or down) to the nearest integer

• For example: ceil(238.08) => 239; or floor(238.08)=>238;

• At last, to multiple it by 10

• For example: 239 * 10 => 2390

• (e) defined in (3) to the list of temperatures.

• You already have the procedure definition so far, but you need to put the value of a and b back to the procedure, if you claim them before the procedure in (c) as I did.

• For example:

• result:=ceil((755/6*t-20855/3)/10)*10; (round up)

• Or floor((755/6*t-20855/3)/10)*10; (round down)