1. read the length of the window
read the width of the window
read delivery choice - y or n
area = length * width
glass cost = area * .50
perimeter = 2 * (length + width)
frame cost = perimeter * .75
total cost = glass cost + frame cost
if delivery == 'y'
total cost = total cost + 50.00
print length, width, total cost
2. This algorithm shows a simple "if" statement. The test data for this
particular problem needs to include the window information (length and
width) and the delivery option. You would need to make up at least TWO
sets of test data; one when the delivery is Yes and one when the delivery
is No.
length width delivery
10 40 n
15 20 y
3. read the length of the window
read the width of the window
area = length * width
perimeter = 2 * (length + width)
if area <= 1000
glass cost = area * .50
frame cost = perimeter * .75
else
glass cost = area * .60
frame cost = perimeter * .80
total cost = glass cost + frame cost
if delivery == 'y'
total cost = total cost + 50.00
print length, width, total cost
4. As you make up test data for this problem, you must realize that
smaller windows are 1 - 1000 square inches and larger windows are 1001
square inches or more. The 1000 and the 1001 become what is referred to
as the "boundary" numbers. They are the numbers that determine the
difference between the two prices. Test data ALWAYS involves using the
boundary numbers.
length width delivery
20 50 n
50 20.05 y
Notice that the area of the first window is 1000, and the area of the
second window is 1001. If the length and width are ints, we cannot choose
data that will give us an area of 1001. In that case we should get as close
to the boundary value as possible, for example, using values like 501 and 2
for the length and the width.
5. read the length of the window
read the width of the window
read the thickness of the window (s/d)
area = length * width
perimeter = 2 * (length + width)
if thickness = 's' // USES TWO-WAY SELECTION
glass cost = area * .50
else
glass cost = area * 1.00
if glass cost < 350.00 // THIS IS ONE-WAY SELECTION INSIDE
glass cost = 350.00 // THE TWO-WAY SELECTION
frame cost = perimeter * .75
total cost = glass cost + frame cost
if delivery == 'y'
total cost = total cost + 50.00
print length, width, total cost
6. This problem has three cases: single pane, double pane with glass cost
less than $350, and double pane with class cost of $350 or more. As in the
previous problem, the best test data uses the boundary values. It's also a
good idea to try a case of double pane with glass cost greater than $350.
length width single/double delivery
10 34.9 d n
10 by 34.9 gives an area of 349, which gives a glass cost of $349
10 35 d y
10 by 35 gives an area of 350, which gives a glass cost of $350
10 15 s n
20 30 d n
7. grand total = 0
read the length of the window
read the width of the window
while length != -1 and width != -1
read the thickness of the window (s/d)
area = length * width
perimeter = 2 * (length + width)
if thickness = 's' // USES TWO-WAY SELECTION
glass cost = area * .50
else
glass cost = area * 1.00
if glass cost < 350 // THIS IS ONE-WAY SELECTION INSIDE
glass cost = 350 // THE TWO-WAY SELECTION
frame cost = perimeter * .75
total cost = glass cost + frame cost
if delivery == 'y'
total cost = total cost + 50.00
print length, width, total cost
grand total = grand total + total cost
read the length of the window
read the width of the window
print grand total
8. For this problem, we can use the test data given for the previous
algorithm. Instead of having to run the program four separate times, we
can use this test data all in one execution. We just need to add -1 -1
to the end of the data to stop the loop.
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