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The delta time for day 1 is 8 hours 1 hour for lunch = 420 minutes; day 2 is 8 hours 30 minutes = 450 minutes. So the productivity for project 1 is 120 LOC/ 870 minutes = 120 LOC/ 2.175 days = 55 LOC/programmer-day (assume 400 minutes per programmer day). The delta times for day 3 and 4 are 7 hours and 4.5 hours = 690 minutes. The productivity is 80 LOC/ 1.725 days = 46.4 LOC/programmer-day. Overall, the programmer averaged 200 LOC/ 3.9 days = 51.3 LOC/programmer-day. 3. Using the following job log, calculate all of the basic measures and the progress indicators. Is the project on schedule Assume that it is currently 5/01/01.

which is the same result obtained in Prob. 7.27(6).

4 12

(a) Using the state space representation, find the unit step response of the system. ( b ) Find the system function H(z).

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Estimated E ort (programmerdays) 50 35 20 40 60 80

From the result from Prob. 7.7 we have q [ n + 11 = A q [ n ] + b x [ n ]

Actual E ort So Far (programmerdays) 70 20 40 40 10 20

Thus, the boat moves downstream in a line making an angle 18 with the direction in which it is headed or making an angle 90 72 with the bank of the river. (See Sec. 4.7 for rounding procedures.)

Setting q[O] = 0 and x [ n ] = u [ n ] in Eq. ( 7 . 2 9 , the unit step response s [ n ] is given by

Estimated Completion Date 1/15/01 2/15/01 2/25/01 4/15/01 6/1/01 7/1/01

+(f)

-,][,]

The component of a vector along a line L is the perpendicular projection of the vector very useful to resolve a vector into two components along a pair of perpendicular lines. on L. It is often

Actual Date Completed 2/1/01 2/15/01 3/1/01 4/1/01

Thus,

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The BCWS is 50+35+20+40=145 programmer-days. The BAC is 50+35+20+40+60+80=285 programmer-days. The planned values (PVs) for the work tasks are 17.5 percent, 12.3 percent, 7.0 percent, 14.0 percent, 21.1 percent, 28.1 percent. The earned value is 17.5 percent + 12.3 percent + 7 percent + 14 percent=50.7 percent. The BCWP for 5/01/01 is the same as BCWS in this example because the scheduled work has been completed. Thus, SPI=145/145=1.

which is the same result obtained in Prob. 4.32(c). (6) By Eq. (7.44) the system function H ( z ) is given by

EXAMPLE 5.5 In Fig. 5.8(a), (b), and (c) the components of R are (1) 4 mi/h in the direction of the current and (2) 12 mi/h in the direction perpendicular to the current. EXAMPLE 5.6 In Fig. 5.9, the force F has horizontal component Fh 30 . Note that F is the vector sum or resultant of Fh and Fv.

Fig. 11-9

Now Thus,

z112] (

Fig. 11-10

The heading of an airplane is the direction (determined from a compass reading) in which the airplane is pointed. The heading is measured clockwise from the north and expressed in degrees and minutes. The airspeed (determined from a reading of the airspeed indicator) is the speed of the airplane in still air. The course (or track) of an airplane is the direction in which it moves relative to the ground. The course is measured clockwise from the north. The groundspeed is the speed of the airplane relative to the ground. The drift angle (or wind-correction angle) is the difference (positive) between the heading and the course.

which is the same result obtained in Prob. 4.32(a).

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