Question 1. The data in table 1 was collected after a two hour infusion of 150 mg/hr. Calculate kel and V.
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Plotting the data on semi-log graph paper gives the green and blue symbols. Semi-log regression through the blue squares give kel = 0.223 hr-1 and Intercept = 5.95 mg/l. Thus CpT = 5.95 x e-0.223 x 2 = 3.8 mg/L (this could have been read from the graph.
Since
Question 2. Fit the data in table 1 using Boomer and the initial estimates calculated above.
Draw the model
Tabulate the model parameters
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kel |
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The .BAT files is included below:
Boomer Batch File
0 wls,bayes,sim,irwls,sim+error,grid
1 Screen, diskfile, printer
HW9911a0
2 Parameter type
kel
0.2230 Parameter value
1 Fixed,adjust,depend1,depend2
0.0000 Lower limit
1.000 Upper limit
0 To
1 From
0 happy or not
18 Parameter type
V
63.70 Parameter value
1 Fixed,adjust,depend1,depend2
1.000 Lower limit
200.0 Upper limit
1 To
Cp
1 From
0 happy or not
0 Parameter type
Duration
2.000 Parameter value
0 Fixed,adjust,depend1,depend2
0 happy or not
3 Parameter type
k0
150.0 Parameter value
0 Start interrupt number
1 Finish interrupt number
0 Fixed,adjust,depend1,depend2
1 To
0 From
0 happy or not
-1 Parameter type
2 Integration method
0.0000 Relative error
0.0000 Absolute error
4 Fitting algorithm
0.0000 PC value
Homework #1 1999 Question 2
1 Data from disk or keyboard
0.0000 X value
0.0000 Y value
1.000 X value
1.950 Y value
2.000 X value
4.000 Y value
3.000 X value
2.900 Y value
4.000 X value
2.500 Y value
5.000 X value
1.800 Y value
6.000 X value
1.600 Y value
7.000 X value
1.280 Y value
-1.000 X value
0 Accept, correct, delete, insert, of
0 Continue or save data
0 Weight type
0 AUC line number
2 Continue,save,plot,both
-2 wls,bayes,sim,irwls,sim+error,grid
This is the .BAt file for the equal weight fit. For the 1/val2 fit change the 0 to a 2 opposite the Weight type above.
The two outputs are:
** FINAL OUTPUT FROM Boomer (v2.7.8) ** 08-Sep-1999 --- 11:07:00 am
Title: Homework #1 1999 Question 2
Input: From hw9911a.BAT
Output: To HW9911a0.OUT
Data for Cp came from keyboard (or ?.BAT)
Fitting algorithm: Simplex Method
Weighting for Cp by Equal weight
Numerical integration method: 2) Fehlberg RKF45
with 1 de(s)
With relative error 0.1000E-03
With absolute error 0.1000E-03
PC = 0.1000E-04
** FINAL PARAMETER VALUES ***
# Name Value S.D. C.V. % Lower <-Limit-> Upper
1) kel 0.22634 0.00 1.0
2) V 63.282 1.0 0.20E+03
AIC = -11.2296 Final WSS = 0.113534
R-squared = 0.9781 Correlation Coeff = 0.9897
** FINAL OUTPUT FROM Boomer (v2.7.8) ** 08-Sep-1999 --- 11:07:00 am
Title: Homework #1 1999 Question 2
Input: From hw9911a.BAT
Output: To HW9911a0.OUT
Data for Cp came from keyboard (or ?.BAT)
Fitting algorithm: DAMPING-GAUSS/SIMPLEX
Weighting for Cp by Equal weight
Numerical integration method: 2) Fehlberg RKF45
with 1 de(s)
With relative error 0.1000E-03
With absolute error 0.1000E-03
DT = 0.1000E-02 PC = 0.1000E-04 Loops = 1
Damping = 1
** FINAL PARAMETER VALUES ***
# Name Value S.D. C.V. % Lower <-Limit-> Upper
1) kel 0.22641 0.145E-01 6.4 0.00 1.0
2) V 63.274 2.32 3.7 1.0 0.20E+03
AIC = -11.2296 Final WSS = 0.113533
R-squared = 0.9781 Correlation Coeff = 0.9897
Model and Parameter Definition
# Name Value Type From To Dep Start Stop
1) kel = 0.2264 2 1 0 0 0 0
2) V = 63.27 18 1 1 0 0 0
3) Duration = 2.000 0 0 0 0 0 0
4) k0 = 150.0 3 0 1 0 0 1
Data for Cp :-
DATA # Time Calculated Observed (Weight) Weighted residual
1 0.0000 0.000000 0.000000 0.000000 0.000000
2 1.000 2.12144 1.95000 1.00000 -0.171441
3 2.000 3.81306 4.00000 1.00000 0.186941
4 3.000 3.04050 2.90000 1.00000 -0.140497
5 4.000 2.42446 2.50000 1.00000 0.755358E-01
6 5.000 1.93324 1.80000 1.00000 -0.133245
7 6.000 1.54155 1.60000 1.00000 0.584487E-01
8 7.000 1.22922 1.28000 1.00000 0.507814E-01
WSS for data set 1 = 0.1135
R-squared = 0.9781 Correlation Coeff = 0.9897
Plots of observed (*) and calculated values (+)
versus time for Cp . Superimposed points (X)
4.000 Linear 4.000 Semi-log
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|_____________________________________ |X____________________________________
0.0000 1.229
0 <--> 7.0 0 <--> 7.0
Plot of Std Wtd Residuals (X) Plot of Std Wtd Residuals (X)
versus time for Cp versus calcd Cp(i) for Cp
1.359 1.359
| X | X
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| X | X
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| X | X
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0X==================================== 0X====================================
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| X X | X X
| X | X
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-1.246 -1.246
0.00 <--> 7.0 0.00 <--> 3.8
** FINAL OUTPUT FROM Boomer (v2.7.8) ** 08-Sep-1999 --- 11:07:35 am
Title: Homework #1 1999 Question 2
Input: From hw9911a.BAT
Output: To HW9911a2.OUT
Data for Cp came from keyboard (or ?.BAT)
Fitting algorithm: Simplex Method
Weighting for Cp by 1/Cp(Obs )^2
Numerical integration method: 2) Fehlberg RKF45
with 1 de(s)
With relative error 0.1000E-03
With absolute error 0.1000E-03
PC = 0.1000E-04
** FINAL PARAMETER VALUES ***
# Name Value S.D. C.V. % Lower <-Limit-> Upper
1) kel 0.21426 0.00 1.0
2) V 66.310 1.0 0.20E+03
AIC = -24.4975 Final WSS = 0.170590E-01
R-squared = 0.9967 Correlation Coeff = 0.9908
** FINAL OUTPUT FROM Boomer (v2.7.8) ** 08-Sep-1999 --- 11:07:35 am
Title: Homework #1 1999 Question 2
Input: From hw9911a.BAT
Output: To HW991a2.OUT
Data for Cp came from keyboard (or ?.BAT)
Fitting algorithm: DAMPING-GAUSS/SIMPLEX
Weighting for Cp by 1/Cp(Obs )^2
Numerical integration method: 2) Fehlberg RKF45
with 1 de(s)
With relative error 0.1000E-03
With absolute error 0.1000E-03
DT = 0.1000E-02 PC = 0.1000E-04 Loops = 1
Damping = 1
** FINAL PARAMETER VALUES ***
# Name Value S.D. C.V. % Lower <-Limit-> Upper
1) kel 0.21420 0.108E-01 5.0 0.00 1.0
2) V 66.321 2.53 3.8 1.0 0.20E+03
AIC = -24.4976 Final WSS = 0.170589E-01
R-squared = 0.9967 Correlation Coeff = 0.9908
Model and Parameter Definition
# Name Value Type From To Dep Start Stop
1) kel = 0.2142 2 1 0 0 0 0
2) V = 66.32 18 1 1 0 0 0
3) Duration = 2.000 0 0 0 0 0 0
4) k0 = 150.0 3 0 1 0 0 1
Data for Cp :-
DATA # Time Calculated Observed (Weight) Weighted residual
1 0.0000 0.000000 0.000000 0.000000 0.000000
2 1.000 2.03590 1.95000 0.512821 -0.440495E-01
3 2.000 3.67924 4.00000 0.250000 0.801894E-01
4 3.000 2.96983 2.90000 0.344828 -0.240789E-01
5 4.000 2.39720 2.50000 0.400000 0.411198E-01
6 5.000 1.93498 1.80000 0.555556 -0.749911E-01
7 6.000 1.56189 1.60000 0.625000 0.238189E-01
8 7.000 1.26073 1.28000 0.781250 0.150518E-01
WSS for data set 1 = 0.1706E-01
R-squared = 0.9967 Correlation Coeff = 0.9908
Plots of observed (*) and calculated values (+)
versus time for Cp . Superimposed points (X)
4.000 Linear 4.000 Semi-log
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|X | X
|_____________________________________ |X____________________________________
0.0000 1.261
0 <--> 7.0 0 <--> 7.0
Plot of Std Wtd Residuals (X) Plot of Std Wtd Residuals (X)
versus time for Cp versus calcd Cp(i) for Cp
1.504 1.504
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| X | X
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0X==================================== 0X====================================
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| X | X
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| X | X
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| X | X
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-1.406 -1.406
0.00 <--> 7.0 0.00 <--> 3.7
Question 3. With a 17 hr half-life and V equal to 19 L what infusion rate would give a steady state concentration of 15 mg/L
What IV bolus dose would you need to reach 15 mg/L quickly:
What IV infusion over 30 minutes would be required to reach 15 mg/L
Sketch the slow infusion on its own
Next the slow infusion with the bolus dose
Finally the fast infusion and the slow infusion
Question 4. Given the model below write the differential equation for Xu and integrate this equation using the Laplace equation table.
Starting with Xp
Continue with Xu
Since there is a repeated term (s2) in the denominator we can't use the finger print method. We must use the Laplace Table. If A = ke•k0 and a = kel
