| Time (hr) | Cp (mg/L) |
| 1 | 7.5 |
| 2 | 5.2 |
| 3 | 2.5 |
| 5 | 0.85 |
| 6 | 0.58 |
Question 1. Graph the data below on semi-log graph paper.
The following data was collected after an I.V. bolus dose of 250 mg.
| Time (hr) | Cp (mg/L) |
| 1 | 7.5 |
| 2 | 5.2 |
| 3 | 2.5 |
| 5 | 0.85 |
| 6 | 0.58 |
Note three cycle graph paper
Using Cp(0) = 13.3 mg/L and Cp(6) = 0.541 mg/L
kel = (ln(13.3) - ln(0.541)) / (6.0 - 0.0) = (2.59 - - 0.614) / 6 = 3.204/6 = 0.534 hr-1
Intercept = Cp(0) = 13.3 mg/L
V = Dose/Cp(0) = 250 / 13.3 = 18.8 L
Question 2. Analyse the same data using Boomer and parameter type 8.
Use Pre-exp value = 13.3 with 1.33 and 133 as upper and lower limit
Use Lambda value = 0.534 with 0.0534 and 5.34 as upper and lower limit
** FINAL OUTPUT FROM Boomer (v3.0) ** 20-Feb-2001 --- 3:31:34 pm
Title: Homework #2
Input: From HW0102.BAT
Output: To HW0102.OUT
Data for Cp came from keyboard (or ?.BAT)
Fitting algorithm: Damping Gauss-Newton Method
Weighting for Cp by 1/Cp(Obs )^2
DT = 0.1000E-02 PC = 0.1000E-04 Loops = 2
Damping = 1
** FINAL PARAMETER VALUES ***
# Name Value S.D. C.V. % Lower <-Limit-> Upper
1) Pre-exp A 13.191 1.13 8.6 1.3 0.13E+03
2) Lambda A 0.53383 0.223E-01 4.2 0.53E-01 5.3
AIC = -13.0821 Final WSS = 0.328295E-01
R-squared = 0.9991 Correlation Coeff = 0.9927
Model and Parameter Definition
# Name Value Type From To Dep Start Stop
1) Pre-exp A = 13.19 8 0 1 0 0 0
2) Lambda A = 0.5338 9 0 0 0 0 0
3) LagTime A = 0.0000 10 0 0 0 0 0
Data for Cp :-
DATA # Time Calculated Observed (Weight) Weighted residual
1 0.0000 13.1911 0.000000 0.000000 0.000000
2 1.000 7.73468 7.50000 0.133333 -0.312913E-01
3 2.000 4.53529 5.20000 0.192308 0.127828
4 3.000 2.65930 2.50000 0.400000 -0.637214E-01
5 5.000 0.914310 0.850000 1.17647 -0.756583E-01
6 6.000 0.536112 0.580000 1.72414 0.756682E-01
WSS for data set 1 = 0.3283E-01
R-squared = 0.9991 Correlation Coeff = 0.9927
Maximum value for Cp is 7.5000 at 1.000
Calculation of AUC and AUMC based on trapezoidal rule
AUC and AUMC for Cp using Observed data
Time Concentration AUC AUMC
0.000000 0.000000
1.00000 7.50000 10.3455 3.75000
2.00000 5.20000 16.6955 12.7000
3.00000 2.50000 20.5455 21.6500
5.00000 0.850000 23.8955 33.4000
6.00000 0.580000 24.6105 37.2650
25.6970 45.8193
MRT = 1.7831
Calculation of AUC and AUMC using Method 9 of R.D. Purves
AUC and AUMC for Cp using Observed data
Time Concentration AUC AUMC
0.000000 0.000000
1.00000 7.50000
2.00000 5.20000 11.7333 13.4667
3.00000 2.50000 15.4200 22.4603
5.00000 0.850000 18.4789 34.1564
6.00000 0.580000 19.1854 38.0193
20.2719 46.5736
MRT = 2.2975
Plots of observed (*) and calculated values (+)
versus time for Cp . Superimposed points (X)
13.19 Linear 13.19 Semi-log
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0.0000 0.5361
0 <--> 6.0 0 <--> 6.0
Plot of Std Wtd Residuals (X) Plot of Std Wtd Residuals (X)
versus time for Cp versus calcd Cp(i) for Cp
1.411 1.411
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-0.8351 -0.8351
0.00 <--> 6.0 0.54 <--> 13.
.BAT as binhexed (Mac) or zipped (Win) files.
Question 3. Use Laplace transform method to integrate the differential equation for Component 2 in the diagram below.
[-------------] [-------------] [------------] [ ] k1 [ ] k2 [ ] [ Component 1 ] ----> [ Component 2 ] ---> [ Component 3] [ ] [ ] [ ] [-------------] [-------------] [------------]
The dose is put into Component 1 and k1 and k2 are first order rate constants.
dX(1)/dt = - k1 x X(1)
dX(2)/dt = k1 x X(1) - k2 x X(2)
dX(3)/dt = k2 x X(2)
Take the Laplace of each equations
s x L(X(1)) - X(1)(0) = - k1 x L(X(1))
s x L(X(2)) - X(2)(0) = k1 x L(X(1)) - k2 x L(X(2))
s x L(X(3)) - X(3(0) = k2 x L(X(2))
Since X(1)(0) = Dose, X(2)(0) = 0, and X(3)(0) = 0
s x L(X(1)) + k1 x L(X(1)) = Dose
s x L(X(2)) + k2 x L(X(2)) = k1 x L(X(1))
s x L(X(3)) = k2 x L(X(2))
the first equation gives L(X(1)) = Dose / (s + k1)
the second equation gives L(X(2)) = k1 x L(X(1)) / (s + k2)
substituting the first (for L(X(1))) into the second gives:
L(X(2)) = k1 x Dose / ((s + k2) x (s + k1))
the third equation gives L(X(3)) = k2 x L(X(2))/s
substituting the first (for L(X(2))) into the third gives:
L(X(3)) = k1 x k2 x Dose / (s x (s+k1) x (s+k2))
with A = k1 x k2 x Dose, a = k1 and b = k2
Using Laplace table gives:
A x (1/(a x b) + exp(-a x t)/(a x (a-b)) - exp(-b x t)/b x (a-b))
or
X(3) = k1 x k2 x Dose x (1/(k1 x k2) + exp(-k1 x t)/(k1 x (k1-k2)) - exp(-k2 x t)/k2 x (k1-k2))