Multiple IV Dosing Pharmacokinetic Research, Bioequivalence Studies, Drug – Drug Interactions Generally all evaluated through single dose studies. But patients receive and most drug therapy is given in a multiple dose regimen Multiple doses generally result in accumulation of drug, … but how much, how fast ?

Problem: A patient with renal dysfunction (CrCl~ 20 mL/min) is given tobramycin IV. The half-life of tobramycin in this patient is 8 hours. Although tobramycin is normally infused over at least 30 minutes, we will administer 100 mg of tobramycin to our patient by IV bolus every 8 hours. Calculate the amount of tobramycin in the body after each dose for the first 10 doses. Dose Time Amount Amount Increase (n) (hr) in the Eliminated in Body Body During Store (mg) Dose Int (mg) 1. Zero 0.00 0.01 100.00 8.00

Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses. Dose Time Amount Amount Increase (n) (hr) in the Eliminated in Body Body During Store (mg) Dose Int (mg) 1. Zero 0.00 0.01 100.00 8.00 2. 8.01 16.0 3. 16.01 24.00

Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses. Dose Time Amount Amount Increase (n) (hr) in the Eliminated in Body Body During Store (mg) Dose Int (mg) 1. Zero 0.00 0.01 100.00 8.00 50.00 50.00 50.00 2. 8.01 150.00 16.0 75.00 75.00 25.00 3. 16.01 175.00 24.00 87.50 87.50 12.50 Increase is difference between dose and what was eliminated

Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses. Dose Time Amount Amount Increase (n) (hr) in the Eliminated in Body Body During Store (mg) Dose Int. (mg) 4. 24.01 187.50 32.00 93.75 93.75 6.25 9. 64.01 199.61 72.0 99.80 99.80 0.20 nth. 200.00 100.00 100.00 zero

Eventually, at some point accumulation stops Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses. Dose Time Amount Amount Increase (n) (hr) in the Eliminated in Body Body During Store (mg) Dose Int. (mg) nth. 200.00 100.00 100.00 zero nth + 1 200.00 100.00 100.00 zero Eventually, at some point accumulation stops This is called steady-state (ss)

At steady state the amount eliminated during an interval is Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses. Dose Amount Amount Increase (n) in the Body Eliminated in Body End of Interval During Store (mg) Dose Int. (mg) 0 0 1 50 50 50 2 75 75 25 3 87.5 87.5 12.5 4 93.75 93.75 6.25 9 99.80 99.8 0.20 nth. 100.00 100.00 = Dose 0.00 At steady state the amount eliminated during an interval is equal to the dose, or with oral dosing the amount of the dose absorbed. IN = OUT

At steady state the amount eliminated during an interval is Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses. Dose Amount Amount Increase (n) in the Body Eliminated in Body End of Interval During Store (mg) Dose Int. (mg) 0 0 1 50 50 50 2 75 75 25 3 87.5 87.5 12.5 4 93.75 93.75 6.25 9 99.80 99.8 0.20 nth. 100.00 100.00 = Dose 0.00 How did SS occur? During each dosing interval (8 hour period) the body appears to eliminate (metabolise, clear) more drug… How is this possible? At steady state the amount eliminated during an interval is equal to the dose, or with oral dosing the amount of the dose absorbed. IN = OUT

body stores increased by half as much as the previous dose… why? Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses. Dose Amount Amount Increase (n) in the Body Eliminated in Body End of Interval During Store (mg) Dose Int. (mg) 0 0 1 50 50 50 2 75 75 25 3 87.5 87.5 12.5 4 93.75 93.75 6.25 9 99.80 99.8 0.20 nth. 100.00 100.00 = Dose 0.00 When did SS occur? Increase in body stores continued with each dose, in fact with each dose body stores increased by half as much as the previous dose… why?

Steady State is considered to have been achieved when conc. (amount) Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses. Dose Amount Amount Increase (n) in the Body Eliminated in Body End of Interval During Store (mg) Dose Int. (mg) 0 0 1 50 50 50 2 75 75 25 3 87.5 87.5 12.5 4 93.75 93.75 6.25 9 99.80 99.8 0.20 nth. 100.00 100.00 = Dose 0.00 When did SS occur? Steady State is considered to have been achieved when conc. (amount) are within 10%. For this patient this occurs between the 3rd and 4th dose.

Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Amount in the Body Body Following a First and SINGLE Dose (mg) 1. Zero 0.00 0.01 100.00 8.00 50.00 16.0 25.00 24.0 12.50 32.0 6.25 40.0 3.12 48.0 1.56 54.0 0.78

Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body after the First Dose (mg) 1. Zero 0.00 0.01 100.00 Amount = Dose 8.00 50.00 Amount = Dose1 * e-Kt = Dose1 * e-Kτ τ (tau) represents dosing interval

What is the concentration immediately following The second dose (8.01 hrs)? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body after the Second Dose (mg) 1. 8.00 50.00 Amount = Dose1 * e-Kt = Dose1 * e-Kτ 2. 8.01 150.00 Amount = Dose2 + Dose1 * e-Kτ = 8hr single dose +0 hr single dose = 50 + 100 = 150 mg =   Time Am’t in (hr) the Body (mg) 0.01 100.00 8.00 50.00 16.0 25.00 24.0 12.50 32.0 6.25   3 = 2 + 1

3 = 2 + 1 Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body after the Second Dose (mg) 1. 8.00 50.00 Amount = Dose1 * e-Kt = Dose1 * e-Kτ 2. 8.01 150.00 Amount = Dose2 + Dose1 * e-Kτ 16.0 75.00 Amount = [Dose2 + Dose1 * e-Kτ] * e-Kτ = Dose2 e-Kτ + Dose1 * e-2Kτ = 8hr single dose +16hr single dose = 50.0 + 25.0 = 75 mg =  Time Am’t in (hr) the Body (mg) 0.01 100.00 8.00 50.00 16.0 25.00 24.0 12.50    3 = 2 + 1

6 = 1 + 2 + 3 Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body after the Third Dose (mg) 2. 16.0 75.00 Amount = Dose2 e-Kτ + Dose1 * e-2Kτ 3. 16.01 175.00 Amount = Dose3 +Dose2 e-Kτ + Dose1 * e-2Kτ = Dose (1 + e-Kτ + e-2Kτ) = 0 hr + 8hr + 16 hr = 100 + 50 + 25 = 175 mg =   Time Am’t in (hr) the Body (mg) 0.01 100.00 8.00 50.00 16.0 25.00 24.0 12.50 32.0 6.25    6 = 1 + 2 + 3

6 = 1 + 2 + 3 Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body following the Third Dose (mg) 3. 16.01 175.00 Amount = Dose3 +Dose2 e-Kτ + Dose1 * e-2Kτ = Dose (1 + e-Kτ + e-2Kτ) 24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ = Dose (e-Kτ + e-2Kτ +e-3Kτ) = 8 hr + 16 hr + 24 hr = 50 + 25 = 12.5 = 87.5 mg =   Time Am’t in (hr) the Body (mg) 0.01 100.00 8.00 50.00 16.0 25.00 24.0 12.50 32.0 6.25    6 = 1 + 2 + 3

Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body Second & Third Doses (mg) 2. 8.01 150.00 Amount = Dose2 + Dose1 * e-Kτ 16.0 75.00 Amount = [Dose2 + Dose1 * e-Kτ] * e-Kτ = Dose2 e-Kτ + Dose1 * e-2Kτ 3. 16.01 175.00 Amount = Dose3 +Dose2 e-Kτ + Dose1 * e-2Kτ = Dose (1 + e-Kτ + e-2Kτ) 24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ = Dose (e-Kτ + e-2Kτ +e-3Kτ)

Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body Third Dose (mg) 3. 16.01 175.00 Amount = Dose3 +Dose2 e-Kτ + Dose1 * e-2Kτ = Dose (1 + e-Kτ + e-2Kτ) 24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ = Dose (e-Kτ + e-2Kτ +e-3Kτ) This is a geometric progression for both the initial concentration following each dose (Cmax) and the end-of interval concentration (Cmin). But what does this show us?

Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body (mg) 3. 24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ = Dose (e-Kτ + e-2Kτ +e-3Kτ) But what does this show us? First, that the concentration in a multiple dose profile can be obtained by adding the conc. from various times following a single dose. For example: [ ]8hr after dose 3 = [3]8hr + [2]16hr + [1]24hr

Can we develop an equation which will predict these amounts / concentrations ? Dose Time Amount (n) (hr) in the Equation for Amount in the Body Body (mg) 3. 24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ = Dose (e-Kτ + e-2Kτ +e-3Kτ) Time Am’t in (hr) the Body (mg) 0.01 100.00 8.00 50.00 16.0 25.00 24.0 12.50 32.0 6.25 For example: [ ]8hr after dose 3 = [3]8hr + [2]16hr + [1]24hr = 50 + 25 + 12.5 = 87.5 mg

Can we develop an equation which will predict these amounts / concentrations ? Second that we have a Geometric Progression Solution to the Geometric Progression Since Cminn = Cmaxn e-Kτ These equations will calculate the concentration for the Cmax and Cmin of any dose (n) in a multiple dosing regimen. Cmax after doses where SS has not been achieved – blue arrows

Can we calculate the concentration at any time after any dose? Can we develop an equation which will predict concentrations at any time? Can we calculate the concentration at any time after any dose? Since Cminn = Cmaxn e-Kτ Then Ct = Cmaxn e-Kt and where n = 6 to concentrations corresponding to the red arrows above.

Can we develop an equation which will predict concentrations at any time? Therefore the more general equation which will calculate concentrations at any time t after any dose (1 through ) is: and when n is large the term e-nKτ approaches zero and then: Predicts for concentrations at steady state.

Can we develop an equation which will predict concentrations at any time? Using our equation: Example: Calculate the concentration 4 hrs after dose 3 in a patient. Dose = 1000 mg, V=10 L; τ = 8 hr T½ = 8 hr (K = 0.0866hr-1) 1000 (1- e-3*0.0866* 8) 10 (1- e-0.0866* 8) Ct = e -0.086*4 Ct = 100*[(1 - 0.1250)/(1-0.50)]*0.7 = 100 * [(0.875)/(.50)] * 0.707 = 100 * 1.75 * 0.707 = 175 * 0.707 = 123.7 mg/L What is 175 ..??

With Multiple Doses Drugs Accumulate. Can we predict the degree of accumulation quickly, without an equation? We have seen that in a patient given 1000 mg Q8H (τ) that when the T½ is 8 hr (K = 0.0866hr-1) and V = 10L, the degree of accumulation is a factor of 2. Dose Amount Amount in the Body in the Body Start of Interval End of Interval (n) (mg) (mg) 1 100 50 2 150 75 3 175 87.5 4 187.5 93.75 9 193.75 99.80 nth. 200.00 100.00

Can we predict how much accumulation WILL occur? Consider giving 100 mg of tobramycin IV, at different dosing intervals, when half-life is 8 hours and Volume is 10 L. Conditions : Dose =100; V = 10 L half-life = 8 hr CMAX Dose Q2H Q4H Q6H Q8H Q12H Q24H 1st. 10.00 10.00 10.00 10.00 10.00 10.00 2nd. 3rd. @ 24. @ 48 @ 96 Using MD 1C-IV Bolus Excel Sheet determine degree of accumulation for Q2, Q4, Q8, Q12, Q16 and Q24 hr regimens. Conditions : Dose = 200; V = 20 L half-life = 8 hr

Can we predict how much accumulation WILL occur? Accumulation depends on the rate (frequency) at which we give the drug (dosing interval). Consider giving 100 mg of tobramycin IV, at different dosing intervals, when half-life is 8 hours and Volume is 10 L. Conditions : Dose =100; V = 10 L half-life = 8 hr CMAX Dose Q2H Q4H Q6H Q8H Q12H Q24H 1st 10.00 10.00 10.00 10.00 10.00 10.00 2nd 18.41 17.07 15.95 15.00 13.54 3rd 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25 @ 48 61.17 33.77 24.44 19.85 15.39 11.41 @ 96 62.83 34.13 24.66 20.00 15.47 11.43

Can we predict how much accumulation WILL occur? Accumulation depends on the rate (frequency) at which we give the drug (dosing interval). Consider giving 100 mg of tobramycin IV, at different dosing intervals, when half-life is 8 hours and Volume is 10 L. Initial concentration is : Dose/ Vd = 100/ 10 = 10 mg/L Dose Q2H Q4H Q6H Q8H Q12H Q24H 1 10.00 10.00 10.00 10.00 10.00 10.00 2. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25 @ 48 61.17 33.77 24.44 19.85 15.39 11.41 @ 96 62.83 34.13 24.66 20.00 15.47 11.43 Notice that at Q8H, degree of accumulation is a factor of 2

Can we predict how much accumulation WILL occur? Repeat the process, but this time use a half-life of 6 hours (patient with a different CrCl). Administer 100 mg of tobramycin IV, at different dosing intervals, when half-life is 6 hours and Volume is 10 L. Conditions : Dose =100; V = 10 L half-life = 6 hr CMAX Dose Q2H Q4H Q6H Q8H Q12H Q24H 1st. 10.00 10.00 10.00 10.00 10.00 10.00 2nd. 3rd. @ 24. @ 48 @ 96 Using MD 1C-IV Bolus Excel Sheet determine degree of accumulation for Q2, Q4, Q8, Q12, Q16 and Q24 hr regimens. Conditions : Dose = 200; V = 20 L half-life = 6 hr

Can we predict how much accumulation WILL occur? Repeat the process, but this time use a half-life of 6 hours (patient with a different CrCl). Administer 100 mg of tobramycin IV, at different dosing intervals, when half-life is 6 hours and Volume is 10 L. Conditions : Dose =100; V = 10 L half-life = 6 hr Dose Q2H Q4H Q6H Q8H Q12H Q24H 1st. 10.00 10.00 10.00 10.00 10.00 10.00 2nd. 17.94 16.30 15.00 13.97 12.50 10.63 3rd. 24.24 20.37 17.50 15.54 13.13 10.66 @ 24. 46.08 25.96 19.38 16.17 13.13 10.66 @ 48 48.36 26.96 19.96 16.56 13.32 10.66 @ 96 48.48 27.03 20.00 16.58 13.33 10.67 When is the degree of accumulation of 2 observed now?

Can we predict how much accumulation WILL occur? Repeat the process, but this time use a half-life of 6 hours (patient with a different CrCl). Administer 100 mg of tobramycin IV, at different dosing intervals, when half-life is 6 hours and Volume is 10 L. Conditions : Dose =100; V = 10 L half-life = 6 hr CMAX Dose Q2H Q4H Q6H Q8H Q12H Q24H 1st. 10.00 10.00 10.00 10.00 10.00 10.00 2nd. 17.94 16.30 15.00 13.97 12.50 10.63 3rd. 24.24 20.37 17.50 15.54 13.13 10.66 @ 24. 46.08 25.96 19.38 16.17 13.13 10.66 @ 48 48.36 26.96 19.96 16.56 13.32 10.66 @ 96 48.48 27.03 20.00 16.58 13.33 10.67 2 x

Can we predict how much accumulation WILL occur? Accumulation depends on the rate at which we give the drug (dosing interval τ) and the rate at which the drug is eliminated (K) When the Dosing Interval and half-life are equal, accumulation is 2x Half-life = 8 hours Dose Q2H Q4H Q6H Q8H Q12H Q24H 1 10.00 10.00 10.00 10.00 10.00 10.00 2. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25 @ 48 61.17 33.77 24.44 19.85 15.39 11.41 @ 96 62.83 34.13 24.66 20.00 15.47 11.43

Can we predict how much accumulation WILL occur? Accumulation depends on the rate at which we give the drug (dosing interval τ) and the rate at which the drug is eliminated (K) When Dosing interval is greater than the half-life, more drug is eliminated during a dosing interval and LESS accumulation occurs Half-life = 8 hours Dose Q2H Q4H Q6H Q8H Q12H Q24H 1 10.00 10.00 10.00 10.00 10.00 10.00 2. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25 @ 48 61.17 33.77 24.44 19.85 15.39 11.41 @ 96 62.83 34.13 24.66 20.00 15.47 11.43

Can we predict how much accumulation WILL occur? Accumulation depends on the rate at which we give the drug (dosing interval τ) and the rate at which the drug is eliminated (K) When Dosing Interval is shorter than the half-life, less drug is eliminated during a dosing interval (because it is short) and MORE accumulation occurs Half-life = 8 hours Dose Q2H Q4H Q6H Q8H Q12H Q24H 1 10.00 10.00 10.00 10.00 10.00 10.00 2. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25 @ 48 61.17 33.77 24.44 19.85 15.39 11.41 @ 96 62.83 34.13 24.66 20.00 15.47 11.43

The ratio of the Peak Concentration at Steady State and the Peak Concentration after the first dose is a direct measure of the degree of accumulation. Cmax ss ----------- = Maximum Accumulation Factor Cmax 1 Dose Q2H Q4H Q6H Q8H Q12H Q24H 1 10.00 10.00 10.00 10.00 10.00 10.00 2. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25 @ 48 61.17 33.77 24.44 19.85 15.39 11.41 @ 96 62.83 34.13 24.66 20.00 15.47 11.43 MAF = 6.28 3.41 2.47 2.00 1.55 1.14

-------- = MAF = ----------- The ratio of the Peak Concentration at Steady State and the Peak Concentration after the first dose is a direct measure of the degree of accumulation. Cmax ss 1 -------- = MAF = ----------- Cmax 1 1 - e -K Cmaxss = Cmax1 x MAF Cminss = Cmin1 x MAF MAF rules of thumb if  = T½… then MAF = 2 if  > T½… then MAF < 2 if  < T½… then MAF > 2

-------- = MAF = ----------- The ratio of the Peak Concentration at Steady State and the Peak Concentration after the first dose is a direct measure of the degree of accumulation. Cmax ss 1 -------- = MAF = ----------- Cmax 1 1 - e -K MAF considers only  … the dosing interval K … the half-life

MAF = ----------- … predicts accumulation Building the Multiple Dose Equation 1 MAF = ----------- … predicts accumulation 1 - e -K 1 x ------------- 1 – e -K Dose V -Kt Ct = ---------- e First Dose Conc. Accumulation Steady State Equation and the Steady State equation predicts ONLY steady state concentrations We need the other piece to calculate concentrations following ANY dose

Building the Multiple Dose Equation Provides for all concentrations from first dose until steady state. 1- e -nK x ------------- 1 – e -K Dose V -Kt C = ---------- e In this equation what happens to the term 1- e -nk when n (number of doses) is large?

Building the Multiple Dose Equation And the term 1- e -nK ------------- 1 – e -K Can convert the single dose IV equation to a multiple dose equation

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the concentration or amount in the body is within 10% of the total amount of drug in the body. For this patient total was 100mg. Dose Amount Amount Increase Cumulative (n) in the Body Eliminated in Body Increase End of Interval During Storein Body Store (mg) Dose Int. (mg) (mg) 0 0 1 50 50 50 50 2 75 75 25 75 3 87.5 87.5 12.5 87.5 4 93.75 93.75 6.25 93.25 9 99.80 99.8 0.20 nth. 100.00 100.00 = Dose 0.00 90% achieved Between 3rd & 4thdose

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. Compare patients with T½ of 6 and 8 hrs for time to reach SS. T½ = 6 hr Dose Time Q2H 1. 0 10.00 2. 2 17.94 4. 6 29.21 10. 14 40.84 11. 16 42.42 12. 18 43.67 3T½ 13. 20 44.66 49. 96 48.48 T½ = 8 hr Dose Time Q2H 1. 0 10.00 2. 2 18.41 4. 6 31.43 10. 22 40.64 11. 24 56.25 12. 26 57.31 3.2T½ 49. 96 62.83 90% of 96 hr [ ] = 43.63 56.55

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 Sometimes, SS is not considered to have been achieved until concentrations exceed 95% of the true steady state concentrations. This occurs between 4 and 5 half-lives. This gives rise to the statement that steady state is achieved in 3-5 half-lives. Each T½ reduces the gap between current concentrations and SS by half. Dosing interval (τ) does not affect time to SS

When is 90% of true (eventual) e-Kt determines proportion lost How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 When is 90% of true (eventual) Steady State Achieved? e-Kt determines proportion lost

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 When is 90% of true (eventual) Steady State Achieved? e-Kt determines proportion lost e-K# determines proportion lost for a set number (#) of half-lives Example: If K = 0.693 and # = 2 T½ = e (-0.693 x 2) = 0.25

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 When is 90% of true (eventual) Steady State Achieved? e-K# determines proportion lost … then 1 - e-K# will determine proportion of steady-state achieved. Example: If K = 0.693 and # = 2 T½ = 1 -e (-0.693 x 2) = 0.75 Or expressed as a % 75% of SS after 2 T½

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 When is 90% of true (eventual) Steady State Achieved? e-K# determines proportion lost and … 1 - e (-0.693 x #) determines proportion of steady-state achieved. and … 100 x (1 - e (-0.693 x #) ) determines percent of steady-state achieved, where # is the number of half-lives. % SS = 100 x (1 - e (-0.693 x #) )

When is 90% of true (eventual) Therefore, 90% of true SS is achieved… How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 When is 90% of true (eventual) Steady State Achieved? % SS = 100 x (1 - e (-0.693 x #) ) Therefore, 90% of true SS is achieved… 90 = 100 x (1 - e (-0.693 x #) ) 0.9 = 1 - e (-0.693 x #) e (-0.693 x #) = 0.1 ln(e (-0.693 x #) = ln(0.1) -0.693 x # = -2.30259 # = 3.322

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 When is 90% of true (eventual) Steady State Achieved? After 3.322 half-lives … frequently stated as 3.3 half lives Or in hours: Time to SS (hr) = 3.322 x T½ If the T½ for your drug is 10 hours it will take (10 * 3.322) hours … or 33.22 hrs to achieve 90% of SS.

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 When is 95% of true (eventual) Steady State Achieved? After how many half-lives (#) is 95% of Steady State Achieved? ln(e (-0.693 x #) = ln(0.05) -0.693 x # = -2.9957 # = 4.322 (95% of SS achieved after 4.322 half-lives) or in hours: Time to SS (hr) = 4.322 x T½

How long does it take to get to steady state? When did SS occur? Steady State is considered to have been achieved when the Concentration is within 10% of the true steady state concentration. 90% of true steady state will occur at ~ 3.3 half-lives. Number Cumulative Of Increase T½ Body Store (mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375 SS 100 When is 90% of SS Achieved? Shown slightly differently, Recall that: the time to eliminate 50% of body stores is T½ = 0.693/K And 0.693147 is ln(0.5) (negative sign omitted) Therefore, the time to reduce the amount of drug in the body to 10% would be Time in hours = ln(0.1)/K = 2.302/K or since K = 0.693/T½, then Time (hr) = (ln(0.1)*T½) / 0.693 or = (3.322)*(T½)

Multiple IV Doses & Concentrations at Steady State In practice, some multiple dose regimens are initiated with a loading dose. Why do we use loading doses? To achieve steady state faster..? Or to achieve a therapeutic concentration sooner? Is a loading dose more useful for a drug with a longer or shorter half-life?

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring It is known that for many drugs, optimal effect is achieved with a certain minimum concentration. It might also be known that toxicity is more likely to occur with concentrations that exceed a certain concentration. Therefore, it becomes the intent of a multiple dose regimen to have peak and trough concentrations fall within this therapeutic range. Half-life is 8 hours, drug is administered every 8 hours. Dose is 400 mg. VD = 40 L. What dose will achieve a SS peak concentration of 30 mg/L?

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring Half-life is 8 hours, drug is administered every 8 hours. Dose is 400 mg. VD = 40 L. What dose will achieve a steady state concentration of 30 mg/L? Dose of 400 mg produces a peak of 400 mg/ 40 L = 10 mg/L Since half-life and Dosing interval are equal … MAF = ??? And the dosing regimen of 400 mg q8h will produce SS [ ] of ? Cpmax ss = Cp1 * MAF = 10 mg/L * 2 = 20 mg/L

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring Half-life is 8 hours, drug is administered every 8 hours. Dose is 400 mg. VD = 40 L. What dose will achieve a steady state concentration of 30 mg/L? A dosing regimen of 400 mg q8h will produce SS [ ] of ? Cpmax ss = Cp1 * MAF = 10 mg/L * 2 = 20 mg/L To acheive a SS peak of 30 mg/L we will need to administer 600 mg every 8 hours in this patient.

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring Half-life is 8 hours, drug is administered every 8 hours. Dose is 400 mg. VD = 40 L. If 600 mg is administered IV every 8 hours, what will the steady-state trough (Cmin) concentrations be? Given Cmaxss = 30 mg/L Since dosing interval and half-life are equal … half the concentration is eliminated during a dosing interval. Cpmaxss = 30 mg/L * 0.5 = 15 mg/L At SS [ ] fluctuate between 15 mg/L and 30 mg/L

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring … tougher Question Following the first 80 mg dose of an aminoglycoside administered by IV bolus*, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations? What do you need to know? What predicts degree of accumulation What is the Half-life? What should the dosing interval be? * Aminoglycosides normally administered by IV infusion over 30-60 min

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring … tougher Question Following the first 80 mg dose of an aminoglycosisde administered by IV bolus, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations? MAF is based on τ and T½ Time [mg/L] 2 4 6 2 T½ = ??? K = 0.1733 hr-1 What do you need to know? What predicts degree of accumulation What is the Half-life? What should the dosing interval be?

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring … tougher Question Following the first 80 mg dose of an aminoglycosisde administered by IV bolus, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations? MAF is based on τ and T½ Dosing interval based on Peak 8 mg/L Trough 1 mg/L 3 T½’s (8,4,2,1) = 12 hr. MAF = 1.143 What do you need to know? What predicts degree of accumulation What is the Half-life? What should the dosing interval be?

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring … tougher Question Following the first 80 mg dose of an aminoglycosisde administered by IV bolus, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations? Peak would be the time zero concentration … What was the concentration at time zero following the first dose? K= 0.17328 hr-1 C2 = 4; C0 = C2e(-Kt) = 4 e(-0.17328 x 2) = 5.66 mg/L

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring … tougher Question Following the first 80 mg dose of an aminoglycosisde administered by IV bolus, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations? What would the current dose (80 mg) yield for a Steady State Peak? CSS-peak = 5.66 mg/L x MAF = 5.66 x 1.1428 = 6.46 mg/L

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring … tougher Question Following the first 80 mg dose of an aminoglycosisde administered by IV bolus, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations? What dose is necessary to produce a peak of 8 mg/L (simple ratio)? Dose = (8.0 / 6.46) x (80 mg) = 1.237 x 80 = 98.99 mg This dose is unreasonable and so should be rounded up to 100 mg.

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring … tougher Question Following the first 80 mg dose of an aminoglycosisde administered by IV bolus, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations? Best Regimen is 100 mg administered every 12 hours. Calculate the SS peak. (use either complete equation or peak1 x MAF CmaxSS = (dose/volume) x MAF = (100 mg / 14.14L) x 1.143 = 8.08 mg/L

Concentrations at Steady State Multiple IV Doses & Concentrations at Steady State Therapeutic Drug Monitoring … tougher Question Following the first 80 mg dose of an aminoglycosisde administered by IV bolus, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations? Best Regimen is 100 mg administered every 12 hours. Calculate the SS Trough (use CmaxSS x e(-Kτ) CmaxSS = 8.08 e(-Kτ) = 8.08 e (-0.1733 x 12) = 1.01 mg/L

Calculation of Parameters at SS – IV dosing Evaluate the following concentrations determined in an patient following the 1st, 2nd, 6th and 15th 100mg doses of a drug given Q8H. Dose Dose Dose Dose 1 2 6 15 Time [ ] Time [ ] Time [ ] Time [ ] (hr) (mg/L) hr) (mg/L) (hr) (mg/L) (hr) (mg/L) 0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00 Steady State?? First Dose Half way 96.9% ~ True SS 1 T½ 5 T½ 14 T½

Calculation of Parameters at SS – IV dosing Evaluate the following concentrations determined in an patient following the 1st, 2nd, 6th and 15th 100mg doses of a drug given Q8H. Dose Dose Dose Dose 1 2 6 15 Time [ ] Time [ ] Time [ ] Time [ ] (hr) (mg/L) hr) (mg/L) (hr) (mg/L) (hr) (mg/L) 0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00 Does the half-life change? Does the volume change? Does clearance change? Does AUC change?

Calculation of Parameters at SS– IV dosing Evaluate the following concentrations determined in an patient following the 1st, 2nd, 6th and 15th 100mg doses of a drug given Q8H. Dose Dose Dose Dose 1 2 6 15 Time [ ] Time [ ] Time [ ] Time [ ] (hr) (mg/L) hr) (mg/L) (hr) (mg/L) (hr) (mg/L) 0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00 Does the half-life change? Does the volume change? Does clearance change? Does AUC change? Protein binding.

Calculation of Parameters at SS – IV dosing Evaluate the following concentrations determined in an patient following the 1st, 2nd, 6th and 15th 100mg doses of a drug given Q8H. Dose Dose Dose Dose 1 2 6 15 Time [ ] Time [ ] Time [ ] Time [ ] (hr) (mg/L) hr) (mg/L) (hr) (mg/L) (hr) (mg/L) 0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00 Does the half-life change? Does the volume change? Does clearance change? Does AUC change? Dose / AUC or K*V

Calculation of Parameters at SS – IV dosing Evaluate the following concentrations determined in an patient following the 1st, 2nd, 6th and 15th 100mg doses of a drug given Q8H. Dose Dose Dose Dose 1 2 6 15 Time [ ] Time [ ] Time [ ] Time [ ] (hr) (mg/L) hr) (mg/L) (hr) (mg/L) (hr) (mg/L) 0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00 Does the half-life change? Does the volume change? Does clearance change? Does AUC change? Let’s calculate AUC

Calculation of Parameters at SS – IV dosing Evaluate the following concentrations determined in an patient following the 1st, 2nd, 6th and 15th 100mg doses of a drug given Q8H. Dose Dose Dose Dose 1 2 6 15 Time [ ] Time [ ] Time [ ] Time [ ] (hr) (mg/L) hr) (mg/L) (hr) (mg/L) (hr) (mg/L) 0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00 Does this make sense? In = Out @ SS AUC (mg*hr/L) AUC(0-t) 57.74 86.61 113.63 115.47 AUC(0-) 115.47 173.21 227.37 230.95

Calculation of Parameters at SS – IV dosing Evaluate the following concentrations determined in an patient following the 1st, 2nd, 6th and 15th 100mg doses of a drug given Q8H. Elimination of 100 mg following the first dose produced an AUC(0-) of 115.47 mg*hr/L. At SS, a dose (100 mg) is eliminated during a dosing interval, in this case 8 hours. At steady state we are eliminating a dose. Therefore, the AUC(0-τ) at SS should be equivalent to the AUC(0-) following the first dose. 1st Dose 2nd Dose 6th Dose 15th Dose Half way 96.9% ~ True SS 1 T½ 5 T½ 14 T½ AUC (mg*hr/L) AUC(0-t) 57.74 86.61 113.63 115.47 AUC(0-) 115.47 173.21 227.37 230.95

Calculation of Parameters at SS – IV dosing Bioequivalence Just as AUC(0-) following the 1st dose can be use to estimate bioequivalence so can AUC(0-τ), as long as you are at steady state. Steady state will require at least 3 T½ of dosing, and the greater the time allowed to achieve steady, the smaller the difference between AUC1(0-) and AUCSS(0-τ). 1st Dose 2nd Dose 6th Dose 15th Dose Half way 96.9% ~ True SS 1 T½ 5 T½ 14 T½ AUC (mg*hr/L) AUC(0-t) 57.74 86.61 113.63 115.47 AUC(0-) 115.47 173.21 227.37 230.95

No change in other parameters Calculation of Parameters at SS – IV dosing No change in other parameters

Multiple Dosing Summary … so far – IV dosing We have seen that: 1. [ ] following single doses can be Summed to create MD profile (Additivity of [ ]) 2. At Steady State In (Dose) = Out (elimination) 3. MAF, predicts the degree of accumulation (K and τ) 4. (1/1-e(-Kτ)), converts a single dose equation to multiple dose 5. (1-e(-nKτ))/(1-e(-Kτ)) allows Calculation of [ ] at any time 6. Time to Reach steady state determine by K (Cl/V) -3.3 T½ AUC(0-τ) over a dosing interval @ SS = AUC(0-) 1st dose. Just as AUC(0-) 1st dose can be use to estimate bioequiavalence so can AUC(0-τ), as long as you are at steady state (min3 T½ dosing) . 9. Using MAF we can design an IV dosing regimen that will achieve desired peak and trough concentrations at SS. At Steady state all other parameters remain unchanged (ClR,ClH, proportion metabolised.)