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A case for Vancomycin : 2

Moving on to step number five to calculate the dosing interval Dosing interval is equal to the natural log of peak concentration over through concentration divided by K the first-order elimination rate constant plus cap T which is the infusion time and a Tau, in this case because we know the therapeutic window is 28 to 17, therefore, the tau turn out to be 10.2 hours and then, of course, we rounded up to 12 hours. So that is the desirable dosing interval in order to constrain the concentration between the peak of 28 and the trough of 17.
So based on the calculated dosing interval, we are going to recommend and maintenance dose and that maintenance dose D is equal to the peak concentration times K times V times cap T times 1 minus e to the minus K tau divided by 1 minus e to the minus K cap T. So this is if we know the peak concentration that we wish to achieve at the dosing interval of tau and we should be able to calculate and maintenance dose in order to maintain the concentration between the therapeutic window. Again here C peak is 28 milligram per liter. K is 0.0542 V is 66.4. The infusion time is one hour and the Tau is 12 hour.
And the result turns out to be 912.9 mg and of course we rounded up to 1000 milligram. So the recommended maintenance dose would be 1000 milligram per hour infusion rate.

Continuing from the Vancomycin case in the previous step, Prof. Lee explains how to calculate dosing interval and maintenance dose this time.

When we get the dosing interval, remember to round up the tau to the actual interval in clinical practice. For instance, we have 10.2 in this case after the calculation, and we usually round it up to 12 then.

Besides, if we wish to achieve the pharmacological effect at the dosing interval of tau, we should be able to calculate the maintenance dose to maintain the concentration between the therapeutic window.

If you have any questions or thoughts, please do not hesitate to leave them below.

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Pharmacokinetics: Drug Dosing in Renal Disease

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