1. Compartmental models divide a system into compartments and model the flow of substances between compartments. Examples include radioactive decay, population growth, pollution in lakes, and drug assimilation. 2. The rate of change of a substance in a compartment equals the inflow rate minus the outflow rate. This is modeled using differential equations. Exponential decay models radioactive elements and is used for radiometric dating. 3. Drug assimilation can be modeled with compartments for the GI tract and bloodstream. The rate of change in each depends on the inflow and outflow between compartments. This models how drugs enter the bloodstream and are removed.

1. Tuesday, 6th February 2018
MATA KULIAH:
PEMODELAN MATEMATIKA
COMPARTMENTAL
MODEL
Lecture:
Heni Widayani, M.Si

2. Compartmental Diagram
Process
input output
Rate in Rate out
Balance Law
𝑛𝑒𝑡 𝑟𝑎𝑡𝑒
𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒
𝑜𝑓 𝑎 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
=
𝑟𝑎𝑡𝑒
𝑖𝑛
−
𝑟𝑎𝑡𝑒
𝑜𝑢𝑡
Example :
1. The decay process of radioactive elements
2. Births and deaths in population
3. Pollution into and our of a lake or river, or the atmosphere
4. Drug asimilation into and removal from the bloodstream

3. Exponential decay and radiometric dating
 Essential to the understanding of our history
(find the approximately date)
RADIOACTIVE
MATERIAL
Emitted particles
Asumption :
1. The amount of an element present is large enough so that we are
justified in ignoring random fluctuations.
2. The process is continuous in time
3. There is a fixed rate of decay for an element
4. There is no increase in mass of the body of material.
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓
𝑟𝑎𝑑𝑖𝑜𝑎𝑐𝑡𝑖𝑣𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡
= −
𝑟𝑎𝑡𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓
𝑟𝑎𝑑𝑖𝑜𝑎𝑐𝑡𝑖𝑣𝑒
𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
𝑑𝑒𝑐𝑎𝑦𝑒𝑑

4. Formulating the differential equation
Let N(𝑡) be the mass (in gram) of radioactive atoms at time 𝑡 and let ∆𝑡
be a small change in time. The change number of atom is proportional
to the number of atom at the start of the time period.
𝑁 𝑡 + ∆𝑡 = 𝑁 𝑡 − 𝑘𝑁 𝑡 ∆𝑡, 𝑁 𝑡0 = 𝑁0
where 𝑘 is a positive constant of proportionality indicating the rate of
decay per atom in unit time (decay constant).
If ∆𝑡 is small enough (∆𝑡 → 0), then
𝑑𝑁
𝑑𝑡
= −𝑘𝑁, 𝑁 𝑡0 = 𝑁0
Initial value
Problem (IVP)
Use the separation variable technique to solve the equation !
𝑵 𝒕 = 𝑵 𝟎 𝒆−𝒌(𝒕−𝒕 𝟎)
Since 𝑡 − 𝑡0 > 0 then
lim
𝑡→∞
𝑁0 𝑒−𝑘(𝑡−𝑡0) = 0
𝑁(𝑡) is a monoton
decreasing function

5. Substance 𝑯𝒂𝒍𝒇 − 𝒍𝒊𝒇𝒆(𝝉)
Xenon-133 5 days
Barium-140 13 days
Lead-210 22 years
Strontium-90 25 years
Carbon-14 5568 years
Plutonium 23103 years
Uranium-235 0.707x109 years
Uranium-238 4.5 x 109 years
Experimentally measured for the half-life (𝝉)
Half-Life (𝜏) is defined as the time taken for
half of a given quantity of atoms to decay. If
𝑁 𝑡0 = 𝑁0 then
𝑁 𝜏 =
𝑁0
2
Find the value of decay constant (𝒌)
for Carbon-14 !
In General,
𝑘 =
ln 2
𝜏

6.  Residence time is defined to be the mean time that an individual
particle is in the compartment.
𝑑𝑁
𝑑𝑡
= −𝑘 𝑁
 𝑘−1 is the residence time for a single compartment where 𝑘 is the
constant rate of 𝑁.
𝑵 𝒕 = 𝑵 𝟎 𝒆−𝒌(𝒕−𝒕 𝟎)
 The fraction of particles remaining in the compartment at time 𝑡 is
given by 𝑒−𝑘(𝑡−𝑡0).
 The quantity 𝐹 𝑡 = 1 − 𝑒−𝑘𝑡
represents the probability an
individual particle has left the compartment by time 𝑡 (cumulative
probability function for the time each particle was in the
compartment)
𝑭 𝒕 = 𝑷𝒓 𝑻 < 𝒕
 𝑇 is the random variable representing the time for each period.
Residence time

7. Study Case :Lascaux Cave Paintings
In the Cave of Lascaux in France there are some ancient wall paintings,
believed to be prehistoric. Using a Geiger counter, the current decay rate of C14
in charcoal fragments collected from the cave was measured as approximately
1.69 disintegrations per minutes per gram of carbon. In comparison,for living
tissue in 1950 the measurement was 13.5 disintegration per minute per gram
of carbon. How long ago was the radioactive carbon formed (the Lascaux Cave
painting painted)?
ANSWER :
Let 𝑁(𝑡) be the amout of C14 per gram in the charcoal at time 𝑡. We know that
𝜏 = 5568 years (the half-life of C14), so we get
𝑘 ≈ 0.0001245 per year
Let 𝑡0 = 0 be the current time. Let 𝑇 be the time that the charcoal was formed,
and thus 𝑇 < 0. For 𝑇 < 𝑡 < 𝑡0, the C14 decays follow the function
𝑁 𝑡 = 𝑁0 𝑒−𝑘𝑇
We don’t have the 𝑁(𝑇) or 𝑁0, but we have 𝑁′
𝑇 =
𝑁′(𝑇)
𝑁′(0)
=
𝑁(𝑇)
𝑁0
Thus, we get
𝑇 = −
1
𝑘
ln
𝑁 𝑇
𝑁0
= −16690 years ago

8. Salt dissolved in a tank
tank
salt inflow salt outflow
𝑟𝑎𝑡𝑒 𝑐ℎ𝑎𝑛𝑔𝑒
𝑜𝑓 𝑠𝑎𝑙𝑡 𝑚𝑎𝑠𝑠
𝑖𝑛 𝑡𝑎𝑛𝑘
=
𝑟𝑎𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑎𝑙𝑡
𝑒𝑛𝑡𝑒𝑟𝑠 𝑡𝑎𝑛𝑘
−
𝑟𝑎𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑎𝑙𝑡
𝑙𝑒𝑎𝑣𝑒𝑠 𝑡𝑎𝑛𝑘
𝑑𝑆
𝑑𝑡
= 10𝑐𝑖𝑛 𝑡 −
1
10
𝑆(𝑡), 𝑆 0 = 𝑠0
Balance law
A large tank contains 100 litres of salt water. Initially 𝑠0 kg of salt is
dissolved. Salt water flows into the tank at the rate of 10 litres per
minute, with the concentration 𝑐𝑖𝑛(𝑡) (kg of salt/litre) of this incoming
water-salt mixture varies with time. We assume that the solution in the
tank is thoroughly mixed and that the salt solution flows out at the
same rate at which it flows in: that is, the volume of water-salt mixture
in the tank remain constant.
Use the technique of integrating factors to solve IVP equation above!
𝑆 𝑇 = 𝑠0 𝑒−𝑇/10
+ 10𝑒−𝑇/10
0
𝑇
𝑐𝑖𝑛(𝑠)𝑒 𝑠/10
𝑑𝑡

9. Lake Pollution Models
Mass of
pollutant in
lake
Mass inflow Mass outflow
Polluted river water, pollution
dump into the lake
Water flows from the lake
carrying some polution with it
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒
𝑝𝑜𝑙𝑙𝑢𝑡𝑎𝑛𝑡 𝑚𝑎𝑠𝑠
𝑖𝑛 𝑙𝑎𝑘𝑒
=
𝑟𝑎𝑡𝑒
𝑜𝑓 𝑝𝑜𝑙𝑙𝑢𝑡𝑎𝑛𝑡
𝑒𝑛𝑡𝑒𝑟𝑠 𝑙𝑎𝑘𝑒
−
𝑟𝑎𝑡𝑒
𝑜𝑓 𝑝𝑜𝑙𝑙𝑢𝑡𝑎𝑛𝑡
𝑙𝑒𝑎𝑣𝑒𝑠 𝑙𝑎𝑘𝑒
Assumption :
1. The lake has a constant volume 𝑉
2. The lake water is continuously well mixed so the pollution is uniform
throughout
Let 𝑀(𝑡) is the mass of the pollutant in the lake
Let 𝐶(𝑡) be the concentration of the pollutant in the lake at time 𝑡.
Let 𝐹 be the rate at which water flows out of the lake in m3/day.
Let 𝑐𝑖𝑛 is the concentration g/m3) of the pollutant in the flow entering lake
𝑑𝑀
𝑑𝑡
= 𝐹𝑐𝑖𝑛 − 𝐹
𝑀 𝑡
𝑉
𝑑𝐶
𝑑𝑡
=
𝐹
𝑉
𝑐𝑖𝑛 −
𝐹
𝑉
𝐶, 𝐶 0 = 𝐶0

10. Problem :
How long it will take for the lake’s pollution level to reach 5% of
its initial level, if only fresh water flows into the lake ?
Implement the result for the cases below
a. Consider Lake Eric with 𝑉 = 458 × 109
m3 and 𝐹 = 480 ×
106 m3/day = 1,75 x 1011 m3/year.
b. Consider Lake Ontario with 𝑉=1636 x 109 m3 and 𝐹 = 572 ×
106
m3/day = 2,089 x 1011 m3/year.
a. t0.05 = 7.8 years
b. t0.05 = 23.5 years
Although the flow rate in and out of Lake Ontario is similar to
Lake Erie, it takes significantly longer to clear the pollution from
Lake Ontario due to the larger volume water in Lake Ontario.

11. Drug Assimilation into the blood
 The drug dissolves in the gastrointestinal tract (GI-tract) and each
ingredient is diffused into bloodstream.
 Drug carried to the locations in which they act and are removed from
the blood by the kidneys and the liver.
GI Tract Blood
Drug intake digestion tissues
𝑟𝑎𝑡𝑒 𝑐ℎ𝑎𝑛𝑔𝑒
𝑜𝑓 𝑑𝑟𝑢𝑔
𝑖𝑛 𝐺𝐼 𝑡𝑟𝑎𝑐𝑡
=
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑑𝑟𝑢𝑔
𝑖𝑛𝑡𝑎𝑘𝑒
-
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑑𝑟𝑢𝑔 𝑙𝑒𝑎𝑣𝑒𝑠
𝐺𝐼 𝑡𝑟𝑎𝑐𝑡
𝑟𝑎𝑡𝑒 𝑐ℎ𝑎𝑛𝑔𝑒
𝑜𝑓 𝑑𝑟𝑢𝑔
𝑖𝑛 𝑏𝑙𝑜𝑜𝑑
=
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑑𝑟𝑢𝑔
𝑒𝑛𝑡𝑒𝑟𝑠 𝑏𝑙𝑜𝑜𝑑
-
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑑𝑟𝑢𝑔
𝑙𝑒𝑎𝑣𝑒𝑠 𝑏𝑙𝑜𝑜𝑑
Let 𝑥(𝑡) be the amount of a drug in the GI-tract at time 𝑡.
Let 𝑦(𝑡) be the amount of a drug in the bloodstream at time 𝑡.

12. A single cold pill
 There is no ingestion of the drug except that which occurs initially.
 Assumption :
1. The output rate of GI-tract is proportional to the drug concentration,
which is proportional to the amount of drug in the bloodstream
2. In the bloodstream, the initial amount of the drugs is zero. The level
increases as the drug diffuses from the GI-tract an decreases as the
kidneys and liver remove it.
𝑑𝑥
𝑑𝑡
= −𝑘1 𝑥, 𝑥 0 = 𝑥0
𝑑𝑦
𝑑𝑡
= 𝑘1 𝑥 − 𝑘2 𝑦, 𝑦 0 = 0
where 𝑥0 is the amount of a drug in the pill, 𝑘1 and 𝑘2 are positive
contant of proportionality.
The cold pill is made up of a decongestant and an antihistamine, which
are define the value of 𝑘1 and 𝑘2.
Decongestant Antihistamine
𝑘1 1.3860/hr 0.6931/hr
𝑘2 0.1386/hr 0.0231/hr
𝑥 𝑡 = 𝑥0 𝑒−𝑘1 𝑡
𝑦 𝑡 =
𝑘1
𝑘1 − 𝑘2
𝑒−𝑘2 𝑡 − 𝑒−𝑘1 𝑡

13. A course of cold pills
 We take a course of pills rather than just one.
 There is a continuous flow of drugs into the GI-tract
𝑑𝑥
𝑑𝑡
= 𝐼 − 𝑘1 𝑥, 𝑥 0 = 𝑥0
𝑑𝑦
𝑑𝑡
= 𝑘1 𝑥 − 𝑘2 𝑦, 𝑦 0 = 0
where 𝐼 is a positive constant represnting the rate of ingestion of the
drug (g/hr).
The analytic solution was
𝑥 𝑡 =
𝐼
𝑘1
1 − 𝑒−𝑘1 𝑡
𝑦 𝑡 =
𝐼
𝑘2
1 −
1
𝑘2 − 𝑘1
𝑘2 𝑒−𝑘1 𝑡
− 𝑘1 𝑒−𝑘2 𝑡
* This solution is valid only if 𝑘1 ≠ 𝑘2.

14. Dull, dizzy, or dead?
 Australian law prohibits driving of vehicles (including boats and
horse) for those with BAL (blood alcohol level) above 0.05. This the
relates to 50mg/100ml alcohol in the bloodstream.
 This restriction is a result of U.S statistics which indicate that a person
with a BAL of 0.15 is 25 times more likely to have a fatal accident than
one with no alcohol. Furthermore, for 41% of Australian men
excessive alcohol leads to confrontational behaviour.
BAL Behavioural effect
5% Lowered alertness, usually good feeling, release of
inhibitors, impaired judgement
Dull and dignified
10% Slowed reaction times and impaired motor function,
less caution
Dangerous and devilish
15% Large consistent increases in reaction time Dizzy
20% Marked depression in sensory and motor capability,
decidedly intoxicated
Disturbing
25% Severe motor disturbance, staggering, sensory
perceptions greatly impaired, smashed
Disgusting and
dishevelled
30% Stuporous but conscious, no comprehension of what’s
going on
Delirious and
disoriented
35% Surgical anaesthesia, minimal level causing death Dead drunk
40% 50 times the minimal level, causing death Dead !

15. • The alcohol intake into the GI-tract is “controlled” by the drinker. The
amount of alcohol subsequently absorbed into the bloodstream
depends on the concentration of alcohol, other liquid and food in the
GI-tract, as well as on the weight and sex of the individual. Alcohol is
removed from the bloodstream at a constant rate by the liver. This is
independent of the body weight, sexm of the individual and
concentration of alcohol in the bloodstream and assumes that the liver
has not been damaged by large doses of alcohol. (Ignoring that a small
percentage leaves through sweat, saliva, breath, and urine. This means
BAL estimate may be slightly above the true value).
• Let 𝐶1(𝑡) be the concentration of alcohol in the GI-tract at time 𝑡.
Let 𝐶2(𝑡) be the concentration of alcohol in the bloodstream at time 𝑡
𝑑𝐶1
𝑑𝑡
= 𝐼 − 𝑘1 𝐶2
𝑑𝐶2
𝑑𝑡
= 𝑘2 𝐶1 −
𝑘3 𝐶3
𝐶2 + 𝑀
• In the case of drinking on an empty stomach, 𝑘1 = 𝑘2
• If drinking occurs together with a meal (or is diluted) then 𝑘1 > 𝑘2