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differential equations report

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Red Wriggler Converion

To begin I will start describing the problem that you asked us to solve. Our group broke it into

three different and distinct phases.

Phase 1 Red Wiggler Conversion

In phase one you stated that the Levittson factory would be able to support twenty million red

wrigglers by November 1st. Both the maximum capacity of the Levittson factory and the date at

which it will be ready are important.

Phase 2 Stocking and Breeding of Red Wrigglers

In phase two you will be stocking 100,000 pounds of worms per week. We aren't really concerned

with how many pounds of worms but how many. But from this we can get a number of worms

using the average worms weighs of about 4 ounces and convert.

100, 000pounds 16ounces worms = 400, 000worms

pound

4ounces

100,000 pounds of worms per week is, on average, 400,000 worms. You also gave us the reproductive

rate of the red wrigglers. According to Dr. Kim they reproduce at a rate of .73% per day.

Phase 3 Harvesting

In your letter you also stated that harvesting could begin at 75% percent capacity, 15,000,000

worms. Also you gave us a date range by which you must reach that amount: January 12-15. So

starting on November 1st we have 73-76 days to stock and breed that many worms.

Explanation of Model

Now that we have all the information relevant to the solution lets find on when at what time 75%

will occur. Our team has elected to use a modified logistics growth equation, which is a very good

model for population growth when we are dealing with a short time frame. The standard logistics

growth equation is as follows:

dP = rP (1 - P )

dt

k

Your chief engineer will be familiar with this notation and maybe even this equation but nonetheless,

I will explain it briefly. dP is the rate of the change of the population. You may also think about

dt

is as how fast the population is changing over time. If you can recall your high school algebra you

might remember the change in y over the change in x? This was the slope of your line. This is

essential the same, but instead we have the change in population over the change in time and our

change is our rates. First we need the growth rate, which we will call r. As we found earlier the

growth rate is the rate at which the worms reproduce .73% per day. Next is the carrying capacity

or population capacity we will call k. Since the Levittson factory can hold 20,000,000 we set k to

that. Also notice what happens when the population reaches the carrying capacity.

dP = r20, 000, 000(1 - 20,000,000)

dt

20,000,000

dP = r20, 000, 000(1 - 1)

dt

dP = r20, 000, 000(0)

dt

dP = 0

dt

The rate of growth goes to zero. This is what we want to happen; this stops the population from

growing at capacity. This model shows the growth of the worms apart from the shipments. We

must account for the shipments of worms. To do this we need realize that our worms reproduce

1

daily, but a shipment only comes once a week. One way to account for the different rates would

be by finding a growth rate per week instead of per day. If we simply convert the growth rate into

weeks we lose a little of the worms growth rate each day since the rate of growth depends on the

worms present.

.73%growth 7days = 5.11%

day

week

week

Similarly we could change the shipment to per day.

400,000worms 1week = 57142.857worms

week

7days

day

But again, you are not getting worms every single day, just every week. In the end, after trying

both ways, the actually difference between each way is negligible, differing only by a fraction of a

day. Our team has elected to show you the solution using days because days is more specific to the

deadline.

Final Equation

Since we decided t will be days, we can now write the equation taking into account the shipment.

To do this we simply add the rate at which we receive worms, about 57142 worms per day. Finally

we can see the last form of the equation by plugging in the proper numbers. Don't let the W

confuse you I am only using it to denote the worm population instead P which is just an arbitrary

population.

dW = .0073 W (1 -

W

) + 57142

dt

20,000,000

Notice also that I used .0073 for the rate because we want the answer in terms of an actually number

and not a percent. Next, using Maple 15, a high-level mathematical programming language, we

can solve this equation to find a solution that gives us the ability to find the population at any

particular time. We can only do this knowing what the starting population is. In this case I will

assume the first shipment of 400,000 worms comes on November 1st. I will spare you details of

this solution because it is just a mess of numbers. However, I will attach the math for when Ms.

Schweinsteiger would like to look over the solution (see end notes). Continuing on, we can set the

equation's population equal to the 15,000,000 worms of 75% of the total capacity to find out on

what exact day the population will reach this number.

2

Solution

At 400,000 worms per week we would reach 75% on 173rd day, April 22, 2012. That puts you about

3 month off of the target day. Shown below is a graph of this. t being the time in days and w(t)

is analogous to w. The w(t) just denotes that population w(t) is with dependent on time. The

top number 1.5 x 107 is the same thing as saying 15,000,000, the target population, it is written

in scientific notation. Also don't mind the red arrows. They just denote other solutions to this

equation such as if we choose a different starting population.

As we can see our graph depicts exactly that, on about the 173rd day we reach the target population.

3

As I stated in my letter, the only way to increase the speed at which the worm population reaches

the target is to increase the shipment. After examining a couple of cases the best suggestion we

can give for you is to triple the worm shipment, 300,000 pounds per week. Given this change, we

now we reach the target by the 68th day or January 8th, 2012. This leaves a few days before the

stockholder's meeting, which is ideal. This gives you a couple of days if the preparations for the

Levittson facility are not quite complete November 1st. Below is the new graph on the same time

scale that accounts for the change in shipment rate. We can see that changing this one factor makes

a big difference.

If you have any questions on the approach to this problem or the math behind it feel free to contact

Dr. Buhl. Also if you have any future issues please don't hesitate to contact us. We enjoy helping

companies, such as yourself, plan for the future. Again we would like to thank you for using

Mathematical Modeling Consultants. Accurate information is power and with this information we

hope that you can find yourself with the power to survive in these hard times.

Notes on attached Maple Worksheet

worms1 in the math refers to the equations using the tripled shipment and worms2 refers to using

the normal shipment. Here we have used dsolve for solving the differential equation. Notice w(0)

is equal to the starting population. Next we used fsolve for solving the by setting the solution of

the population equation to 15,000,000, the population we wanted. Lastly don't be alarmed by the

differences in a negative sign in our dsolve equation and our fsolve. They turn out to be the same

equation.

4

5

To begin I will start describing the problem that you asked us to solve. Our group broke it into

three different and distinct phases.

Phase 1 Red Wiggler Conversion

In phase one you stated that the Levittson factory would be able to support twenty million red

wrigglers by November 1st. Both the maximum capacity of the Levittson factory and the date at

which it will be ready are important.

Phase 2 Stocking and Breeding of Red Wrigglers

In phase two you will be stocking 100,000 pounds of worms per week. We aren't really concerned

with how many pounds of worms but how many. But from this we can get a number of worms

using the average worms weighs of about 4 ounces and convert.

100, 000pounds 16ounces worms = 400, 000worms

pound

4ounces

100,000 pounds of worms per week is, on average, 400,000 worms. You also gave us the reproductive

rate of the red wrigglers. According to Dr. Kim they reproduce at a rate of .73% per day.

Phase 3 Harvesting

In your letter you also stated that harvesting could begin at 75% percent capacity, 15,000,000

worms. Also you gave us a date range by which you must reach that amount: January 12-15. So

starting on November 1st we have 73-76 days to stock and breed that many worms.

Explanation of Model

Now that we have all the information relevant to the solution lets find on when at what time 75%

will occur. Our team has elected to use a modified logistics growth equation, which is a very good

model for population growth when we are dealing with a short time frame. The standard logistics

growth equation is as follows:

dP = rP (1 - P )

dt

k

Your chief engineer will be familiar with this notation and maybe even this equation but nonetheless,

I will explain it briefly. dP is the rate of the change of the population. You may also think about

dt

is as how fast the population is changing over time. If you can recall your high school algebra you

might remember the change in y over the change in x? This was the slope of your line. This is

essential the same, but instead we have the change in population over the change in time and our

change is our rates. First we need the growth rate, which we will call r. As we found earlier the

growth rate is the rate at which the worms reproduce .73% per day. Next is the carrying capacity

or population capacity we will call k. Since the Levittson factory can hold 20,000,000 we set k to

that. Also notice what happens when the population reaches the carrying capacity.

dP = r20, 000, 000(1 - 20,000,000)

dt

20,000,000

dP = r20, 000, 000(1 - 1)

dt

dP = r20, 000, 000(0)

dt

dP = 0

dt

The rate of growth goes to zero. This is what we want to happen; this stops the population from

growing at capacity. This model shows the growth of the worms apart from the shipments. We

must account for the shipments of worms. To do this we need realize that our worms reproduce

1

daily, but a shipment only comes once a week. One way to account for the different rates would

be by finding a growth rate per week instead of per day. If we simply convert the growth rate into

weeks we lose a little of the worms growth rate each day since the rate of growth depends on the

worms present.

.73%growth 7days = 5.11%

day

week

week

Similarly we could change the shipment to per day.

400,000worms 1week = 57142.857worms

week

7days

day

But again, you are not getting worms every single day, just every week. In the end, after trying

both ways, the actually difference between each way is negligible, differing only by a fraction of a

day. Our team has elected to show you the solution using days because days is more specific to the

deadline.

Final Equation

Since we decided t will be days, we can now write the equation taking into account the shipment.

To do this we simply add the rate at which we receive worms, about 57142 worms per day. Finally

we can see the last form of the equation by plugging in the proper numbers. Don't let the W

confuse you I am only using it to denote the worm population instead P which is just an arbitrary

population.

dW = .0073 W (1 -

W

) + 57142

dt

20,000,000

Notice also that I used .0073 for the rate because we want the answer in terms of an actually number

and not a percent. Next, using Maple 15, a high-level mathematical programming language, we

can solve this equation to find a solution that gives us the ability to find the population at any

particular time. We can only do this knowing what the starting population is. In this case I will

assume the first shipment of 400,000 worms comes on November 1st. I will spare you details of

this solution because it is just a mess of numbers. However, I will attach the math for when Ms.

Schweinsteiger would like to look over the solution (see end notes). Continuing on, we can set the

equation's population equal to the 15,000,000 worms of 75% of the total capacity to find out on

what exact day the population will reach this number.

2

Solution

At 400,000 worms per week we would reach 75% on 173rd day, April 22, 2012. That puts you about

3 month off of the target day. Shown below is a graph of this. t being the time in days and w(t)

is analogous to w. The w(t) just denotes that population w(t) is with dependent on time. The

top number 1.5 x 107 is the same thing as saying 15,000,000, the target population, it is written

in scientific notation. Also don't mind the red arrows. They just denote other solutions to this

equation such as if we choose a different starting population.

As we can see our graph depicts exactly that, on about the 173rd day we reach the target population.

3

As I stated in my letter, the only way to increase the speed at which the worm population reaches

the target is to increase the shipment. After examining a couple of cases the best suggestion we

can give for you is to triple the worm shipment, 300,000 pounds per week. Given this change, we

now we reach the target by the 68th day or January 8th, 2012. This leaves a few days before the

stockholder's meeting, which is ideal. This gives you a couple of days if the preparations for the

Levittson facility are not quite complete November 1st. Below is the new graph on the same time

scale that accounts for the change in shipment rate. We can see that changing this one factor makes

a big difference.

If you have any questions on the approach to this problem or the math behind it feel free to contact

Dr. Buhl. Also if you have any future issues please don't hesitate to contact us. We enjoy helping

companies, such as yourself, plan for the future. Again we would like to thank you for using

Mathematical Modeling Consultants. Accurate information is power and with this information we

hope that you can find yourself with the power to survive in these hard times.

Notes on attached Maple Worksheet

worms1 in the math refers to the equations using the tripled shipment and worms2 refers to using

the normal shipment. Here we have used dsolve for solving the differential equation. Notice w(0)

is equal to the starting population. Next we used fsolve for solving the by setting the solution of

the population equation to 15,000,000, the population we wanted. Lastly don't be alarmed by the

differences in a negative sign in our dsolve equation and our fsolve. They turn out to be the same

equation.

4

5

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