Mathematical Modeling-optimisation

Mathematical

Modeling

for

Optimisation

In many business activites,you have to optimise certain variables . Suppose you own a pick up truck which you have been using for a few years. You may like to replace the old one for a new truck.You may ask yourself whether it is economical to replace it or continue to use the old one. This is called "Replacement Theory". This is an optimisation problem. In any optimisation problem, may be opposing to each other

we have more than one factor. The factors with respect to cost or time or some factor like usage.

Truck

Problem

Replacement

Here the total cost has two components. Capital cost of owning the truck which keep decreasing with th emileage run or time .This is decreasing function--right. The other factor is the increasing operating and maintenance costs {O&M Cost]. As time passes by,the capital cost is almost zero.But the O&M cost will go up. A time comes when you want to sell the truck and get a new one. Let us model first in simple formulas: Say, capital cost C decreases linearly with mileage: If you bought the truck for $30000. Let the capital cost decrease 3000 dollars for every 10000 kms run. We write : C= 30000 3000 * X Equation (1) where x is the mileage run in 10000 kms. This is a straight line decrease----called linear depreciation"----Accountant like to use such depreciation equations. The book value for the truck becomes zero after 10 years according to accountants.! We will use a better model later. Again, for O&M costs, let us use a linear model: O&M Cost= 1000 + 5000 * X Equation (2) You see that O&Mcost keeps increasing with miles run as it should. Now the

Total

Cost TC=

C + O&M Cos cost

An easy way to understand this is to plot both th the X. Let us forma table before we plot the X

C 0 1 2 3 4 5 6

Now we plot

30000 27000 24000 21000 18000 15000 12000

costs as lines:

a function of

O&M Cost 1000 6000 11000 16000 21000 26000 31000

the two lines

in Excel

spreadsheet.

Replacement models 32500 30000 27500 25000 22500

Cost

20000 17500

C O&M Cost

15000 12500 10000 7500 5000 2500 0 0

1

2

3

4

5

6

miles run x 10000 kms

Now we note that Beyond that The total cost is Solving algebraically

the two costs are equal when miles run is O&M costs will increase and the total cost will go up. a minumum at this point where the wo lines intersect.! 30000 - 3000*X= 0*X = 1000 + 5000*X Solving

X =29000/8000 X= 3.8 X = 38000 kms

IMPROVED MODEL

The model given above is too simplistic , because we used straight line functions and for O&M costs… real life is more complicated. The capital cost decreases rapidly as any vehicle owner knows. The capital cost decreases as follows: C=k/X For our model ,let k=24000 This function called "rational expression' in Algebra gebra. It is a better model for

capital cost. Now the O&M cost

is not linear with

miles run;it t increases much faster;let

O&M Cost= 5000*(1.1)^X I have used a sort of 'compound interest fromula' The total cost Let us

TC =

for this cost.

24000/X + 5000*(1.1)^X

form a table for

these numbers:

X

C 1 2 3 4 5 6 7 8

O&M cost 24000 5500 12000 6050 8000 6655 6000 7320.5 4800 8052.55 4000 8857.81 3428.57 9743.59 3000 10717.94

25000 22500 20000

Cost

17500 15000 12500

C O&M cost

10000 7500 5000 2500 1

2

3

4

5

6

7

8

miles run (10000 km)

The graph shows that the two curves cut around 3.5 or 35000 km run as the repalcement time. This example Thermal

is illustrative of many such optimisation modes used in business.

Insulation Problem Here is a typical problem solved by engineers. You want to insulate a cylindrical pipe with outer diameter D. With each layer of insualtion tape wound over the pipe, the path of heat flow inc is reduced by increased diameter of D + X where X is the thickness of tape layer.Since heat flow is proport temp difference across this layer divided by thickness X we write: heat energysaving S= k/ X where k is a constant to convert energy saved in terms of cost of saving. As thickness of insulation layer increases, the cost of added insulation is proportioanl portional to thickness tothickness squared., since c pi * [(D+X)^2 - D*D} .Expanding we will get: pi* {D*X + X*X] Cost of Insulation C= n * [D*X + X*X] where n is a constant to convert the quantity in Total Cost= S+C

You can optimize

by finding

the value of X at which TC is a minimum.

Note that any optimization problem involves one increasing function and one decreasing function for a common va Do it yourself Exercise 1 John wantsto buy a diesel

engine for his yacht. The cost of engine increases with reliability or M C= a+b * MTBF MTBF is a common measure of relaibility and is expressed in hours. 16000 The maintenance cost would decrease with increased reliability or MTBF. 15000 14000 Let maintenanace nance cost M=8000-100(MTBF/1000)^2 13000 12000 For a = 5000 b=1 MTBF is in the range 1000 to 10000 hours. 11000 10000 Solve for the optimum value for MTBF 9000 MTBF C M Total 8000 7000 1000 6000 7900 13900 6000 5000 2000 7000 7600 14600 4000 3000 4000 9000 6400 15400 2000 6000 11000 4400 15400 1000 1000 2000 3000 4000 5000 6000 8000 12000 1600 13600 Interpret the results from the table and the graphs.

NOTE

2 Mike, a chemical enginer wants to build a cylindrical reactor vessel to hold hold1000 100 cubic meters. a height and diameter for the vessel such that the volume will be 100 cu.meter but the surface area will b he wants to optimise the surface area for two reasons;1 to reduce the cost of sheet required to build the 2 to reduce the heat loss from the reactor vessel. Volume= pi*r*r* h Surface area SA = 2*pi*r*h + 2*pi*r*r= 2*pi*(r*h+r*r) Try to solve for the ratio h/r for maximum of volume/SA ratio and get V=100. See my other articles on mathematical modeling for basic concepts relavant to this article in www.pdfcoke.com This article is based on the experience of the author in mathematical modeling for aircraft systems and r The numbers given are indicative of the real world numbers for certain systems.

C O&M Cost

6

about 38000 kms.

for capital decreases rapidly as any

is a better model for

C O&M cost

8

a cylindrical pipe with the path of heat flow increase increases and and heat loss .Since heat flow is proportional to

squared.

t to convert the quantity into cost.

function for a common variable.

creases with reliability or MTBF {mean time between Failure]

C M Total

0

2000

3000

4000

5000

6000

7000

8000

cu.meters. He wants to choose but the surface area will be minimum. He wants to optimise the surface area for two reasons:1 to reduce sheet area required to build the ve heet required to build the vessel;

for aircraft systems and realted work.

area required to build the vessel 2.to reduce heat loss from the re actor vessel.