classifying optimization problems by the independent variables: –integer optimization --- integer...

Classifying optimization problems

By the independent variables:– Integer optimization --- integer variables– Continuous optimization – real variables

By the problem: – Linear optimization, Nonlinear optimization, Dynamic optimization, ….

By constraints– Unconstrained optimization – Constrained optimization

1 2Optimize ( , , , )mf x x x

1 2

1 2

Optimize ( , , , )

Subject to

( , , , ) for all in an index set

m

l m l

f x x x

g x x x b l I

Minimize the cost of delivery and storage

Scenario: – You have been hired as a consultant by a chain of gasoline stations to

determine how often and how much gasoline should be delivered to the various stations. After some questioning, you determine that each time gasoline is delivered the stations incur a charge of d dollars, which is in addition to the cost of the gasoline and is independent of the amount delivered.

– Costs are also incurred when the gasoline is stored. One such cost is capital tied up in inventory – money that is invested in the stored gasoline and that cannot be used elsewhere. The cost is normally computed by multiplying the cost of the gasoline to the company by the current interest rate for the period the gasoline was stored. Other costs include amortization of the tanks and equipment necessary to store the gasoline, insurance, taxes, and security measures.


– The gasoline stations are located near interstate highways, where demand is fairly constant throughout the week. Records indicating that gallons sold daily are available for each station.

Problem identification– The firm wishes to maximize its profits and that demand and price are

constant in the short run– Total revenue is constant, total profit can be maximized by minimizing the

total costs.– There are many components of total costs

• Such as overhead and employee costs which are not affected by the amount of gasoline and the timing of the deliveries

• Storage and delivery costs.


– Focus problem -- Minimize the average daily cost of delivery and storing sufficient gasoline at each station to meet consumer demand!

– Intuitively we expect such a minimum to exist. – Two special cases:

• If the delivery charge is very high and the storage cost very low, we should expect very large orders of gasoline delivered infrequently.

• If the delivery cost is very low and the storage costs very high, we should expects small orders of gasoline delivered very frequently.

Assumptions– We consider factors important to be deciding how large an inventory to

maintain– Obvious factors:


• Delivery costs• Storage costs• Demand rate for the product• Perishability of the product being stored, especially when the gasoline level

gets lower and lower in the tank• The market stability of selling price of the product • The cost of raw materials• The stability of the demand for the product by the consumer• ……..

– The inventory decision is not an easy one!!!– For short-term plan, we restrict our initial model to the following variables:

average daily cost = (storage costs, delivery costs, demand rate)f


The Sub-models– Storage costs -- how the storage cost per unit varies with

the number of units being stored. • Renting space and receiving a discount when storage exceeds

certain levels• Rent the least expensive storages first (adding more spaces as

needed)• Rent an entire warehouse or floor first, the per unit prices is likely to

decrease as the quantity stored increases until another warehouse or floor needs to be rented.

• In our model, we take per unit storage as a constant!!


– Delivery costs – in many cases the delivery charge depends on the amount delivered

• In our model, we consider a constant delivery charge independent of the amount delivered

– Demand rate


• Daily demand• Frequency of each demand level• We assume the daily demand as being constant and we take a

continuous submodel for demand.

Model formulation– Variables

• s --- storage costs per gallon per day• d -- delivery cost in dollars per delivery• r --- demand rate in gollons per day• q – quantity of gasoline in gallons• t --- time in days


– Problem setup: • An amount of gasoline, say q, is delivered at time t=0 and the gasoline is used

up after t days. The same cycle is repeated. • The problem is to determine an order quantity q* and a time between orders t*

that minimizes the delivery and storage costs.• We seek an expression for the average cost, so consider the delivery and

storage cost for a cycle of length t days. • The delivery costs are the constant amount d because only one delivery is

made over the single time period.• To compute the storage costs, take the average daily inventor q/2, multiply by

the number of days in storage t, and multiply that by the storage cost per item per day s.


– Mathematical formulation• Cost per cycle = d + s t q/2• Objective function --- cost per day

– Model solution

0

( , ) ( )2 2

min ( )

q rt

t

d sq d srtc t q c t

t tc t

1/2

2

2( ) 0 * & * *

2

d sr dc t t q r t

t sr


– Model interpretation• Given a (constant) demand rate r, a proportionality between the

optimal period t* and . Intuitively we would expect t* to increase as the delivery cost d increases and to decrease as the storage costs s increase. The model at least makes common sense!!

• More mathematically way to compute the storage cost for one cycle as an integral:

• We neglect the cost of the gasoline in the analysis. • Question: Does the cost of gasoline actually affect the optimal order

quantity and period???

1/2( / )d s

2

0

( )2 2

t rt sqts q rx dx s qt


– Model implementation• Run out of stock problem--- the model assumes the entire inventory is

used up in each cyclic period, yet all demands are supposed to be satisfied immediately. Note that this assumption is based on an average daily demand of r gallons per day. Thus over the long run, for roughly half of the time cycles the stations will run out of stock before the end of the period and the next delivery time, and for the other half of the time cycles the stations will still have some gasoline left in the storage tanks when the next delivery arrives!!! Such a situation won’t do good for the credibility as a gasoline station!!

• Solution – recommendation a buffer stock to help prevent the stock-outs.

Computational methods

For one variable minimization

Numerical methods– Find critical points– Lagrange multiplier

min ( , ) with ( , ) ( ) [ ( ) ]

( , ) ( ) ( ) 0

( , ) ( ) 0x

F x F x f x g x b

F x f x g x

F x g x b

max ( ) subject to

( )

f x

g x b

Computational methods

Multi-variables minimization– Newton’s method– Interior point method

Linear programming– Simplex method

Nonlinear programmingConic programmingDynamic or stochastic programming

max ( ) subject to

0

f x

Ax b

An example: A space shuttle water container

The problem – Consider a space shuttle and an astronaut’s water container is to be stored within the shuttle’s wall. – The container is made in the form of a sphere surmounted by a cone (see figure) – The radius of the sphere is restricted to exactly 6 ft and a surface area of 450 ft2 is all that is allowed in the design – Find the dimensions x1 & x2 such that the volume of the container is a maximum!!


Problem identification – maximize the volume of the water container for the astronauts while meeting the design restrictions

Assumptions – the design of the water container followed the shape in the figure including dimensions, volume, surface area and radius of the sphere

Model formulation– Variables

• Volume of the conical top • Volume of the cut sphere

2 21

(2 )

3c

x r xV x

3 22 2

4 1(3 )

3 3sV r x r x


• Volume of the water container• Surface area of the cone -----• Surface of the sphere --------------------------------• Total surface area

– The model

w c sV V V 2 21 2 2, (2 )cS b x b b x r x

224 2sS r rx T c sS S S

3 22 21 2 1 2 2

2 2 21 2 2 2 1 2 2 2

(2 ) 4 1 Max ( , ) 3

3 3 3Subject to

( , ) : (2 )( 2 ) 4 2 450 0

x r xf x x x r x r x

g x x x r x x rx x r x


Model solution– Method of Lagrange Multipliers

– Critical points

– Plugging the parameters– The solution by numerical method

1 2 1 2 1 2( , , ) ( , ) ( , )L x x f x x g x x

1 2 1 2 1 2

1 2

( , , ) ( , , ) ( , , )0, 0, 0

L x x L x x L x x

x x

6 & 3.14r

31 2 1 21.95ft, 1.56ft, =1.83, ( , ) 898.72fx x f x x t

classifying optimization problems by the independent variables: –integer optimization --- integer...

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