data structures course assignment

8/8/2019 Data Structures Course Assignment

1/8

Faculty Of Engineering

Department of Electronics & Electrical Communications

Data Structure Course Contest One

Time: Two Days

M obil e T echnol ogy S chedul ing

A l gor ithms

In a mobile t elecommunicat ions company; it 's requ ired t o inc rease t he overall

t hroughput [ D at a rat e] of t he wireless coverage.

For each mobile cell t here are n slot s t hat t he n users could be assigned.

[ A ssuming tha t the nu mber o f user s = the nu m ber o f the fr equ ency s lot s ]

i .e. User number one could be operated on slot one or two or three or any

combinations of them

If w e assume that the user one could be operated in both slots one , two

and three then the user one data rate = the summation of the data rate in

each slot.

There's a metri c coefficient to indicate the datarate corresponds to a specific

user assuming certain operating frequency slot.

I f user number one operat es on t he f requency slot number one t hen t he

dat arat e =10 M bps and i f t he same user operat es on t he f requency slot


2/8

number t wo t hen the dat a rat e =15 M bps and if t he same user operat es on t he

f requency slot number t hree t he dat arat e cou ld be =12M bps

But why f or t he same user if he operat es on dif f erent slot t he dat arat e is

di f ferent?

The answer is relat ed t o t he communicat ions t heory; t here are many f act ors

relat es t he user locat ion (distance bet ween t he user and t he cell base st at ion)

, t he t ransmit t ing power of t he mobile , t he f requency t hat t he mobile

operat ing in [ The f requency slot select ed when init iat ing t he communicat ion

process] ,...et c

These f act ors are analyzed in a st ochast ic f ashion [ Probabilist ic model] t o

calculat e t he data rat e of each user if s/he operat es on a cert ain f requency

slot .

We cou ld develop a mat rix called t he met ric mat rix it s rows represent t he

user number and it s columns represent t he f requency slot number.

You are requ ired t o develop a c++ program t hat makes a channel assignment .

so t hat we could decide which channel t he user number j should use t o get

t he maximum overall syst em t hroughput .

This problem is known in t he algor it hms wor ld as a sched u ling pr oblem .

Given the required algorithms to be implemented by your program; and the

algorithm exp lanation


3/8

A lso t he scheduling shou ld be implement ed under some assumpt ions

1-Each user must operat es only [ at most and at least ] on a single

channel(f requency slot ).

[ To make t he f airness as t he number of channels = t he number of users; if a

user operat es on more t han a single channel t hen t here's anot her u ser t hat

wouldn't f ind any channel t o operat e on and it 's not f air]

2-Each channel couldn't be assigned t o more t han a single user.

(A nd i t s realist ic t he user sends t he dat a as an analog signal

[ D igit al inside t he mobile or t he comput er] and af t er t he t ransmit t er + t he

ant enna; it 's convert ed t o analog signal.

The analog signal is a sinusoidal signal of a cert ain f requency band.

(A f t er a proc ess known as t he modulat ion process)

The f inal deduc t ion:

Each user shou ld operat es on only one channel

Each channel must only assigned t o a single user.

M ore pract ical:

The user could operat e on more t han a channel.

[ But st ill each channel couldn't be assigned t o more t han a single user at t he

same t ime] .


4/8

A ssuming t hat t he met ric mat rix as t he f ollowing

We will give each met r ic mat rix element a symbol as which means t he

met r ic of t he user number i if s/he uses t he channel (f requency slot ) number c

So is t he met r ic (dat a rat e) of t he f irst user if s/he uses t he f irst channel

(f requency slot ) which is equal 380 M bps in our example.

A nd is t he met ric (dat a rat e) of t he second user if s/he uses t he t hird

channel (f requency slot ) which is equal 1390 M bps in ou r example.

The algor it hm procedure:

Init ially we pick t he largest dat a rat e f rom t he mat rix (1530) t hen delet e all t he

dat a on t he same row (f irst row) of t he picked dat a rat e

(As the sam e row means th e same user a nd as the u ser i s ass igned to the

chan ne l then th e r es t chann e ls cou ldn 't be ass igned to th e user )

and also delet e all t he dat a on t he same column(t hird column) of t he picked dat a rat e


5/8

( as th e same colum n m eans the sam e chann e l and a s the chan ne l is ass igned

to the u ser t hen i t couldn ' t be ass igned to an y other user s)

So the resultant matrix is as the following:

S t ep 1

From t he result ant dat a pick t he largest dat a rat e (810) and repeat t he process

unt il t he mat r ix is empt y and all t he users are assigned t o t he required

channel.

S tep 2

S tep 3 Each t ime t he maximum select ed element is t he t arget dat a rat e.

Which is meaning t hat we f inally choose the f lowing dat a rat es:

1530,810 and 300


6/8

Or which is meaning t hat t he user one will use t he

channel number t hree and t he user t wo will use t he channel number t wo and

t he user t wo will choose channel number one

The t ot al syst em thr oughput =1530+810+300=2640 M bps


7/8

0- Init ialize t he t ot al t hroughput t o zero and a count er t o zero.1- St ore all of t he mat r ix dat a int o an in-order linked list2- Inc rement t he count er3- Get t he f irst element of t he in-order linked list and st ore it in a variable

(SD R[ count er] an acr onym f or Select ed_dat a_rat e)

=>specif y it s row number and column number

=>Get all t he dat a on t he same row (same user) and t he same column

(same channel ) and delet e t hem f rom t he in-order linked list .

4- A dd t he SD R[ i] t o t he t ot al t hroughput and st ore t he result in t he t ot althroughput.

5- I f t he result ed linked list is not empt y go t o st ep 2 and r epeat6- A f t er t he linked list becomes empt y t hen t he number in t he t ot al

t hroughput is t he result ed Overall syst em t hroughput .

7- The numbers on t he arr ay SD R represent s t he select ed dat a rat es t o beassigned .

8- Out put whic h channel is assigned t o eachuser .

[ Repeat t he previous 8 st eps but using a linear in-order list ] compare bet ween

t he memory ut ilizat ion f or bot h implement at ions and f or a given number of

users.

N ot es:

1-N umber of user [ =number of channels] cou ld be bet ween 2 and t ill t o 1000


8/8

2-The cu rr ent algorit hm is not t he opt imized one as f or t he given met ric

mat rix

I f we choose t he f ollowing met r ics: 1530, 730 and 650

We sat isf ied t he condit ions t hat each user t akes only one channel and in t he

same t ime maximizing t he t ot al t hr oughpu t 1530+730+650=2910M bps

We will learn in t he next sessions more opt imized methods but need a new

dat a st ruct ures t o be implement ed like t he binary t ree, Quad-t ree..et c.

For a bet t er memory ut ilizat ion st ore t he mat rix int o a dynamically t wo

dimensional ar ray [ D ouble Point er] .

In t he beginning of t he program ask t he program user f or t he number of t he

users and t he number of t he f requency slot s. [They must be t he same]. and

t hen allocat e t he cor responding memory blocks

I f you use a st at ic t wo d imensional array 1000*10000 t hen in t he worst case

You will lose 1000*1000 -2*2 memory loc at ions [ You now are not a programmer]

The same concept s are applied when using t he in-order linear linked list rat her

t han t he in-order linear st at ic list .

G ood L uck"A bdelrhman M ohamed A bot aleb" 30 August 2010

data structures course assignment

Documents