Topology, Capacity and Flow Assignment Computer Communication - - PowerPoint PPT Presentation

Topology, Capacity and Flow Assignment

Computer Communication Networks: Analysis and Design (Klei. Vol. 2, Chap. 5)

Original Material Prepared by: Professor James S. Meditch Lecturer:Prof. Massimo Tornatore Typesetter: Dr. Anpeng Huang

1

The problem

• The analysis of stochastic flows in packet

networks is extremely complex

• We study some the more important

problems arising in the design process

• Design variables:

– Routing procedure (flow assignment) – Channel capacity – Topological configuration

– Queuing discipline, packet numbering and sequencing, error control, etc..

2

Packet vs. Circuit Switching

fA fX fY fZ Lh/f0 Lp/f0 L/f0 Spazio Tempo Ritardo end-to-end

Propagazione Trasmissione Elaborazione Attesa A B X Y Z

Packet Switching

Propagation Transmission Processing Queuing Time Space End-to-end delay

Packet vs. Circuit Switching

Circuit Switching

X Y Z A B

Spazio Instaurazione Dati Rilascio Ritardo end-to-end Propagazione Trasmissione Elaborazione Tempo

End-to-end delay Propagation Transmission Processing Data Setup phase Release phase Time Space

• A. Evolution of network structures

A.a. Private/most expensive (star network)

5

A.b. Least cost/slowest (minimal spanning tree) Multidrop lines – polling or contention

6

A.c. Compromise/multiplexing or concentrating (statistical multiplexer)

7

A.d. Large network – combinations of (c)

8

• B. Network modeling

M channels, N nodes, message (packet) switching Assumptions: (i) Links and nodes perfectly reliable (combinatorial in nature) (ii) Nodal processing time negligible (for reading, error checking) (iii) Nodal storage infinite (iv) Fixed (or random) routing (v) Offered traffic Poisson, packet length negative exponential

9

B.1. Arriving traffic

• Poisson with avg. rate [msg/sec]
• =

total average flow entering in the system

jk

γ

∑∑

= =

=

N j N k jk 1 1

γ γ

k j ≠

10

B.2. Messages

• Message lengths exponentially distributed with

mean bits µ 1

B.3. Channel model

M/M/1 queue Channel I ,

• = channel capacity in bits/sec (bps)
• = avg. channel flow in msg/sec
• Note(1): Total average flow inside

the network (≠γ !)

• Note(2): propagation time Pi

i

C

i

λ

i

λ

i

C

11

∑

=

=

M i i 1

λ λ

B.4. Channel cost (in $)

• d is a generic cost function

12

∑

=

=

M i i i C

d D

1

) (

B.5. Average message delay

• T = E [message delay]
• E [message delay for message from j to k]

fraction of traffic over j–k node pair

≅

jk

Z

∑∑ ∑∑

= = = =

= =

N j N k jk jk N j N k jk jk

Z Z T

1 1 1 1

1 γ γ γ γ

13

• C. Network design problems

C.1. Given: node locations, , traffic matrix {γjk} C.2. Objective function T C.3. Parameters: Ci , λi, topology τ C.4 Constraint 4 optimization problems can be defined that differ

• nly in the set of permissible design variables ->

µ 1

14

D C d

M i i i

=

∑

=1

) (

C.3. The 4 problems are:

Problem Given Minimize w.r.t. s.t. CA λi, τ T Ci D FA Ci, τ T λi CFA

τ

T Ci, λI D TCFA

• T

Ci, λI, τ D

i i

C µ λ < ≤

Increasing complexity

C.4. Dual problems

Swap D with T No dual for FA

15

• D. Delay analysis

D.1. Model for T Little’s result applied: : one-way or two-way message flow

First take-away: global T can be expressed as sum of local Ti

∑

=

= =

M i i iT

T N

1

λ γ

∑

=

=

M i i iT

T

1

1 λ γ

i

λ

16

D.2. Independence assumption (a) and channel model (b)

D.2.a. Message lengths at nodes are independent at each node, and have exponentially distributed length b with p.d.f.

• avg. message length = bits
• In fact, message lengths are not independent at each node since a message

enters the network with a given length and retains that length from source to

• destination. This independence assumption is based on (i) large numbers of

messages passing through each node and (ii) the moderate connectivity of the network, which support the assumption that packet lengths can be approximated as independent, exponentially-distributed message lengths.

b

e b p

µ

µ

−

= ) (

≥ b

µ 1

17

D.2.b. M/M/1 channel model and T

Poisson arrivals: messages/sec. Exponential service: sec/message

T

i

C

i

λ

i i

C x µ 1 =

i i i i i i i i i i i

C C C x x x T λ µ µ λ µ λ ρ − = − = − = − = 1 1 1 1 1

18

[ ]

sec : message C nb

i =

µ

It follows that Other results (easily obtainable):

∑

=

− =

M i i i i

C T

1

1 λ µ λ γ

i i i i i i

C T N λ µ λ λ − = =

i i qi

N ρ ρ − = 1

2

i i i i i

C C W λ µ µ λ − =

19

i i i

C µ λ ρ =

Cont’d:

D.2.c. Other effects

Note: so far we neglected: 1. control traffic, 2.propagation and nodal processing delay D.2.c.1. Average data message delay when control traffic is present = avg. flow of data messages = avg. length of data message = avg. flow of all traffic = avg. length of all messages = avg. data message delay = avg. service time for data messages = avg. waiting time for all messages

i

λ

µ 1

' i

λ

'

1 µ

i

T

i

x

i

W

20

nb: all traffic = data + control

i i i

x W T + =

' ' ' ' i i i i i

C C W λ µ µ λ − =

i i

C x µ 1 =

i i i i i i

C C C T µ λ µ µ λ 1

' ' ' '

+ − =

            + − = ∑

= i i i i i M i i

C C C T µ λ µ µ λ γ λ 1

' ' ' ' 1

21

The avg. system time (with control traffic) can be now easily calculated as

D.2.c.2. Propagation delay and nodal processing delay

Pi = propagation delay [sec/msg]

Depends upon the medium and the length of the link, e.g., ground station to geosynchronous

satellite is on the order of 120 to 135 msec.

K = avg. nodal processing time [sec/msg]

            + + + − + =

∑

=

K P C C C K T

i i i i i i M i i

µ λ µ µ λ γ λ 1

' ' ' ' 1

22

A simple example

23

Question: How should be split in

• rder to minimize T?

Routing: all of via 1 of via and 5 of via all of via msg/sec msg/sec msec

12

γ

1

C

23

γ

13

γ

3

C

2

C

1

C

13

γ

2

C

4

1 =

λ 2

2 =

λ 5

3 =

λ

10

23 13 12

= + + = γ γ γ γ

      − + − + − = − = ∑

=

5 9 5 2 6 2 4 12 4 10 1 1

3 1 i i i i

C T λ µ λ γ

225 = T

13

γ

24

D.2.d. Delay T vs Offered load γ

Fixed routing

D.2.d.1. [Low load] Decrease γ such that ∀ i So, no queuing in the network, all delay is due to service time (i.e., transmission delay)

∑

=

− =

M i i i i

C T

1

1 λ µ λ γ

i i

C µ λ <<

∑

=

→

M i i i

C T

1

1 µ λ γ

∑

=

≅

M i i i

C T

1

1 µ λ γ

25

D.2.d.1. [High load] Increase γ until some channel saturates, i.e., .

For that channel ,

and, therefore so does T≅Ti .

Let γ* be the

corresponding value of γ.

i i

C µ λ →

∞ →

i

T

26

D.2.e. Average path length

= avg. path length

• no. of links through which a message passes in

proceeding from its source to its destination averaged

• ver all source-destination node pairs

n

27

But Hence, where

14 13 12

γ γ γ γ + + =

γ γ γ γ 3 . 2 . 1 .

14 13 12

+ + = n

γ γ γ γ γ γ γ

14 14 13 14 13 12

) ( ) ( + + + + + =

14 13 12 1

γ γ γ λ + + =

14 13 2

γ γ λ + =

14 3

γ λ =

∑

=

= + + =

3 1 3 2 1

1

i i

n λ γ γ λ λ λ

γ λ = n

∑

=

≅

3 1 i i

λ λ

28

In general, “path” from j to k

• no. of links in

=

jk

π

=

jk

n

∑∑

= =

=

N j N k jk jkn

n

1 1

γ γ

k j ≠

jk

π

29

But, Since msg/sec will traverse njk links in passing through the network Hence, where Using this result, we get for no-load delay with fixed routing.

∑∑ ∑

= = =

= ≅

N j N k jk jk M i i

n

1 1 1

γ λ λ

jk

γ

γ λ = n

∑

=

=

M i i 1

λ λ

∑

=

=

M i i i

C n T

1

µ λ λ

30

See formula at pp. 22

Example: Same network as in delay analysis example above

; , , So, Also, using either expression for , we have msec This is the avg. msg. delay when the network is very lightly loaded.

10 = γ

4

1 =

λ

2

2 =

λ

5

3 =

λ

1 . 1 10 11 = = n

T

2 , 122

0 ≈

T

31

• E. Capacity Assignment (CA) Problem

Given: node locations, , , , ; min w.r.t. s.t. N.B.: the actual problem requires selection of Ci from a finite set, e.g., Ci ∈{100Mbps, 1Gbps, etc.}, but, to find closed form solution, we have to relax this requirement

jk

γ

µ 1 τ

{ }

i

λ

T

{ }

i

C

( )

D C d

M i i i

=

∑

=1

32

E.1. Linear costs

and it can solve using the Lagrangian:

M , 1,2,...... ) ( = = i C d C d

i i i i

D C d C T

M i i i M i i i i

= − =

∑ ∑

= = 1 1

s.t. 1 min λ µ λ γ

      − + =

∑

= M i i i

D C d T J

1

β       − + − =

∑ ∑

= = M i i i M i i i i

D C d C

1 1

1 β λ µ λ γ M 1,2,... = = ∂ ∂ j C J

j

) (

2

= + − − = ∂ ∂

j j j j j

d C C J β λ µ γ µ λ

33

Let us assume: Then the optimization problem can be written as

( )

j i j j

d C βγ µλ λ µ = −

2

j j j j

d C λ βγµ µ λ 1 + =

∑ ∑ ∑

= = =

+ = =

M i i i M i i i M i i i

d d C d D

1 1 1

1 λ βγµ µ λ

∑ ∑ ∑

= = =

≅ − =

M i i i e M i i i M i i i

d D d d D

1 1 1

1 λ λ µ λ βγµ

34

Solving the Lagrange’s equations: We can then find β by «forming the constraint»:

«Excess dollars»

j = 1, 2,……, M (*) {min. required capacity to satisfy flow reqmnt.} + {sq. root allocation of normalized excess capacity} If we substitute Cj in the o.f. Note: from (*)

∑

=

+ =

M i i i j j j e j j

d d d D C

1

λ λ µ λ

=

j

C

∑

=

− =

M i i i i

C T

1

1 λ µ λ γ

∑

=

= −

M j j j i i i e i i

d d d D C

1

λ λ µ λ µ

35

Cont’d:

∑ ∑

= =

=

M i M j j j i i i e i

d d d D T

1 1 min

1 λ λ µ λ γ

              =

∑ ∑

= = M j j j M i i i e

d d D

1 1

1 λ λ γ µ

36

Cont’d:

2 1 2 1

1         =       =

∑ ∑

= = M i i i e M i i i e

d D d D λ λ γ µ λ λ γ µ

2 1 min

        =

∑

= M i i i e

d D n T λ λ µ

Note: De must be greater than 0!

Special case: for all i Hence, is the constraint Using the definition

• f De at pp. 31

1 =

i

d

C C D

M i i =

= ∑

=1

37

Cont’d:

λ µ λ µ 1 1

1

− = − =

∑

=

C C D

M i i e

∑

=

+ =

M i i j e j j

D C

1

λ λ µ λ

2 1 min

1       =

∑

= M i i e

D T λ γ µ

Example:

3

10 11 1 × = − = λ µ C De

11 1 1 2 3 2 2

6 1

= + + + + + =

∑

= i i

λ

bps C

j j j

11 10 11 1000

3

λ λ × + =

kbps) (in

j j j

C λ λ + = →

kbps C C C 6

4 2 1

= = =

kbps C 12

3 =

kbps C C 2

6 5

= =

( )

( )

2 3 2 6 1 min

11 16 10 11 1 1000 1 1 × =               =

∑

= i i e

D T λ γ µ

sec 5 . 687 sec 16 11 m = →

4

4 2 1

= = = λ λ λ

9

3 =

λ 1

6 5

= = λ λ

16 = → γ

kbps C C

i i

34 = =∑

23 = → λ

38

FLOWS

1 step: calculate CAPACITIES

BUDGET (di=1) D=34 (=C)

2 step: calculate AVG DELAY

E.2. Other cost functions

Approximation of actual cost structures

a. Logarithmic cost (quantity discount) CA is proportional

• b. Power law (ARPANET study - 1970)

β = Lagrange multiplier (iterative solution required)

∑

=

=

M i i i

C d D

1

logα

i i i i

a b C λ λ µ λ = + =

∑

=

=

M i i iC

d D

1 α

1 < <α

44 . ≅ α

( )

2 1

= − −

−α

µ λ

i i i i

C g C

2 1

        =

i i i

d g µγαβ λ

39

E.3. Other forms of capacity assignment

• Heuristic approaches based on the notion of the network

being “balanced” in some physically meaningful way E.3.a. All links in the network have the same avg. utilization Fix “ ” by specifying the desired avg. utilization, e.g., Proportional CA

a Ci

i i

= = µ λ ρ

1 < < a

6 . =

i

ρ

µ λi

i

a C 1 =

) 1 ( a Ma T − = γ

40

∑

=

− =

M i i i i

C T

1

1 λ µ λ γ

a

E.3.b. All links in the network have the same link delay

Simply, let us fix “ τ ” by specifying the desired avg. delay for each link, (e.g., to Ti msec), then:

τ λ µ = − =

i i i

C T 1

µτ µ λ 1 + =

i i

C

τ n T =

41

Example:

Find capacities Ci so that all Ti are equal and T=110 msec

3

12 =

γ

6

13 =

γ

1

23 =

γ

bits 200 1 = µ

4

1 =

λ

2

2 =

λ

5

3 =

λ

1 . 1 = = γ λ n

sec 1 . sec 100 = = = m n T τ

2000 1 = µτ kbps bps C 8 . 2 2800

1

= =

kbps bps C 4 . 2 2400

2

= =

kbps bps C . 3 3000

3

= =

42

[msg/sec] [msg/sec]

• F. Flow Assignment (FA) problem

F.1. Given τ and {Ci}, minimize T with respect to {λi} subject to capacity constraints, 0≤ λi ≤Ci, external traffic requirements {γjk}, and flow conservation

∑

=

− =

M i i i i

C T

1

1 λ λ γ

43

F.2. FA properties

1. Multicommodity-flow non-linear optimization problem 2. Closed-form solution not possible 3. Use iterative computational algorithm 4. Solution shall give both the amount of flows {λi} and the corresponding

• ptimal (minimum delay) routing for all commodities

5. T=T(λ1,λ2 …. λM) is a convex function of the λi’s 6. Capacity constraints, 0≤ λi ≤Ci, are convex 7. Fundamental theorem

1. Any minimum of a convex function over a convex set is the global minimum

8. Best known and most effective algorithm for solving this problem is the Flow Deviation (FD) algorithm 9. FD algorithm based on two basic notions:

• a. “Shortest path” flows
• b. Flow deviation which reduces minimization of T=T(λ1,λ2 …. λM)
• ver all the λi simultaneously to minimization of T(a) where a is a scalar,

0≤ a≤1

44

F.3. Shortest path flow part of FD

F.3.1. Link “length” , F.3.2. Problem statement subject to capacity, total flow, and commodity flow constraints (a) Linear programming problem (b) Many efficient algorithms available

i

l

i

C

i

λ

( )

2 >

− = ∂ ∂ =

i i i i i

C C T l λ γ λ

∑

= M i i i

l

i

1

min φ

φ

i

l

45

F.4. FD Algorithm

F.4.1. FD algorithm n=iteration index f(n)=(λ1

n,λ2 n …. λM n) = flow vector on nth iteration

f(0)=initial feasible flow (see Kleinrock vol. II, pp. 344-345) a. Set n=0 b. Solve shortest-path flow problem with weigths li and let c. Set and find such that is minimized

c. Any search method is ok, e.g, Fibonacci search

• d. If where is the accuracy

threshold, stop. Otherwise, set and return to Step 2.

( ) ( ) ( ) ( )

[ ]

n M n n n

φ φ φ φ ,..., ,

2 1

=

( )

( )

( ) ( )

n n n

a f a f φ + − =

+

1

1

1 ≤ ≤ a

( )

( )

1 + n

f T

( )

( )

( )

( )

ε ≤ −

+1 n n

f T f T

> ε

1 + = n n

46

F.4.2. FD algorithm flow chart

Test ( accuracy threshold) (A) (B)

= ε

( )

( ) ( )

ε > −

+1 n n

f T f T

( )

( ) ( )

ε ≤ −

+1 n n

f T f T

47

F.5. FA Problem

F.5.1. Plausibility of FD algorithm Assume given and let Then, and Therefore, to minimize to within first

• rder in a, it is necessary to choose the

such that is minimized.

( )

n

f

( )

( )

n n n

a f a f φ + − =

+

1

1

( )

( )

( )

( )

( ) ( ) ( )

( )

n i n i M i n i n n

f f f T f T f T − ∂ ∂ + ≅

+ = +

∑

1 1 1

( )

( )

( )

( )

( ) ( ) ( )

( )

n i n i M i n i n n

f l a f T f T − + ≅

∑

= +

φ

1 1

( )

( )

1 + n

f T

( )

n i

φ

( ) ( )

∑

= M i n i n i

l

1

φ

48

F.5.2. Problem set-up example

Conservation of total avg. flow: If there were linearly independent, then the problem would be over.

( ) ( ) ( )

∑

=

− =

4 1

1

i n i i n i n

C T λ λ γ

13 12

γ γ γ + =

13 12 2 1

γ γ λ λ + = +

12 4 3 1

γ λ λ λ = + −

13 4 3 2

γ λ λ λ = − +

13 12 2 1

γ γ λ λ + = +

12 4 3 1

γ λ λ λ = + −

49

• a. Shortest path flow subproblem

where subject to ( refers to ) and where for example

• b. FD algorithm gives the and ,but not the

routing

∑

= 4 1

min

i i i

l

i

φ

φ

( )

2 i i i i

C C l λ γ − =

i

φ

i

λ

13 12 2 1

γ γ φ φ + = +

12 4 3 1

γ φ φ φ = + −

i i

C β φ ≤ ≤

99 . = β

i

λ

min

T

50

• c. Extension to include commodity flow

Link 1: , Link 2: , Link 3: Link 4: Conservation of commodity flow: 6 eqns. of which 4 are linearly independent

12 12

u

13 12

u

12 13

u

12 32

u

13 23

u

13 13

u

13 12 12 12 1

u u + = λ

13 13 12 13 2

u u + = λ

13 23 3

u = λ

12 32 4

u = λ

12 12 13 12 12

γ = + u u

13 13 23 13 13

γ = + u u

13 13 13 13 12

γ = + u u

12 12 32 12 12

γ = + u u

13 23 13 12

u u =

12 13 12 32

u u =

51

• c. Extension to include commodity flow (cont.)

Shortest path flow subproblem where Subject to and FD algorithm gives , ,

13 12 12 12 1

u u + = φ

13 13 12 13 2

u u + = φ

13 23 3

u = φ

12 32 4

u = φ

12 12 13 12 12

γ = + u u

13 13 23 13 13

γ = + u u

13 13 13 13 12

γ = + u u

12 12 32 12 12

γ = + u u

∑

= 4 1

min

i i i

l

i

φ

φ

( )

2 i i i i

C C l λ γ − =

i i

C β φ ≤ ≤

i

λ

min

T

jk ab

u

52

• G. CFA Problem

G.1. Given τ, minimize T with respect to {Ci} and {λi} subject to and 0≤ λi ≤Ci

where

Note: When we combine CA and FA, we are no longer able to give globally optimal solutions, but rather describe procedures that find local minima for T

∑

=

=

M i i i

D C d

1

∑

=

− =

M i i i i

C T

1

1 λ λ γ

53

G.2. Approach

• a. Combine CA and FA algorithms
• b. Since {Ci}’s have to be assigned, it can be

shown that fixed routing is optimal.

– This implies setting a=1 in the FD algorithm

• c. It can also be shown that shortest path routing

gives only local minima for T.

54

– This means that several initial feasible flows f(0) must be tried

G.3.Flow chart of CFA algorithm

55

• G. 4. Algorithm properties

a. Converges for each f(0) since the number of possible shortest-path flows is finite b. Since li≥0, there are no negative loops in the shortest- path flow algorithm c. The form of li that results from the “CA part” of algorithm is such that

– See Kleinrock’s Vol. 2, Eq. 5.45

– It implies that once λi and Ci become zero, they remain there

d. Since only local minima are possible, one has to be clever in the f(0) that are used

∞ =

→ i

l

i

lim

λ

56

) / ( ) / (

2 / 1 1 2 / 1 1

>               +       = ∂ ∂ =

∑ ∑

= = M j i i e i i i e M j i i i

d D d d D n T l λλ µ λλ µ λ µ λ

• H. TCFA Problem (Dual Form)

H.1. Given node locations and {γjk}, minimize

with respect to τ, {Ci} and {λi} subject to and connectivity constraints.

Note: M = no. of links = variable! Note(2): in the slides µ is not reported (..normalized)

( )

∑

=

=

M i i i C

d D

1 MAX M i i i i

T C T ≤ − = ∑

=1

1 λ λ γ 1)

• N(N

M 1,2,...M ≤ < = < ≤ i Ci

i

λ

57

H.2. Approach

a. Real design requires selection of discrete capacities b. Heuristic solution procedure via iterative use of CFA and choices of τ c. Only local minima possible d. di(Ci)= termination cost + (line cost) x (length of line) expressed in $/month

( )

α i i i i i

C d a C d + =

58

H.3. TCFA2 Problem

CBE (Concave Branch Elimination) Algorithm a. Select τ(0), e.g., fully-connected b. For each channel, assume a power-law approximation and choose di and αi in c. Execute the CFA2 (minimize D subject to constraint T=TMAX). At each iteration in CFA2, use a linearized value for capacity about the current value of flow. If at any step in CFA2, the connectivity constraint is violated, proceed to Step (d) with the previous step CFA2 results; otherwise, let CFA2 run to completion d. Discretize the {Ci } to the nearest actual line capacities available such that λi<Ci and T≤TMAX

τ

∑

=

=

M i i i

i

C d D

α

59

CBE Algorithm (continued)

• e. Refine the flow assignment using the dual form of the FD

algorithm where and adjust the {Ci } accordingly so that λi<Ci and T≤TMAX

• f. Repeat Steps 3-5 for various feasible f(0)
• g. Repeat Steps 1-6 for a number of different τ(0)

            + = ∂ ∂ ≅

∑

= i i MAX M j j j i i i

d T d d D l λ γ λ λ

1

1

60

ARPANET CBE Study

a. γjk 1kbps over all possible jk pairs gives γ=650kbps

• b. TMAX=200 msec
• c. Connectivity = 2

d.

∑

=

+ =

M i i i i

i

C d c D

1 α 61

• e. Channel capacities and costs

62

• f. Results

1) 30 initial topologies studied 2) CBE algorithm required 1 to 2 seconds CPU time on 360/91 per topology 3) 0.8≤αi ≤ 1.0 provided good fit to capacity costs and CBE worked well for these values 4) Two best solutions on next slides

63

Best Solution of ARPANET CBE

64

a. Distribution of capacities vs. link lengths b. τ(0) was a fully-connected topology c. Solution has 61 channels d. Cost = $89,580/month

65

Best Solution of ARPANET CBE

Second Best Solution of ARPANET CBE

66

• a. τ(0) was a low-connected topology
• b. Both initial and final topologies had 29

channels

• c. Uses higher speed lines (50 and 100kbps)

for medium and long distances; 230kbps for short distances

• d. Cost = $ 94,288/month

Second Best Solution of ARPANET CBE (cont.)

67

Convex Sets

• A set S is said to be a convex set if, for any two

points in the set, the line joining those two points is also in the set

• Mathematically, S is a convex set if for any two

vectors x(1) and x(2) in S, the vector

is also in S for any number between 0 and 1

Examples: Figures A.1 and A.2 represent convex sets, while Figure A.3 is not a convex set.

( ) ( )

2 1

) 1 ( x x x λ λ − + =

68

Theorem 1 The set of all feasible solutions to a linear programming problem is a convex set Theorem 2 The intersection of convex sets is a convex set (Figure A.4) Theorem 3 The union of convex sets is not necessarily a convex set (Figure A.4) Theorem 4 A hyperplane is a convex set Theorem 5 A half-space is a convex set

69

Theorems

• A convex combination of vectors x(1), x(2),…, x(k) is

a vector x such that

• An “extreme point” or “vertex” of a convex set is a

point in the set that cannot be expressed as the midpoint of any two points in the set. For example, consider the convex set . This set has four extreme points given by (0,0), (0,2), (2,0), and (2,2)

( ) ( ) ( )

k kx

x x x λ λ λ + + + = 

2 2 1 1

1

2 1

= + + +

k

λ λ λ  for ≥

i

λ

k i , , 2 , 1  =

( )

{ }

2 , 2 ,

2 2 1

≤ ≤ ≤ ≤ = x x x x S

i

70

Vertex

• A hyperplane is the set of all points x satisfying

cx=z for a given vector c≠0 and a scalar z

– The vector c is called the normal to the hyperplane. – For example, is a hyperplane.

• A half-space is the set of all points x satisfying

cx≤z or cx≥z for a given vector c≠0 and a scalar z

( )

{ }

5 3 2 , ,

3 2 1 3 2 1

= + − = x x x x x x H

71

Hyperplane & Half Space

Convex Function

A function of n variables f(x) defined on a convex set D is said to be a convex function if and only if for any two points x(1) and x(2)∈ D and 0≤ λi ≤1,

( )

( ) ( )

[ ]

( )

( ) (

)

( )

( )

2 1 2 1

1 1 x f x f x x f λ λ λ λ − + ≤ − +

The figure illustrates the definition of a convex function of a single variable.

72

Properties of Convex Functions

a. The chord joining any two points on the curve always falls entirely on or above the curve between those two points b. The slope or first derivative of f(x) is increasing or at least non-decreasing as x increases c. The second derivative of f(x) is always non-negative for all x in the interval d. The linear approximation of f(x) at any point in the interval always underestimates the true function value e. For a convex function, a local minimum is always a global minimum

73

Figure A.6 illustrates property 4

74

Figure A.6 illustrates property 4

The linear approximation of f(x) at the point x0, denoted by , is obtained by ignoring the second and other higher

• rder terms in the Taylor’s series expansion

For a convex function, property 4 implies that for all x The gradient of a function f(x1,…xn) is given by The Hessian matrix of a function f(x1,…xn) is an nxn symmetric matrix given by

( )

; ~ x x f

( ) ( ) ( )( )

; ~ x x x f x f x x f − ∇ + =

( )

( ) ( )( )

x x x f x f x f − ∇ + ≥

( )

      = ∇

n n

x f x f x f x x f δ δ δ δ δ δ ,..., , ,...,

2 1 1

( )

f x x f x x H

j i n f 2 2 1,...,

∇ =         = δ δ δ

75

Backup slides

76

Example:

Require all Ti to be the same and that msec

3

12 =

γ

6

13 =

γ

1

23 =

γ

bits 200 1 = µ

4

1 =

λ

2

2 =

λ

5

3 =

λ

110 = T

77

[msg/sec] [msg/sec]

1 . 1 = = γ λ n

sec 1 . sec 100 = = = m n T τ

2000 1 = µτ

kbps bps C 8 . 2 2800

1

= =

kbps bps C 4 . 2 2400

2

= =

kbps bps C . 3 3000

3

= =

FA Problem Example:

Ground network and satellite system Determination of T includes 0.5 sec. Assumed propagation delay for each of the three satellite links.

3

42 =

C

2

34 =

C

2

32 =

C

3

24 =

C

5 . 2

13 =

C

3

12 =

C

51 . 1

12 =

λ 17 .

42 =

λ 66 .

34 =

λ 83 .

32 =

λ 51 . 1

24 =

λ 49 .

13 =

λ

51 . 1 = T

78