On Network Simplification for Gaussian Half-Duplex Diamond Networks

0.2 MB·

by  Martina Cardone

On Network Simplification for Gaussian
Half-Duplex Diamond Networks
Martina Cardone†, Christina Fragouli†, Daniela Tuninetti∗
† UCLA, Los Angeles, CA 90095, USA, Email: {martina.cardone, christina.fragouli}@ucla.edu
∗ University of Illinois at Chicago, Chicago, IL 60607, USA, Email: [email protected]
Abstract—This paper investigates the simplification problem capacity one can achieve by smartly selecting a subset of k
in Gaussian Half-Duplex (HD) diamond networks. The goal is relays out of the N possible ones in a Gaussian HD diamond
6 to answer the following question: what is the minimum (worst- network. As a first step in this direction, we here provide
1 case) fraction of the total HD capacity that one can always a worst-case (i.e., independent of the channel parameters)
0 achieve by smartly selecting a subset of k relays, out of the N
2 possible ones? We make progress on this problem for k = 1 approximate (i.e., up to a constant gap) capacity guarantee
and k = 2 and show that for N = k + 1, k ∈ {1,2} at in terms of achievable fraction for k = 1 and k = 2 when
ul least k+k1 of the total HD capacity is always approximately (i.e., N = k+1 and N ≫ 1. We also present network examples
J up to a constant gap) achieved. Interestingly, and differently forwhichthebestk-relaysubnetworkapproximatelyachieves
from the Full-Duplex (FD) case, the ratio in HD depends on
4 theworst-casefractionofthetotalHDcapacity,henceshowing
N, and decreases as N increases. For all values of N and k
for which we derive worst case fractions, we also show these to thatthederivedworst-performanceguaranteesareindeedtight.
]
T be approximately tight. This is accomplished by presenting N- Moreover, we find significant differences of the wireless
I relayGaussianHDdiamondnetworksforwhichthebestk-relay simplificationproblemforHDandFDnetworks.Forexample,
. subnetworkhasanapproximateHDcapacityequaltotheworst-
s (i) in HD the fraction of the achieved capacity dependson N
casefractionofthetotalapproximateHDcapacity.Moreover,we
c and decreases as N increases, (ii) the worst-case networks in
provideadditionalcomparisonsbetweentheperformance of this
[
simplificationproblemforHDandFDnetworks,whichhighlight HD and FD are not necessarily the same and (iii) the best

2 their different natures. k-relay subnetworks in HD and FD might be different.
v
1 I. INTRODUCTION Related Work. The capacity characterizationof the Gaussian
6 Wireless network simplification, first introduced in [1] in HD relay network is a long-standing open problem. The
1 thecontextofGaussianFull-Duplex(FD)diamondnetworks1, tightestupperboundonthecapacityisthewell-knowncut-set
5
shows the surprising result that a significant fraction of the upper bound. In [2], this bound was evaluated by using the
0
. capacity can be achieved by using only a small subset of the approach first proposed in [3] and shown to be achievable to
1
availablerelays.Inthispaper,weseektoanswerthefollowing within1.96(N+2)bitsperchanneluse (independentlyofthe
0
question:howdotheseresultsextendtoGaussianHalf-Duplex channel parameters), by noisy network coding [4].
6
1 (HD) diamond networks? Ingeneral,theevaluationofthecut-setupperboundrequires
: The wireless simplification approach offers a number of anoptimizationover2N listen/transmitstates.Recently,in[5]
v
i benefits. First, it promises energy savings since only the theauthorsprovedasurprisingresult:atmostN+1states(out
X power of the active relays is used to transmit information; of the 2N possible ones) suffice for capacity characterization
r the power available at the relays that are kept silent is saved. (uptoaconstantgap)foraclassofHDrelaynetworks,which
a
Then, it simplifies the synchronization problem as only the includestheGaussiannoisenetwork.However,theproblemof
selected relays have to be synchronized for transmission. finding which are the N +1 active states might still require
Finally, for HD networks, simplification offers benefits in the use of all the N relays.
terms of scheduling. Indeed, in an N-relay HD network,
This work presents partial results on the wireless simplifi-
a capacity-achieving scheme requires a global optimization
cation problem for the HD case, whose FD counterpart was
over the 2N possible listen/transmit configuration states. This
solved in [1]. In particular, in [1] it was shown that, by
approach, as N increases, quickly becomes computationally
selectingkrelays(outoftheN possibleones),onecanalways
prohibitive. Thus, selecting a small subset of the relays leads approximatelyachieve at least a fraction k of the capacity.
to a significant complexity reduction in the scheduling. k+1
This result was proved to be tight, i.e., there exist N-relay
Inthispaper,motivatedbythenumerousbenefitsofwireless
Gaussian FD diamond networks for which the best k-relay
simplification, we seek to understand how much of the HD
subnetwork approximately achieves this fraction of the total
FD capacity. A polynomial-time algorithm to discover these
The work of M. Cardone and C. Fragouli was partially funded by NSF
underawardnumber1514531.TheworkofD.Tuninettiwaspartiallyfunded high-capacity k-relay subnetworks was also proposed. In [6]
byNSFunderawardnumber1218635.M.Cardonewouldliketoacknowledge the authors proved that, by selecting k = 2 relays out of the
insightful discussions withYahyaH.Ezzeldin.
N possible ones in a Gaussian HD diamond network, and by
1AnN-relaydiamondnetworkisarelaynetworktopologywherethesource
cancommunicatewiththedestinationonlythroughN non-interferingrelays. operating them only in a complementary fashion (i.e., when
one relay listens, the other transmits), at least 1 of the total The capacity2 C¯HD of the channel in (1) is not known, but
2
HD capacity is approximately achieved. can be upper and lower bounded as
From the result in [1], it directly follows that in HD, by
CHD−G ≤C¯HD ≤CHD+G , (3a)
selecting k relays, one can always approximately achieve a NF 1 NF 2
fisraaccticoonm2p(lkik+sh1e)dobfythoepHerDatcinagpatchietysuobfntheetwwohrkoliennoentwlyor2k.stTahteiss CHNDF :=mλaxAmF⊆inNFs∈X[0:1]Nλs(cid:18)i∈AmFa∩xRsℓi+i∈AmcFa∩xRcsri(cid:19), (3b)
(outofthe2k possibleones)ofequalduration:thefirstwhere
all the k relays listen and the second where all the k relays where: (i) G and G are both O(N) and independent of the
1 2
transmit. However, as we show in this paper this capacity actual value of the channel parameters; (ii) λ = [λ ] with
s
guarantee is not tight in general. Moreover, differently from λ := P[S = s] ∈ [0,1] : λ = 1; (iii) R
s [1:N] s∈[0:1]N s s
the work in [6], in this paper we do not restrict the selected containstherelaysthat,instates∈P[0:1]N,arereceiving,i.e.,
relays to operate only in certain states. This leads to better among the relays ‘on the side of the destination’ (indexed by
performance guarantees in terms of achievable fraction. A ) only those in receive mode matter, and similarly, among
F
the relays ‘on the side of the source’ (indexed by Ac) only
Paper Organization. Section II describes the N-relay Gaus- F
those in transmit mode matter.
sian HD diamond network and summarizes known capacity
In the next section we will use the result in (3) to prove
results. Section III proves our main result, i.e., it provides
that a significant fraction of CHD can always be achieved by
a tight worst-case guarantee in terms of achievable fraction NF
selecting k=1 and k=2 relays out of the N possible ones.
of the capacity for k = 1 and k = 2 when N = k + 1
and N ≫ 1. Section III also highlights differences between
III. ACHIEVING A FRACTION OF THEHD CAPACITY
the simplification problem in HD and FD networks. Finally,
Section IV concludes the paper and discusses future research In this section we prove our main result, i.e., we derive a
directions. Some of the proofscan be found in the Appendix. worst-case fraction guarantee for any N-relay Gaussian HD
diamondnetworkwith k =1 andk =2 whenN =k+1 and
when N ≫ 1. Moreover, for each case we provide network
II. SYSTEMMODELAND KNOWN RESULTS
examples which approximately achieve the derived fraction
With [n :n ] we denote the set of integers from n to of the total HD capacity, hence showing that our worst-case
1 2 1
n ≥n ; ⌊a⌋ and ⌈a⌉ are the floor and ceiling functions of performance guarantees are indeed tight. Our main result is
2 1
a∈R, respectively.Calligraphiclettersdenotesets; |A| is the presented in the following theorem.
cardinality of A, A\B is the set of elementsthat belongto A
Theorem 1. In the Gaussian HD N-relay diamond network,
but notto B and Ac is the complementof A. A⊆B indicates
byselectingk ≤N relaysandbykeepingtheremainingN−k
that A is a subset of B, A∪B is the union of A and B and onessilent,wecanachieve,uptoagap,arateCHD suchthat
k,N
A∩B is the intersection of A and B; ∅ is the empty set.
The Gaussian HD diamond network consists of a source 1 k =1, N =2
communicating with a destination only through N non- CHk,DN ≥ 124 k =1, N ≫1 . (4)
irneltaetrifoenrisnhgiprefolarythsitsatnioentwsoorkpeirsating in HD. The input/output CHNDF  213 kk ==22,, NN =≫31
2

Moreover, the bound in (4) is tight up to a constant gap.
Y =(1−S )h X +Z , ∀i∈[1:N], (1a)
i i i 0 i
N Before going into the technical details of the proof of
YN+1 = SigiXi+ZN+1, (1b) Theorem 1 we make a couple of remarks.
Xi=1
Remark1. Theorem1showsthat,forafixedvalueofk ∈[1:
where: (i) X0 denotes the channel input at the source and 2], the fraction CHk,DN decreases as N increases. In particular,
CHD
YN+1 the channel output at the destination; (ii) the channel NF CHD
inputsaresubjecttotheaveragepowerconstraintE |X |2 ≤ for N = k+1 we have k,N = k as in FD [1, Theorem
i CHD k+1
1,i ∈ [0 : N]; (iii) Si,i ∈ [1 : N], is the binar(cid:2)y rand(cid:3)om 1]. Differently from FD whNeFre the fraction does not depend
variable, which represents the state (either listening or trans- on N, in HD the fraction decreases as N increases.
mitting) of the i-th relay, with i ∈ [1 : N], i.e., if S = 0
i Remark 2. The lower bound on CHD for k =2 and N ≫1
then it is receiving,while if Si =1 then it is transmitting[3]; k,N
(iv) the channel parameters (h ,g ) ∈ C2,∀i ∈ [1 : N], are is the one derived in [6, Theorem 2.2] where the k = 2
i i
selectedrelaysareconstrainedtooperateinacomplementary
constant and therefore known to all terminals; (v) the noises
fashion. However, for k = 2 and N = 3, the result in
are independent. In the following we shall indicate
Theorem 1 improves over [6, Theorem 2.2], hence showing
ℓ :=log(1+|h |2), ∀i∈[1:N], (2a) the importance of optimizing over all possible states.
i i
r :=log(1+|g |2), ∀i∈[1:N]. (2b)
i i 2Weusestandarddefinitions forcodes,achievable ratesandcapacity.
In what follows we let C⋆NHiD,i ∈ [1 : 2] be the HD 0.65 k=1
achievable rate of the subnetwork Ni when operated with the k=2
0.6
‘natural’ schedule derived from λ⋆ (the optimal schedule of
0.55
thewholenetwork).ClearlywehaveC⋆HD ≤CHD.Theproof
ofTheorem1makesuseofthefollowinNgilemmaN,iwhoseproof HDCk,NHDCNF00.4.55
can be found in Appendix A. 0.4
0.35
Lemma 2. For any Gaussian HD diamond network N with
F 0.3
N relays, we have
0.25
0 100 200 300 400 500 600 700 800
CHD ≤C⋆HD+C⋆HD, (5) Fig. 1: CHk,DN for k∈[1:2]Nversus different values of
NF N1 N2 CHD
where{N ,N }isapartitionofthefullnetworkN =[1:N]. N ∈[1N:F800] for the network in (6) with c=1.
1 2 F
A. The case k =1 and N =2
CHD
For the network in (6) in Fig. 1 we plot 1,N (see solid line)
From the result in Lemma 2 we obtain CHD
versusdifferentvaluesofN ∈[1:800]forNcF=1.FromFig.1
CHNDF=CH{1D,2}≤C⋆{H1}D+C⋆{H2}D≤CH{1D}+CH{2D}≤2im∈[1a:x2]nCH{iD}o, we observe that for N ≫ 1 we have CH1,DN = 41CHNDF, i.e., the
bound in Theorem 1 for k = 1 and N ≫ 1 is precisely met.
wNh=ich2imwhpelirees, CbyH1,D2se≥le12ctCinHNgDF.thWeebensotwreplraeys,ewnteaanpeptrwoxoirmkawteitlhy In particular, for N ≫1, we have CHNDF =CFNDF =c.
achieve 1 of the HD capacity of the whole network. C. The case k=2 and N =3
2
Example. Let ℓ =r ,i∈[1:2] and ℓ =ℓ . For this network WehereprovethatineveryGaussianHDdiamondnetwork
i i 1 2
from [7] we have CHD=ℓ and CHD=ℓ1, i.e., CHD=1CHD. withN =3relays,therealwaysexistsasubnetworkofk =2
NF 1 1,2 2 1,2 2 NF
relayssuchthatitsapproximateHDcapacityislowerbounded
B. The case k =1 and N ≫1
by 2 of the approximate HD capacity of the whole network.
3
From the result in [1, Theorem 1] for k =1 we obtain Tothisend,wemakeuseoftheresultsinLemma2,Lemma3
1 1 (whose proof can be found in Appendix B) and Lemma 4.
CHD ≤ CFD ≤CFD,
2 NF 2 NF 1
Lemma 3. Consider a Gaussian HD diamond network N
F
where we used the notation (i) CFD (which indicates the with N = 3. Let A⋆ ⊆ N = {i,k} with |A⋆| = a⋆ and
1 i i i i
approximate FD capacity of the k = 1 selected relay) to A⋆ ⊆N ={i,j}with A⋆ =a⋆ be themincutsfortheHD
CFD j j j j
highlightthat in FD the ratio 1 doesnot dependon N and networks N and N , re(cid:12)spec(cid:12)tively, so that a⋆−a⋆ 6=2. Let
(2i-ir)eClaFNyDFnteotwinodrikc.aItet itshenaoptpdriofxfiCicFNmuDFlattetoFDseecatphaactitCyF1oDft=he2wC˜hH1,oDNle, λtCh⋆HeDonf=etthwCeoHrwkDhisoNleiwnaeentwjhdaovNrke(cid:12)jCoHN(cid:12)pDFerwatiethwNith=th3e(cid:12)(cid:12)reoilpatyims.ajTl(cid:12)(cid:12)hsecnh,ewduitlhe
where C˜HD is the approximate HD capacity of the k = 1 NF {1,2,3}
1,N
selected relay when it receives for 1 of the time and it CHD+C⋆HD ≤C⋆HD +C⋆HD. (8)
transmits for 1 of the time. Thus, 2 NF {i} {i,j} {i,k}
2 Lemma 4. In any 3-relay Gaussian diamond network (both
1CHD≤1CFD≤CFD=2C˜HD ≤2CHD =⇒CHD ≥1CHD. FD and HD) let A⋆1 ⊆N1 ={1,2}, A⋆2 ⊆N2 ={1,3} and
2 NF 2 NF 1 1,N 1,N 1,N 4 NF A⋆⊆N ={2,3}be themin cutsofthe networksN , N and
3 3 1 2
Wenextprovideanetworkexampleforwhich(uptoaconstant N , respectively with |A⋆|=a⋆,∀i ∈ [1 : 3]. Then, ∃(i,j) ∈
3 i i
gap) CHD = 1CHD for N ≫1. [1:3]2 withi6=j forwhichA⋆ andA⋆ satisfy a⋆−a⋆ 6=2.
1,N 4 NF i j i j
Example. We let (cid:12) (cid:12)
Proof: Assume, without loss of ge(cid:12)nerality,(cid:12) that
N+2 N+2 c |a⋆−a⋆| = 2, i.e., either {A⋆,A⋆} = {∅,{1,3}} or
ℓ = 2 2 , i∈[1:N], (6a) 1 2 1 2
i (cid:6)N+2 (cid:7)+(cid:4) N+(cid:5)2 i {A⋆1,A⋆2} = {{1,2},∅}. Let {A⋆1,A⋆2} = {∅,{1,3}} (the
2 2 samereasoningalsoholdsfor{A⋆,A⋆}={{1,2},∅}).Then,
(cid:6) N+(cid:7)2 (cid:4)N+2 (cid:5) c 1 2
r = 2 2 , i∈[1:N], (6b) • if A⋆ =∅, then |a⋆−a⋆|=06=2;
i (cid:6)N2+2 (cid:7)+(cid:4) N2+(cid:5)2 N −i+1 • if A⋆33={2}, then |1a⋆1−3a⋆3|=16=2 and |a⋆2−a⋆3|=16=2;
wherec∈(cid:6)R .N(cid:7)otic(cid:4)ethat(cid:5)inthisnetworkwe haveℓ ≥ℓ ≥ • if A⋆={3}, then |a⋆−a⋆|=16=2 and |a⋆−a⋆|=16=2;
+ 1 2 3 1 3 2 3
...≥ℓN andr1 ≤r2 ≤...≤rN.Moreover,itisnotdifficult • if A⋆3={2,3}, then |a⋆2−a⋆3|=06=2.
to see that the vectorof the relay-destinationcapacitiesis just Thanks to the result in Lemma 4 we know that ∃i∈[1:3]
theflippedversionofthevectorofthesource-relaycapacities. such that (8) is satisfied. Without loss of generality,let i=1.
For the network in (6) all the single-relay capacities are the Moreover, from Lemma 2 we have CHD ≤ C⋆HD +C⋆HD .
NF {1} {2,3}
same and, up to a constant gap, evaluate to By summing this with (8) evaluated in i = 1 we obtain
CH1,DN = (cid:6)NN2+2+22(cid:7)+(cid:4)NN2+2+2(cid:5)2 Nc+1 N=≫1 4c. (7) 32CmHNaDFx≤nCCH{⋆{1DH1,,2D2}},+CH{C1D,⋆{3H1},,D3C}H{+2D,C3}⋆{oH2,,D3}w≤hiCcH{h1D,i2m}p+lieCsH{1DC,3H2},D3+≥CH{322DC,3HN}DF≤.
(cid:6) (cid:7) (cid:4) (cid:5)
CHD
We nowprovidea networkexamplewithN=3 where,byse- • Worst-case ratio with respect to FD, i.e., k,N. The
CFD
lecting the best subnetworkof k=2 relays, we approximately problem of finding the capacity of HD relay netwNForks is
achieve 2 of the HD capacity of the whole network.
3 computationallyexpensive,asit requiresan optimizationover
Example.Letℓ1=13, ℓ[2:3]=1, r1=r, r[2:3]=32 withr→∞. 2N cuts each of which depends on 2N listen/transmit states.
For this network CHNDF=1 and CH2,D3 =32, thus CH2,D3 =32CHNDF. On the contrary, the cut-set upper bound in FD can be more
easily evaluated. Thus, one can think of comparing the HD
D. The case k =2 and N ≫1
capacity of the k-relay selected subnetwork with respect to
The lower bound in Theorem 1 for this case is simply the the FD capacity of the whole network. By doing so, we get
one derived in [6, Theorem 2.2], valid for the case when the results presented in the next theorem,whose proofcan be
the k = 2 selected relays are allowed to operate only in a found in Appendix C.
complementary fashion. We next provide a network example
for which CHD =1CHD for N≫1, up to a constant gap. Theorem 5. In the Gaussian HD N-relay diamond network,
2,N 2 NF byselectingk ≤N relaysandbykeepingtheremainingN−k
Example.Weagainconsiderthenetworkdefinedin(6).Since
onessilent,wecanachieve,uptoagap,arateCHD suchthat
for this network all the single-relay capacities, given in (7), k,N
arethe same, thenCH2,DN ≤2CH1,DN. Itis notdifficulttosee that 1 k =1, N =1
by selecting the first and the last relays we exactly get CHk,DN ≥ 123 k =1, N =2 . (9)
CH2,DN =2CH1,DN =2 (cid:6)NN+2+22(cid:7)+(cid:4)NN2++2(cid:5)2 Nc+1 N=≫1 2c. CFNDF  1142 kk ==12,, NN ≫∈[21:3]
(cid:6) 2 (cid:7) (cid:4) 2 (cid:5) Moreover, the boundin (9) is tight up to a constant gap.
This, with the fact that for N ≫ 1 we have CHD = c, gives
the claimed ratio as we observe from Fig. 1 (daNsFhed line). Although, for the case k =2, Theorem 5 provides a lower
boundforN ∈[2:3],webelievethatCHD ≥ 1CFD fork =2
2,N 2 NF
E. HD versus FD simplification andN ≥2.Ifthisconjectureisprovedtobetrue,thenitwould
We here discuss differences between the selection perfor- imply CHk,DN ≥ 21CFNDF for k > 1 and N ≥ k, as by selecting
more relays (i.e., by increasing k) the performance guarantee
mances in HD and FD networks. We believe that the reason
cannot decrease. By comparing Theorem 1 and Theorem 5,
for this different behavior is that in HD the schedule plays a
two conclusions can be drawn:
keyrole:removingsomeoftherelaysmightchangetheglobal
schedule of the network. 1) For N=k+1, k∈[1 : 2], Theorem 1 provides a better
CHD
•InHDtheratio k,N decreasesasN increases.Asalready guaranteewithrespecttoTheorem5,becauseoftheuse
CHD
NF of a tighter upper bound to the HD performance.
highlightedinRemark1thisrepresentsasurprisingdifference
2) For N ≫ 1 and k = 1 the two bounds in Theorem 1
withrespectto FD andshowsthatFD andHD relaynetworks
and Theorem 5 coincide. This may indicate that, as N
have a different nature.
increases, the schedule optimization is less crucial.
• Worst-case networks in HD and FD are not necessarily
the same. Consider the network exampleℓ1=ℓ2=r1=r2=1 IV. CONCLUSIONS ANDFUTURE WORK
andsupposewewanttoselectk=1relay.We alreadyshowed
We studied the simplification problem in an N-relay Gaus-
beforethat,byselectingeitherthefirstorthesecondrelay,we
sianHDdiamondnetwork.We showedthat,whenN =k+1,
get CHD= 1CHD. Now, suppose we operate this network in
1,2 2 NF by selecting k ∈ [1 : 2] relays one can achieve (to within a
FD. Then,itisnotdifficultto see that,byselectinganyofthe
constantgap)atleast k ofthetotalHDcapacity.Differently
N =2 relays, we get CFD = CFD, which is greater than 1, k+1
1 NF 2 from the FD case, this fraction decreases as N increases.
i.e., the worst-case ratio proved in [1, Theorem 1].
The extension of these results to k>2 and the design of a
• The best HD and FD subnetworks are not necessarily
polynomial-timealgorithmwhich efficientlydiscoversa high-
the same. Consider the network in (6). When the N relays
capacity k-relay subnetwork are interesting open problems,
operate in HD, they all have the same single capacity given
which are object of current investigation.
in (7). This means that by selecting any of the relays (i.e., at
random) we get the same performance guarantee. Differently, APPENDIX A
when the N relays operate in FD, only the ⌊N+2⌋-th relay PROOF OF LEMMA2
2
(when N is an odd number) and the relays number ⌊N+2⌋
2 We hereprovetheresultin Lemma2. Tothis endwe make
and number⌊N+2⌋−1 (whenN is an evennumber)give the
2 use of the following lemma.
performanceguaranteeof[1,Theorem1].Asanotherexample
consider a Gaussian 2-relay diamond network with ℓ = 1, Lemma 6. For any Gaussian FD diamond network, we have
1
ℓ2 = 52, r1 = 21 and r2 = 154 and suppose we want to select CFD ≤CFD+CFD, (10)
thebestrelay.Itisnotdifficulttoseethatiftherelaysoperate NF N1 N2
inFD,thenthefirstrelayisthebestanditachievesCFD = 1, where {N ,N } is a partition of the full network N , with
1 2 1 2 F
while if the relays operate in HD then the second relay is the
CFD = min I ,I := maxℓ + max r ,
best giving CH1,D2 = 270. NF AF⊆NF AF;NF AF;NF i∈AF i i∈NF\AF i
TABLE I: Proof of Lemma 7.
A1 A2 f(A1,A2) AF AS g(AF,AS)
∅ ∅ max{ri,rj}+max{ri,rk} ∅ ∅ max{ri,rj,rk}+ri
∅ {i} max{ri,rj}+ℓi+rk ∅ {i} max{ri,rj,rk}+ℓi
∅ {k} max{ri,rj}+ℓk+ri {k} ∅ ℓk+max{ri,rj}+ri
{i} ∅ ℓi+rj+max{ri,rk} ∅ {i} max{ri,rj,rk}+ℓi
{i} {i} 2ℓi+rj+rk {i} {i} 2ℓi+max{rj,rk}
{i} {k} ℓi+rj+ℓk+ri {i,k} ∅ max{ℓi,ℓk}+rj+ri
{i} {i,k} ℓi+rj+max{ℓi,ℓk} {i,k} {i} max{ℓi,ℓk}+rj+ℓi
{j} ∅ ℓj+ri+max{ri,rk} {j} ∅ ℓj+max{ri,rk}+ri
{j} {i} ℓj+ri+ℓi+rk {j} {i} ℓj+max{ri,rk}+ℓi
{j} {k} ℓj+2ri+ℓk {j,k} ∅ max{ℓj,ℓk}+2ri
{j} {i,k} ℓj+ri+max{ℓi,ℓk} {i,j,k} ∅ max{ℓi,ℓj,ℓk}+ri
{i,j} {i,k} max{ℓi,ℓj}+max{ℓi,ℓk} {i,j,k} {i} max{ℓi,ℓj,ℓk}+ℓi
CFD= min I ,I :=maxℓ + max r ,j∈[1:2]. N = {1,2,3} with (i,j,k) ∈ [1 : 3]3 and i 6= j 6= k. Let
Nj Aj⊆Nj Aj;Nj Aj;Nj i∈Aj i i∈Nj\Aj i AF ⊆ N with |A | = a , and A ⊆ N with |A | = a .
1 1 1 1 2 2 2 2
Proof: We have DefineN =N ∩N .Clearly N =N ∪N .If|a −a |6=
S 1 2 F 1 2 1 2
2, then there exist A ⊆N , A ⊆N such that
I +I =maxℓ +maxℓ + max r + max r F F S S
A1;N1 A2;N2 i∈A1 i i∈A2 i i∈N1\A1 i i∈N2\A2 i maxℓ + max r +maxℓ + max r
t t t t
≥ max ℓi+ max ri t∈A1 t∈N1\A1 t∈A2 t∈N2\A2
i∈A1∪A2 i∈(N1\A1)∪(N2\A2) ≥ maxℓ + max r +maxℓ + max r . (13)
t t t t
(=a) max ℓ + max r t∈AF t∈NF\AF t∈AS t∈NS\AS
i i
i∈A1∪A2 i∈(N1∪N2)\(A1∪A2) Proof: We define
=I ≥ min I =CFD, (11)
A1∪A2;NF B⊆NF B;NF NF f(A1,A2):=tm∈aAx1ℓt+t∈mN1a\xA1rt+tm∈aAx2ℓt+t∈mN2a\xA2rt,
where the equality in (a) follows since N1∩A2=∅ and N2∩ g(AF,AS):= maxℓt+ max rt+maxℓt+ max rt.
A1=∅ and (B\A)∪(C\A)=(B∪C)\A. The result in (11) is t∈AF t∈NF\AF t∈AS t∈NS\AS
valid ∀A1⊆N1 and ∀A2⊆N2, hence also for the minimum Table I proves that, ∀A1 ⊆ N1 and ∀A2 ⊆ N2 for which
cuts of the networks N1 and N2, i.e., CFND1+CFND2≥CFNDF. |a1−a2|6=2, then f(A1,A2)≥g(AF,AS).
We now show how the result in (11) extends to HD. We Notice that we did not consider the cases {A1,A2} =
notice that CHNDF in (3b) can be equivalently written as e{q{ui,ivja}l,e{nit}t}oa{nAd {,AA1,}A=2}{{=i}{,{{ii,,kj}},}{kan}d} tsoin{ceAth,eAse}a=re
1 2 1 2
CHD = min λ⋆ maxℓ′ +maxr′ , {{j},{i,k}}, respectively. Therefore, Table I covers all the
NF AF⊆NFs∈X[0:1]N s(cid:18)i∈AF i,s i∈AcF i,s(cid:19) cases except {A1,A2} = {∅,{i,k}} and {A1,A2} =
{{i,j},∅} since for these cases |a −a |=2.
1 2
where
We are now ready to prove the result in Lemma 3. Let,
ℓ′ = ℓi if i∈Rs , r′ = ri if i∈Rcs . (12) without loss of generality, i = 1, j = 2 and k = 3, i.e.,
i,s (cid:26) 0 otherwise i,s (cid:26) 0 otherwise consider A⋆ ⊆ N = {1,3} and A⋆ ⊆ N = {1,2} with
1 1 2 2
|a⋆−a⋆|6=2. With this we have
From (11), ∀A1 ⊆ N1 and ∀A2 ⊆ N2, with N1,N2 being a 1 2
partition of NF, we have that C⋆HD +C⋆HD
{1,2} {1,3}
λ⋆ maxℓ′ +maxℓ′ + max r′ + max r′ = λ⋆ maxℓ′ + max r′ +maxℓ′ + max r′
s∈X[0:1]N s(cid:20)i∈A1 i,s i∈A2 i,s i∈N1\A1 i,s i∈N2\A2 i,s(cid:21) s∈X[0:1]Ns(cid:18)i∈A⋆1 i,s i∈N1\A⋆1 i,s i∈A⋆2 i,s i∈N2\A⋆2 i,s(cid:19)
≥s∈X[0:1]Nλ⋆s(cid:18)im∈AaxFℓ′i,s+im∈AaxcFri′,s(cid:19)≥CHNDF, (≥as)∈X[0:1]Nλ⋆s(cid:18)im∈AaxFℓ′i,s+i∈mNFa\xAFri′,s+im∈AaxSℓ′i,s+i∈mNSa\xASri′,s(cid:19)
for A1∪A2 =AF, whiAchPPimENplDieIXs CBHNDF ≤C⋆NH1D+C⋆NH2D. =s∈X[0:1]Nλ⋆s(cid:18)im∈AaxFℓ′i,s+i∈mNFa\xAFri′,s(cid:19)
PROOF OF LEMMA3
+ λ⋆ maxℓ′ + max r′
We hereprovethe resultin Lemma3. To thisendwe make s∈X[0:1]N s(cid:18)i∈AS i,s i∈NS\AS i,s(cid:19)
use of the following lemma.
LLeetmNm1a=7. C{io,njs}idaerndaFND2 d=iam{oi,nkd}nebtewotwrkoNsuFbwneitthwNork=s o3f. ≥AmF⊆inNFs∈X[0:1]Nλ⋆s(cid:18)im∈AaxFℓ′i,s+i∈NmFa\xAFri′,s(cid:19)
 
problem in (15), which can be equivalently rewritten as
+ min  λ⋆ maxℓ′ + max r′ 
AS⊆NSs∈X[0:1]N s(cid:18)i∈AS i,s i∈NS\AS i,s(cid:19) min t
=CHNDF +C⋆{H1}D,  sa.ntd. ℓℓℓ112+rrr121 ≤≤tt,,
whereℓ′i,s andri′,s aredefinedin(12)andwheretheinequality and ℓℓ12+≥r2c, ℓ2+r1 ≥c, max{r1,r2}≥c, (16)
in (a) follows from Lemma 7. and ℓ1 ≥ℓ2,
and min{r ,ℓ ,r ,t}≥0.
1 2 2
APPENDIX C We now analyze two cases:
PROOF OF THEOREM5 • Case (i): r ≥r ; then, the problem in (16) becomes
1 2
A. The case N =k, k ∈[1:2] min ℓ1r1
and ℓℓ1+≥r1c, ℓ +r ≥c, r ≥c,
We here prove the result in Theorem 5 for the case N = 1 2 1 1 (17)
and ℓ ≥ℓ , r ≥r ,
k, k ∈[1:2]. For this case, we trivially have 1 2 1 2
and min{ℓ ,r }≥0.
2 2
CHk,DN = CHk,Dk = CHk,Dk ≥ C˜Hk,Dk = 1, Since we have ℓ1 ≥ c and r1 ≥ c and the objective function
CFD CFD 2C˜HD 2C˜HD 2 in (17) is increasing in ℓ and r , we get that the optimal
NF NF k,k k,k solution is f ≥ 1c, which1implies1CHD ≥ 1CFD.
1 2 1,2 2 NF
whereC˜HD istheHDcapacityofthek-relaysubnetwork(with • Case (ii): r <r ; then, the problem in (16) becomes
k,k 1 2
k ∈ [1 : 2]) when the relays receive for 1 of the time and
2 min t
transmitfor 1 ofthetime.Wenextprovideanetworkexample
2 s. t. ℓ1r1 ≤t,
for which CHD = 1CFD, with k ∈[1:2]. ℓ1+r1
k,k 2 NF and ℓ2r2 ≤t,
Efoxratmhipslen.eFtwoorrNk w=e hkav=e 1C,FcDon=si1dearnℓd1C=HDr1== ℓ11;r1cle=ar1ly; and ℓℓ12+≥r2c, ℓ2+r1 ≥c, r2 ≥c, (18)
hence CHD = 1CFD. For NN=F k = 2, co1n,1sider ℓℓ1+r=1 r 2= and ℓ1 ≥ℓ2, r1 <r2,
1,ℓ =r1,=1 0; c2leaNrFly for this network we have CFD1=11and and min{r1,ℓ2,t}≥0.
2 2 NF
CHD= ℓ1r1 =1; hence CHD=1CFD. Itisnotdifficulttoseethatifr >c,thentheoptimalsolution
2,2 ℓ1+r1 2 2,2 2 NF wouldbe f > 1c since ℓ ≥c1, which impliesCHD > 1CFD.
B. The case k =1 and N =2 Thus,we n1owfo2cusonth1e case r1 ≤c. Forthis1c,a2se, w2eNcaFn
set ℓ =c−r , without loss of optimality (since ℓ2r2 is an
We here prove the result in Theorem 5 for the case k = 1 2 1 ℓ2+r2
increasing function in ℓ ); with this we obtain
and N = 2. To this end, we let CFD = c, for some given 2
c∈R and we assume, without losNsFof generality, that ℓ ≥ min t
+ 1
ℓ . As such,provingthe lowerboundon CHD is equivalentto s. t. ℓ1r1 ≤t,
so2lve the following problem 1,2 and ℓ(c1−+rr11)r2 ≤t, (19)
and ℓc−r≥1+cr,2r ≥c, r <r , r ≤c
1 2 1 2 1
fs.1t:.=min mc=axmi∈i[n1:2{]ℓn1,ℓℓℓii+2rri+ior1,max{r1,r2}}, (14) Now,intheparnodblemmiinn{(1r91,)tw}e≥ca0n. setℓ =c,withoutlossof
and ℓ ≥ℓ , 1
1 2 optimality (since ℓ1r1 is an increasing function in ℓ ); with
and min{r1,ℓ2,r2}≥0. this we obtain ℓ1+r1 1
Notice that the solution f1 of the problem in (14) is greater min t
than the solution f2 of the following problem as.ntd. c(cc+r−r1r11)≤r2t≤, t, (20)
fs.2t:.=min mℓ1a≥xi∈c,[1:ℓ22]n+ℓℓiri+1rri≥ioc, max{r1,r2}≥c, aanndd rmc−2irn≥1{+crr,12,rt1}<≥r02., r1 ≤c
and ℓ ≥ℓ ,
1 2
and min{r1,ℓ2,r2}≥0. Now,in theproblemin (20) wecansetr2 =c(since ℓℓ22+rr22 is
(15) an increasing functionin r2), without loss of optimality;with
this we obtain
The reason why we have f ≥f is because we are minimiz-
1 2 min t
ing the objective function and in the problem in (15) we are s. t. cr1 ≤t,
tiihnnicsr(e1iam4s)ip.nlgiTehsthutesh,asptifaaclwesoeoffarthe≥easbe1laecr.ctHhoewsnhictohew,rwetsehpaentcotfwt2oft≥ohceu31spcroo,nbtlhtehemne aanndd 0(c2c+−≤cr−r1r1r1)1c<≤ct., (21)
1 3
0.34 D. The case k =2 and N =3
We here prove the result in Theorem 5 for the case k = 2
0.32
and N ≫1. To this end we make use of the following upper
bound
0.3
HDC1,NFDCNF CHD ≤CFD,
0.28 NF NF
and we let C¯HD be the HD capacity of the subnetwork of
jk
0.26 relays j and k with (j,k)∈[1:N]2 and j 6=k.
Without loss of generality, we let CFD =c and we assume
0.24F0 ig.120:0 CH12,0DN0 fo30r0the400net5wN00ork60i0n (670)0wi8t0h0 c9=001.1000 ℓ1 ≥ℓ2 ≥ℓ3. With these assumptionsNwFe need to have
CFNDF c=min{ℓ1,ℓ2+r1,ℓ3+max{r1,r2},
max{r ,r ,r }}. (24)
It is not difficult to see that if r ≥ c, then cr1 ≥ (c−r1)c, 1 2 3
leading to 1 2 c+r1 2c−r1 Since we are interestedin findingthe worstpossible ratio,the
above constraint can be relaxed as
min cr1
and cc+≤r1r <c, (22) ℓ1 ≥c, (25a)
2 1 ℓ +r ≥c, (25b)
2 1
w13ChiFNcDFh.hSaismfil1ar=ly,13icf ra1s ≤opt2cim, tahlesnolcuc+rtir1o1n≤. T(h2c−ics−rr1im1)cp,llieeasdCinH1g,D2to= ℓm3a+xm{ra1x,r{2r,1r,3r}2}≥≥c.c, ((2255dc))
min (c−r1)c Thereasonwhywecanrelaxtheconstraintin(24)asthosein
and 02≤c−rr1 ≤ 1c, (23) (25)isbecausein(25)weareenlargingthespaceofthesearch
1 2
and because we are interested in the worst possible ratio. It is
which has f = 1c as optimal solution. This implies CHD = notdifficultto see that, since the capacityis a non-decreasing
1 3 1,2
1CFD. This concludes the proof that CHD ≥ 1CFD. We next function of the channel parameters, we can set, without loss
p3roNviFde a network example for which C1H1,2,D2 =331CNFNFDF. of generality, ℓ1 =c. With this, we need to have
Example. Consider ℓ1 = 1,ℓ2 = r1 = 12 and r2 = 1; clearly ℓ +r ≥c, (26a)
for this network we have CFD = 1 and CHD = ℓ1r1 = 1; 2 1
hence CHD = 1CFD. NF 1,2 ℓ1+r1 3 ℓ3+max{r1,r2}≥c, (26b)
1,2 3 NF max{r ,r ,r }≥c. (26c)
1 2 3
C. The case k =1 and N ≫1
In what follows we will show by contradiction that at least
We here prove the result in Theorem 5 for the case k = 1 one of the three possible subnetworks of k = 2 relays has a
and N ≫ 1. To this end we make use of the result derived HD capacity greater than or equal to 1CFD.
in [1, Theorem 1] for this case, namely Case (i): r1r2 ≥ cℓ2; with this we o2btaNinF that the capacity
C¯HD of the first pair (i.e., relay 1 and relay 2) is given by [7]
12
1
CF1D ≥ 2CFNDF, C¯HD = cℓ2r1−ℓ22r2+cℓ2r2+cr1r2,
12 (ℓ +r )(c−ℓ +r )
2 2 2 1
where we used the notation CFD (which indicates the FD
capacity of the k = 1 selected r1elay) to highlight that in FD which implies C¯H12D = cℓ2r1−ℓ22r2+cℓ2r2+cr1r2. We now want
the ratio CF1D does not depend on N. It is not difficult to see to showthat, inCoFNrDFderto mce(eℓt2+thre2)(cco−nℓd2i+tiro1n)sin (26), there are
CFD
that CFD =NF2C˜HD, where C˜HD is the HD capacity of the no networks for which
k =11selected re1,lNay when it r1e,cNeives for 21 of the time and it C¯H12D = cℓ2r1−ℓ22r2+cℓ2r2+cr1r2 < 1,
transmits for 1 of the time. Thus, CFD c(ℓ +r )(c−ℓ +r ) 2
2 NF 2 2 2 1
1 CHD 1 which is equivalent to show that there are no networks for
CFD ≤CFD =2C˜HD ≤2CHD =⇒ 1,N ≥ . which
2 NF 1 1,N 1,N CFD 4
NF cℓ r −2ℓ2r +3cℓ r +cr r −ℓ c2+cℓ2−c2r <0,
2 1 2 2 2 2 1 2 2 2 2
We next providea network example for which CH1,DN = 14CFNDF ⇒c(ℓ2+r2)(r1+ℓ2−c)+2ℓ2r2 (c−ℓ2) <0.
for N ≫1.
Example. We consider the network defined in (6) for which ≥0 from (26a) ≥0 since c=ℓ1≥ℓ2
allthesingle-relayHDcapacitiesarethesameandgivenin(7) Thus, there ar|e no c{hzanne}l conditions|for{wzh}ich C¯HD < 1c,
12 2
and CFD = c. With this we obtain the claimed ratio CHD = i.e., by letting relay 1 and relay 2 operate we always achieve
NF 1,N
1CFD for N ≫1 as we also observe from Fig. 2. (to within a constant gap) at least half of the capacity of the
4 NF
corresponding FD network. which implies C¯H13D = cℓ3r3−cr12+cr1r3+ℓ3r1r3. We now want
Case (ii): r1r2 ≤ cℓ2 and r1 ≥ r2; with this we obtain that to showthat, inCoFNrDFderto mce(ec+ttrh1e)(ℓc3o+nrd3i−tiro1n)sin (26), there are
the capacity C¯H12D of the first pair (i.e., relay 1 and relay 2) is no networks for which
given by [7], i.e., C¯HD cℓ r −cr2+cr r +ℓ r r 1
13 = 3 3 1 1 3 3 1 3 < , (28)
cℓ r −ℓ r2+cr r +ℓ r r CFD c(c+r )(ℓ +r −r ) 2
C¯HD = 2 1 2 2 1 2 2 1 2, NF 1 3 3 1
12 (ℓ2+r2)(c+r1−r2) which is equivalent to show that there are no networks for
which implies C¯H12D = cℓ2r1−ℓ2r22+cr1r2+ℓ2r1r2. We now want which
to show that,inCoFNrDFderto mce(eℓ2t+thr2e)(cco+nrd1−itiro2)nsin (26), there are 2cℓ3r3−cr12+cr1r3+2ℓ3r1r3
no networks for which −c2ℓ3−c2r3+c2r1−cr1ℓ3 <0
C¯H12D = cℓ2r1−ℓ2r22+cr1r2+ℓ2r1r2 < 1, ⇒ℓ3(c+r1) (r3−c) +cr3(ℓ3+r1−c)
CFNDF c(ℓ2+r2)(c+r1−r2) 2 ≥0 from (26c) ≥0 from (26b)
+cr (c−|r {)z +}ℓ r r <| 0. {z }
which is equivalent to show that there are no networks for 1 1 3 1 3
which ≥0 since r1<c
Thus, there are no c|ha{nznel}conditions for which C¯HD < 1c,
cℓ r −2ℓ r2+cr r +2ℓ r r −ℓ c2 13 2
2 1 2 2 1 2 2 1 2 2 i.e., by letting relay 1 and relay 3 operate we always achieve
+ℓ r c−r c2+r2c<0
2 2 2 2 (to within a constant gap) at least half of the capacity of the
⇒2ℓ r (r −r ) +cr (r +ℓ −c) corresponding FD network.
2 2 1 2 2 1 2
Case (iii): r r ≤ cℓ and r ≤ r ; with this we obtain that
≥0 since r1≥r2 ≥0 from (26a) the capacity1C¯H2D of t2he first1pair (2i.e., relay 1 and relay 2) is
+c ℓ r| −{zℓ c}+r2 <|0. {z } 12
2 1 2 2 given by [7], i.e.,
(cid:0) (cid:1)
≥0 if r1≥c cℓ r −cr2+cr r +ℓ r r
| {z } C¯HD = 2 2 1 1 2 2 1 2, (29)
Thus,asufficientconditionfortheabovequantitytobealways 12 (c+r )(ℓ +r −r )
1 2 2 1
positive is r ≥ c, i.e., under these channel conditions we
never have C¯1H12D < 12c. We now analyze the case r1 < c, which implies CC¯FNH12DDF = cℓ2cr(2c−+crr112)+(ℓc2r+1rr22+−ℓr21r)1r2. We now want
which,becauseoftheconstraintin(26c),impliesr ≥r ≥r . to showthat, in orderto meetthe conditionsin (26), there are
3 1 2
Case (ii-a): cℓ ≤r r ; with this we obtain that the capacity no networks for which
3 1 3
C¯HD of the pair relay 1 and relay 3 is given by [7], i.e., C¯HD cℓ r −cr2+cr r +ℓ r r 1
13 12 = 2 2 1 1 2 2 1 2 < ,
C¯HD = cℓ3r1−ℓ23r3+cℓ3r3+cr1r3, CFNDF c(c+r1)(ℓ2+r2−r1) 2
13 (ℓ +r )(c−ℓ +r ) which is equivalent to show that there are no networks for
3 3 3 1
which
which implies C¯H13D = cℓ3r1−ℓ23r3+cℓ3r3+cr1r3. We now want
to show that,inCoFNrDFderto mce(eℓt3+thre3)(cco−nℓd3i+tiro1n)sin (26), there are 2cℓ2r2−cr12+cr1r2+2ℓ2r1r2
−c2ℓ −c2r +c2r −cr ℓ <0
no networks for which 2 2 1 1 2
⇒c (r −r ) (ℓ +r −c)+ℓ r c+2r r −c2 <0.
C¯HD cℓ r −ℓ2r +cℓ r +cr r 1 2 1 2 1 2 2 1 2
13 = 3 1 3 3 3 3 1 3 < , (cid:0) (cid:1)
CFNDF c(ℓ3+r3)(c−ℓ3+r1) 2 ≥0|sinc{ezr2≥}r1|≥0 fro{mz (26a)} | ≥0 if r{2z≥c+c22r1 }
which is equivalent to show that there are no networks for Thus,asufficientconditionfortheabovequantitytobealways
which wpoesintievveerishra2ve≥C¯Hc+Dc22r<1, 1i.ce..,Tuhnudse,rwtheenseowchaannanleylzecotnhdeitcioanses
cℓ3r1−2ℓ23r3+3cℓ3r3+cr1r3−ℓ3c2+cℓ23−c2r3<0 r < c2 , which12,becau2seofthe constraintin (26c), implies
⇒c(ℓ3+r3)(r1+ℓ3−c)+2ℓ3r3 (c−ℓ3) <0. r23 ≥rc2+2≥r1r1.
Thus, there are|≥n0ofcro{hmzan(2n6eb)l}condition≥s0fs|oinrce{wzc=hiℓc}1h≥ℓC¯3HD < 1c, CC¯H2a3sDeo(ifiit-hae):pℓa2iℓr3re≤lary2r23a;nwditrheltahyis3wiesogbivtaeinntbhyat[7th],eic.ea.p,acity
13 2 ℓ ℓ r −ℓ2r +ℓ ℓ r +ℓ r r
i.e., by letting relay 1 and relay 3 operate we always achieve C¯HD = 2 3 2 3 3 2 3 3 2 2 3, (30)
(to within a constant gap) at least half of the capacity of the 23 (ℓ3+r3)(ℓ2−ℓ3+r2)
corresponding FD network. whichimplies C¯H23D = ℓ2ℓ3r2−ℓ23r3+ℓ2ℓ3r3+ℓ2r2r3.Wenowwant
CC¯HaDseo(fii-tbh)e:pcaℓi3r≥relra1yr31;awnidthrethlaiys w3eisogbitvaeinntbhyat[t7h]e, ic.ea.p,acity to showthat, inCFNoDFrderto mc(eℓe3t+trh3e)(cℓ2o−nℓd3i+tiro2n)sin (26), there are
13 no networks for which
C¯H13D = cℓ3r(3c−+crr11)2(+ℓ3c+r1rr33+−ℓr31r)1r3, (27) CC¯H2FN3DDF = ℓ2ℓ3cr(2ℓ−3+ℓ23rr33)+(ℓ2ℓ2−ℓ3ℓr33++rℓ22)r2r3 < 12,
which is equivalent to show that there are no networks for from≥(26b) 3c(c−ℓ3)−c2 +ℓ −c=ℓ3(c−2ℓ3) c≥ℓ>3>2c 0.
which 2(c−ℓ3) 3 2(c−ℓ3)
2ℓ2ℓ3r2−2ℓ23r3+2ℓ2ℓ3r3+2ℓ2r2r3−cℓ2ℓ3+cℓ23 Thus, there are no channel conditions for which C¯H23D < 12c,
−cℓ r −cr ℓ +cr ℓ −cr r <0. (31) i.e., by letting relay 2 and relay 3 operate we always achieve
3 2 3 2 3 3 2 3 (to within a constant gap) at least half of the capacity of the
It is notdifficultto see that the Left-HandSide (LHS) of (31) corresponding FD network.
is always increasing in ℓ ; the derivative of the LHS of (31) Case(iii-b):ℓ ℓ ≥r r ;withthisweobtainthatthecapacity
2 2 3 2 3
with respect to ℓ is, in fact, given by C¯HD of the pair relay 2 and relay 3 is given by [7], i.e.,
2 23
2ℓ r +2ℓ r +2r r −cℓ −cr ℓ ℓ r −ℓ r2+ℓ r r +ℓ r r
3 2 3 3 2 3 3 3 C¯HD = 2 3 3 2 2 2 2 3 3 2 3, (32)
=2ℓ3r2+r2r3+ℓ3 (r3−c) +r3(ℓ3+r2−c)≥0. 23 (ℓ2+r2)(ℓ3+r3−r2)
≥0 from (26c) ≥0 from (26b) whichimplies C¯H23D = ℓ2ℓ3r3−ℓ2r22+ℓ2r2r3+ℓ3r2r3.Wenowwant
Hence, from (31) we obta|in {z } | {z } CFNDF c(ℓ2+r2)(ℓ3+r3−r2)
to showthat, in orderto meetthe conditionsin (26), there are
2ℓ ℓ r −2ℓ2r +2ℓ ℓ r +2ℓ r r −cℓ ℓ +cℓ2 no networks for which
2 3 2 3 3 2 3 3 2 2 3 2 3 3
−cℓ3r2−cr3ℓ2+cr3ℓ3−cr2r3 C¯H23D = ℓ2ℓ3r3−ℓ2r22+ℓ2r2r3+ℓ3r2r3 < 1,
ℓ2≥≥ℓ3 2ℓ2r +2ℓ r r −cℓ r −cr r CFNDF c(ℓ2+r2)(ℓ3+r3−r2) 2
3 2 3 2 3 3 2 2 3
=r (r +ℓ )(2ℓ −c) . which is equivalent to show that there are no networks for
2 3 3 3
which
≥0 if ℓ3≥2c
| {z } 2ℓ ℓ r −2ℓ r2+2ℓ r r +2ℓ r r −cℓ ℓ −cℓ r
Thus,asufficientconditionfortheabovequantitytobealways 2 3 3 2 2 2 2 3 3 2 3 2 3 2 3
npeovsietrivheaviseC¯ℓ3HD≥<2c,1ci..eT.,huusndweer nthoewseancahlaynzneetlhecocnadseitiℓons<wce. +cℓ2r2−cr2ℓ3−cr2r3+cr22 <0. (33)
23 2 3 2 It is not difficult to see that the LHS of (33) is always
We startbynoticingthattheLHSof(31) isalwaysincreasing
increasinginr ;thederivativeoftheLHSof(33)withrespect
in r ; the derivative of the LHS of (31) with respect to r is 3
2 2 to r is, in fact, given by
in fact given by 3
2ℓ ℓ +2ℓ r −cℓ −cr =(ℓ +r )(2ℓ −c)≥0. 2ℓ2ℓ3+2ℓ2r2+2ℓ3r2−cℓ2−cr2
2 3 2 3 3 3 3 3 2
=ℓ (ℓ +r −c)+ ℓ ℓ −cr
Notice that the fact that ℓ ≥ c follows since: (i) from (26b) 2 3 2 2 3 2
ℓ < c impliesr > c and2(ii)2sincer ≤r < c2 implies ≥0 from (26b) ≥0 since ℓ2ℓ3≥r2r3 and r3≥c
r3 <2c from (262a) w2e must have ℓ >1 c.2Hencc+e,2rf1rom (31) +ℓ2r|2+2{ℓz3r2 ≥}0. | {z }
1 2 2 2
we obtain
Hence, from (33) we obtain
2ℓ2ℓ3r2−2ℓ23r3+2ℓ2ℓ3r3+2ℓ2r2r3−cℓ2ℓ3+cℓ23 2ℓ ℓ r −2ℓ r2+2ℓ r r +2ℓ r r −cℓ ℓ −cℓ r
−cℓ r −cr ℓ +cr ℓ −cr r 2 3 3 2 2 2 2 3 3 2 3 2 3 2 3
3 2 3 2 3 3 2 3 +cℓ r −cr ℓ −cr r +cr2
r2≥≥c−ℓ3 ℓ2ℓ3c−2ℓ2ℓ23−2ℓ23r3+ℓ2r3c+2cℓ23 r3≥≥cℓ 2ℓ 2c−2ℓ2r32+3ℓ2r3c+ℓ2r c−c2ℓ −c2r +cr2
−c2ℓ3+2cr3ℓ3−c2r3 =c(ℓ2 +3 r )(ℓ2 +2 r −2c)2+2ℓ 3r 2 (c−r2) ≥20. 2
2 2 3 2 2 2 2
=ℓ (c−2ℓ ) (r −c) +ℓ ℓ (c−2ℓ )
3 3 3 2 3 3
≥0 from (26b) ≥0 since r2≤c
+c>r0|(siℓn{cez+ℓ3ℓ<}2c−≥0c|)si.n{cze r3}≥c >0|sin{cezℓ3<}2c Thus, there are|no ch{aznnel }conditions |for{wzhi}ch C¯H23D < 21c,
3 2 3 i.e., by letting relay 2 and relay 3 operate we always achieve
We next show that the term (ℓ +ℓ −c) is also always (to within a constant gap) at least half of the capacity of the
2 3
positive. We recall that we are considering the regime r c+ corresponding FD network. This concludes the proof of the
2
2r r − c2 < 0 since otherwise, as we already proved, lower bound in (9). We next provide a network example for
1 2
C¯HD ≥ 1c. We have which CHD=1CFD.
12 2 2,3 2 NF
Example. Consider ℓ =r =1,ℓ =r =0; clearly for
r2c+2r1r23r−cc2−<c20from=⇒(26a)3r2c−2r2ℓ2−c2 <0 tChHisDn=et1wCoFrkD.we have1CFNDF1=1 an[d2:3C]H2,D3 [=2:3ℓ]ℓ11+rr11 = 12; hence
=⇒ℓ > 2 . 2,3 2 NF
2 2r
2
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