SOLUTIONS MANUAL for Optical Networks 1st Edition by Debasish Datta

Solution Manual (March 2021)

Optical Networks Debasish Datta

Oxford University Press

Preface

This solution manual has been prepared chapter-wise starting from Chapter 2, after submission of the book manuscript. Special attention has been given to the holistic design issues and performance features in all segments of optical networks, calling for the crosslayer aspects therein, and hence in most of the chapters some of the exercise problems need adequate knowledge in the technologies (Chapter 2) and transmission impairments in optical networks (Chapter 10). The instructors may please address this issue and encourage the students to appreciate this aspect for developing a practical perspective of optical networks.

Moreover, while solving the problems after submission of the book manuscript, some changes on the numerical data have been incorporated in the exercises of the book to make them more worthwhile, along with a correction in an expression in Chapter 6. All these modifications are included in the errata presented in the following page.

The instructors are most welcome to contact me by email (ddatta@ece.iitkgp.ac.in), if they need any clarifications, and they may please write to me if they come across any error in the book.

Debasish Datta

i

Errata

Context

Present version in the book

Corrected/changed version

5.27 nm

527 nm

Gd = L/[2(M – 1) + L]

Gd = L/[2(M – 1 + L)]

Page 572, Exercise 14.6.

Γ = 0 24 40 50 24 0 24 40 24 24 0 0 50 0 40 0.

Γ = 0 50 25 60 25 0 50 60 25 30 0 30 25 50 30 0.

Page 593, Exercise 15.7

0.1 µs

0.8 µs

Page 130, Exercise 2.7 Page 248, expression for Gd below Eq. 6.5.

ii

Exercise Problems and Solutions for Chapter 2 (Technologies for Optical Networks) 2.1 A step-index multi-mode optical fiber has a refractive-index difference Δ = 1% and a core refractive index of 1.5. If the core radius is 25 µm, find out the approximate number of propagating modes in the fiber, while operating with a wavelength of 1300 nm.

Solution: Δ = 0.01, n1 = 1.5, a = 25 μm, w = 1300 nm, and the number of modes Nmode is given by 𝑉𝑉 2

𝑁𝑁𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 2 , with 𝑉𝑉 =

The numerical aperture NA is obtained as

2𝜋𝜋𝜋𝜋 𝑁𝑁𝑁𝑁. 𝑤𝑤

𝑁𝑁𝑁𝑁 = �𝑛𝑛12 − 𝑛𝑛22 ≈ 𝑛𝑛1 √2∆ = 1.5√0.02.

Hence, we obtain V parameter as, 2𝜋𝜋 × 25 × 10−6 × �1.5√0.02� = 25.632, 1300 × 10−9 leading to the number of modes Nmode , given by 𝑉𝑉 =

𝑁𝑁𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 =

25.6322 ≈ 329.

2

2.2 A step-index multi-mode optical fiber has a cladding with the refractive index of 1.45. If it has a limiting intermodal dispersion of 35 ns/km, find its acceptance angle. Also calculate the maximum possible data transmission rate, that the fiber would support over a distance of 5 km.

Solution: The cladding refractive index n2 =1.45, and the intermodal dispersion Dmod = 35 ns/km. Dmod is

expressed as 𝐷𝐷𝑚𝑚𝑚𝑚𝑚𝑚 ≈

𝑛𝑛1 Δ 𝑛𝑛1 𝑛𝑛1 − 𝑛𝑛2 𝑛𝑛1 − 𝑛𝑛2 = � �= = 35 ns/km. 𝑐𝑐 𝑛𝑛1 𝑐𝑐 𝑐𝑐

Hence, (n1 – n2) = cDmod = (3 × 105) × (35 × 10-9) = 0.0105, and n1 = n2 + 0.0105 = 1.4605. Therefore, we obtain NA as 𝑁𝑁𝑁𝑁 = �𝑛𝑛12 − 𝑛𝑛22 = �1.46052 − 1.452 = 0.174815,

and the acceptance angle is obtained as θA = sin-1(NA) = sin-1(0.174815) = 10.068o.

The pulse spreading due to dispersion should remain ≤ 0.5/r, with r as the data-transmission rate, implying that r ≤ 0.5/(Dmod L). Hence, we obtain the maximum possible data transmission rate rmax over L = 5 km as 𝑟𝑟𝑚𝑚𝑚𝑚𝑚𝑚 =

0.5 = 2.86 Mbps.

35 × 10−9 × 5

2.3 Consider that a step-index multi-mode optical fiber receives optical power from a Lambertian source with the emitted intensity pattern given by I(θ) = I0 cosθ, where θ is the angle subtended by an incident light ray from the source with the fiber axis. The total power emitted by the source is 1 mW while the power coupled into the fiber is found to be - 4 dBm. Derive the relation between the

2.1

launched power and the numerical aperture of the optical fiber. If the refractive index of the core is 1.48, determine the refractive index of the cladding.

Solution: Transmit power PT = 1 mW, and the power coupled into fiber PC = - 4 dBm = 10- 0.4 W = 0.3981 mW. For a Lambertian source, the coupled power PC = NA2 × PT X (for derivation, see Cherin 1983). Hence, 𝑃𝑃 NA2 = PC/PT = 0.3981. Further, 𝑁𝑁𝑁𝑁2 = 𝑛𝑛12 − 𝑛𝑛22 = 𝑃𝑃 𝐶𝐶 implying that n22 = n12 – PC/PT . 𝑇𝑇

Thus, we obtain n2 as

𝑛𝑛2 = √1.482 − 0.3981 = 1.34.

2.4 Consider a 20 km single-mode optical fiber with a loss of 0.5 dB/km at 1330 nm and 0.2 dB/km at 1550 nm. Presuming that the optical fiber is fed with an optical power that is large enough to force the fiber towards exhibiting nonlinear effects, determine the effective lengths of the fiber in the two operating conditions. Comment on the results.

Solution: With L = 20 km, first we consider the case with fiber loss αdB = 0.5 dB/km. So, the loss α in neper/km is determined from αdB = 10log10[exp(α)] as

α = ln (10αdB/10) = ln(100.05) = 0.1151.

Hence, we obtain the effective fiber length as Lef f = [1 – exp(-αL)]/α = [(1 – exp(-0.1151 × 20)]/0.1151 = 7.82 km.

With αdB = 0.2 dB/km, we similarly obtain Lef f = 13.06 km, which is expected because with lower attenuation, the power decays slowly along the fiber and thus the fiber nonlinearity effects can take place over longer fiber length.

2.5 Consider an optical communication link operating at 1550 nm over a 60 km optical fiber having a loss of 0.2 dB/km. Determine the threshold power for the onset of SBS in the fiber. Given: SBS gain coefficient gB = 5 ×10-11 m/W, effective area of cross-section of the fiber Aeff = 50 µm2, SBS bandwidth = 20 MHz, laser spectral width = 200 MHz. Solution: With αdB = 0.2 dB/km at 1550 nm, we obtain α = ln (10αdB/10) = ln(100.02) = 0.0461.

Hence, for L = 60 km, we obtain Lef f as Lef f = [1 – exp(-αL)]/α = [1 – exp(-0.0461 × 60)]/0.0461 = 20.327 km.

With Aeff = 50 μm , gB = 5 × 10 m/W, and assuming the polarization-matching factor to be ηp = 2, we obtain the SBS threshold power as 2

𝑃𝑃𝑡𝑡ℎ (𝑆𝑆𝑆𝑆𝑆𝑆) =

-11

21 𝜂𝜂𝑝𝑝 𝐴𝐴𝑒𝑒𝑒𝑒𝑒𝑒 𝛿𝛿𝛿𝛿 21 × 2 × 50 × 10−12 200 �1 + �= �1 + � = 22.73 mW. −11 3 𝑔𝑔𝐵𝐵 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 5 × 10 × 20.327 × 10 Δ𝜔𝜔𝐵𝐵 20

2.6 Consider an optical communication link operating at 1550 nm over a 60 km optical fiber having a loss of 0.2 dB/km. The effective area of cross-section of the fiber Aeff = 50 µm2, where an optical power of 0 dBm is launched. Determine the nonlinear phase shift introduced by SPM in the fiber.

Given: ñ(ω) = 2.6 × 10-20 m2/W.

2.2

Solution: The transmit power PT = 1 mW and the effective area of cross-section of the fiber is Aeff = 50 μm2 Hence, the intensity of launched light I = PT/Aeff = 10-3/(50 × 10-12) = 2 × 107 W/m2. With αdB = 0.2 dB/km., α = 0.0461 and Leff = 20.327 km (see Exercise 2.5). Further, the nonlinearity parameter 𝜉𝜉of the fiber is given by 𝜉𝜉 =

𝑛𝑛� (𝜔𝜔)𝜔𝜔 𝑐𝑐

=

2𝜋𝜋 �𝑛𝑛 (𝜔𝜔) 𝑤𝑤

with ñ(ω) = 2.6 × 10-20 m2/W and w = 1550 nm.

Therefore, the phase shift Δ𝜙𝜙𝑆𝑆𝑆𝑆𝑆𝑆 introduced by SPM in the optical fiber is obtained as Δ𝜙𝜙𝑆𝑆𝑆𝑆𝑆𝑆 = 𝐿𝐿𝑒𝑒𝑒𝑒𝑒𝑒 𝜉𝜉 𝐼𝐼 = (20.327 × 103 ) ×

2𝜋𝜋×2.6×10 −20 1.55×10 −6

× (2 × 107 ) = 0.04285 radian = 2.455o

2.7 Determine the Bragg wavelength of an integrated-optic grating with a period of 527 nm (earlier it was 5.27 nm in the book) and an effective refractive index of 1.47. Sketch a block schematic using Bragg grating and other relevant components for an optical add-drop multiplexer for three wavelengths at a given node in a WDM ring network.

Solution: The period of grating is Λ = 527 nm, and the effective refractive index is neff = 1.47. Hence the Bragg wavelength wb of the integrated-optic grating is obtained as 𝑤𝑤𝑏𝑏 = 2𝑛𝑛𝑒𝑒𝑒𝑒𝑒𝑒 Λ = 2 × 1.47 × 0.527 = 1549.38 nm.

Using Bragg grating one can realize an add-drop multiplexer (OADM) for three wavelengths w1, w2,

w3 (w2 as the drop wavelength) as shown in the following.

Circulator Bragg grating

w1, w2, w3 w2

Coupler

w1, w2, w3

w2

OADM using Bragg grating

2.8 A laser cavity (i.e., an active layer in DH configuration) has a length L with a medium loss α dB per unit length, which is pumped with a gain g per unit length using an appropriate forward bias current. The walls on two ends of the cavity are designed with reflectivities r1 and r2, and the cavity has a confinement factor of Γ. Determine the condition to be satisfied by the pumped cavity to function as a laser. Give a typical sketch of g as a function of wavelength, and explain its impact on the laser spectrum.

Solution: The DH laser cavity operates as shown in the following diagram with the two reflecting walls (shaded parts).

r1

Cavity (α, g, Γ)

r2

L Total loss-cum-gain ρLG in the cavity must be ≥ 1, with ρLG given by 𝜌𝜌𝐿𝐿𝐿𝐿 = 𝑟𝑟1 𝑟𝑟2 𝑒𝑒𝑒𝑒𝑒𝑒{−(𝛼𝛼 − 𝑔𝑔Γ)𝐿𝐿)}.

2.3

Hence, the condition that the cavity must satisfy to function as a laser is expressed as 𝑟𝑟1 𝑟𝑟2 𝑒𝑒𝑒𝑒𝑒𝑒{−(𝛼𝛼 − 𝑔𝑔Γ)𝐿𝐿)} ≥ 1,

leading to the expression for the minimum cavity gain, given by 𝑔𝑔𝑚𝑚𝑚𝑚𝑚𝑚 =

𝛼𝛼 ln (𝑟𝑟 𝑟𝑟 ) − 𝐿𝐿Γ1 2 . Γ

g (w)

w

g(w) shapes the spectrum and thus reduces the number of modes in a laser.

2.9 What is the fundamental difference between the spectral spreads in lasers due to the phase noise and modulation? Determine the spectral spread of a 1550 nm laser transmitting at 10 Gbps with the unmodulated linewidths of (i) 200 MHz and (ii) 0.08 nm.

Solution: The spectral spread due to phase noise is an internal phenomenon in a laser, arising from the underlying spontaneous noise and finite bandwidth of the laser cavity. On the other hand, the spectral spread due to the modulation is an external effect from the phase, frequency, amplitude, intensity modulation (governed by the modulation scheme) by the input data stream and hence is related to the bit rate of the data stream. (i) Laser linewidth BL = 200 MHz, modulation bandwidth BM = 10 GHz (considering intensity modulation). Hence the total spectral spread BT of the laser after modulation (3-dB bandwidth) is obtained as 𝐵𝐵𝑇𝑇 = �𝐵𝐵𝐿𝐿2 + 𝐵𝐵𝑀𝑀2 = �(200 × 106 )2 + (10 × 109 )2 = √100.04 × 1018 ≈ 10 GHz.

(ii) Laser linewidth is given in terms of wavelength spread δw = 0.08 nm, which leads to a frequency spread BL, given by 𝐵𝐵𝐿𝐿 =

𝑐𝑐𝑐𝑐𝑐𝑐 3×10 8 ×0.08×10 −9 = (1.55×10 −6 )2 = 9.99 GHz ≈ 10 GHz. 2 𝑤𝑤

Using this value for BL, we obtain the total laser spectral spread as

𝐵𝐵𝑇𝑇 = �𝐵𝐵𝐿𝐿2 + 𝐵𝐵𝑀𝑀2 = �(10 × 109 )2 + (10 × 109 )2 = 14.14 GHz.

2.10 The threshold current density of a stripe-geometry AlGaAs laser is 3000 A/cm2 at 15oC. Estimate the required threshold current at 50oC, when the laser characteristic temperature T0 = 170oK and the contact stripe of the laser has a size (area) of 20 μm × 100 μm.

Solution: The threshold current Ith (T) in laser has a strong dependence on temperature, given by

Let T1 = 50oC and T2 = 15oC.

𝑇𝑇 𝑇𝑇0

𝐼𝐼𝑡𝑡ℎ (𝑇𝑇) ∝ exp � �.

At T = T2, the current density Jth (T2) = 3000 A/cm2 = 3 × 107 A/m2. Using the above relation between Ith with T (and hence between Jth with T), we obtain 𝐼𝐼𝑡𝑡ℎ (𝑇𝑇1 ) 𝐽𝐽 (𝑇𝑇 ) exp (𝑇𝑇 /𝑇𝑇 ) 𝑇𝑇 −𝑇𝑇 = 𝑡𝑡ℎ (𝑇𝑇1 ) = exp (𝑇𝑇1 /𝑇𝑇0 ) = exp � 1𝑇𝑇 2 � 𝐼𝐼𝑡𝑡ℎ (𝑇𝑇2 ) 𝐽𝐽 𝑡𝑡ℎ 2 2 0 0

2.4

Therefore, we obtain Jth (T1) as 𝑇𝑇 −𝑇𝑇

50−15 � A/m2 .

170

𝐽𝐽𝑡𝑡ℎ (𝑇𝑇1 ) = 𝐽𝐽𝑡𝑡ℎ (𝑇𝑇2 )exp � 1𝑇𝑇 2 � = 3 × 107 × exp � 0

Using the above expression, we therefore obtain Ith (T1) as the product of Jth (T1) and the stripe area Ast (= 20 μm × 100 μm), given by 50−15 � × (20 × 100 × 10−12 ) = 73.72 A. 170

𝐼𝐼𝑡𝑡ℎ (𝑇𝑇1 ) = 𝐽𝐽𝑡𝑡ℎ (𝑇𝑇1 )𝐴𝐴𝑠𝑠𝑠𝑠 = 3 × 107 × exp �

2.11 Consider that a binary optical signal is incident from a fiber onto a photodetector. Presuming that the incident light has a duality in its nature (particle and wave), give examples for the manifestation of both the forms of light on the performance of a digital optical receiver.

Solution: One of the manifestations of the wave nature of light takes place in the form of fiber dispersion mechanisms leading to the pulse spreading at the receiving end. On the other hand, the particle nature of light is reflected from the fact that, the photocurrent (number of electrons per unit time) is proportional to the number of photons (light particles) arriving per unit time at the receiver.

2.12 An APD operates at a wavelength of 900 nm with 95% quantum efficiency. Consider that an incident light with a power of -30 dBm has produced a photocurrent of 15 μA at the APD output.

Determine the mean avalanche gain of the APD. Solution: The APD operates at 900 nm with a quantum efficiency of η = 0.95. The incident power Pin = -30 dBm = 10-6 W, which has generated a photocurrent IAPD = 15 μA. Hence, the primary photocurrent Ip in the APD is obtained as 𝐼𝐼𝑝𝑝 = 𝑅𝑅𝑤𝑤 𝑃𝑃𝑖𝑖𝑖𝑖 =

𝜂𝜂𝜂𝜂 𝜂𝜂𝜂𝜂𝜂𝜂 0.95×1.6×10 −19 ×900×10 −9 𝑃𝑃𝑖𝑖𝑖𝑖 = 𝑃𝑃𝑖𝑖𝑖𝑖 = × 10−6 = 0.6878 μA. 6.63×10 −34 ×3×10 8 ℎ𝑓𝑓 ℎ𝑐𝑐

Using the above result, we obtain the APD mean gain MAPD as 𝑀𝑀𝐴𝐴𝐴𝐴𝐴𝐴 =

𝐼𝐼𝐴𝐴𝐴𝐴𝐴𝐴 15 = = 21.8. 𝐼𝐼𝑝𝑝 0.6878

2.13 If the received optical power in a PIN-based optical receiver is -20 dBm and the preamplifier needs a voltage swing of 1 mV at its output, calculate the value of the feedback resistance needed for the preamplifier. Given: optical transmitter uses a laser with perfect extinction, PIN diode responsivity = 0.8 A/W. Solution: The received optical power PR = -20 dBm = 10-5 W and the PIN responsivity = 0.8 A/W. Hence, the photocurrent Ip = 0.8 × 10-5 A. The preamplifier output voltage swing needs to be Vs = 10-3 V which is the product of the input photocurrent Ip and the transimpedance RTI, implying Vs = Ip RTI. Hence we obtain the transimpedance as 𝑅𝑅𝑇𝑇𝑇𝑇 =

𝑉𝑉𝑠𝑠 10−3 = = 125 ohm. 𝐼𝐼𝑝𝑝 0.8 × 10−5

2.14 Draw a block schematic for a strictly nonblocking 8 × 8 optical switch, employing 2 × 2 electrooptic switching elements as the building block in Spanke's architecture. Estimate the total insertion loss of a similar N × N (with N = 2k) optical switch in terms of its number of stages and the losses incurred in all the passive devices during the traversal of a lightpath from an input port to an output port.

Solution: 2.5

Considering the Spanke’s switch architecture in Fig. 2.58, we note that the first (i.e., input) column will have N (1:N) electro-optic switches, with k = log2N. Similarly, the second (output) column will have N (N:1) electro-optic switches. Using this observation, the readers are instructed to draw the full switch configuration.

Hence, the insertion loss Lsw incurred by a lightpath in the above switch will be the sum of the losses in the two columns of the switches along with the connector losses Lc the input and output sides of each 1:N and N:1 switch. Thus, we express Lsw as 𝐿𝐿𝑠𝑠𝑠𝑠 = 2𝑘𝑘𝐿𝐿𝑒𝑒𝑒𝑒 + 4𝐿𝐿𝑐𝑐 = 2(log 2 𝑁𝑁 × 𝐿𝐿𝑒𝑒𝑒𝑒 + 2𝐿𝐿𝑐𝑐 ),

with Leo as the insertion loss in each (2 × 2) electro-optic switch, used as (1 × 2) and (2 × 1) switches at the input and output columns, respectively.

2.15 Using the results on gain saturation in EDFA (Eq. 2.119), estimate the decrease in EDFA gain when its input power increases from -20 dBm to -10 dBm. Given: Q = 10 dBm. Discuss how the gain saturation in EDFAs can affect the performance of an optical link in a WDM network and suggest some possible remedy.

Solution: Using Eq. 2.119, EDFA gains with the input power levels of -20 dBm and -10 dBm are estimated as 27.5 dB and 22.4 dB respectively, implying that the gain falls by 27.5 – 22.4 = 5.1 dB with the 10 dB increase in the input power level, thereby affecting the optical SNR at the destination receivers.

Gain control: By using a separate wavelength in each fiber link and controlling its own power, one can bring down the variation in the total optical power (from all wavelengths) handled by each EDFA along the fiber links across the network.

2.6

Exercise Problems and Solutions for Chapter 3 (Optical LANs/MANs and SANs) 3.1 Discuss with suitable illustrations why the fiber-optic bus or ring topology cannot use traditional one-packet-at-a-time MAC protocols, such as CSMA/CD or token-ring protocol, as used in IEEE standards for copper-based LANs. Also indicate the difficulty in using CSMA/CD protocol in passive-star-based optical LANs. Solution: In optical LANs, the packet sizes shrink with high speed while the network size increases. This leaves a large space (and hence bandwidth) in the optical fiber as unused resource. Further, in bus, the progressive power losses of each packet along the optical fiber, makes collision detection difficult when the packets overlap with dissimilar power levels. Passive-star-based optical LANs don’t suffer from the progressive loss, but the ratio of the packet duration to the round-trip propagation delay becomes much smaller with large network size.

3.2 Obtain an expression for the progressive power loss in an optical fiber bus connecting N nodes to set up a LAN, with each node tapping power from and transmitting (coupling) power into the fiber bus using a transmit-receive coupler (Fig.3.2). Using this formulation, evaluate the dynamic range needed in the receivers used for the network nodes connected to the fiber bus with N = 5 and 10. If the receiver dynamic range cannot be more than 10 dB, comment on the feasibility of the LANs in the two cases. Given: connector loss = 1 dB, power tapping for the receiver in the transmit-receive coupler = 10%, insertion loss in the waveguide of each transmit-receive coupler = 0.3 dB, loss at the transmitter-coupling point in each transmit-receive coupler = 0.2 dB, fiber loss = 0.2 dB/km, distance between two adjacent nodes = 1 km.

Solution: RX

TX Each node employs a passive transmit-receive coupler, as shown above, with two taps for transmit and receive operations, and four connectors on input/output and transmit/receive ports. The necessary parameters associated with the nodes and the network are defined in the following for the analysis.

• • • • • • • • •

Ptx: transmitted power from each node in mW (including the loss of isolator/circulator). d: distance between two successive nodes in km. All nodes are assumed to be equispaced. lf : loss in a fiber segment (d km) in neper/km between each adjacent node pair with, lf = exp(αf d) and lf (dB) = 10 log10(lf ). lc: connector loss (ratio). lcin: insertion loss (loss) inside the coupler. lrtap: tapping ratio for the receiving port. ltxc: loss (ratio) at the transmitting port. lrxc: loss (ratio) at the receiving port due to power tapping (= 1 – lrtap). ltcp: loss (ratio) while coupling the optical signal from the transmitter into the transmit-receive coupler

The power at node 1 output is given by 𝑃𝑃1 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 ,

and the power received at node 2 from node 1 is given by 3.1

𝑃𝑃𝑅𝑅12 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 × 𝑙𝑙𝑓𝑓 𝑙𝑙𝑐𝑐 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑐𝑐 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐4 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑓𝑓 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 .

Following the same approach, we obtain the power received at node N as the power received at the input of node N multiplied by 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 . Moreover, the power loss at each transmit-receive coupler is given by 𝑙𝑙 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝑙𝑙𝑐𝑐2 (𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑖𝑖𝑛𝑛 ), and the optical signal from node 1 to node N has to pass through N-2 transmit-receive couplers with this loss. Therefore, we obtain the power received at the receiver of node N as 𝑁𝑁−1 2 𝑃𝑃𝑅𝑅1𝑁𝑁 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙 𝑁𝑁−2 𝑙𝑙𝑐𝑐 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 . 𝑇𝑇𝑇𝑇𝑇𝑇 𝑙𝑙𝑓𝑓

Substituting for lTRC in the above expression, we obtain

𝑃𝑃𝑅𝑅1𝑁𝑁 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 {𝑙𝑙𝑐𝑐2 (𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 )}𝑁𝑁−2 𝑙𝑙𝑓𝑓𝑁𝑁−1 𝑙𝑙𝑐𝑐2 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

𝑃𝑃𝑅𝑅1𝑁𝑁 = 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2𝑁𝑁 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑓𝑓𝑁𝑁−1 (𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 )𝑁𝑁−2 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 .

or,

From the above results, the dynamic range DR of the bus can be expressed as 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐4 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑓𝑓 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑃𝑃𝑅𝑅12 � 𝐷𝐷𝐷𝐷 = 10log10 � 1𝑁𝑁 � = 10log10 � 𝑃𝑃𝑅𝑅 𝑃𝑃𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐2𝑁𝑁 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑓𝑓𝑁𝑁−1 (𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 )𝑁𝑁−2 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

= 10log10 �

1

𝑁𝑁−2 � = �𝑙𝑙𝑐𝑐2 𝑙𝑙𝑓𝑓 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 �

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 (𝑁𝑁 − 2)�2𝑙𝑙𝑐𝑐𝑑𝑑𝑑𝑑 + 𝑙𝑙𝑓𝑓𝑑𝑑𝑑𝑑 + 𝑙𝑙𝑟𝑟𝑟𝑟𝑟𝑟 + 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 �.

For N = 10, d = 1 km, αfdB = 0.2 dB/km, lcdB = 1 dB, ltxcdB= 0.2 dB, lcindB = 0.3 dB, and lrxc = 1 – lrtap = 1 – 0.1 = 0.9 with lrxcdB = - log10(0.9) = 0.46 dB, we obtain DR as 𝐷𝐷𝐷𝐷 = 8[2 × 1 + 0.2 + 0.46 + 0.2 + 0.3] = 8 × 3.16 = 25.28 dB,

which is not acceptable for the given optical receiver. However with N = 5, DR comes down to 3 × 3.16 = 9.48 dB (<10 dB), making the design realizable.

3.3 Sketch a block schematic for a possible realization of an 8×8 transmissive passive-star coupler using 2×2 3-dB power splitters as the building block. Assume that, each 2×2 building block has an insertion loss of 1.5 dB (including connectors) apart from its 3-dB power-splitting loss and the LAN diameter is 5 km. Estimate the end-to-end signal loss in a passive-star-based optical LAN using this device. If each node transmits a power of 0 dBm and its receiver sensitivity is -23 dBm, check whether the network would offer a feasible power budget. Given: fiber loss = 0.2 dB/km, connector loss = 1 dB. Solution:

Input ports

Output ports

3.2

The above figure shows the 8×8 passive star coupler (PSC) using 2×2 optical power splitters as the building block. The 8×8 PSC requires log28 = 3 columns, each column consisting of 4 2×2 powersplitting blocks. Each 2×2 block has a power-splitting loss of 3 dB along with an insertion loss of 1.5 dB including connectors. Hence the total loss in each 2×2 power splitter is Lsp = 3 + 1.5 = 4.5 dB. So the total end-to-end loss Lpsc across the 8×8 PSC would be 𝐿𝐿𝑝𝑝𝑝𝑝𝑝𝑝 = 3 × 4.5 = 13.5 dB.

With the above PSC configuration, the end-to-end power loss LT in the PSC-based LAN is obtained as the sum of the losses in the PSC, the interconnecting optical fibers with a length of 5 km (2.5 × 2 km as the network diameter) and a loss of 0.2 dB/km and four connectors (at two nodes and PSC ports) each with a loss of 1 dB. We therefore obtain LT as 𝐿𝐿 𝑇𝑇 = 𝐿𝐿𝑝𝑝𝑝𝑝𝑝𝑝 + 5 × 0.2 + 4 × 1 = 13.5 + 1 + 4 = 18.5 dB.

With the given transmit power PT = 0 dBm and the receiver sensitivity PR = -23 dBm, the total loss LT must be well within the power budget, i.e., 𝑃𝑃𝑇𝑇 − (𝑃𝑃𝑅𝑅 + 𝑃𝑃𝑀𝑀 ) ≥ 𝐿𝐿 𝑇𝑇 ,

with PM as the power margin. With PT – PR = 23 dB, which is higher than LT by 23 – 18.5 = 4.5 dB, the design is therefore realizable with a reasonable value for PM (= 4.5 dB).

3.4 Consider that a passive-star optical LAN operates at 1 Gbps with a fixed packet size of 1500 bytes. Calculate the maximum possible diameter of the network that would support CSMA/CD as an acceptable MAC protocol. Repeat the calculation for 10 Gbps transmission and comment on the result. (Given: refractive index of fiber core = 1.5).

Solution: Packets with 1500 bytes transmitted at 1 Gbps leads to a packet duration τp = 1500×8×10-9 = 12 × 10-6 sec = 12 μs. If the network diameter is dN, then the CSMA/CD protocol would require the packet duration τp ≥ round-trip delay, i.e., 𝜏𝜏𝑝𝑝 ≥

2𝑑𝑑 𝑁𝑁 , 𝑣𝑣

withv as the velocity of light in the fiber, given by v = c/n, n being the refractive index of the fiber core. Hence, with the above value of τp and v = 3×108/1.5 = 2 × 108 m/sec, the LAN can have a maximum diameter, given by 𝑣𝑣𝜏𝜏𝑝𝑝 2 × 108 × 12 × 10−6 = = 1.2 km.

2 2 As the bit rate is increased to 10 Gbps, the packets shrink by a factor of 10, thereby bringing the network diameter down to 120 m only. 𝑑𝑑𝑁𝑁 (𝑚𝑚𝑚𝑚𝑚𝑚) =

3.5 Consider a passive-star optical LAN operating with pure ALOHA as the MAC protocol with a large number of nodes. Assuming that the packet arrival process in the network follows a Poisson distribution with a traffic Λ, determine the expression for the probability that a successful transmission takes place with k attempts. Using this result, determine the average number of attempts to be made for a successful packet transmission.

Solution: With the total LAN traffic as Λ, throughput of the pure-Aloha protocol is S = Λ exp(-2Λ) with the probability of zero arrival during a packet transmission time as exp(-2Λ). Hence, the probability of successful transmission at the kth attempt is given by 𝑃𝑃𝑆𝑆 (𝑘𝑘) = exp(−2Λ){1 − exp(−2Λ)}𝑘𝑘−1 .

Hence, the average value for k is obtained as

∞ 𝑘𝑘−1 . 𝑘𝑘� = ∑∞ 0 𝑘𝑘 𝑃𝑃𝑆𝑆 (𝑘𝑘) = ∑0 𝑘𝑘 exp(−2Λ){1 − exp(−2Λ)}

3.3

3.6. Consider two identical passive-star optical LANs as in Exercise 3.5 (LAN-A and LAN-B, say), each of them working independently using pure ALOHA with a traffic Λ. Next, connect optically one free port of the passive star coupler used in LAN-A to one free port of the passive-star coupler of LAN-B using an optical fiber. Determine the expression for the throughput of the combined network and compare the same with the throughput of one single LAN (i.e., LAN-A or LAN-B).

Solution: When not interconnected, the throughput in LAN-A and also in LAN-B is S = Λ exp(-2Λ).

However, after the two LANs are interconnected, the traffic seen by each LAN increases from Λ to 2Λ with increased collisions. This changes the overall throughput to SAB = 2Λ exp(-4Λ).

3.7. Assume that, in Exercise 3.6 the two free ports of the two LANs are now connected through an active device (as in a bridge), which can convert the optical packets coming from both LANs into electrical domain and store them in a buffer, with the capability to read each packet header from LAN A and LAN B and switch the received packet from one LAN to another, if necessary, and ignore the packets that are destined to some node in the same LAN. What will be the throughput of the combined network?

Solution: When kept connected by a bridge, each LAN has a traffic of Λ, however with a collision domain that is half of the collision domain experienced with passive interconnection between the two LANs. Thus, each LAN operates with a traffic of Λ plus a fraction α of Λ coming from the other LAN. Assuming that the inter-LAN traffic is same in both directions, the throughput in each LAN becomes Λ(1+ α) exp{-2Λ(1+ α)}, leading to the overall throughput SAB, given by 𝑆𝑆𝐴𝐴𝐴𝐴 = 2Λ(1 + α) exp{−2Λ(1 + α)}.

3.8 Consider a 10-km long DQDB network with 31 equally distant nodes, where the 1st and 31st nodes have negligible distances from their respective nearest head-ends. Estimate the amount of unfairness (in terms of the time to access the bus) between the 15th and 27th nodes for packet transmission along the unidirectional bus operating from the 1st node to the 31st node. What will be the impact on unfairness between these two nodes (i.e., between 15th and 27th nodes), if we accommodate 30 more nodes in the same network (maintaining equal separation between nodes and same bus lengths)? Given: refractive index of fiber core = 1.5.

Solution: Inter-node distance in the 31-node DQDB network is 10/30 = 1/3 km. Hence the distance between the nodes 15 and 27 is d = (27-15)/3 = 4 km. Therefore, the propagation delay between the two nodes gives the measure of the unfairness τuf in respect of getting access to a free slot in the DQDB bus.

Thus, we obtain τuf as 𝜏𝜏𝑢𝑢𝑢𝑢 = 𝑑𝑑/𝑣𝑣 = 4 × 5 = 20 𝜇𝜇s, with v as the velocity of light in fiber (1/v = 5 μs/km with a core refractive index of 1.5).

With 61 nodes, d reduces to 1/6 km, which halves the propagation delay between the two nodes and hence the unfairness reduces to 10 μs.

3.9 In a DQDB network, the terminating end of the upper bus has become defective and hence taken out. Discuss the impact of removal of the termination on each bus and on the overall network operation.

Solution: Removal of the termination in one of the two buses will cause reflections of data packets, thereby making the bus inoperative. Since any given bus in a DQDB network works for both buses, i.e., supports the transmission of its own data packets and also carries the transmission requests on the 3.4

other bus in the opposite direction, the removal of the termination in any one of the two buses will jeopardize the operation of the entire network.

3.10 Consider an optical LAN based on the ring topology with 100 nodes. Each node in the LAN employs OEO conversion and requires a processing time of 1 μs for each packet that passes by. If the ring operates at a transmission speed of 100 Mbps, and each packet consists of 2000 bytes, determine a suitable value of the ring circumference in km, for which the standard token-based MAC protocol (i.e., IEEE 802.5) will be able to operate efficiently. Given: refractive index of the fiber core = 1.5.

Solution: The total processing time at all nodes of the ring is Tproc = 100 × 1 = 100 μs, and the duration of 2000-byte packets (transmitted at 100 Mbps) is Tp = 2000 × 8 × 10-8 sec. = 160 μs.

The circumference of the ring CR should be such that the propagation time around the ring Tprop + Tproc ≤ Tp. Hence, we obtain the maximum allowable value for CR as 𝐶𝐶𝑅𝑅 (max) ≤ 𝑇𝑇𝑝𝑝 − 𝑇𝑇𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 . 𝑣𝑣 Using the above condition, we obtain the value for CR (max) as 𝑇𝑇𝑝𝑝𝑟𝑟𝑜𝑜𝑝𝑝 =

𝐶𝐶𝑅𝑅 (max) = 𝑣𝑣�𝑇𝑇𝑝𝑝 − 𝑇𝑇𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 � = 2 × 108 × (160 − 100) × 10−6 = 12 × 103 m = 12 km.

3.11 Estimate the capacity utilization and maximum delay incurred in an FDDI ring with the following values of ring length L (circumference), number of nodes N, and TTRT setting: a) L = 4 km, N = 20, TTRT = 5 ms and 50 ms, b) L = 50 km, N = 100, TTRT = 5 ms and 50 ms.

Given: processing time at each node Tproc = 1 μs.

Solution: Case (a): L = 4 km, N = 20 (i)

TTRT = 5,000 μs.

Delay around the FDDI ring D = L/v + NTproc = 4 × 5 + 20 × 1 = 40 μs.

Hence, FDDI efficiency η is obtained as 𝜂𝜂 =

TTRT − 𝐷𝐷 5000 − 40 = = 99.16%.

TTRT + 𝐷𝐷/𝑁𝑁 5000 + 40/20

The maximum delay around the ring is obtained as

𝐷𝐷(max) = (𝑁𝑁 − 1)TTRT + 2𝐷𝐷 = 19 × 5000 + 2 × 40 = 95.08 ms.

(ii) TTRT = 50,000 μs.

𝜂𝜂 =

TTRT − 𝐷𝐷 50,000 − 40 = = 99.92%.

TTRT + 𝐷𝐷/𝑁𝑁 50,000 + 40/20

𝐷𝐷(max) = (𝑁𝑁 − 1)TTRT + 2𝐷𝐷 = 19 × 50,000 + 2 × 40 = 950.08 ms

Case (b): L = 50 km, N = 100

(i) TTRT = 5,000 μs.

D = 50 × 5 + 100 × 1 = 350 μs.

3.5

𝜂𝜂 =

TTRT − 𝐷𝐷 5000 − 350 = = 92.93%.

TTRT + 𝐷𝐷/𝑁𝑁 5000 + 350/100

𝐷𝐷(max) = (𝑁𝑁 − 1)TTRT + 2𝐷𝐷 = 99 × 5000 + 2 × 350 = 495.7 ms.

(ii) TTRT = 50,000 μs.

𝜂𝜂 =

50,000 − 350 TTRT − 𝐷𝐷 = = 99.29%.

50,000 + 350/100 TTRT + 𝐷𝐷/𝑁𝑁

𝐷𝐷(max) = (𝑁𝑁 − 1)TTRT + 2𝐷𝐷 = 99 × 50,000 + 2 × 350 = 4.95 s.

3.12 Discuss why the hubs (along with UTP cables) were introduced in 10-Mbps Ethernet in place of coaxial cables as the transmission lines. Indicate the limitations faced subsequently by the hub-based Ethernet architecture and explain how this difficulty was overcome by using switched-Ethernet architecture with bridges/switches.

Solution: See the discussions in Section 3.5.1.

3.13 Consider a 100BaseT (i.e., hub-based) fast Ethernet, wherein one has set up by mistake the twisted-pair links between the computers of end-users and the hub with the link lengths ≥ 150 m.

Discuss the problems (with appropriate calculations) that the LAN will face in running the CSMA/CD MAC protocol with this network setting.

Solution: See the discussions in Section 3.5.1.

3.14 The IEEE standard committee had to introduce 40/100 GbE architectures together, though the earlier Ethernet upgrades (e.g., 100-Mbps, 1-GBps and 10-Gbps versions) took place one at a time with the speed-enhancement factor of 10. Discuss the underlying issues that had driven the committee to make such a choice.

Solution: See the discussions in Section 3.5.4.

3.6

Exercise Problems and Solutions on Chapter 4 (Optical Access Networks) 4.1 Discuss the various candidate topologies for realizing optical access networks and indicate why PON has evolved as the favoured option. Also discuss the basic differences between the EPON and GPON architectures.

Solution: See the discussions in Sections 4.3 and 4.4.

4.2 Consider a PON operating at 1 Gbps (in both directions) with the maximum reach of 20 km, using single-stage RN. Assume that the transmitters in the PON transmit an optical power of 3 dBm and the receiver sensitivity is -27 dBm. The optical splitter in the RN is realized using multiple stages of 2×2 splitters as the building block, with each having an insertion loss of 0.3 dB. Calculate the received power levels for three different values of the number of ONUs in the PON: (a) 32, (b) 64, and (c) 128.

Indicate which value(s) for N will offer a feasible power budget for the PON. Given: fiber loss = 0.2 dB/km, connector loss = 1 dB. Solution: (a) Number of ONUs N = 32. Hence the PON needs a 1:16 power splitter/combiner at RN, needing k = log232 = 5 stages of (2×2) power splitters, each having an insertion loss Lsp = 0.3 dB. Hence the loss in the RN, including splitting and insertion losses, is obtained as 𝐿𝐿𝑅𝑅𝑅𝑅 = 10log10 2𝑘𝑘 + 𝑘𝑘𝐿𝐿𝑠𝑠𝑠𝑠 = 10log10 32 + 5 × 0.3 = 16.5 dB.

Further, the transmit power PT = 3 dBm, receiver sensitivity is -27 dBm for 1 Gbps, fiber loss αf = 0.2 dB/km, and connectors have losses Lc = 1 dB each. Hence, the received power at an ONU at 20-km distance is obtained as 𝑃𝑃𝑅𝑅 = 𝑃𝑃𝑇𝑇 − 20𝛼𝛼𝑓𝑓 − 𝐿𝐿𝑅𝑅𝑅𝑅 − 4𝐿𝐿𝑐𝑐 = 3 − 20 × 0.2 − 16.5 − 4 × 1 = −21.5 dBm,

which is higher than the receiver sensitivity (-27 dBm). Hence the design is feasible in respect of power budget. (b) N = 64. Hence k = log264 = 6.

𝐿𝐿𝑅𝑅𝑅𝑅 = 10log10 64 + 6 × 0.3 = 19.8 dB. 𝑃𝑃𝑅𝑅 = 3 − 20 × 0.2 − 19.8 − 4 × 1 = −24.8 dBm, which also offers a feasible power budget. However, to have a higher system margin, the transmit power can be increased to about 5 dBm. (c) N = 64. Hence k = log2128 = 7. 𝐿𝐿𝑅𝑅𝑅𝑅 = 10log10 128 + 7 × 0.3 = 23.1 dB. 𝑃𝑃𝑅𝑅 = 3 − 20 × 0.2 − 23.1 − 4 × 1 = −28.1 dBm, which doesn’t offer a feasible power budget.

4.3 Consider the PON with the largest number of ONUs from Exercise 4.2, that can offer a feasible power budget. For the same PON, assume that the average grant size requested from each ONU is 5000 bytes and all the ONUs are equidistant (i.e., 20 km) from the OLT. Calculate the average cycle time of the PON. Given: guard time = 1 μs, velocity of light in optical fiber = 2 × 108 m/s.

Solution: The PON with N = 64 in the previous example is the one with feasible power budget with largest N. We therefore evaluate the cycle time for this PON. 4.1

The RTT for the PON with 20-km span is (20 × 5) × 2 = 200 μs, and the grant size is 5000 bytes at 1 Gbps leading to a grant duration τ = 5000 × 8 × 10-9 s = 40 μs. Further, the PON uses a guard time τg = 1 μs between consecutive transmissions. Therefore, the cycle time TC for this PON is obtained as 𝑇𝑇𝐶𝐶 = 𝑅𝑅𝑅𝑅𝑅𝑅 + 𝑁𝑁𝜏𝜏𝑔𝑔 + 𝑁𝑁𝑁𝑁 = 200 × 10−6 + 64 × 10−6 + 64 × 40 × 10−6 s = 2.824 ms.

4.4 A PON operates at 1 Gbps both ways with three ONUs. The REPORT and GATE messages use 60 bytes. The processing time for reading these messages is 3 ns. Initially, the OLT sends GATE messages using broadcast transmission to all ONUs at time t (t = 0, say) with zero grant sizes. In the ensuing cycle, the OLT polls only after REPORT from the last ONU arrives. Assume that the ONUs 1, 2, and 3 are located at 10 km, 15 km, and 20 km from the OLT and have the queue sizes of 8,000, 10,000, and 20,000 bytes, respectively, before sending their REPORT messages. Having sent the REPORT messages, the ONUs wait for their GATE messages from the OLT, and start data transmission immediately after receiving the respective GATE messages. The OLT employs DBA using interleaved-polling scheme with a guard time of 1 μs, starting with the nearest ONU. Find out the scheduling times for the GATE messages at the OLT to the three ONUs and the cycle time for the following grant-sizing schemes: a) fixed scheme, granting 10,000 bytes to each ONU, b) gated scheme.

Given: velocity of light in optical fiber = 2 × 108 m/s.

Solution: First we define the following parameters/variables.

• • • • • •

Ti: duration of the grant.

τG: GATE duration. τR: REPORT duration (we assume τR = τG).

RTT(i): RTT for ONUi.

τproc: message processing time. τg: guard time.

The RTT for the three ONUs are: RTT(1) for ONU1 = 10×2×5 = 100 μs, RTT(2) for ONU2 = 15×2 ×5 = 150 μs, and RTT(3) for ONU3 = 20×2×5 = 200 μs. Further, τR = τG = 60×8×10-9 = 0.48 μs, τproc = 0.003 μs, and τg = 1 μs a) Fixed grant size = 10,000 bytes, leading to a grant duration T = 10,000 × 8 × 10-9 s = 80 μs. With t = 0 as the reference time, the OLT waits for the response of the farthest ONU (ONU3) and then schedules its first transmission to the nearest ONU (ONU1) at time t1, given by 𝑡𝑡1 = 0 + 𝑅𝑅𝑅𝑅𝑅𝑅(3) + 𝜏𝜏𝐺𝐺 + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 + 𝜏𝜏𝑅𝑅 + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 0 + 200 + 0.48 + 0.003 + 0.48 + 0.003 = 200.966 μs

Thereafter, the OLT sends a GATE message to ONU1 and estimates the time of arrival of the granted packets at the OLT. From this instant of time going further by the message processing time and the guard time, the OLT goes back in time domain by the time required for the granted packets to arrive at the OLT itself, leading to the scheduling time t2 for ONU2, given by 𝑡𝑡2 = �𝑡𝑡1 + 𝜏𝜏𝐺𝐺 + 𝑅𝑅𝑅𝑅𝑅𝑅(1) + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 + 𝑇𝑇 + 𝜏𝜏𝑅𝑅 + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 + 𝜏𝜏𝑔𝑔 � − �𝑅𝑅𝑅𝑅𝑅𝑅(2) + 𝜏𝜏𝐺𝐺 + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 � = {200.966 + 0.48 + 100 + 0.003 + 80 + 0.48 + 0.003 + 1} − {150 + 0.46 + 0.003} 4.2

= 232.449 μs.

Similarly, we obtain the scheduling time t3 at the OLT for ONU3 as 𝑡𝑡3 = �𝑡𝑡2 + 𝜏𝜏𝐺𝐺 + 𝑅𝑅𝑅𝑅𝑅𝑅(2) + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 + 𝑇𝑇 + 𝜏𝜏𝑅𝑅 + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 + 𝜏𝜏𝑔𝑔 � − �𝑅𝑅𝑅𝑅𝑅𝑅(3) + 𝜏𝜏𝐺𝐺 + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 � = {232.499 + 0.48 + 100 + 0.003 + 80 + 0.48 + 0.003 + 1} − {200 + 0.46 + 0.003} = 263.932 μs.

Using the above results, we obtain the cycle time TC as

𝑇𝑇𝐶𝐶 = �𝑡𝑡3 + 𝜏𝜏𝐺𝐺 + 𝑅𝑅𝑅𝑅𝑅𝑅(3) + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 + 𝑇𝑇 + 𝜏𝜏𝑅𝑅 + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 � − �𝑡𝑡1 + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 � = {263.932 + 0.48 + 200 + 0.003 + 80 + 0.48 + 0.003} − {200.966 + 0.003} = 343.929 μs.

b) Gated grant size: T1 = 8,000 × 8 × 10-9 = 64 μs, T2 = 10,000 × 8 × 10-9 = 80 μs, T3 = 20,000 × 8 × 10-9 = 160 μs. All other parameters remain same. Using the same method, we obtain the scheduling times for the three ONUs at the OLT and the cycle time as 𝑡𝑡1 = 200.966 μs, 𝑡𝑡2 = 216.449 μs, 𝑡𝑡3 = 247.932 μs, 𝑇𝑇𝐶𝐶 = 407.929 μs

4.5 Consider again the PON of the above exercise (Exercise 4.4) and assume that the DBA employs: a) limited scheme with maximum grant size of 9,000 bytes, b) limited scheme with excess distribution (using maximum grant size of 9,000 bytes).

Find out the grant sizes, scheduling times for the GATE messages at the OLT, and the cycle time for the two schemes.

Solution: a) Grant sizes of ONU2 and ONU3 will be pruned down from 10,000 and 20,000 bytes to 9,000 bytes, while ONU1 will get its reported size, i.e., 8,000 bytes. Using this observation, and following the method used for Exercise 4.4, one can calculate the grant-scheduling times and the cycle time in a straightforward manner. b) Grant size of ONU1 is 8000 bytes, which is less than the upper limit by 9,000 – 8,000 = 1,000 bytes, leading to an excess capacity.

On the other hand, ONU2 needs 10,000 bytes, i.e., 10,000 – 9,000 = 1,000 bytes more than the limited quota. Similarly, ONU3 needs 20,000 bytes, i.e., 20,000 – 9,000 = 11,000 bytes more than the limited quota. So, ONU2 and ONU3 are granted demand-driven excess capacity (i.e., in proportion to the respective excess demands).

Thus, ONU2 gets 8,000 + 1,000 × {1,000 / (1,000 + 11,000)} = 8,083 bytes, and ONU3 gets 8,000 + 1,000 × {11,000 / (1,000 + 11,000)} = 8917 bytes. With this limited excess distribution, one can again follow the same method as used for Exercise 4.4 to obtain the grant-scheduling times and the cycle time.

4.6 Consider a DBA framework where the ONUs can be polled only after receiving the REPORT from the last ONU in a cycle. It is understood that a walk-time of at least an RTT is wasted between the grant cycles. Apart from polling the ONU with the least RTT first, what could be a possible grantsizing strategy to reduce the delay experienced by the worst-affected packets due to long walk time?

4.3

Solution: One possible scheme is to ensure that the packets experiencing maximum delay satisfy the delay bound DB. In a polling-based DBA scheme, wherein the REPORT message is sent after data, the packet which arrives while the REORT is being sent experiences the maximum delay, as it can only be transmitted in the next cycle.

In order to guarantee the specified delay bound, a viable protocol can be adapted in the OLT, which would keep track of the packets (critical packets) which are going to experience delay exceeding DB and design a DBA scheme accordingly. For example, by using a limited-grant scheme, the OLT can adjust the cycle time to be at the most (DB - 0.5 RTT)/2, so that the worst-case packets can get around this problem, even when taking two cycle times to reach the OLT from the ONU end. The readers are encouraged to sketch the space-time diagram for the DBA and check how this scheme would work to keep the packet delay within DB. 4.7 Consider a PON employing IPACT-based DBA with the gated grant-sizing scheme for N ONUs. Following parameters are defined to model the gated grant-sizing scheme. τj: RTT/2 for ONUj, λ: average arrival rate in bits per second at an ONU from the user-end, T: bit interval, bR: size (bits) of REPORT message, bG: size (bits) of GATE message, τg: guard time, j ti : time of arrival of the ith transmission from ONUj at the OLT, qij: queue length (bits) of ONUj after completion of the ith data transmission, gij: grant size (bits) allocated by the OLT to ONUj for the ONU's ith transmission (including the REPORT message).

Using the above definitions, determine the expressions for: j

j

a) grant size gi in terms of qi-1 , which is the queue length after completion of the (i -1)th data transmission, j j j j b) queue length qi j in terms of λ, ti , ti-1, and gi , gi-1 for i > 1. j j-1 c) initial transmission time t1 in terms of t1 , j j j d) transmission time ti in terms of ti-1 and gi-1 if the PON has only one ONU. Solution: a) qi-1 j = queue length after the completion of (i-1)th transmission at ONUj. This is exactly the number of bytes reported through the REPORT message and will be cleared in the ith transmission window (since it is a gated scheme). Hence, the OLT would set the grant size gi j as 𝑗𝑗

𝑗𝑗

𝑔𝑔𝑖𝑖 = 𝑞𝑞𝑖𝑖−1 + 𝑏𝑏𝑅𝑅 ,

𝑗𝑗

with 𝑔𝑔1 = 𝑏𝑏𝑅𝑅 .

j

b) Further, the average queue size qi can be obtained as 𝑗𝑗

𝑗𝑗

𝑗𝑗

𝑗𝑗

𝑗𝑗

𝑗𝑗

𝑗𝑗

𝑗𝑗

𝑗𝑗

𝑞𝑞𝑖𝑖 = 𝜆𝜆�(𝑡𝑡𝑖𝑖 − 𝜏𝜏 𝑗𝑗 ) + 𝑇𝑇�𝑔𝑔𝑖𝑖 − 𝑏𝑏𝑅𝑅 � − �𝑡𝑡𝑖𝑖−1 − 𝜏𝜏 𝑗𝑗 + 𝑇𝑇�𝑔𝑔𝑖𝑖−1 − 𝑏𝑏𝑅𝑅 ��� = 𝜆𝜆�(𝑡𝑡𝑖𝑖 − 𝑡𝑡𝑖𝑖−1 ) + 𝑇𝑇�𝑔𝑔𝑖𝑖 − 𝑔𝑔𝑖𝑖−1 ��.

c) The transmission time depends on the arrival of the GATE message at the ONU. For the first transmission, the OLT sends GATE for all ONUs with the grant size equal to the REPORT message 4.4

size bR and the transmission of ONUj is scheduled ahead of t1 j-1 appended by a guard interval τg, or j the sum of all the j GATE messages plus RTTj, whichever is higher. Hence, we express ti as 𝑗𝑗

𝑗𝑗 −1

𝑡𝑡1 = 𝑚𝑚𝑚𝑚𝑚𝑚�𝑡𝑡1

+ 𝑇𝑇𝑏𝑏𝑅𝑅 + 𝜏𝜏𝑔𝑔 , 𝑗𝑗𝑗𝑗𝑏𝑏𝐺𝐺 + 2𝑑𝑑 𝑗𝑗 � j

d) With single ONU, the transmission time ti will depend only on the end of the previous j j transmission time ti-1 and gi-1 , as the GATE message can be sent only after receiving the REPORT message along with (i-1)th data transmission. Once the REPORT message arrives, the OLT sends the GATE message which reaches the ONUj after RTTj/2 (= dj) and the ith transmission can start after j RTTj/2. Hence, ti can be expressed as 𝑗𝑗

𝑗𝑗

𝑗𝑗

𝑡𝑡𝑖𝑖 = 𝑡𝑡𝑖𝑖−1 + 𝑇𝑇𝑔𝑔𝑖𝑖−1 + 𝑇𝑇𝑏𝑏𝐺𝐺 + 2𝑑𝑑 𝑗𝑗 + 𝜏𝜏𝑔𝑔 .

4.8 Consider a single-ONU PON, where the average delay Dav incurred by a packet from the instant of arrival at the ONU from the user-end until its delivery to the OLT is estimated by an M/G/1 queuing system, with the assumption that the RTT is relatively small compared to Dav. If the packet arrival at the ONU from the user-end follows Poisson distribution with a mean arrival rate of λ and the arriving packets have an average length Lav and a variance σL2, determine the expression for Dav. Ignore guard time.

Solution: λ = packet arrival rate at the ONU from the user-end, Lav = average packet size in bits, σL = standard deviation of the packet size, and r = 1/T = bit rate.

Hence, using the M/G/1 queuing model, we express the average delay D incurred by a packet at the ONU as 𝜎𝜎 2 𝐿𝐿2 𝜆𝜆 � 𝐿𝐿2 + 𝑎𝑎𝑎𝑎 � 𝑟𝑟 𝑟𝑟 2 𝐷𝐷 = , 2(1 − 𝜌𝜌) where ρ represents the traffic, expressed as 𝜆𝜆 𝜌𝜌 = = 𝜆𝜆𝑇𝑇𝑇𝑇𝑎𝑎𝑎𝑎 . 𝜇𝜇

4.5

Exercise Problems and Solutions for Chapter 5 (SONET/SDH, OTN, and RPR)

5.1 What are the different levels of synchronization in a telecommunication network and in what sequence they are carried out in a receiving node? Discuss critically the differences between the bit synchronization in a point-to-point communication link and the network synchronization.

Solution: See the discussions in Section 5.2.

5.2 What are the typical network synchronization schemes used in telecommunication networks?

Which scheme is used in SONET/SDH networks and why?

Solution: See the discussions in Sections 5.3 and 5.4.

5.3 What are the different clock references in SONET? With a suitable example, illustrate how the clock references are used in a SONET ring and how does the network try to get around typical network failures.

Solution: See the discussions in Section 5.4.

5.4 Consider two PCM bit streams with the frame durations of 125 μs arriving from two independent sources (nodes) at a third node, where the incoming bit streams have to be multiplexed. The system requirement dictates that the two incoming data streams must not slip from each other with an upper bound of one slip a day. Determine the clock-inaccuracy limit in terms of slips/frame for the incoming PCM tributaries.

Solution: In a digital telecommunication network, one slip a day implies a timing misalignment between clocks in 24 hrs = 24 × 60 × 60 = 86,400 s. This leads to a difference in clock rates Δrck between the two incoming PCM bit streams arriving at the multiplexing node, given by 1

∆𝑟𝑟𝑐𝑐𝑐𝑐 = 86,000 slips/s.

125-μs PCM frames arrive at a rate of rF, given by 1

𝑟𝑟𝐹𝐹 = 125×10 −6 = 0.008 × 106 = 8,000 frames/s

Hence, the slip per frame ρF is obtained as

𝜌𝜌𝐹𝐹 =

∆𝑟𝑟 𝑐𝑐𝑐𝑐 𝑟𝑟 𝐹𝐹

=

1/86,400 8,000

= 1.44676 × 10−9 slips/frame.

Thus, the clocks of the two PCM streams must be accurate and stable enough to ensure that the slip between them doesn’t exceed 1.44676 slips/frame.

5.5 Assume that, in a two-channel PDH system, each multiplexed frame accommodates at the most five bits from each incoming data channel, along with two additional bits for realizing the pulsestuffing mechanism. Sketch the multiplexed frame obtained using pulse-stuffing. Calculate the percentage variation in the incoming data rates (around the average per-channel data rate), that can be accommodated in the given PDH system.

Solution: For the sketch see Fig. 5.3. In the time-multiplexed PDH frame, each tributary can carry either 4 or 5 bits, with 4.5 bits as the average number of bits per frame.

5.1

Hence, the clock rate can vary by 0.5 bit in a frame with an average length of 4.5 bits. Thus, the given PDH system has a clock-jitter tolerance σj , given by 𝜎𝜎𝑗𝑗 =

0.5 × 100% = 11.11%.

4.5

5.6 Discuss a suitable scheme, which can be used in a SONET/SDH-based ADM to sense the transmission rates of the incoming tributaries with respect to the local clock rate for deciding the actions of the pointer H3.

Solution: The relative transmission rate of the incoming tributary is assessed by using an elastic buffer wherein the incoming bit stream stored by the recovered clock and emptied out by using the local clock. Thus, the buffer length dynamics (growing or decaying) shows which clock is faster and helps in deciding the value of the pointer in SONET frame.

5.7 Sketch a typical SONET connection over a path involving several links and sections along with terminal nodes, add-drop multiplexers, and digital crossconnect. Using this diagram, show how the various overheads for path/line/section are utilized in maintaining the connections.

Solution: See the discussions in Section 5.5.

5.8 Give the maximum number of VTs that can be accommodated in an SPE of the basic SONET (STS-1) frame for VT1.5, VT2, VT3, and VT6, and determine the effective transmission rates (i.e., excluding the overhead bytes) for each VTx. Solution: As STS-1 frame can carry 7 VT groups (VTGs), and each VTG can be populated one type of VT only. In particular, one VTG can carry 4 VT1.5 (DS1/T1), 3 VT2 (E1), 2 VT3 (DS1C), and 1 VT6 (DS2) tributaries/tributary.

These VTx’s occupy specific number of columns (c) and rows (r) in the STS-1 frame as follows: • • • •

VT1.5: each VT1.5 uses r = 9 and c = 3, with 4 such VT1.5’s in a VTG, leading to 4×7 = 28 VT1.5’s in a VTG. VT2: each VT2 uses r = 9 and c = 4, with 3 such VT2’s in a VTG, leading to 3×7 = 21 VT2’s in a VTG. VT3: each VT3 uses r = 9 and c = 6, with 2 such VT3’s in a VTG, leading to 2×7 = 14 VT3’s in a VTG. VT6: each VT6 uses r = 9 and c = 12, with 1 such VT6 in a VTG, leading to 1×7 = 7 VT6’s in a VTG.

The effective transmission rate rVTx of each VTx in an STS-1 frame is obtained as 𝑟𝑟𝑉𝑉𝑉𝑉1.5 = 𝑟𝑟𝑉𝑉𝑉𝑉2 = 𝑟𝑟𝑉𝑉𝑉𝑉3 =

8𝑟𝑟𝑟𝑟 8×9×3 = = 1.728 × 106 bps = 1.728 Mbps (original bit rate = 1.544 Mbps), 125×10 −6 125×10 −6

8𝑟𝑟𝑟𝑟 8×9×4 = = 2.304 × 106 bps = 2.304 Mbps (original bit rate = 2.048 Mbps), 125×10 −6 125×10 −6 8𝑟𝑟𝑟𝑟 8×9×6 = = 3.456 × 106 bps = 3.456 Mbps (original bit rate = 3.152 Mbps), 125×10 −6 125×10 −6

𝑟𝑟𝑉𝑉𝑉𝑉1.5 =

8𝑟𝑟𝑟𝑟 8×9×12 = = 6.912 × 106 bps = 6.912 Mbps (original bit rate = 6.312 Mbps).

125×10 −6 125×10 −6

5.2

5.9 Calculate the capacity-utilization efficiency for DoS connections with and without VCAT for the following three Ethernet clients and justify the benefits of VCAT: a) 10 Mbps Ethernet, b) 100 Mbps Ethernet (fast Ethernet), c) 1 Gbps Ethernet (Gigabit Ethernet).

Solution: a)10 Mbps Ethernet: Without VCAT, 10 Mbps data is carried with an effective capacity of 48.384 Mbps in a STS-1 (capacity including overhead bytes = 51.84 Mbps). Thus the capacity-utilization efficiency η is obtained as 10×100

𝜂𝜂 = 48.384 % = 20.67%.

With VCAT using VT1.5-7v scheme, 7 VT1.5’s are transmitted in one SPE, with the bandwidth consumed being 7 × (1.728 – rPA × 0.064), with rPA (=2) consuming two preamble bits corresponding to a capacity of rPA × 64 kbps = 2 × 0.064 Mbps. Hence, the improved capacity-utilization efficiency η is obtained as 10×100

b)100 Mbps Ethernet:

𝜂𝜂 = 7×(1.728−2×0.064) % = 89.29%.

Without VCAT, 100 Mbps data is carried over 3 STS-1 frames, each with an effective capacity of 48.384 Mbps, giving the capacity-utilization efficiency η as 100 × 100 % = 64.3%.

3 × 48.384 With VCAT, using VT1.5-64v scheme along with two preamble bits, the improved capacityutilization efficiency η is obtained as 𝜂𝜂 =

100×100

c) 1 Gbps Ethernet:

𝜂𝜂 = 64×(1.728−2×0.064) % = 97.67%.

Without VCAT, 1000 Mbps data is carried over STS-48 frames, each with an effective capacity of 48.384 Mbps, giving the capacity-utilization efficiency η as 1000 × 100 % = 43.06%.

48 × 48.384 With VCAT, using STS-1-21v scheme along with two preamble bits, the improved capacityutilization efficiency η is obtained as 𝜂𝜂 =

1000 ×100

𝜂𝜂 = 21×(48.384−2×0.064) % = 98.68%.

Thus, in each case, VCAT offers a significant increase in bandwidth utilization.

5.10. Consider the following client networks for OTN: a) b) c) d) e) f)

GbE (1 Gbps), OC-48 (2.488 Gbps), OC-192 (9.953 Gbps), 10GbE (10 Gbps), 40GbE (40 Gbps), 100GbE (100 Gbps).

5.3

Describe the basic mapping process and sketch the overall mapping schemes to obtain the respective OTUi’s from each client network.

Solution: Basic mapping schemes are as follows: • • • •

Two 1 Gbps Ethernet (GbE) or one OC-48 (2.488 Gbps) are/is encapsulated and encoded for FEC (using RS coding) into 2.67 Gbps.

One 10GbE or one OC-192 (9.953 Gbps) is encapsulated and RS-coded into 10.71 Gbps.

One 40GbE or one OC-768 (39.813 Gbps) is encapsulated and RS-coded into 43.02 Gbps.

One 100GbE or one OC-1920 (99.533 Gbps) is encapsulated and RS-coded into 111.81 Gbps.

Overall mapping and multiplexing schemes are illustrated in the following diagram.

1 GbE

OC-48

Mapper

ODU0 Multiplexer (x 2)

OTU1

Mapper ODU1 Multiplexer (x 4)

10 GbE OC-192

Mapper ODU2 Multiplexer (x 4)

40 GbE

100 GbE

OTU2

OTU3 OTU3

Mapper

OTU4

Mapper

5.11 What are the limitations of FDDI, DQDB, and SONET, and how are these limitations addressed in RPR-based metro ring networks. Justify that, with uniform traffic pattern in the network, RPR with its spatial-reuse feature of fiber rings can achieve a capacity utilization that is eight-times higher than that in an FDDI ring.

Solution: See the discussions in Section 5.10.1.

5.4

Exercise Problems and Solutions for Chapter 6 (WDM LANs) 6.1. Consider a single-hop WDM LAN using PC over 16 wavelengths. Assume the following: control packets consist of five bytes for source MAC address, five bytes for destination MAC address, log2M bits for wavelength indexing, with M as the number of wavelengths, and two bytes for miscellaneous purposes. a) Find the sizes of control and data packets, if the ratio of the data packet to the control packet sizes L = 100. b) Consider two designs of the above network with L = 100 and 30. Calculate for both cases the maximum achievable per-channel throughput and the corresponding data-channel traffic and delay for the following MAC protocols: i) Aloha/Aloha, ii) slotted-Aloha/Aloha.

Solution: a) Control packet consists of 5 bytes for source address, 5 bytes for destination address and log232 bits for the wavelength indexing, along with 2 additional bytes for miscellaneous purpose. Hence, the control packet size nCP is given by 𝑛𝑛𝐶𝐶𝐶𝐶 = 5 × 8 + 5 × 8 + log 2 16 + 2 × 8 = 100 bits.

Hence the data packet size nDP is given by 𝑛𝑛𝐷𝐷𝐷𝐷 = 100 × 𝐿𝐿 = 10,000 bits. b) Performance of Aloha/Aloha and slotted Aloha/Aloha protocols for L = 100 and 30: L = 100 (i) Aloha/Aloha protocol: Maximum per-channel throughput is obtained as 100 𝐿𝐿 = = 0.15995, 2𝑒𝑒(𝑀𝑀 − 1 + 𝐿𝐿) 2𝑒𝑒(16 − 1 + 100) which takes place at the optimum data traffic Gd (opt), obtained as (the typo in the expression for the optimum Gd in the book has been corrected here) 100 𝐿𝐿 𝐺𝐺𝑑𝑑 (opt) = = = 0.43478.

2(𝑀𝑀 − 1 + 𝐿𝐿) 2(16 − 1 + 100) The delay experienced at this traffic is D = e = 2.718. (ii) Slotted Aloha/Aloha protocol: Maximum per-channel throughput is obtained as 100 𝐿𝐿 = = 0.17191, 𝑆𝑆𝑑𝑑 (max) = 𝑒𝑒{𝑀𝑀 + 2(𝐿𝐿 − 1)} 𝑒𝑒{16 + 2(100 − 1)} which takes place at the optimum data traffic Gd (opt), obtained as 100 𝐿𝐿 𝐺𝐺𝑑𝑑 (opt) = = = 0.46729. 𝑀𝑀 + 2(𝐿𝐿 − 1) 16 + 2(100 − 1) The delay experienced with this traffic is D = e = 2.718. 𝑆𝑆𝑑𝑑 (max) =

L = 30 (i) Aloha/Aloha protocol: Maximum per-channel throughput is obtained as 6.1

30 𝐿𝐿 = = 0.12263, 2𝑒𝑒(𝑀𝑀 − 1 + 𝐿𝐿) 2𝑒𝑒(16 − 1 + 30) which takes place at the optimum data traffic Gd (opt), obtained as 30 𝐿𝐿 = = 0.3333. 𝐺𝐺𝑑𝑑 (opt) = 2(𝑀𝑀 − 1 + 𝐿𝐿) 2(16 − 1 + 30) The delay experienced at this traffic is D = e = 2.718. 𝑆𝑆𝑑𝑑 (max) =

(ii) Slotted Aloha/Aloha protocol: Maximum per-channel throughput is obtained as 30 𝐿𝐿 = = 0.14914 𝑒𝑒{𝑀𝑀 + 2(𝐿𝐿 − 1)} 𝑒𝑒{16 + 2(30 − 1)} which takes place at the optimum data traffic Gd (opt), obtained as 100 𝐿𝐿 = = 0.40541. 𝐺𝐺𝑑𝑑 (opt) = 𝑀𝑀 + 2(𝐿𝐿 − 1) 16 + 2(30 − 1) The delay experienced at this traffic is D = e = 2.718. 𝑆𝑆𝑑𝑑 (max) =

6.2 From the analytical models for the single-hop WDM LANs developed for the three PC-based protocols, obtain an estimate for the upper-bound on the tuning delay τTD in tunable devices (transceivers) in the network, such that τTD doesn't affect the network latency significantly.

Solution: Assume that the single-hop WDM LAN uses M wavelengths wi’s with equal channel spacing δw, i.e., 𝑤𝑤𝑖𝑖 = 𝑤𝑤𝑖𝑖−1 + 𝛿𝛿𝛿𝛿.

Hence, the maximum tuning range is (M-1) δw. If the tuning delay to change wavelength from wi to wi+1 is δτTD, then the maximum tuning delay τTD (max) incurred by the LAN can be expressed as 𝜏𝜏 𝑇𝑇𝑇𝑇 (max) = (𝑀𝑀 − 1)𝛿𝛿𝜏𝜏 𝑇𝑇𝑇𝑇 .

In order that, the tuning time doesn’t affect seriously the LAN performance, one should therefore ensure that τTD (max) is less than the delay DMAC (i.e., network latency) incurred in the MAC by a reasonable margin (K, say, with K > > 1), i.e., the LAN should satisfy 𝜏𝜏 𝑇𝑇𝑇𝑇 (max) ≤

𝐷𝐷𝑀𝑀𝑀𝑀𝑀𝑀 . 𝐾𝐾

Using the above criterion, one can obtain the conditions needed for the respective PC-based singlehop WDM LANs as follows.

Aloha/Aloha:

Slotted Aloha/Aloha:

𝜏𝜏 𝑇𝑇𝑇𝑇 (max) ≤

Slotted Aloha/M-server switch:

𝜏𝜏 𝑇𝑇𝑇𝑇 (max) ≤

1 2𝐺𝐺𝐺𝐺 𝑀𝑀 − 1 exp � �1 + ��. 𝐾𝐾 𝑀𝑀 𝐿𝐿 1 2𝐺𝐺𝐺𝐺 𝑀𝑀 − 2 exp � �1 + ��. 𝐾𝐾 𝑀𝑀 2𝐿𝐿 1

𝜏𝜏 𝑇𝑇𝑇𝑇 (max) ≤ 𝐾𝐾

exp (𝐺𝐺)

. [𝐿𝐿𝐿𝐿 exp (−G )]𝑀𝑀 /𝑀𝑀 !

1− 𝑀𝑀 𝑖𝑖 ∑𝑖𝑖=0 [𝐿𝐿𝐿𝐿 exp (−G )] /𝑖𝑖!

Note that, the above conditions need to be fulfilled for the entire range of G expected in the LAN. Moreover, a large value for K (say 10) might make the design infeasible for the LANs, and in such cases, the used optical spectrum can be divided into multiple segments, and each segment can be assigned one tunable laser and filter in each node.

6.2

6.3 Consider a spanning tree representing the minimum-hop paths from a source to all the destinations in a multihop 31-node WDM-LAN with a logical nodal degree of two. Evaluate the full-tree height and the expected number of hops.

Solution: The number N of nodes in a spanning tree with a nodal degree p and a height H is given by 2𝐻𝐻+1 − 1 𝑁𝑁 ≤ , 𝑝𝑝 > 1. 𝑝𝑝 − 1 Hence with N = 31 and p = 2, we obtain H as 2𝐻𝐻+1 − 1 31 ≤ , 2−1 leading to the estimate of H as 𝐻𝐻 = ⌈𝑙𝑙𝑙𝑙𝑔𝑔2 (31 + 1) − 1 ⌉ = 4.

The expected number of hops is therefore obtained as 2 − 24+1 + 32 × 4 + 4 𝐸𝐸[ℎ] = = 3.29.

32 − 1

6.4 Consider a six-node passive-star multihop WDM LAN. For this network, form a unidirectional logical ring topology and determine the required number of wavelengths. Estimate the expected number of hops needed in this network to communicate between any two nodes. If each wavelength operates at 10 Gbps, evaluate the network throughput in Gbps.

Solution: The 6-node multihop WDM LAN with the unidirectional (clockwise) ring as its logical topology is shown in the following.

1 6

2

5

3 4

With 6 nodes, the LAN needs M = 6 wavelengths, wherein the connections are set up with node 1 as the source node (note that hop requirements will be the same for each node due to the circular topology). The number of hops (h) for each connection would be as follows • Node 1 to 2: h = 1 • Node 1 to 3: h = 2 • Node 1 to 4: h = 3 • Node 1 to 5: h = 4 • Node 1 to 6: h = 5.

Hence, the expected number of hops is obtained as 1+2+3+4+5 = 3. 𝐸𝐸[ℎ] = 6−1 Using E[h], we obtain the network throughput in Gbps (for 10 Gbps as the transmission speed) as 6.3

𝑀𝑀 × 10 Gbps 6 × 10 = = 20 Gbps 𝐸𝐸[ℎ] 3 6.5 Consider a (p,k) Shufflenet with p = 3, k = 2. a) Draw the (3,2) Shufflenet topology using the node numbering convention used for self routing. b) How many wavelengths should this network have to set up concurrent transmissions for all possible connections? c) Find out the number of hops from node 0 to all the other nodes in the network. d) Calculate the expected number of hops in the network. e) Trace a shortest-path route from node 0 to the highest-numbered node in the network. How many such shortest-path routes exist? Identify all of these paths. 𝜂𝜂 =

Solution: a) The number of nodes in the (3,2) ShuffleNet is N = kpk = 2×32 = 18. So the first column of the ShuffleNet has 9 nodes numbered from top to bottom as (using p = 3 as the radix for the row numbering): (1,00), (1,01), ... , (1,22). Similarly the 9 nodes in the second column are numbered as: (2,00), (2,01), ... , (2,22), while the wrapped-up last column is the replica of the first column. Using this numbering scheme, sketch the interconnection for the ShuffleNet as illustrated in Section 6.4.1. b) The number of wavelengths M = kpk+1 = 2×33 = 54. c) The numbers of hops needed to reach from node 0 to all other nodes are presented below in tabular form. h = 1: Route from node 0 Number of hops (h) Node 0 → 9, 10,11 1 h = 2: Route from node 0 Node 0 → 9 → 1,2 Node 0 → 10 → 3,4,5 Node 0 → 11 → 6,7,8

Number of hops (h) 2 2 2

Route from node 0 Node 0 → 9 → 1 → 12,13,14 Node 0 → 9 → 2 → 15,16,17

Number of hops (h) 3 3

Alternate paths with h = 3: Route from node 0 Node 0 → 10 → 3 → 9,10,11 Node 0 → 10 → 4 → 9,10,11 Node 0 → 10 → 5 → 12,13,14 Node 0 → 11 → 6 → 12,13,14 Node 0 → 11 → 7 → 15,16,17 Node 0 → 11 → 8 → 15,16,17

Number of hops (h) 3 3 3 3 3 3

h = 3:

d) The expected number of hops E[h] is obtained as 6.4

𝑘𝑘𝑝𝑝𝑘𝑘 (𝑘𝑘 − 1)(3𝑘𝑘 − 1) − 2𝑘𝑘(𝑝𝑝𝑘𝑘 − 1) 𝐸𝐸[ℎ] = 2(𝑝𝑝 − 1)(𝑘𝑘𝑝𝑝𝑘𝑘 − 1) =

2×32 (3−1)(3×2−1)−2×2�32 −1� 2(3−1)(2×32 −1)

= 2.1765.

e) The shortest path between node 0 to node 17 is: 0 → 9 → 2 → 17. (other options: 0 → 11→ 7 → 17 and 0 → 11 → 8 → 17).

6.6 Consider a (3,4,5) GHCNet and find out the following: a) number of nodes in the network, b) number of wavelengths in the network for collision-free transmission, c) logical nodal degree, d) nodes reachable in one hop from node (3,2,1) (in mixed-radix representation), e) path from node (0,0,0) to (4,3,2) using the GHCNet routing algorithm, f) expected number of hops for this network.

Solution: (3,4,5) GHCNet: a) Number of nodes N: 𝑟𝑟

𝑁𝑁 = � 𝑚𝑚𝑘𝑘 = 3 × 4 × 5 = 60. 𝑘𝑘=1

b) Logical nodal degree p: 𝑟𝑟

𝑝𝑝 = �(𝑚𝑚𝑘𝑘 − 1) = (3 − 1) + (4 − 1) + (5 − 1) = 9. 𝑘𝑘=1

c) Number of wavelengths M:

𝑀𝑀 = 𝑁𝑁𝑁𝑁 = 60 × 9 = 540. d) Nodes reachable in one hop from node (3,2,1): i.

Decrease or increase the third digit: (3,2,1) → (3,2,0), (3,2,2). ii.

Decrease or increase the second digit: (3,2,1) → (3,0,1), (3,1,1), (3,3,1). iii.

Decrease or increase the first digit: (3,2,1) → (0,2,1), (1,2,1), (2,2,1), (4,2,1). e) Path from node (0,0,0) to (4,3,2): (0,0,0) → (0,0,2) → (0,3,2) → (4,3,2). f) Expected number of hops E[h]: ∑𝑟𝑟ℎ=1 ℎ𝑛𝑛ℎ . 𝑁𝑁 − 1 In the given GHCNet, r = 3. Hence, we need to evaluate n1, n2 and n3. 𝑛𝑛1 = ∑3𝑖𝑖=1(𝑚𝑚𝑖𝑖 − 1) = (3 − 1) + (4 − 1) + (5 − 1) = 9.

Similarly, 𝐸𝐸[ℎ] =

𝑛𝑛2 = ∑3𝑖𝑖=1 ∑3𝑗𝑗 ≠𝑖𝑖=1(𝑚𝑚𝑖𝑖 − 1)�𝑚𝑚𝑗𝑗 − 1� = 26,

3

Hence,

3

𝑛𝑛3 = � �

3

�

(𝑚𝑚𝑖𝑖 − 1)�𝑚𝑚𝑗𝑗 − 1�(𝑚𝑚𝑘𝑘 − 1) = 24.

𝑖𝑖=1 𝑗𝑗 =1 𝑘𝑘=1(𝑖𝑖≠𝑗𝑗 ≠𝑘𝑘)

6.5

𝐸𝐸[ℎ] =

∑𝑟𝑟ℎ=1 ℎ𝑛𝑛ℎ 1 × 9 + 2 × 26 + 3 × 24 = = 2.2542. 𝑁𝑁 − 1 60 − 1

6.7 Consider a (2,4) dBNet and draw its topology. Find out the following: a) number of nodes in the network, b) number of wavelengths in the network for collision-free transmission, c) logical nodal degree, d) route between two extreme nodes according to the self-routing algorithm, e) number of hops from node (0000) to all other nodes, f) expected number of hops for this network.

Solution: (p,D) dBNet with p = 2, D = 4. a) Number of nodes N = pD = 16. b) Number of wavelengths M = Np – p = 32 – 2 = 30. c) Logical nodal degree is p + 1 for N – p nodes, and p for p nodes. Hence 14 nodes have a logical degree of 3 and 2 nodes have a logical degree of 2 due to self-looping. d)Route between two extreme nodes (node (0,0,0,0) and node (1,1,1,1): Using the routing scheme (shift and merge), we get Adds Shift Merge Routing 0000 0000 1111 1111 00001111 0000 → 0001 → 0011 → 0111 → 1111. e) Hop counts from node (0000) : 1 hop 0000 to 0001: Adds Shift Merge Routing 0000 0000 0001 0001 00001 0000 → 0001.

2 hops 0000 to 0010: Adds Shift 0000 0000 0010 0010 0000 to 0010: Adds Shift 0000 0000 0011 0011

Merge

Routing

000010

0000 → 0001. → 0010.

Merge

Routing

000011

0000 → 0001. → 0011..

Similarly, for 3 and 4 hops, we obtain the following results: 3 hops 6.6

0000 to 0101: 0000 → 0001→ 0010 → 0101, 0000 to 0110: 0000 → 0001→ 0011 → 0110, 0000 to 0111: 0000 → 0001→ 0011 → 0111.

4 hops 0000 to 1000: 0000 → 0001→ 0010 → 0100 → 1000, and so on for the remaining 7 (below) 0000 to 1001 0000 to 1010 0000 to 1011 0000 to 1100 0000 to 1101 0000 to 1110 0000 to 1111 f) Expected number of hops E[p] is obtained as 𝐸𝐸[ℎ] =

𝑝𝑝 − 𝑝𝑝𝐷𝐷+1 + 𝑝𝑝𝐷𝐷 𝐷𝐷(𝑝𝑝 − 1)2 + 𝐷𝐷(𝑝𝑝 − 1) 2 − 24+1 + 24 × 4(2 − 1)2 + 4(2 − 1) = = 2.5333. (𝑝𝑝𝐷𝐷 − 1)(𝑝𝑝 − 1)2 (24 − 1)(2 − 1)2

6.8 Draw the topology of a (2,3) ShuffleNet with the number of nodes starting from 0. For this network, a) trace the route from node 0 to node 23 following the self-routing scheme of ShuffleNet, b) if both the outgoing links from the nodes 9 and 11 fail, then check whether node 23 would be reachable from node 0 - justify your answer by tracing the routes, c) mention one possible situation, when node 18 would be unreachable from node 0 and justify your answer from the topology.

Solution: a, b) The (2,3) ShuffleNet with column-row-based addressing for the 24 nodes (0 to 23) is shown below: Columns → 0 1 2 0 • Following routes/paths are shown by the respective Rows ↓ --------------------------arrows in the network diagram on the left side.

000 | 0 8 16 0 • Actual path obtained using the self-routing scheme based on the column-row-based addresses of the 001 | 1 9 17 1 nodes: 010 | 2 10 18 2 0 → 9 → 19 → 7 → 15 → 23 011 | 3 11 19 3 • If outgoing links of node 9 are down, still there will 100 | 4 12 20 4 exist the following alternate shortest paths: 101 | 5 13 21 5 0 → 8 → 17→ 3 → 15 → 23, 110 | 6 14 22 6 0 → 8 → 16→ 1 → 11 → 23.

111 | 7 15 23 7 c) Having drawn the full topology of the network, it becomes evident that, node 18 will become unreachable from node 0, if the outgoing links from the nodes 9 and 13 are down.

6.7

Exercise Problems and Solutions for Chapter 7 (WDM Access Networks) 7.1 Consider three types of PONs using WDM transmission: (a) WDM PONs with single-stage RN using AWG, (b) TWDM PON using two stages of RN with AWG as the first-stage RN and multiple optical splitters at the second stage, and (c) TWDM PON using splitters only at RNs. Make an exhaustive comparison between these three options in respect of the power budget, cycle time, and hardware complexity of the OLT/ONUs. Solution: See the discussions in Sections 7.2, 7.3, and 7.9.

7.2 A long-reach WDM PON has to be set up to serve 64 ONUs over a span of 100 km by using AWG-based RN. If the transmit power of the optical source in the OLT is 3 dBm for each wavelength and the per-channel receiver sensitivity of the ONUs is -28 dBm for 1 Gbps downstream transmission, examine the feasibility of the power budget for downstream connection. Given: fiber loss = 0.2 dB/km, AWG insertion loss = 4 dB, connector loss = 1 dB. Solution: OLT transmit power PT = 3 dBm, receiver sensitivity PR (min) = -28 dBm, fiber loss = 0.2 dB/km, AWG loss lawg = 4 dB, and connector loss lc = 1 dB. Power budget: PR = PT – (100 × αf + lawg + 4 × lc) = 3 – (100 × 0.2 + 4 + 4 × 1) dBm = 3 – (20 + 4 + 4) = - 25 dBm > PR (min) (= -28 dBm).

Hence, the power budget is feasible, though there will be some variation of power at the AWG output ports, which should be taken care of by the dynamic range of the receivers at the ONUs. 7.3 A TWDM PON has to be set up to serve 64 ONUs over a span of 40 km at 1 Gbps using 8 wavelengths, and the network designer is given two options: (a) a static TWDM PON using AWGOSP configuration, and (b) dynamic TWDM PON using OSP-only configuration. The OSPs are realized with multiple stages of 2 × 2 optical splitters as building block, with each of them having an insertion loss of 0.3 dB. If the transmit power of the optical source at the OLT is 3 dBm over each wavelength and the per-channel receiver sensitivity of the ONUs is -28 dBm, examine the feasibility of the downstream power budget in each case. Given: fiber loss = 0.2 dB/km, AWG insertion loss = 4 dB, connector loss = 1 dB. Solution: Number of ONUs N = 64, maximum distance L = 40 km, transmit power PT = 3 dBm, receiver sensitivity PR (min) = -28 dBm, bit rate r = 1 Gbps, connector loss lc = 1 dB, AWG insertion loss = 4 dB. Power budget with AWG and power splitter: Power splitter loss = 10 log10N + 0.3 × log2N. The number of wavelengths M = 8, and hence the AWG (with the insertion loss of 4 dB) needs 1:8 configuration, with each wavelength needing 64/8 = 8 splitters, each with a split ratio of 1:8. Between the OLT and each ONU, 6 connectors are needed with a loss of 6lc = 6 dB. Hence, the power budget is obtained as PR = PT – [40 × 0.2 + 4 + (10log108 + 0.3 × log28) + 6] dBm = PT - [8 + 4 + 9 + 0.9 + 6] dBm = 3 – 27.9 = -24.9 dBm,

7.1

which is larger than the receiver sensitivity (= - 28dBm) with a 3.1 dB as system margin. Hence, this design offers a feasible power budget.

Power budget with two stages of power splitters: In this case, end-to-end connection between the OLT and each ONU requires 6 connectors with a loss of 6 dB. Hence, we obtain PR as PR = PT – [40 × 0.2 + (10 × log108 + 0.3 × log28) + (10 × log108 + 0.3 × log28) + 6] = 3 – [8 + (9 + 0.9) + (9 + 0.9) + 6] = 3 - 33.8 = -30.8 dBm, which is less than the receiver sensitivity. Hence this design doesn’t offer a feasible power budget.

7.4 Consider a static TWDM PON with AWG-OSP configuration. The PON transmits at 1 Gbps (both ways) and supports 128 ONUs over 40 km with a fixed grant size of 10,000 bytes with a guard time of 1 μs. There are two options for the PON design in respect of the AWG and OSPs: a) 1 × 16 AWG, 1 × 8 OSP, b) 1 × 8 AWG, 1 × 16 OSP. Determine the cycle time of a TDMA group for each design. Compare the two designs. Given: propagation time in optical fiber = 5 μs/km.

Solution: Maximum distance L = 40 km, grant size G =10,000 bytes, bit rate r = 1 Gbps RTT = 40 × 2 × 5 = 400 μs, guard time τg = 1 μs. a) 16 × 8 (1:16 AWG, 1:8 power splitter) Number of wavelengths M = 16, number of ONUs in each TDM group = 8 Cycle time TC = RTT + 8τg + 8G/r = 400 × 10-6 + 8 × 10-6 + 8 × 80,000 × 10-9 s = 400 + 8 + 640 = 1048 μs = 1.048 ms. b) 8 × 16 (1:8 AWG, 1:16 power splitter) Number of wavelengths M = 8, number of ONUs in each TDM group = 16 Cycle time TC = RTT + 16τg + 16G/r = 400 × 10-6 + 16 × 10-6 + 16 × 80,000 × 10-9 sec = 400 + 16 + 1280 = 1696 μs = 1.696 ms.

In case (a), the power budget is (obviously) better, as well as, the cycle time is smaller, however needing AWG. In case (b), more flexible DBA scheme can be realized due to larger number of ONUs sharing each wavelength, thereby offering higher bandwidth utilization.

7.5 Consider the design of a 128 ONU TWDM PON using two stages of RN using OSPs only. The TWDM PON needs to cover a distance of 40 km and transmit at 1 Gbps both ways, with a transmit power of 3 dBm. The receivers have a sensitivity of -28 dBm while operating at 1 Gbps, and the OLT allocates a fixed grant size of 10,000 bytes for all ONUs and uses a guard time of 1 μs. If the design needs to ensure that the cycle time for the TDMA groups of ONUs remains confined within 1.5 ms, then estimate the number of wavelengths to be used and check whether any EDFA is needed to realize a feasible power budget. Given: fiber loss = 0.2 dB/km, connector loss = 1 dB. Solution: The design uses splitters only, while the cycle time TC < 1∙5 ms. The number of ONUs N = 128, distance L = 40 km, and receiver sensitivity PR (min) = - 28dBm, fiber loss αf = 0.2 dB/km, connector loss lc = 1 dB. For the ease of deployment (fiber laying), we assume that, the 1:128 splitting would be realized in two stages, e.g, with M × K = 128, with the second stage of splitting having K splitters. This configuration 7.2

will divide the ONUs into M groups, each group of K ONUs sharing typically one of the M wavelengths. However, large K will lead to long cycle time. Thus, one needs to design judiciously the values for K and M. Cycle time in each group of ONUs: Grant size duration τ = 10,000 × 8 × 10-9 s = 80 μs. Hence, the cycle time is expressed as TC ≥ RTT + Kτg + Kτ. with τg as the guard time. Hence, 1500 ≥ 400 + K × 1 + K × 80 or, K ≤ 1100/81 = 13.58 → 13, thereby needing 1:16 splitters at the second stage (with 2n-splitting using 2 × 2 splitters as the building block), with two ports remaining unused.

With K = 13, we therefore obtain 𝑀𝑀 = �

128 � = 10.

13

Hence, the first-stage splitter will also employ the nearest available, i.e., the 1:16 splitter configuration, wherein 6 ports will remain unused. Thus, with both stages of power splitting needing the 1:16 configuration, overall power-splitting will increase from 1:128 to 1:162 (= 1:256), due to the unused ports in both stages, thereby making the power budget more critical.

Power budget: We obtain the power budget with total splitting ratio of 1:256 (with 6 connectors) as PR = PT – (αfL + 6lc + 10log10256 + log2256 × 0.3) = 3 – [0.2 × 40 + 6 × 1 + (24 + 2.4)] = -37.4 dBm. Hence, the TWDM PON needs to employ EDFAs in both directions with a gain Gedfa given by Gedfa ≥ {-28 – (-37.4)} + PM = 9.4 dB + PM, with PM as the system power margin. With a PM ≈ 5 dB, an EDFA with a gain of 15 dB would make the design feasible in the physical layer.

7.6 A WDM PON uses one 1:N AWG, which has a non-uniform power distribution P(θ) across its output ports, following a Gaussian distribution, given by 𝑃𝑃(𝜃𝜃) =

1

exp �

−θ2 �, 2σ2

√2𝜋𝜋𝜎𝜎 2 where θ is the angle made by the output ports with respect to the longitudinal axis passing through the input port. The angle θ takes up discrete values across the output ports and σ is determined by the overall AWG construction. Determine the expression for the power ratio PR (in dB) of the signal power levels at the central output port (with N assumed to be odd) and the outermost output port on one of two sides. Using this result, the network provider decides to connect the farthest ONU to the central AWG port, while the closer ONUs are connected to the outer ports of the AWG. If dmax and dmin are the distances of the farthest and closest ONUs from the RN respectively, determine the expression for the revised PR (as seen at the ONUs) with the above connection strategy for the ONUs. Solution: Power received at the outermost output port of AWG (i.e., θ = θmax) expressed as 𝑃𝑃(𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 ) =

1

√2𝜋𝜋𝜎𝜎 2

−θ2

exp � 2σmax 2 �.

Similarly, the power received at the central port with θ = 0 is given by 𝑃𝑃(0) =

1

√2𝜋𝜋𝜎𝜎 2

.

The ratio of the two power levels at the central and outermost ports can be expressed in dB as PR, given by

7.3

θ2max 𝑃𝑃(0) 𝑃𝑃𝑃𝑃 = 10log10 = 10log10 exp � 2 �. 𝑃𝑃(𝜃𝜃𝑚𝑚𝑚𝑚𝑚𝑚 ) 2σ

Hence, the ONU receivers should have a dynamic range DR in dB, such that the combined effect of PR and the varying distances of ONUs from the OLT keeps the received-power ratio at ONUs below the receiver DR. In order to address this issue, the farthest ONU (at a distance d = dmax) from the AWG is connected to the central AWG output port and the nearest ONU (at a distance d = dmin) is connected to the outermost AWG output port (one of the two on two sides), such that the receivedpower ratio PR′ at the ONUs gets reduced as dmax �� 𝑃𝑃𝑃𝑃 ′ = �𝑃𝑃𝑅𝑅 − 10log10 � dmin θ2

d

max min = �10log10 exp �� 2σ ��� 2 � �d

with α = dmin/dmax.

max

αθ2max = �10log10 exp �� 2σ 2 ���,

7.7 Develop a pseudo code to simulate the EFT and EFT-VF schemes for DWBA in a dynamic TWDM PON with N ONUs and compare the performance of the two schemes using computer simulation for a set of practical network specifications.

Solution: Consider Poisson arrivals of data traffic at ONUs from the user-ends and implement the respective DWBA schemes (EFT and EFT-VF) as described in Section 7.9.5 for a TWDM PON with typical system/device parameters.

7.4

Exercise Problems and Solutions for Chapter 8 (WDM Metro Networks)

8.1. Discuss using an appropriate layered architecture, how the packet- and circuit-switched traffic flows are circuit-multiplexed in SONET-over-WDM ring networks. What is the limitation of the circuit-switched transmission of packet-switched traffic and why does this practice continue to exist?

Solution: See the discussions in Section 8.2.

8.2. Consider an N-node mesh-configured WRON having bidirectional fiber links with a given set of lightpaths, where a node pair is allocated only one lightpath in each direction. Assume that a cut-set Ci bisects the network into two parts with N-ni and ni nodes. If the cut-set consists of fi fiber pairs and li lightpaths, then find out the expression for the lower-bound for the number of wavelengths Mmin needed in the entire network. Using this model, show that for an N-node ring (i.e., two-connected mesh) with N as an odd integer, Mmin = (N2 – 1)/8.

Solution: The N-node WRON mesh is bisected into two parts with N-ni and ni nodes with a cut-set Ci, as shown below.

N - ni nodes

ni

...

nodes

Cut-set Ci Cut-set Ci consists of fi fiber pairs and li lightpaths. Hence to maintain the connectivity between the two sides, one needs at least M wavelengths, given by 𝑙𝑙𝑖𝑖 𝑀𝑀 ≥ . 𝑓𝑓𝑖𝑖

Hence, the lower-bound on M for all possible cut-sets is obtained as 𝑙𝑙𝑖𝑖 𝑀𝑀𝐿𝐿𝐿𝐿 = �𝑚𝑚𝑚𝑚𝑚𝑚𝑖𝑖 �. 𝑓𝑓𝑖𝑖

For a WDM ring with N as an odd number, (N-1)/2 and (N+1)/2 are almost the half of N, with fi = 2.

Hence, we obtain MLB for the WDM ring as 𝑁𝑁 2 − 1 1 (𝑁𝑁 − 1) (𝑁𝑁 + 1) 𝑀𝑀𝐿𝐿𝐿𝐿 = � � × . �� = 2 2 2 8

Note that, N being odd ≥ 3, N is also odd and hence N -1 is always divisible by 8.

2

2

8.1

8.3. Repeat the above exercise for WRON rings with even number of nodes and determine the lowerbound for the number of wavelengths in the network.

Solution: Following the same approach as in the previous exercise, we divide the network into two sets for N as an even number, each having N/2 nodes. Hence, we express MLB as 𝑁𝑁 2 1 𝑁𝑁 𝑁𝑁 𝑀𝑀𝐿𝐿𝐿𝐿 = � � × �� = � �.

8 2 2 2

In this case, the ceiling function is needed to estimate MLB as N 2 is not always divisible by 8.

8.4. Consider a six-node bidirectional ring with the following traffic matrix (integer multiples of OC12 connection). Γ=

077256 905976 740546 886049 375403 967540

Find out the numbers of ADMs and wavelengths by using a multihop heuristic design to support the traffic demand with a grooming capacity of 4 in each wavelength.

Solution: Use the heuristic method presented in Section 8.3.3 to design the WDM ring.

8.5. Repeat the above exercise using a multihop LP-based method. Compare the results obtained from the LP formulation and the heuristic scheme.

Solution: Use the LP-based design presented in Section 8.3.3 to design the WDM ring, and compare the results from the heuristic and LP-based methods, as illustrated using the case-study results (Fig. 8.14).

8.6. Consider that a WRON-based ring with five nodes employs single-hub configuration, where the numbers of OC-48 lightpaths needed from the various nodes to the hub node have been estimated from the traffic matrix for the WRON as follows (node 0 is the hub): a) b) c) d)

from node 4 to hub: 2, from node 3 to hub: 4, from node 2 to hub: 1, from node 1 to hub: 3.

Determine the number of wavelengths needed in the network. Also calculate the number of OC-48 transceivers to be provisioned in the hub.

8.2

Solution: Consider the 5-node single-hub WRON ring, as shown below with node 0 as the hub.

Hub node

0 1

4 WRON ring

2

3

The number of lightpaths li0 needed from a node to the hub on either side of the hub are as follows: • • • •

l40 = 2, l30 = 4, l20 = 1, l10 = 3.

Hence, the number of OC-48 lightpaths needed on each side of the ring (lightpaths on two sides of the hub being link-disjoint) can be expressed as the sum of the lightpaths needed from each node to the hub on one of the two sides. Thus, we obtain the number M of wavelengths necessary for the ring as 1 1 𝑀𝑀 = � (𝑙𝑙10 + 𝑙𝑙20 + 𝑙𝑙30 + 𝑙𝑙40 )� = � (3 + 1 + 4 + 2)� = 5.

2 2

The number of transceivers required in the hub is 2M = 10.

8.7. Consider a WRON-based ring with six nodes, whose virtual topology looks like a circle with several arcs as lightpaths representing the end-to-end connections between all non-adjacent node pairs. With this network-setting, sketch the schematic diagram of an OADM for the add-drop and passthrough operations.

Solution: The 6-node WRON ring is shown below, wherein the dashed lines are used to show some end-to-end connections between non-adjacent nodes. However, all such connections are not shown to avoid 0 5

1 2

4 3

8.3

clumsiness. The number of nodes being even, we express the number M of wavelengths as 𝑀𝑀 = �

𝑁𝑁 2 𝑁𝑁 𝜌𝜌 + �� �, 8 4 𝑁𝑁 − 1

with ρ as the ingress traffic at each node (OADM), which is distributed uniformly between the

remaining (N-1) nodes. Assuming ⌈𝜌𝜌/(𝑁𝑁 − 1)⌉ = 1 (which can be increased proportionately corresponding to the actual traffic) and N = 6, we obtain M as 62

6 + � × 1 = 6.

8 4

𝑀𝑀 = �

Consider one half of the node operating for the clockwise traffic over the shortest paths. In order to construct the configuration of node 0 for the clockwise traffic, we consider the relevant transit as well as add/drop lightpaths (with shortest-path routing) as follows: 1.

2.

3.

4.

5.

6.

7.

8.

9.

node 4 to node 1 (transit lightpath on wavelength w1), node 5 to node 2 (transit lightpath on wavelength w2), node 5 to node 1 (transit lightpath on wavelength w3), node 3 to node 0 (drop lightpath on wavelength w4), node 4 to node 0 (drop lightpath on wavelength w5), node 5 to node 0 (drop lightpath on wavelength w6), node 0 to node 3 (add lightpath on wavelength w4), node 0 to node 2 (add lightpath on wavelength w5), node 0 to node 1 (add lightpath on wavelength w6).

Using the above add/drop/transit lightpaths, the clockwise part of the OADM at node 0 is constructed as shown below.

Optical demux w1

w1,w2, ..., w6

Optical mux

w2 w3

w1,w2, ..., w6

w4 w5 w6 5-0 4-0 3-0

0-1

0-2

0-3

RX TX

RX TX

8.4

RX TX

8.8. In the six-node WRON ring of the above exercise, assume that the optical transmitters transmit 15 dBm of optical power at 2.5 Gbps for each lightpath and the receivers have a sensitivity of -23 dBm, and the ring has a circumference of 30 km with equal distances between the adjacent nodes.

Trace the longest lightpath in the ring and check the feasibility of the power budget for the same.

Given: connector loss = 1 dB, demultiplexer loss = 3 dB, multiplexer loss = 3 dB, fiber loss = 0.2 dB/km.

Solution: As shown in the WRON ring in the previous exercise, one of the longest lightpaths runs from node 3 to node 0.

0 With equal distance between the adjacent nodes, the distance between any adjacent node pair is 30/6 = 5 km.

1 2 3

With the above observation, we therefore consider the lightpath from node 3 to node 0, and estimate its power budget. For this purpose, we enumerate in the following the losses incurred by the lightpath from node 3 to node 0.

•

Loss at node 3: Connector loss = 2lc = 2 × 1 = 2 dB, optical multiplexer (OMUX) loss lmx = 3 dB, Hence, total loss = 2 + 3 = 5 dB. • Losses at nodes 2 and 1 (same loss at both nodes): Connector loss = 4lc = 4 dB, optical demultiplexer (ODMUX) loss ldm = 3 dB, OMUX loss lmx = 3 dB. Hence, total loss at node (and at node 2) = 4 + 3 + 3 = 10 dB. • Loss at node 0: Connector loss = 2lc = 2 dB, ODMUX loss ldm = 3 dB, Hence, total loss = 2 + 3 = 5 dB. Further, the transmit power PT = 15 dBm and loss in each 5-km link = 0.2 ×5 = 1 dB. We therefore express the received power PR as 𝑃𝑃𝑅𝑅 = 𝑃𝑃𝑇𝑇 − (losses in 4 nodes + losses in 3 links)

= 15 − {(5 + 10 + 10 + 5) + 3 × 1} = −18 dBm,

which is larger than the receiver sensitivity (= -23 dBm) with a 5-dB power margin. Hence, the design is feasible.

8.9. Discuss critically the realizability of the various packet-switched WDM metro networks, particularly in respect of their scalability of physical size, node hardware and latency.

8.5

Solution: See the discussion on packet-switched WDM rings in Section 8.4.

8.10. Consider a five-node packet-switched WDM metro ring based on HORNET architecture. With the HORNET-node configuration presented in Fig. 8.28, formulate the power budget for the data packets with a ring circumference of 15 km. If the transmit power from the lasers is 15 dBm and the receiver sensitivity for the data packets is -27 dBm at 1 Gbps, estimate the worst-case power budget and comment on its feasibility. Given: insertion and connecting loss in the input splitter (in addition to the 10% tapping loss) = 1 dB, insertion loss in FBG (including circulator) = 2 dB, insertion loss in the output combiner = 2 dB, coupling and insertion loss in the output combiner = 1.5 dB, connector loss = 1 dB. Assume equal distance between adjacent nodes.

Solution: In order to examine the power budget of HORNET, we consider an example scenario as shown in the figure below.

Circulator

Circulator Con

Splitter

Con

FBG Combiner

10%

Splitter

FBG Combiner

3 intermediate nodes (without details Add CP Drop HORNET node (destination node)

Add CP Drop HORNET node (source node)

The figure shows the source and destination nodes along with three intermediate nodes for the longest connection span in the 5-node ring, with the add and drop ports (shown in bold face) of the source and destination nodes, being under consideration for the transmission and reception, respectively, between a given farthest node pair. In the figure, CP represents the tapped signal for the control packets, FBG stands for fiber Bragg grating, Con represents the connectors, while Add and Drop ports are for the data packets. For evaluation of the path loss, we define the necessary parameters as follows: lf = fiber loss in dB/km, lc = connector loss, lcom = combiner loss, lcoup = coupling loss in combiner, lfbg = FBG loss (including circulator), lsp = 1 + 10log10(10/9) = 1.456 dB = splitter loss for passthrough signal including the connector losses integrated with the device. Using these physical-layer settings, we express the received power in the farthest node as (e.g., node 0 to node 3 with 3 intermediate nodes with 4 fiber links each with a length = 15/5 = 3 km) 𝑃𝑃𝑅𝑅 = 𝑃𝑃𝑇𝑇 − ��𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑙𝑙𝑐𝑐 � + 3�𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑠𝑠𝑠𝑠 + 𝑙𝑙𝑓𝑓𝑓𝑓𝑓𝑓 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑙𝑙𝑐𝑐 � + 4 × 3 × 𝑙𝑙𝑓𝑓 + �𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑠𝑠𝑠𝑠 + 𝑙𝑙𝑓𝑓𝑓𝑓𝑓𝑓 + 𝑙𝑙𝑐𝑐 ��

= 15 − [(1 + 2 + 1) + 3{1 + 1.456 + 2 + 2 + 1} + 12 × 0.2 + {1 + 1.456 + 2 + 1}]

= 15 − [4 + 3 × 7.456 + 2.4 + 5.456] = −19.224 dBm,

which is larger than the receiver sensitivity (= - 27 dBm), thereby making the design feasible with a system margin of 7.776 dB.

8.6

Exercise Problems and Solutions for Chapter 9 (WDM Long-haul Networks) 9.1 As illustrated in the example shown in Fig.9.1, sketch some other candidate topological interconnections between multiple WDM rings to form mesh-like topology and specify the requirements of the interconnecting nodes in the proposed interconnected-ring networks.

Solution: MDR(4)

MDR(4)

MDR(4)

MDR(4)

MDR(4)

MDR(4)

MDR (3)

MDR(6)

MDR(5) MDR(3)

MDR(3)

MDR(5)

MDR(5) MDR(4)

MDR(4)

• • • •

Shaded boxes: multi-degree ROADMs with degree n (MDR(n)).

White boxes: ROADMs with a degree of 2.

Dashed lines: to interconnect separate rings. In-line EDFAs are needed when these links exceed a limit (≈ 60 km, typically).

All fiber links are bidirectional using WDM transmission and all nodes (ROADMs) are capable of wavelength routing.

9.1

9.2 Consider an MD-ROADM used in Fig. 9.1 at node nAB and sketch its configuration using optical demultiplexers, multiplexers, and switches. Sketch alternate configurations for the same MDROADM by using (i) optical splitters and WSSs and (ii) WSSs only. Comment on the basic differences between these configurations.

Solution: See the figures and the related discussions in Sections 2.16 and 9.2.

9.3 Draw the configuration for a MEMS-based OXC that can support three input and three output fibers with 16 wavelengths on each fiber, such that any four wavelengths from/to each fiber can be dropped/added through tunable transceivers (i.e., with logical nodal degree of four). Specify the sizes of the optical demultiplexers and multiplexers, optical switch (MEMS) and the number of tunable transceivers needed.

Solution: 3 × 3 OXC with Δ = 4 (tunable transceivers), number of wavelengths M = 16:

ODMUX

OMUX

.. .

16

.. 2 . 16

1 2

1

w1,w2, ..., w16

w1,w2, ..., w16

w1,w2, ..., w16

.. ..

52 × 52 . 32 MEMS 32 .

17 18

17 18

.. .

switch 33 34

33 34

48

48

49 50 51 52

49 50 51 52

4 tunable TXs

.. .

w1,w2, ..., w16

w1,w2, ..., w16

w1,w2, ..., w16

4 tunable RXs

9.4. Consider an N-node mesh-configured WRON offering a set C of connections {cs,d} with s and d representing the source and destination nodes for the connection cs,d. Assume that the connection cs,d employs l(s,d) fiber links. If the WRON has E bidirectional fiber links, determine the expression for the lower-bound on number of wavelengths needed in the network in terms of the number of nodes,

9.2

total inter-nodal fiber links used for the connections in C and the average physical degree of a node in the network.

Solution: The N-node WRON mesh employs C = {cs,d} connections, each connection having l(s,d) fiber links.

Hence, the total number of intermodal fiber links LT can be expressed as 𝐿𝐿 𝑇𝑇 = � 𝑙𝑙(𝑠𝑠, 𝑑𝑑). 𝑠𝑠,𝑑𝑑

On the other hand, the network with E edges has 2E links (each edge being bidirectional), and each link carries M wavelengths. Hence, total capacity of the network is 2EM, which turns out to be the upper-bound for LT, i.e., 2𝐸𝐸𝐸𝐸 ≥ 𝐿𝐿 𝑇𝑇,

leading to the lower limit for the number of wavelengths to support the network as 𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚 =

𝐿𝐿 𝑇𝑇 ∑𝑠𝑠,𝑑𝑑 𝑙𝑙(𝑠𝑠, 𝑑𝑑) = .

2𝐸𝐸 2𝐸𝐸

Using E and N, the average physical degree Dav of the nodes in the WRON can be expressed as

Thus, Mmin can be expressed using Dav as

𝐷𝐷𝑎𝑎𝑎𝑎 =

𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚 =

2𝐸𝐸 . 𝑁𝑁

∑𝑠𝑠,𝑑𝑑 𝑙𝑙(𝑠𝑠, 𝑑𝑑) . 𝐷𝐷𝑎𝑎𝑎𝑎 𝑁𝑁

9.5 Consider a 3 × 3 OXC for four wavelengths using MEMS in Spanke’s switch architecture and sketch the OXC configuration. Estimate the gain of an EDFA as the loss-compensating amplifier placed after the OXC. Given: connector loss = 1 dB, loss in optical demultiplexer/multiplexer = 3 dB, loss in a MEMS element = 1 dB. Solution: Consider the OXC shown in Fig. 2.80 with MEMS-based optical switch fabric using Spanke’s architecture, wherein we assume that there are 3 input/output ports operating with 4 wavelengths.

Further, we define the loss parameters as follows: loxc = OXC loss, lc = connector loss, LM = MEMS loss in one stage (it uses two stages in Spanke’s architecture), ldm = demultiplexer loss, lmx = multiplexer loss. Using these definitions, we obtain the OXC loss for a passthrough lightpath as a sum of three components (demultiplexer, switch, and multiplexer units) as 𝑙𝑙𝑜𝑜𝑜𝑜𝑜𝑜 = (𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑑𝑑𝑑𝑑 + 𝑙𝑙𝑐𝑐 ) + (𝑙𝑙𝑐𝑐 + 2𝑙𝑙𝑀𝑀 + 𝑙𝑙𝑐𝑐 ) + (𝑙𝑙𝑐𝑐 + 𝑙𝑙𝑚𝑚𝑚𝑚 + 𝑙𝑙𝑐𝑐 ) = 5 + 4 + 5 = 14 dB.

Hence the OXC should be followed by an EDFA with a gain of 14 dB to fully compensate for the OXC loss for a passthrough lightpath.

9.6 Consider an N-node WRON mesh with a given traffic matrix. The network vendor can afford ΔT transceivers for the entire network and decides to provision them for the nodes in accordance with the traffic handled by each node. Using this provisioning scheme, develop an expression for the number 9.3

of transceivers Δi for each node, while ensuring that each node has at least one transceiver to remain connected with the network.

Solution: Each node in the WRON must have at least one incoming and one outgoing lightpath to remain connected with the rest of the network. Besides this constraint, the logical in/out degree of a node is made proportional to the traffic handled by the node.

Hence, if ΔT represents the total number of transceivers available in the WRON, then (ΔT - 1× N) transceivers should be proportionately distributed to all nodes determined by the traffic handled therein. Assuming the network traffic matrix is mostly symmetric, and denoting λ(s,d) as the traffic from node s to node d, we therefore obtain the number Δi of transmitters/receivers in node i as ∆𝑖𝑖 = (∆ 𝑇𝑇 − 𝑁𝑁)

∑𝑑𝑑 𝜆𝜆(𝑖𝑖, 𝑑𝑑) + ∑𝑠𝑠 𝜆𝜆(𝑠𝑠, 𝑖𝑖) + 1. ∑𝑠𝑠 ∑𝑑𝑑≠𝑠𝑠 𝜆𝜆(𝑠𝑠, 𝑑𝑑)

9.7 Consider the seven-node WDM mesh network shown in Fig. 9.19 with two tunable transmitters and two tunable receivers per node and two wavelengths per fiber. Assume that the nodes don’t have wavelength converters. The traffic matrix Γ for node pairs in the network is given below (in arbitrary units): Γ=

0344521 3042210 3308641 2210220 5735077 1484102 0 1 4 2 2 3 0.

Determine a set of lightpaths to be constructed using an algorithm maximizing the single-hop traffic with the largest traffic considered first, where if two traffic elements have the same value, the first traffic element obtained in the row-wise sequence (i.e., beginning with the topmost row from left to right and then moving downward to the next row) is selected. Show the RWA for the lightpaths obtained by using the shortest-path and first-fit schemes for routing and WA, respectively.

Solution: Using the heuristic scheme on the WRON with the given traffic matrix, the RWA solution is obtained and presented in a tabular form as follows.

Serial No.

Lightpath route

Cost (distance)

Wavelength (wi)

1.

2.

3→ 4 6 → 2 →3

100 390

w1 w1

3.

5 →3 →2

420

w1

4.

5→6

400

w1

5.

3→5

120

w1

6.

1 →2 → 3 → 5

620

w2

9.4

7.

4 →5 → 6 → 7

650

w2

8.

1→2→3

500

w2

9.

6→2

90

w2

10.

2→ 1

200

w1

11.

7→6

160

w1

12.

2→6→7

250

w1

13.

4→3→2→1

600

w2

14.

7→6→5→4

650

w2

The direct lightpaths for other (s,d) pairs cannot be allocated, and the connections for these node pairs need to be accommodated with concatenated (multihop) lightpaths by using traffic grooming.

9.8 Evaluate the wavelength utilization ratio in a WRON mesh using 10 wavelengths with full and without wavelength conversion for five-hop connections for a blocking probability of 0.001. Using these results, estimate also the gain in wavelength utilization that can be achieved by using full wavelength conversion in a network which currently doesn't employ any wavelength conversion.

Solution: Number of wavelengths M = 10, number of hops H = 5, and the probability of blocking Pb due to resource exhaustion should be ≤ 0.001. Hence, the wavelength utilization without wavelength conversion is obtained as 1 𝐻𝐻

1/𝑀𝑀

𝛾𝛾𝑛𝑛𝑛𝑛 = − ln �1 − 𝑃𝑃𝑏𝑏

1 5

� = − ln�1 − 0.0011/10 � = 0.13910.

With full wavelength conversion at all nodes, the wavelength utilization improves to 𝑃𝑃𝑏𝑏 1/𝑀𝑀 0.001 1/10 𝛾𝛾𝑓𝑓𝑓𝑓 = =� � = 0.42668. 𝐻𝐻 5 Thus, with full wavelength conversion, one can obtain a gain in wavelength utilization given by 𝐺𝐺𝑤𝑤𝑤𝑤 =

0.42668 = 3.06743.

0.13910

9.9 In a given WRON, two client nodes (IP routers) are connected to two OXCs at two different WRON nodes, with four intermediate nodes between them. In the WRON, each fiber between any two nodes supports four wavelengths. Consider that 0.7 is the probability that a wavelength is being used on a fiber for an ongoing connection. a) Find the probability that a new connection request is blocked when the nodes are not equipped with wavelength converters. b) Find the probability that a new connection request is blocked when the nodes are equipped with wavelength converters, having full wavelength-conversion capability.

Solution: OXC

OXC

OXC

OXC

OXC

OXC

OXC Client

Client 9.5

Number of wavelengths M = 4, number of hops H = 5, wavelength utilization γ = 0.7. a) Without wavelength conversion: 𝑃𝑃𝑏𝑏 = [1 − (1 − 𝛾𝛾)𝐻𝐻 ]𝑀𝑀 = [1 − (1 − 0.7)5 ]4 = 0.99032.

b) With full wavelength conversion:

𝑃𝑃𝑏𝑏 = 1 − [1 − 𝛾𝛾 𝑀𝑀 ]𝐻𝐻 = 1 − [1 − 0.74 ]5 = 0.74661.

9.10 Figure 9.20 shows a five-node linear WRON with six all-optical connection requests in the form of VPs, needing WA. Form an auxiliary graph using the VPs (VPi's) shown in the given network, where these VPs would be represented by the equivalent nodes. Using the auxiliary graph, assign wavelengths (i.e., colors) to the VPs by using a heuristic scheme, where the nodes in the auxiliary graph with the largest degree are considered first in every step to assign the wavelengths.

Solution: Using the given linear network and the set of VPs assigned therein (i.e., VP1, VP2, VP3, VP4, VP5, and VP6), the auxiliary graph (with its nodes representing the VPs) is drawn, as shown below.

VP1

VP2

VP3

VP6 VP4

VP5

With the above auxiliary graph, the wavelengths indexed as w1, w2, ... and so on, we start coloring the the nodes, in descending order of the nodal degrees Di’s and obtain the wavelength assignment to the VPs (transforming them into lightpaths) as follows (nodes with the same degree are chosen arbitrarily in the assignment process).

• • • • • •

VP6 (D6 = 4): assigned wavelength = w1, VP2 (D2 = 4): having a common edge with VP6, cannot be assigned w1. Hence, VP2 is assigned the next available wavelength in the list, i.e., w2.

VP5 (D5 = 4): having common edges with VP2 and VP6, VP5 shouldn’t be assigned w1, w2.

Hence, VP5 is assigned the next wavelength w3.

VP4 (D4 = 4): having common edges with VP2, VP3, VP5 and VP6, VP4 shouldn’t be assigned w1, w2, and w3. Hence VP4 is assigned the next wavelength w4.

VP1 (D1 = 2): having common edges with VP2 and VP5, VP1 shouldn’t be assigned w2, w3.

Hence, VP1 is assigned the wavelength available with lowest index from the list, i.e., w1.

VP3 (D3 = 2): similarly, having common edges with VP6 and VP4, VP3 is assigned w2.

The readers are encouraged to repeat the above WA problem using ILP as well.

9.6

Exercise Problems and Solutions for Chapter 10 (Transmission Impairments and Power Consumption in Optical Networks) 10.1 How in practice is the power/energy-consumption efficiency defined in telecommunication networks? Discuss how the efficiency has evolved over the last three centuries. Also, discuss the conflicting roles between the ways to enhance power-consumption efficiency and spectrum utilization in the high-speed WDM networks.

Solution: See the discussion in Section 10.1.

10.2. Indicate the major differences between the shot and thermal noise components in optical receivers. Write down the expression for the shot-noise variance in an APD-based optical receiver and discuss how one can optimize the receiver performance (in respect of SNR) in the presence of APD excess noise.

Solution: Shot noise: signal-dependent noise, with its route cause lying in the optical source.

Thermal noise: doesn’t depend on signal and is typically a localized noise process in the receiver.

The shot-noise variance in an APD is expressed as 2 𝜎𝜎𝐴𝐴𝐴𝐴𝐴𝐴 = 2𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅 𝑀𝑀2 𝐹𝐹(𝑀𝑀)𝐵𝐵𝑒𝑒 ,

with the excess noise factor F(M) ≈ M x, x being in the range of [0,1]. Hence, the signal-to-noise ratio (SNR) in an APD-based receiver is expressed as 𝑆𝑆𝑆𝑆𝑆𝑆 =

(𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅 𝑀𝑀)2 .

2𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅 𝑀𝑀2+𝑥𝑥 𝐵𝐵𝑒𝑒 + 4𝑘𝑘𝑘𝑘𝐵𝐵𝑒𝑒 /𝑅𝑅

To maximize the SNR with an optimum APD gain Mopt, we equate the two terms in the denominator with M = Mopt as or

leading to Hence, we obtain Mopt as

2+𝑥𝑥 𝐵𝐵𝑒𝑒 = 2𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅 𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜

2+𝑥𝑥 𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜 =

4𝑘𝑘𝑘𝑘𝐵𝐵𝑒𝑒 , 𝑅𝑅

2𝑘𝑘𝑘𝑘 1 × = 𝑍𝑍 (say), 𝑅𝑅 𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃𝑅𝑅

(2 + 𝑥𝑥)log10 𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜 = log10 𝑍𝑍. 𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜 = 10{log 10 𝑍𝑍/(2+𝑥𝑥)}

10.3 Consider that, a PIN-based binary optical receiver, operating at 10 Gbps with an incoming optical signal at 1.55 μm, offers a BER of 10-9 with a received optical power of -18 dBm. Determine how inferior it is (in dB) with respect to an equivalent receiver operating in quantum limit. Discuss how one can improve the BER of this receiver. (Given: Planck's constant = 6.63 × 10-34 Joulesecond).

10.1

Solution: In quantum limit, the received optical power at 10 Gbps and 1550 nm for a BER = 10-9 is obtained as

Λ 𝑎𝑎𝑎𝑎 ℎ𝑓𝑓 Λ 𝑎𝑎𝑎𝑎 ℎ𝑐𝑐 10.5 × 6.63 × 10−34 × 3 × 108 𝑃𝑃𝑅𝑅 = = = = 13.474 nW = −48.73 dBm, 𝑇𝑇 𝑇𝑇𝑇𝑇 10−10 × 1550 × 10−9

with Λav as the average photon count received and T = 1/r is the bit interval. The given PIN-based receiver requires -18 dBm, and hence the receiver is inferior to the receiver in quantum limit by -18 – (-48.73) = 30.73 dB. In order to improve the receiver sensitivity, one needs to employ APD as the photodetector with appropriate APD gain.

10.4. Consider an optical communication link operating between two cities using WDM transmission over C band of a single-mode fiber. If the link uses 32 wavelengths, each operating at 10 Gbps, calculate the effective transmission capacity (in Gbps) of the link. Estimate the maximum allowable value of chromatic dispersion for each wavelength, if the link length is 100 km and the laser linewidth is 100 MHz. Ignore polarization-mode dispersion.

Solution: Number of wavelengths M = 32 and per-channel bit rate r = 10 Gbps. Hence, the effective transmission capacity of the WDM link is 𝐶𝐶𝑊𝑊𝑊𝑊𝑊𝑊 = 𝑀𝑀𝑀𝑀 = 32 × 10 = 320 Gbps.

The effective spectral width of the laser due to the inherent linewidth and modulation is estimated as (Eq. B.57) 𝐵𝐵𝐿𝐿𝑀𝑀 = �𝑟𝑟 2 + 𝛿𝛿𝛿𝛿 2 = �(1010 )2 + (107 )2 ≈ 10 GHz

The corresponding value for wavelength spread δw (with the central wavelength being ≈ 1550 nm in C band), is obtained as 𝛿𝛿𝛿𝛿 =

𝑤𝑤 2 𝑀𝑀 (1550 × 10−9 )2 𝐵𝐵 = × 1010 = 0.08 nm. 𝑐𝑐 𝐿𝐿 3 × 108

With Dch as the chromatic dispersion, the pulse spread over L = 100 km link is expressed as Δ𝜏𝜏𝑐𝑐ℎ = 𝐷𝐷𝑐𝑐ℎ 𝛿𝛿𝛿𝛿𝐿𝐿 = 𝐷𝐷𝑐𝑐ℎ × 0.08 × 100 ps,

which should be ≤ T with T = 100 ps (for per-channel 10 Gbps transmission). Thus, the link should satisfy the condition, given by

thereby needing to satisfy

𝐷𝐷𝑐𝑐ℎ × 0.08 × 100 < 100, 𝐷𝐷𝑐𝑐ℎ <

1 = 12.5 ps/nm/km.

0.08

10.5. A 1550-nm optical communication link transmits over 70 km of single-mode optical fiber at 10 Gbps without any optical regeneration. The laser at the transmitting end emits a power of 3 dBm and the fiber has a loss of 0.2 dB/km. The optical fiber link needs to splice multiple fiber spools, each having a length of 30 km. Each splicing incurs a splicing loss of 0.1 dB and each of the two connectors at the transmitting/receiving ends has a loss of 1 dB. If the receiver has a sensitivity of -18 dBm, estimate the system-power margin, if any.

Solution: 70

Transmit power PT = 3 dBm, fiber loss αf = 0.2 dB/km, number of splices 𝑛𝑛𝑠𝑠𝑠𝑠 = � � = 2, splice 30

loss nsp = 0.1 dB, and connector loss lc = 1 dB. We therefore express the received power PR as 10.2

𝑃𝑃𝑅𝑅 = 𝑃𝑃𝑇𝑇 − �70 × 𝛼𝛼𝑓𝑓 + 𝑛𝑛𝑠𝑠𝑠𝑠 × 0.1 + 2 × 𝑙𝑙𝑐𝑐 � = 3 − (70 × 0.2 + 2 × 0.1 + 2 × 1) = −13.2 dBm.

Hence, with a receiver sensitivity of -18 dBm, the system power margin is obtained as 𝑃𝑃𝑀𝑀 = −13.2 − (−18) = 4.8 dB.

10.6. Consider the optical fiber communication link of the last exercise, and estimate its system risetime. Comment on the result. Given: fiber dispersion = 8 ps/nm/km, transmitter rise time = 10 ps, laser linewidth = 100 MHz.

Solution: Chromatic dispersion in 100-km fiber is Δ𝜏𝜏𝑐𝑐ℎ = 𝐷𝐷𝑐𝑐ℎ 𝛿𝛿𝛿𝛿𝐿𝐿 with δw = 0.08 nm (see the solution of Exercise 10.4). Hence, Δ𝜏𝜏𝑐𝑐ℎ = 8 × 0.08 × 70 = 44.80 ps. Transmitter rise time tTX = 10 ps and receiver rise time tRX = 0.35/(10×109) = 35 ps. Hence, we obtain the system rise time tsys as 𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 = �∆𝑐𝑐ℎ 2 + 𝑡𝑡𝑇𝑇𝑇𝑇 2 + 𝑡𝑡𝑅𝑅𝑅𝑅 2 = �44.82 + 102 + 352 = 57.73 ps.

With the bit rate r = 10 Gbps, bit interval T = 100 ps, and thus the above design keeps more than 30% of the bit interval (i.e., 100 – 57.73 = 42.27 ps out of 100 ps) free to absorb the clock-timing jitter.

This makes the system feasible in respect of the system rise-time budget.

10.7. Discuss the needs of preamble bits in the data bursts transmitted in a PON. Discuss the pros and cons of using long preamble fields in the data bursts in respect of the power penalty in BMRs, clock jitter and bandwidth utilization.

Solution: See the discussions in Section 10.3.1.

10.8. Consider a TDM PON with the maximum distance between the OLT and ONUs as 20 km, where the RN uses an optical splitter with a splitting ratio of 1:32, which is realized using 2 × 2 optical splitters as the building block. Assuming that the upstream transmission takes place at 1 Gbps, verify whether the BMR at the OLT would ensure a BER ≤ 10-9. Given: PIN responsivity = 0.8 A/W, receiver input resistance = 50 ohms, receiver temperature = 270C, BMR power penalty = 2 dB, transmit power from ONU = 5 dBm, laser extinction = 5%, insertion loss in 2 × 2 optical splitter = 0.3 dB. Solution: The RN in the PON uses a 1:32 optical splitter, bit rate r = 1 Gbps, transmit power PT = 5 dBm, receiver input resistance R = 50 ohm, PON span L = 20 km, receiver temperature θ = 270C, fiber loss αf = 0.2 dB/km. The optical splitter loss lsp = 10log1032 + log232 × 0.3 = 16.5 dB. Using these parameters, we express the received power at a BMR as 𝑃𝑃𝑅𝑅 = 𝑃𝑃𝑇𝑇 − �𝐿𝐿𝛼𝛼𝑓𝑓 + 𝑙𝑙𝑠𝑠𝑠𝑠 + 4𝑙𝑙𝑐𝑐 � = 5 − (20 × 0.2 + 16.5 + 4) = −19.5 dBm

The received power is the average of the power levels P1 and P0 received for binary 1 and 0 transmitted from the source node, respectively. Hence, BER at the receiver is expressed as 1 2

𝑅𝑅 (𝑃𝑃 −𝑃𝑃 )

1 2

𝑄𝑄

𝐵𝐵𝐵𝐵𝐵𝐵 = erfc � 𝑤𝑤2(𝜎𝜎 1+𝜎𝜎0 )� = erfc � 2�. √

1

0

√

With the laser extinction ratio ε = 0.05, we obtain P1 and P0 as follows.

10.3

1 1 𝑃𝑃𝑅𝑅 = (𝑃𝑃1 + 𝑃𝑃0 ) = (𝑃𝑃1 + 𝜖𝜖𝜖𝜖1 ).

2 2

Hence, we obtain

2𝑃𝑃𝑅𝑅 = and 𝑃𝑃0 = 𝜖𝜖𝑃𝑃1 .

1 + 𝜖𝜖 Due to the power penalty ΔP in BMR, PR will get reduced effectively by the same amount, leading to the revised values of P1 and P0. With ΔP = 2 dB, the equivalent received power would be revised as PR = -19.5 – 2 = -21.5 dBm = 10-2.15 × 10-3 = 7.08 × 10-6 W. Hence, we obtain P1 and P0 from the revised PR as 𝑃𝑃1 =

2𝑃𝑃

𝑃𝑃1 = 1+𝜖𝜖𝑅𝑅 =

2×7.08 × 10 −6 = 13.4857 𝜇𝜇W, 1+0.05

𝑃𝑃0 = 𝜖𝜖𝑃𝑃1 = 0.05 × 13.4857 = 0.6743 𝜇𝜇W

Using P1 and P0, we express σ12 and σ02 (both having shot and thermal noise components) as 4𝑘𝑘𝑘𝑘𝐵𝐵𝑒𝑒 , 𝑅𝑅 4𝑘𝑘𝑘𝑘𝐵𝐵𝑒𝑒 . 𝜎𝜎02 = 2𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃0 𝐵𝐵𝑒𝑒 + 𝑅𝑅 Using the above formulas, we obtain the values of σ12 and σ02 as 𝜎𝜎12 = 2𝑞𝑞𝑅𝑅𝑤𝑤 𝑃𝑃1 𝐵𝐵𝑒𝑒 +

10 9

𝜎𝜎12 = 2 × 1.6 × 10−19 × 0.9 × 13.4857 × 10−6 × 0.7 × 109 + 4 × 1.38 × 10−23 × 300 × 0.7 × 50 = 2345.59 × 10−16 𝐴𝐴2 .

𝜎𝜎02 = 2 × 1.6 × 10−19 × 0.9 × 0.6743 × 10−6 × 0.7 × 109 + 4 × 1.38 × 10−23 × 300 × 0.7 × = 2319.76 × 10−16 𝐴𝐴2 .

10 9 50

Hence, we obtain the Q factor of BMR receiver as 𝑄𝑄 =

𝑅𝑅𝑤𝑤 (𝑃𝑃1 − 𝑃𝑃0 ) √2(𝜎𝜎1 + 𝜎𝜎0 )

=

0.9(13.4857 − 0.6743) × 10−6

1.414 × �√2345.59 × 10−16 + √2319.76 × 10−16 � -9

= 8.44.

The value of Q required to ensure a BER of 10 is 5.99, which is lower than the present value and hence the BMR would offer a feasible power budget.

10.9 Consider a building complex, which needs to be provisioned either by Ethernet or GPON for about 2000 end-users. As shown in Fig. 10.11, the GPON design supports two OLTs connected to one core Ethernet switch from the top and each OLT can support 40 GPON ports and each port can connect to 32 ONUs using OSPs. However, the Ethernet design employs three levels of switching: core, distribution, and access, with two core switches, two distribution switches and 16 access switches, with each access switch being able to support 150 switched ports. Thus, both designs employ a degree of fault tolerance at the switching levels, where the Ethernet design can accommodate at the most 16 × 150 = 2400 end-users, and the GPON design can support at the most 2 × 40 × 32 = 2560 ONUs, thereby ensuring that both designs can serve close to but not less than 2000 end-users. The power consumptions in the respective network elements are as follows: Ethernet core switch: 836 W, Ethernet distribution switch: 725 W, Ethernet access switch: 970 W, GPON OLT: 2500 W, GPON ONU: 10 W. 10.4

Calculate the power consumptions in both networks and explain using the results why GPON would be preferred as a green architecture by the service providers.

Solution: Ethernet consumptions: Core switch power consumption Pcore = 836 W, Distribution switch power consumption Pdist = 725 W, Access switch power consumption Pacc = 970 W Total power consumption in the Ethernet is obtained as 𝑃𝑃𝑇𝑇 (𝑒𝑒𝑒𝑒ℎ𝑒𝑒𝑒𝑒) = 2𝑃𝑃𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 2𝑃𝑃𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 16𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = 2 × 836 + 2 × 725 + 16 × 970 = 18,642 W. Power consumption for the service provider is obtained as 𝑃𝑃𝑆𝑆𝑆𝑆 (𝑒𝑒𝑒𝑒ℎ𝑒𝑒𝑒𝑒) = 𝑃𝑃𝑇𝑇 (𝑒𝑒𝑒𝑒ℎ𝑒𝑒𝑒𝑒) = 18,642 W. GPON consumptions:

Core switch power consumption Pcore = 836 W, OLT power consumption Polt = 2500 W, ONU power consumption Ponu = 10 W Total power consumption in GPON is obtained as 𝑃𝑃𝑇𝑇 (𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔) = 2𝑃𝑃𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 2𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 + 2000 × 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 = 2 × 836 + 2 × 2500 + 2000 × 10 = 26,672W. Power consumption for the service provider is obtained as

𝑃𝑃𝑆𝑆𝑆𝑆 (𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔) = 2𝑃𝑃𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 2𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 = 2 × 836 + 2 × 2500 = 6,672 W.

Thus, even though the total power consumption is more in GPON, the service providers will have lower power consumption at their end in GPON and hence would prefer GPON as compared to Ethernet.

10.10 A TDM PON operates with 32 ONUs, with each ONU operating in deep-sleep LPM to save power consumption. If the power-saving and time-saving ratios in the ONUs are 20% and 40%, respectively, determine the overall power-saving ratio in the PON. Repeat the calculation when the values of power-saving and time-saving ratios are reversed.

Solution: Power-consumption efficiency in a TDM PON operating in deep-sleep mode is given by 𝑁𝑁−1 − 𝜌𝜌𝑇𝑇 � (1 − 𝜌𝜌𝑃𝑃 ), 𝑁𝑁

𝜂𝜂𝑝𝑝𝑝𝑝𝑝𝑝 = �

with ρT and ρP representing the time and power-saving ratios, respectively, and N as the number of ONUs. For the PON under consideration, N = 32 ONUs. Power-saving ratio ηP = 20%, time-saving ratio ηT = 40%: 31 𝜂𝜂𝑝𝑝𝑝𝑝𝑝𝑝 = � − 0.4� (1 − 0.2) = 45.5%.

32 Power-saving ratio ηP = 40%, time-saving ratio ηT = 20%: 31 𝜂𝜂𝑝𝑝𝑝𝑝𝑝𝑝 = � − 0.2� (1 − 0.4) = 46.1%.

32 Thus, both (time/power) saving schemes have comparable effect on the power consumption, though one needs to choose them judiciously with due consideration to the hardware design aspects of the ONUs and the overall network performance.

10.5

Exercise Problems and Solutions for Chapter 11 (Survivability in Optical Networks) 11.1. The problem of survivability against failures in access and metro segments is generally addressed using protection-based designs. On the other hand, the long-haul mesh networks can employ protection and/or restoration schemes to ensure network survivability. Justify this practice with reference to the basic features and constraints of the respective network segments.

Solution: See the discussions in Section 11.2.

11.2. Discuss the pros and cons of the UPSR and BLSR2/4 schemes in SONET rings. Explain how the upper-bound on recovery time (50 ms) influences the design of SONET rings. By using a suitable illustration, describe a scheme to recover from a node failure in the BLSR2 scheme.

Solution: See the discussions in Section 11.3.

11.3. Consider that two SONET rings using UPSR-based protection scheme are interconnected using dual access, thereby involving four nodes (with two nodes from each ring). Sketch the two interconnected rings and illustrate the protection scheme that can be used for inter-ring connections.

Discuss how the devices used for the protection scheme might influence the scalability of the ring network.

Solution: Protection in two interconnected UPSR rings (Ring A and Ring B) is realized with dual access between the two rings involving two nodes from each by using drop-and-continue scheme, as shown below. Optical protection switches (OPSs), optical splitters (OSPs), and drop-and-continue (DC) devices (which are also basically splitters) are utilized for implementing the scheme. Note that, the use of power splitters (in OPSs and DCs) would reduce the scalability of ring sizes on both sides.

B2

A2 A3

B3 OPS

A1

B1

DC DC

OSP

OPS

OPS OSP

OPS

OPS

OPS DC

DC

A4

B4

A5

B5

Ring A

Ring B

11.1

11.4. As indicated in the section on PON protection, the survivability of a PON can be assessed in terms of the unavailability of its connections to the ONUs, by using the concept of asymptotic unavailability of the network components, given by Ux = τR × failure rate, with x representing the various PON elements/devices. Using this model and the following values of Ux's, estimate the worstcase availability of the ONU connections in the unprotected TDM PON shown in Fig. 4.2 (c). Assume that the PON has a feeder length of 15 km and the distribution fibers are all within 5 km.

Asymptotic unavailabilities: OLT: UOLT = 5.12 × 10-7, ONU: UONU = 5.12 × 10-7, optical fiber: UF = 3.42 × 10-6/km, remote node (RN) using optical splitter (OSP): URN = UOSP = 10-7, optical switch (OS): UOS = 4 × 10-7 (used in the next problem).

Solution: The PON given in Fig. 4.2 (c) is an unprotected configuration, and hence all the relevant asymptotic unavailability entities will appear in series. Thus, all the asymptotic unavailability values will add up to give the overall unavailability of connections in the PON. For the fibers used in the feeder and distribution segments with the length of LF = LFF + LDF (= 15 + 5 = 20km), the overall unavailability of fiber will be equal to (LF × UF). Thus, we obtain the PON unavailability as 𝑈𝑈𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑈𝑈𝑂𝑂𝑂𝑂𝑂𝑂 + 𝑈𝑈𝐹𝐹 𝐿𝐿𝐹𝐹 + 𝑈𝑈𝑅𝑅𝑅𝑅 +𝑈𝑈𝑂𝑂𝑂𝑂𝑂𝑂

= 5.12 × 10−7 + (3.42 × 10−6 × 20) + 10−7 + 5.12 × 10−7

= 6.9524 × 10−5 .

11.5. Repeat the above problem with a PON using the Type-A protection scheme as shown in Fig.11.9.

Solution: In this case, the feeder fiber segment is kept protected with parallel fibers, while the distribution fibers remain unprotected. Hence, for the feeder fiber segment the unavailability value will be (UFLFF)2, while the distribution fiber will have the unavailability value = UFLDF. Hence, the unavailability of the PON will improve as 𝑈𝑈𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑈𝑈𝑂𝑂𝑂𝑂𝑂𝑂 + (𝑈𝑈𝐹𝐹 𝐿𝐿𝐹𝐹𝐹𝐹 )2 + 𝑈𝑈𝐹𝐹 𝐿𝐿𝐷𝐷𝐷𝐷 + 𝑈𝑈𝑅𝑅𝑅𝑅 + 𝑈𝑈𝑂𝑂𝑂𝑂𝑂𝑂

= 5.12 × 10−7 + (3.42 × 10−6 × 15)2 + (3.42 × 10−6 × 5) + 10−7 + 5.12 × 10−7 = 1.8224 × 10−5 .

11.6. Compare the two-fiber OCh-DP-Ring and the four-fiber OCh-SP-Ring protection schemes used for WDM ring networks, in respect of the recovery time and differential delay in full-duplex connections. Discuss the implications of these differences while setting up a WDM ring network.

Solution: See the discussion in Section.11.5.

11.2

11.7. Consider a long-haul WRON with its lightpaths carrying data at 10 Gbps across the network, where each fiber can carry 40 wavelengths. Imagine that the network has been bisected using a cut-set C generating two parts having 10 fiber links in the cut-set. With this bisected view, one finds that the WRON has suffered a link failure, which happens to be one of the 10 fiber links in the cut-set. The total traffic following across the cut-set before the link failure was found to be 1000 Gbps, which needs to be accommodated with the 9 out of 10 fiber links due to the link failure. In this scenario, if the average usage of the wavelengths is 60%, estimate the upper-bound for the traffic scale-up factor in the network after the link failure.

Solution: Cut-set C

.. .

In the cut-set C, there are k = 10 links, and one of them fails. Hence, after the failure, the cut-set will have (k – 1) = 9 fiber links. Each fiber carries M = 40 wavelengths. Before the failure, the traffic flow between the two parts (shaded areas) of the network was rcs = 1000 Gbps. After the failure, the same traffic would flow over 9 fiber links with a bandwidth usage of wavelengths u = 0.6..

Hence, after the failure, the effective data flow in a wavelength across the cut-set is obtained as 𝜌𝜌𝑐𝑐 =

𝑟𝑟𝑐𝑐𝑐𝑐 1000 = = 4.63 Gbps.

0.6 × 40 × 9 𝑢𝑢𝑢𝑢(𝑘𝑘 − 1)

Thus, the traffic can be scaled up in the network after the failure by a factor αSU , given by 𝛼𝛼𝑆𝑆𝑆𝑆 =

10 10 = = 2.16. 𝜌𝜌𝑐𝑐 4.63

11.3

Exercise Problems and Solutions for Chapter 12 (Optical Network Control and Management) 12.1. Using a suitable illustration, describe how the SONET overhead bytes carry out the tasks of OAM. Solution: See the discussion in Section 12.4.

12.2. Discuss the pros and cons of the distributed and centralized control planes in telecommunication networks.

Solution: See the discussion in Section 12.7.

12.3. Considering the coexistence of the control (distributed) and data planes in WRONs, as shown in Fig. 9.6, propose and draw a suitable 3 × 3 node configuration for an OXC-based mesh-connected WRON with a logical nodal degree (in/out) of 2. Each node needs to perform the operations of both the planes, with the control information being handled by using one unique wavelength out of 8 wavelengths available in the network. Assume that the control-plane topology follows the physical topology of the network. Also indicate how the node configuration will change, if the control plane uses a ring topology.

Solution: A 3 × 3 WRON node with the symmetrical control plane (i.e., the control information flows over all links in the physical topology) is shown below. Note that, the control wavelength w0 after demultiplexing at each input port visits the control module, wherein OEO conversion is carried out.

After the processing of the control information in electrical domain, appropriate control signal is passed onto the MEMS switch for switch reconfiguration, while the output control information are forwarded to the multiplexer inputs at the control wavelength w0.

Control module w0 w0, w1, ..., w7 w0, w1, ..., w7 w0, w1, ..., w7

w0

...

w1

w1

w7

w7

...

w1

.. .

w1

w1

w7

w7

w7

MEMS switch

w1 w7

.. .

w0, w1, ..., w7

...

w0, w1, ..., w7

.. .

w0, w1, ..., w7

TR TR

TT TT

12.1

If the above 3 × 3 WRON node uses a control plane designed with a ring topology, then only one out of the three input/output ports of the control module will be active and carry the control plane information.

12.4. Mention the driving forces behind the development of the MPLS protocol and explain the role of FEC in MPLS protocol. Consider an MPLS network with five ingress packets arriving with the destination (IP) addresses a.b.c.3, a.b.c.5, a.b.c.7, a.b.d.4, and a.b.d.9 at one of its edge nodes (i.e., at an LER). Explain with a suitable illustration how the LER will treat them by using their CoSs and destination addresses to form the respective FECs and set up the corresponding LSPs. Solution: See the discussion in Section 12.5.1 12.5. Describe the role of suggested labels in setting up unidirectional LSPs in a GMPLS network.

Discuss a suitable method to set up bidirectional LSPs in GMPLS networks.

Solution: See the discussion in Section 12.5.3.

12.6. Describe how ASON is used to interconnect multiple domains of GMPLS networks and roles of the necessary ASON interfaces therein. Illustrate the interdomain routing between various GMPLS domains using ASON interfaces.

Solution: See the discussion in Section 12.6.

12.7. Describe with a suitable illustration how the centralized control operation is realized in SDN. Explain the roles of the flow tables in OpenFlow protocol in controlling the flows of circuit as well as packet-switched traffic.

Solution: See the discussion in Section 12.7.2.

12.8. Consider a cluster of N SDN nodes operating with one control node (see Fig. 12.17). Each SDN node has an ingress arrival rate of λ, of which on average 30% packets are forwarded to the control node. The control node serves these requests from the SDN nodes with a service rate of μ. Using the concept of open network of queues with feedback (see Section 12.7.5), determine the maximum possible value of N that can be accommodated in the SDN cluster, so that the control node can remain in the steady state.

Solution: Refer to Fig. 12.17. Each SDN node in the figure forwards on the average 30% of its incoming packets (with an arrival rate of λ) to the controller. Hence, the total traffic coming to the SDN controller from N nodes is given by Γc = 𝑁𝑁𝑁𝑁 × 0.3.

In order that the controller remains in steady state, one must limit Γc below the service rate μ of the

controller, i.e.,

Γc < 𝜇𝜇.

12.2

Substituting for Γc in the above expression, we obtain leading to

𝑁𝑁𝑁𝑁 × 0.3 < 𝜇𝜇,

𝜇𝜇 𝜇𝜇 is not an integer, � when 0.3𝜆𝜆 0.3𝜆𝜆 𝜇𝜇 𝜇𝜇 − 1 when is an integer. 𝑁𝑁 ≤ 0.3𝜆𝜆 0.3𝜆𝜆

𝑁𝑁 ≤ �

12.9. Explain how SDN orchestration helps in realizing multi-domain networks and mention the basic differences between the horizontal and vertical frameworks of SDN orchestration. Describe, with suitable illustration, how SDN operates through a multi-domain network using a horizontal framework.

Solution: See the discussion in Section 12.7.4.

12.10. Explain, with suitable diagram, how the legacy WDM nodes can be retro-fitted in an SDON. Construct an example SDON schematic for setting up circuit- and packet-switched connections.

Solution: See the discussion in Section 12.7.6.

12.3

Exercise Problems and Solutions for Chapter 13 (Datacenters)

13.1. Consider a DCN using a 6-ary fat tree architecture and find out the following: a) b) c) d)

number of servers, number of ToR switches, number of aggregate switches, number of core switches.

Taking the two extreme ToR switches at the two ends, trace the path between them and justify that the connection is non-blocking.

Solution: For the 6-ary fat tree (k = 6), the numbers of servers and the switches at three levels are obtained as: a) number of servers = k3/4 = 63/4 = 54, b) number of ToR switches = k2/2 = 62/2 = 18, c) number of aggregate switches = k2/2 = 62/2 = 18, d) number of core switches = k2/4 = 62/4 = 9.

Sketch the DCN following the method used in Fig. 13.3, and trace the path from two extreme ports of the two extreme ToR switches. The connection should be non-blocking in nature as the fat-tree topology is a folded version of Clos architecture. Prove this by observation.

13.2. Consider that a DCN employs Aspen tree as its physical topology with four levels of switching with each switch having six ports. Calculate the following: a) number of ToR switches for the three FTVs: (0,0,0), (0,2,0) and (2,2,2), b) number of servers for the three FTVs: (0,0,0), (0,2,0) and (2,2,2).

Sketch the Aspen tree topology with four levels and six-port switches with the FTV of (2,2,2) and illustrate how the number of servers is significantly reduced to achieve high level of fault tolerance.

Solution: For the Aspen tree with n = 4 levels and k = 6 ports in switches, the numbers of ToR switches and servers for different values of FTV are as follows: a) number of ToR switches (S) 𝐹𝐹𝐹𝐹𝐹𝐹 = (𝜙𝜙𝑛𝑛 , 𝜙𝜙𝑛𝑛−1 , … , 𝜙𝜙2 ),

𝐶𝐶𝐶𝐶𝐶𝐶 = (𝜙𝜙𝑛𝑛 + 1)(𝜙𝜙𝑛𝑛 + 1) … (𝜙𝜙2 + 1), 63

𝑘𝑘 𝑛𝑛 −1

𝑆𝑆 = 2𝑛𝑛 −2 ×

FTV = (0,0,0): CEF = 1×1×1 = 1, 𝑆𝑆 = 22 × 1 = 54.

63

1

FTV = (0,2,0): CEF = 1×3×1 = 3, 𝑆𝑆 = 22 × 3 = 18.

63

1

FTV = (2,2,2): CEF = 3×3×3 = 27, 𝑆𝑆 = 22 × 27 = 2.

13.1

1 . 𝐶𝐶𝐶𝐶𝐶𝐶

b) number of servers (NS) 𝑘𝑘 𝑛𝑛

𝑁𝑁𝑆𝑆 = 2𝑛𝑛 −1 ×

64

FTV = (0,0,0): CEF = 1, 𝑆𝑆 = 23 × 1 = 162.

64

1 . 𝐶𝐶𝐶𝐶𝐶𝐶

1

FTV = (0,2,0): CEF = 3, 𝑆𝑆 = 23 × 3 = 54.

64

1

FTV = (2,2,2): CEF = 27, 𝑆𝑆 = 23 × 27 = 6.

Aspen tree with n = 4, k = 6 and FTV = (2,2,2)

Level 4 Level 3

Switch

Switch

Level 2

Switch

Switch

Level 1

Switch

Switch

Servers

Note that, due to high fault-tolerance (i.e., with FTV (2,2,2)), all three uplink ports in a switch at a given level (below level 4) are used up for one single higher-level switch, thereby bringing down the number of servers to only 6 as compared to the one without any protection (i.e., 162 with FTV (0,0,0)).

Switch

S

S

S

S

S

S

13.3 A server-centric DCN has to be designed following DCelln topology with the basic DCell network (i.e., DCell0) having one mini-switch connected to five servers. Design and draw the DCell0 and DCell1 topologies. If the overall DCelln topology needs to accommodate at least 800,000 servers, determine the value for n.

Solution: DCell0 : 1 miniswitch and 5 servers MS S: server MS: miniswitch

S

S

S

13.2

S

S

DCell1: 5 + 1 = 6 miniswitches and 5 × (5 + 1) = 30 servers.

MS

MS

MS S S

S

S

S

S

S

S

S

S

S

S

MS

S

S

S

S

S

S

S

S

S

S

MS

MS S

S

S

S

S

S

S

S

Similarly, DCell2 will have 30 × (30 + 1) = 930 servers, and DCell3 will finally be able to accommodate 930 × 931 = 865,830 servers, thereby reaching the 800,000 mark.

13.4. A DCN using fat-tree topology employs 48-port Ethernet switches as the building block.

Assume that the links in the DCN carry an average traffic of 0.75 Erlang, and it is desirable that the probability of packet loss in the switches, each of which uses a shared buffer for all ports, remains below 10-6. Assuming that the packet arrivals and service durations follow Poisson and exponential distributions, respectively, estimate the needed buffer size in the switches. Also indicate how the estimated buffer size would perform, when the arrival processes at the switch ports become nonPoissonian with heavy-tail distributions. Given: average packet size = 1500 byte.

Solution: Each up/downlink port of a switch is assumed to follow the M/M/1/K queuing model. Hence, each 48port switch with a shared buffer would have a buffer size of 48 (K-1), with K chosen at a value so that the probability of packet loss PL ≤ 10-6.

The probability of packet loss in an M/M/1/K queue is expressed as

With ρ = 0.75, K = 43, we get PL as

𝑃𝑃𝐿𝐿 =

(1 − 𝜌𝜌)𝜌𝜌𝐾𝐾 .

1 − 𝜌𝜌𝐾𝐾+1

(1 − 0.75)0.7543 𝑃𝑃𝐿𝐿 = = 1.06−6 .

1 − 0.7544

Using K = 43, 48 switch ports and an average packet size of 1500 byte, we obtain the full buffer size BS of a switch (with a PL ≈ 10-6) as 𝐵𝐵𝑆𝑆 = 48 × (𝐾𝐾 − 1) × 1500 = 48 × 42 × 1500 = 3.024 MB. 13.3

Note that, with non-Poissonian heavy-tail distribution of the arrival processes, the buffer size will have to be increased beyond the one obtained in the above.

13.5. Explain why the two sets of addresses, location and application addresses (LA and AA), are used in a VL2 network. Justify how the valiant load-balancing (VLB) in VL2 addresses the challenges of the traffic agility and network failures.

Solution: See the discussion in Section 13.2.1.

13.6. Why does Helios employing HyPac switching classify the traffic flows into mouse and elephant categories and distribute these traffic streams between the optical and electrical switching units?

Discuss how one should decide the threshold between the mouse and elephant traffic flows.

Solution: See the discussion in Section 13.2.2.

13.7. Consider the OSA architecture for an all-optical DCN, and determine the end-to-end path loss for an all-optical connection between any two ToR switches. Using this estimate, determine the transmit power needed from each transmitter if the receiver sensitivity for each wavelength is -18 dBm at 10 Gbps. Given: loss in optical multiplexer/demultiplexer = 3 dB, loss in WSS = 3 dB, loss in circulator = 2 dB, loss in combiner = 2 dB, loss in MEMS switch = 4 dB, connector loss = 1 dB. Solution: Receiver sensitivity = -18 dBm, multiplexer loss lmx = 3 dB, demultiplexer loss ldm = 3 dB, WSS loss lwss = 3 dB, circulator loss lcir = 1 dB, MEMS switch loss lsw = 4 dB, combiner loss lcomb = 2 dB, connector loss lc = 1 dB, bit rate r = 10 GBps. Note that, an end-to-end connection will have to go through 16 connectors altogether, with 16 × 1 dB = 16 dB loss.

Hence, the needed transmit power PT (including the power margin PM = 4 dB) is obtained as 𝑃𝑃𝑇𝑇 = −18 + (𝑙𝑙𝑚𝑚𝑚𝑚 + 𝑙𝑙𝑤𝑤𝑤𝑤𝑤𝑤 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑙𝑙𝑠𝑠𝑠𝑠 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑙𝑙𝑑𝑑𝑑𝑑 + 16𝑙𝑙𝑐𝑐 ) + 𝑃𝑃𝑀𝑀 = −18 + (3 + 3 + 1 + 4 + 1 + 2 + 3 + 16) + 4 = 19 dBm.

13.8. Consider an OSPS-FT architecture of DCN, where the lowest level of the network uses 4-ary fat trees in electrical domain and a MEMS-based optical switch at the top. Below the MEMS switch, for the three levels of PSC-based multihop WDM LANs, the DCN uses three types of ShuffleNet as follows: • • •

(2,3) ShuffleNets at level 1 (bottom-most level of PSCs), (4,2) ShuffleNets at level 2, one (2,3) ShuffleNet at level 3 (top-most level of PSCs).

13.4

Using the above design guideline, construct and sketch the entire OSPS-FT architecture and evaluate the number of servers the DCN can support, the number of PSCs at levels 1 and 2 and the size of the MEMS switch. Also specify the sizes of the PSCs needed at each level and check whether the power budget of the network would be feasible, if each optical link has a length ≤ 15 m and all transmitters transmit 10 dBm of optical power at 10 Gbps with a receiver sensitivity of -18 dBm. Given: connector loss = 1 dB, fiber loss = 0.2 dB/km, insertion loss in each 2 × 2 optical splitter in a PSC = 0.5 dB. Solution: The given OSPS-FT architecture uses one MEMS switch, 3 levels of WDM LAN and 4-ary fat trees.

The following parameters are defined for the design: • • • •

number of level-3 PSC (L3-PSC): 1, L3-PSC uses (2,3) ShuffleNet with NL3SN = kpk = 3×23 = 24 nodes. number of level-2 PSC (L2-PSC): N2, each L2-PSC uses (4,2) ShuffleNet with NL2SN = kpk = 2×42 = 32 nodes. number of level-1 PSC (L1-PSC): N1, each L1-PSC uses (2,3) ShuffleNet with NL1SN = kpk = 3×23 = 24 nodes.

MEMS size: K × K, K = N1 × N2.

Each 4-ary fat tree has the following features: • •

number of servers S = 43/4 = 16, number of core switches NCSFT = 42/4 = 4, and each core switch must be connected to an L1PSC.

L1-PSC employs a (2,3) ShuffleNet, with the capacity NL1SN = 3×23 = 24 nodes, and the corresponding PSC (nearest available) uses 32 ports. Each L1-PSC needs to connect with (i) MEMS switch using 1 port, (ii) one L2-PSC using 1 port, and the remaining ports should connect to the core switches of the underlying fat trees, with each fat tree needing NCSFT = 4 ShuffleNet ports for its core switches.

Thus, out of 24 nodes of ShuffleNet (from 32 ports of L1-PSC), 2 nodes are used for the connections to MEMS switch and L2-PSC, and hence the remaining 24 - 2 = 22 nodes can accommodate L fat trees, given by 𝑆𝑆𝑆𝑆 24 − 2 𝑁𝑁𝐿𝐿1 −2 𝐿𝐿 = � � = 5. 𝐶𝐶𝐶𝐶 � = � 4 𝑁𝑁𝐹𝐹𝐹𝐹

Hence, each L1-PSC can support 5 4-ary fat trees, with each 4-ary fat tree supporting 16 servers, leading to L × S = 5 × 16 = 80 servers, while some of the L1-PSC ports remain unused. On the other hand, L3-PSC has N2 = NL3SN = 24 nodes (out of 32 PSC ports) connected to 24 L2-PSCs, while each L2-PSC can utilize all of its 32 ports and connect downward to N1 = NL2SN -1 = 32 – 1 = 31 L1-PSCs (since one of its ports must connect to L3-PSC). With these observations, we obtain the following design: • • •

total number of L1-PSCs N = N2 × N1 = 24 × 31 = 744, size of MEMS switch (N × N): 744 ×744, total number of servers Stotal = N LS = 744 × 5 × 16 = 59,520.

Next, we examine the power budget for the largest (worst-case) PSC; however in this example all PSCs have 32 ports and hence 1:32 power splitting.

13.5

The power loss lpsc in the 32-port PSC would have two components, one due to the power-splitting and the other due to the insertion loss. Thus, lpsc is obtained as 𝑙𝑙𝑝𝑝𝑝𝑝𝑝𝑝 = 10log10 32 + log 2 32 × 0.5 = 17.5 dB.

With the total fiber loss lf = 0.2 × 0.015 = 0.003 dB, connector loss lc = 1 dB, and the receiver sensitivity PR(min) = -18 dBm, we obtain the required transmit power PT (with a system power margin PM = 5 dB) as 𝑃𝑃𝑇𝑇 = 𝑃𝑃𝑅𝑅 (min) + 𝑙𝑙𝑝𝑝𝑝𝑝𝑝𝑝 + 𝑙𝑙𝑓𝑓 + 4𝑙𝑙𝑐𝑐 + 𝑃𝑃𝑀𝑀 = −18 + 17.5 + 0.003 + 4 × 1 + 5 = 8.503 dBm.

Hence, 10-dBm transmitters at 10 Gbps would work for the given OSPS-FT architecture, wherein the additional losses in circulators (used in the WDM LAN nodes to separate transmitted and received signals) get also accounted for.

OSPS-FT design: MEMS switch (744 × 744)

...

1

744

L3-PSC (32 ports) 24-node ShuffleNet 1

24

L2-PSC (32 ports) (32-node ShuffleNet) 1

5 × 4 = 20

...

L2-PSC (32 ports) ... (32-node ShuffleNet) 1

31

FTG-1

FT-1

... ... ...

FT-2

FT-5

31 FTG-744

L1-PSC (32 ports) (24-node ShuffleNet)

...

...

...

FT: fat tree (16 servers), FTG: fat-tree group (5 FTs), Note that, the OEO blocks are not shown explicitly here (see Fig. 13.17 for the placement of the OEO blocks)

5 x 16 = 80 servers 13.9. In a long-haul WRON mesh, 6 contents/objects need to be replicated at multiple DCN locations, governed by their respective popularity indices. Following Zipf's distribution, estimate the probabilities of accessing the objects using their respective popularity indices, i.e., 1 through 6.

Assume that there are six DCNs placed at as many locations across the WRON with an ordered list, where the DCN locations are arranged in ascending order of their mean distances from the other nodes. Using the popularity indices of the objects and the location-based ordered list of the DCNs, determine the locations of the DCNs, where each object has to be placed/replicated.

Solution: Zipf’s distribution for the access probability φi is given by 13.6

𝛽𝛽

𝜙𝜙𝑖𝑖 = 𝑖𝑖 =

With the number of objects Q = 6, we obtain β as 𝛽𝛽 =

𝑄𝑄

1 −1 𝑖𝑖

�∑𝑖𝑖=1 � 𝑖𝑖

1

1 1 1 1 1 1+2+ + + + 3 4 5 6

.

= 0.4082.

Using the value of β we next obtain the values of φi for the six objects as • • • • • •

φ1 = β/1 = 0.4082, φ2 = β/2 = 0.2041, φ3 = β/3 = 0.1361, φ4 = β/4 = 0.1021, φ5 = β/5 = 0.0817, φ6 = β/6 = 0.0680.

In the given problem the number of datacenters K = 6 = Q. With K = Q, the numbers of replications Ri (including the original) for the object i is given by 𝑅𝑅𝑖𝑖 = �𝜙𝜙

(𝐾𝐾−1)𝜙𝜙 𝑖𝑖

𝑚𝑚𝑚𝑚𝑚𝑚 −𝜙𝜙 𝑚𝑚𝑚𝑚𝑚𝑚

Hence, we obtain the values of Ri’s for i = 1 to 6, as • • • • • •

𝑅𝑅1 = �

(6−1)×0.4082

0.4082−0.0680 (6−1)×0.2041

𝑅𝑅2 = � 𝑅𝑅3 = � 𝑅𝑅4 = � 𝑅𝑅5 = � 𝑅𝑅6 = �

0.4082−0.0680 (6−1)×0.1361 0.4082−0.0680 (6−1)×0.1021 0.4082−0.0680 (6−1)×0.0817 0.4082−0.0680 (6−1)×0.0680 0.4082−0.0680

�.

� = ⌈5.99⌉ = 6,

� = ⌈2.99⌉ = 3,

� = ⌈2.003⌉ = 3, � = ⌈1.5⌉ = 2, � = ⌈1.2⌉ = 2,

� = ⌈. 999⌉ = 1.

The objects would be placed in 6 locations, L1 through L6, which are ordered with the increasing value of the average distance from other nodes in the network (i.e., L1 has the least average distance and L6 has the highest average distance). With the heuristic that the most popular object will be placed in all locations, and the least popular one will get only one location, and hence L1, we obtain the placements of the objects as follows.

1.

2.

3.

4.

5.

6.

Object 1 (most popular): with R1 = 6, the object is placed at the locations L1 through L6.

Object 2: with R2 = 3, the object is placed at the locations L1 through L3.

Object 3: with R3 = 3, the object is placed at the locations L1 through L3.

Object 4: with R4 = 2, the object is placed at the locations L1 and L2.

Object 5: with R5 = 2, the object is placed at the locations L1 and L2.

Object 6 (least popular): with R6 = 1, the object is placed at location L1.

13.7

Exercise Problems and Solutions for Chapter 14 (Elastic Optical Networks) 14.1. Consider the versions of OOFDM transceivers that need to employ coherent optical transmission. Justify the need of coherent transmission in the respective transceivers and identify the components that would need special care to ensure coherent operation of the transceivers.

Solution: See the discussions in Section 14.3.

14.2. Consider an all-optical OOFDM receiver (Fig. 14.7) to be used in EON. Indicate the roles of the components that are used to carry out the task of signal processing in optical domain for OFDM demodulation.

Solution: The optical devices used in the fi-planes (Fig. 14.7) would in effect perform the Fourier transform, and hence any deviation from the designed values for these devices (delay lines and phase shifters) will affect the accuracy of the signal processing. Further, the timing jitter in the gating instants in optical gates (OGs) following the combiner needs to be controlled to avoid timing offsets in picking up the signal at the right instants of time.

14.3. You are given four transmission systems for optical WDM links using a 50 GHz transmission grid, which have to operate at 10 Gbps, 40 Gbps, 100 Gbps, and 400 Gbps. Grade them as Nyquist, quasi-Nyquist, super-Nyquist, or non-Nyquist systems and justify your statement using the basic criterion of Nyquist transmission.

Solution: The channel bandwidth Δf = 50 GHz. Hence, the class of transmission system is decided by the ratio ρ of the channel bandwidth Δf to the bit rate r as follows. a) r = 10 Gbps: ρ = Δf/r = 50/10 = 5 > 1. Hence, this is a non-Nyquist transmission system, b) r = 40 Gbps: ρ = Δf/r = 50/40 = 1.25. In this case, ρ being greater than but close to 1, this is a quasi-Nyquist transmission system. c) r = 100 Gbps: ρ = Δf/r = 50/100 = 0.5 <1. This is a super-Nyquist transmission system. d) r = 400 Gbps: ρ = Δf/r = 50/400 = 0.125 <1. This is a super-Nyquist transmission system.

14.4. In a given EON, five connection requests have been made with the speeds r1 =10 Gbps, r2 = 20 Gbps, r3 = 40 Gbps, r4 = 100 Gbps, and r5 = 400 Gbps. Each FS has the width of 12.5 GHz and the width of the guard bands is the same as one FS. Further, BPSK modulation is used for each of the FSs, which are combined using OFDM when multiple FSs are to be transmitted. Estimate the number of FSs needed for the five connection requests.

Solution: With BPSK modulation, the number of FSs for a connection request i is given by 𝑟𝑟𝑖𝑖 𝑛𝑛𝑖𝑖 = � � + 𝑛𝑛𝐺𝐺 , Δ𝑓𝑓 with nG = 1 representing one FS for the guard band.

Hence, the required numbers of FSs for the five connection requests are obtained as: 14.1

10 � + 1 = 2, 12.5 20 𝑛𝑛2 = � � + 1 = 3, 12.5 40 𝑛𝑛3 = � � + 1 = 5, 12.5 100 𝑛𝑛4 = � � + 1 = 9, 12.5 400 𝑛𝑛5 = � � + 1 = 33.

12.5

a) 𝑛𝑛1 = � b) c) d) e)

14.5. Consider the above exercise with QPSK-OFDM transmission and estimate the number of FSs needed for the five connection requests.

Solution: With QPSK modulation, the number of FSs for a connect request i is given by 𝑟𝑟𝑖𝑖/2 𝑛𝑛𝑖𝑖 = � � + 𝑛𝑛𝐺𝐺 , Δ𝑓𝑓 with nG = 1 representing one FS for the guard band.

Hence, the required numbers of FSs for the connection requests are obtained as: 10/2 � + 1 = 2, 12.5 20/2 𝑛𝑛1 = � � + 1 = 2, 12.5 40/2 𝑛𝑛1 = � � + 1 = 3, 12.5 100/2 𝑛𝑛1 = � � + 1 = 5, 12.5 400/2 𝑛𝑛1 = � � + 1 = 17.

12.5

a) 𝑛𝑛1 = �

b) c) d) e)

14.6. Consider the four-node EON, as shown in Fig. 14.27, with the link lengths indicated beside the respective links in arbitrary units. Assume that the width of each FS is 12.5 GHz, and each connection uses a guard band of one FS, appended at the end of its allocated FS(s). The network needs to set up all connections as end-to-end all-optical lightpaths. Carry out an offline RSA for the EON using the heuristic scheme described in Section 14.6, where each link can carry at the most eight FSs in both directions. The traffic matrix Γ in terms of transmission rates (in Gbps) is given by: Γ=

0 50 25 60 25 0 50 60 25 30 0 30 25 50 30 0.

From the spectral assignment determine the spectral utilization in the network and the probability of blocking due to resource exhaustion.

Note that, while working on the solution manual (after the submission of the book manuscript), the traffic matrix Γ given in the book exercise (which was a sparse matrix with some of the off-diagonal matrix elements being zero) has been replaced by a full traffic matrix (i.e., with all non-zero offdiagonal elements) to make the RSA problem more worthwhile and interesting, which is evident explicitly for the next exercise problem (Exercise 14.7) with connection blocking, where the traffic matrix is scaled up by a factor of 1.5.

14.2

Solution: Following the approach used in the last two exercises, the given traffic matrix (in Gbps) is transformed into the equivalent number of FSs (without any guard band) as follows: ΓFS = {γ(sd)} = 0 4 2 5 2045 2303 2 4 3 0.

Next, we tabulate below the above requests ri’s of FSs in the decreasing order of number of FSs along with one FS as guard band.

Ordered request (ri)

r1

r2

r3

r4

r5

r6

r7

r8

r9

r10

r11

r12

Connection

C14

C24

C12

C23

C42

C32

C34

C43

C13

C21

C31

C41

5+1 =6

5+1 =6

4+1 =5

4+1 =5

4+1 =5

3+1 =4

3+1 =4

3+1 =4

2+1 =3

2+1 =3

2+1 =3

2+1 =3

(Csd) γ(sd) + 1

For a given connection Csd, the necessary fiber links are allocated along the shortest path with the requested FSi’s following the heuristic scheme described in Section 14.6. Using this method with the number of available slots M = 8 in each fiber link, we obtain the offline RSA, as tabulated below.

FSi

FS1 FS2 FS3

FS4

FS5 FS6

C14

r1

r1

r1

r1

r1

C24

r2

r2

r2

r2

r2

C12

r3

r3

r3

r3

C23

r4

r4

r4

C42

r5

r5

C32

r6

C34 C43

FS7

FS8

Route

Number of FSs used (excluding guard bands)

r1

1→4

5×1

r2

2→4

5×1

r3

1→2

4×1

r4

r4

2→1→3

4×1×2 (for 2 links)

r5

r5

r5

4→2

4×1

r6

r6

r6

3→2

3×1

r7

r7

r7

r7

3→4

3×1

r8

r8

r8

r8

4→3

3×1

Csd

C13

r9

r9

r9

1→3

2×1

C21

r10

r10

r10

2→1

2×1

C31

r11

r11

r11

3→1

2×1

C41

r12

r12

r12

4→1

2×1

14.3

The number of FSs used (excluding guard bnds) = (2 × 5) + (4 × 4) + (3 × 3) + (4 × 2) = 43. Total number of available FSs in all (6) fiber links (bidirectional) with 8 FSs in each link = 8 × (6 × 2) = 96.

Hence, spectrum utilization ηspec in the network is 𝜂𝜂𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

43 × 100% = 44.79%, 96

without any blocking due to resource unavailability.

14.7. Scale up the traffic in the above exercise by a factor of 1.5 and carry out the offline RSA again.

Compare the results obtained in the two cases.

Solution: The traffic in the network of Exercise 14.6 is scaled up by a factor of 1.5. With the scaled up traffic, the traffic matrix is changed as Γ=

0 75.0 37.5 90.0 37.5 0 75 0 90.0 37.5 45.0 0 45.0 37.5 75.0 45.0 0, leading to ΓFS = {γ(sd)} = 0 6 3 8 3068 3404 3 6 4 0.

Next, as before, we tabulate below the above requests ri’s of FSs in the decreasing order of number of FSs along with one FS as guard band Ordered request (ri)

r1

r2

r3

r4

r5

r6

r7

r8

r9

r10

r11

r12

Connection

C14

C24

C12

C23

C42

C32

C34

C43

C13

C21

C31

C41

8

8

6+1 =7

6+1 =7

6+1 =7

4+1 =5

4+1 =5

4+1 =4

3+1 =4

3+1 =4

3+1 =4

3+1 =4

(Csd) γ(sd) + 1

Note that for C14 and C24 with 8 FSs, no guard band is necessary (as total number of slots in a link is M = 8). Again, for a given connection Csd, the necessary fiber links are allocated along the shortest path with the requested FSi’s. Using this method, we obtain the offline RSA, as tabulated below.

FSi

FS1

FS2

FS3

FS4

FS5

FS6

FS7

FS8

Route

Number of FSs used (excluding guard bands)

C14

r1

r1

r1

r1

r1

r1

r1

r1

1→4

7×1

C24

r2

r2

r2

r2

r2

r2

r2

r2

2→4

7×1

C12

r3

r3

r3

r3

r3

r3

r3

1→2

6×1

C23

r4

r4

r4

r4

r4

r4

r4

2→1→3

6×1×2

C42

r5

r5

r5

r5

r5

r5

r5

4→2

6×1

Csd

14.4

C32

r6

r6

r6

r6

r6

3→2

4×1

C34

r7

r7

r7

r7

r7

3→4

4×1

C43

r8

r8

r8

r8

r8

4→3

4×1

Blocked

0

C13 C21

r10

r10

r10

r10

2→1

3×1

C31

r11

r11

r11

r11

3→1

3×1

C41

r12

r12

r12

r12

4→1

4×1

Total number of FSs utilized = 56, leading to ηspec as 𝜂𝜂𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

60 × 100% = 62.5%, 96

however with the blocking of 1 out of 12 connections, implying a blocking probability of (1/12) × 100 = 8.33% due to resource exhaustion. Thus, with a traffic scale-up factor of 1.5, the blocking due to resource exhaustion has increased from zero to 8.33%.

14.8. Consider the six-node EON shown in Fig. 14.28 with the following connection requests.

Node pair | FSs requested | Selected path 1,2

2

1→2

1,3

2

1→2→3

4,6

2

4→5→6

2,4

1

2→3→4

1,4

1

1 → 2→ 3 → 4

Each connection in the EON uses one FS as the guard band, appended after the requested FS(s). The total number of FSs available in each link (bidirectional) is 10. a) Carry out the spectral allocation and illustrate the same on the network topology, b) Calculate the spectral fragmentation for each fiber link and the average spectral fragmentation across the entire network.

Solution: The EON is shown below with FS allocations on the respective links as per the given RSA. Node pair | FSs requested | Selected path

FSi’s allotted (with guard band)

1,2

2

1→2

FS1, FS2, FS3

1,3

2

1→2→3

FS4, FS5, FS6

4,6

2

4→5→6

FS1, FS2, FS3

2,4

1

2→3→4

FS7, FS8

1,4

1

1 → 2→ 3 → 4

FS9, FS10

14.5

FSs: 4,5,6(G),7,8(G),9,10(G) FSs: 1,2,3(G),4,5,6(G), 9,10(G)

2

700

200

100

1

FSs: 4,5(G),7,8(G),9,10(G)

3

800 200

900 700

6

700

5

4 150

FSs: 1,2,3(G),4,5(G),9,10(G)

FSs: 1,2,3(G),4,5(G),F,7,8(G)

Fragmentation F(l) in a link l is defined as

𝐹𝐹𝑖𝑖𝑖𝑖 =

𝑇𝑇 𝐿𝐿𝐿𝐿 (𝑙𝑙) (𝑙𝑙)− 𝑁𝑁𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑁𝑁𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑇𝑇 (𝑙𝑙) 𝑁𝑁𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

.

𝑇𝑇 𝐿𝐿𝐿𝐿 (𝑙𝑙) and 𝑁𝑁𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (𝑙𝑙) representing the total number of free FSs and the number of FSs in the with 𝑁𝑁𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

largest free band of FSs in the same link. Hence the fragmentation on the links are obtained as follows: • •

F(12) = 0, F(23) = 0, F23 = (3-2)/3 = 1/3, F45 = 0, F56 = 0 Other links have no FS allocation, hence fragmentation = 0 for them.

Therefore average fragmentation in the network Fav is obtained as

𝐹𝐹𝑎𝑎𝑎𝑎 =

1/3 = 1.85%.

9×2

14.6

Exercise Problems and Solutions for Chapter 15 (Optical Packet and Burst-Switched Networks) 15.1. Consider a 1200 byte optical packet to be switched at an OPS node with a bit rate of 40 Gbps.

What should be the maximum allowable switching time for the OPS node? Comment on the suitability of the various possible optical-switching technologies for the OPS node.

Solution: 1200-byte optical packets transmitted at a bit rate of 40 Gbps would have a packet duration τops,

given by 1 s = 240 ns.

40 × 109 The maximum allowable switching time in an OPS node must be one order lower than τops. Hence, the switching time of the OPS node should be ≤ τops /10 = 24 ns. For the candidate optical-switching technologies for OPS nodes, see the discussions in Section 15.2.3. 𝜏𝜏𝑜𝑜𝑜𝑜𝑜𝑜 = 1200 × 8 ×

15.2. Consider an OPS node operating with the configuration shown in Fig. 15.1 and express its total insertion loss in terms of the losses incurred in their constituent optical devices. Assuming the given losses in various constituent building blocks as given below, estimate the insertion loss in the node and indicate whether one should use post-amplifier following the node, and if so, what should be the amplifier gain. Given: insertion loss in demultiplexer/multiplexer = 3 dB, power tapping for header processing = 10%, insertion loss in power tap = 0.5 dB, insertion loss in FDL (for largest delay) = 5 dB, insertion loss in the optical switch = 12 dB, insertion loss in optical combiner = 2 dB, connector loss = 1 dB. Assume that the TWCs (using SOAs) don't contribute to the node insertion loss.

Solution: Insertion loss in an OPS node will have the following components: demultiplexer loss ldm, loss at the optical tap ltap, loss in FDL lfdl, loss in TWC ltwc, loss in switch lsw, loss in combiner lcomb, loss in multiplexer lmx, and the losses lc’s in all connectors used to interconnect all the elements in the node and the input/output fibers (14 connectors). Hence, by tracing along the path of a transit signal, we express the total insertion loss in an OPS node as 𝑜𝑜𝑜𝑜𝑜𝑜

𝑙𝑙𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑙𝑙𝑑𝑑𝑑𝑑 + 𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑙𝑙𝑓𝑓𝑓𝑓𝑓𝑓 + 𝑙𝑙𝑠𝑠𝑠𝑠 + 𝑙𝑙𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑙𝑙𝑚𝑚𝑚𝑚 + 14 × 𝑙𝑙𝑐𝑐.

The total loss in an optical tap will depend on the tapping ratio and the insertion loss in the device, given by 𝑜𝑜𝑜𝑜𝑜𝑜

Hence, we obtain 𝑙𝑙𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 as 𝑜𝑜𝑜𝑜𝑜𝑜

10 � + 0.5 = 0.96 dB. 9

𝑙𝑙𝑡𝑡𝑡𝑡𝑡𝑡 = 10log10 �

𝑙𝑙𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 = 3 + 0.96 + 5 + 12 + 2 + 3 + 14 × 1 = 39.96 dB. With this loss (indeed a large loss), one needs to use a post-amplifier with a gain of 40 dB. 15.3. Consider a WDM network operating with both WRON-based optical circuit-switching (OCS) as well as OBS, where for the control operation one wavelength is kept reserved, and the OCS and OBS parts of the network are assigned three and two wavelengths, respectively. At each node, the control wavelength from the adjacent nodes undergoes OE-EO conversions for sharing OCS/OBS control information with the rest of the node. Draw a suitable configuration for a 3 × 3 node with the given network-setting, that will be able to cater to the needs of both OCS and OBS operations. Discuss on the feasibility of independent control operations of the OCS and OBS segments of the network with 15.1

one control wavelength and the switching technologies that would be appropriate for the OCS and OBS operations.

Solution: The block schematic of the OCS-cum-OBS node is shown below with w0 as the control wavelength for both OCS and OBS, w1, w2, and w3 as the wavelengths for the OCS operation, and w4 and w5 as the wavelengths for the OBS operation. The OCS switches can use MEMS as the switching technology, while the OBS switches need to be agile enough such that its switching takes place with an adequate time margin before the burst arrivals. Both, the OCS and OBS parts of the node, have two tunable transceivers. The control wavelength uses subcarrier multiplexing with two subcarriers, for the OCS and OBS control operations. Tunable transceivers are employed for OCS and OBS to make use of the available wavelengths for transmission.

w0 w0,w1, ..., w5 w0,w1, ..., w5 w0,w1, ..., w5

Control OCS

OBS

w1 w2 w3

w1 w2 w3

OCS switch

w1 w2 w3

w1 w2 w3

w1 w2 w3 w1 w2 w3

w0 w0,w1, ..., w5 w0,w1, ..., w5 w0,w1, ..., w5

Control

w4 w4 w5 w5 w4 OBS w4 w5 w5 w4 switch w4 w5 w5

Tunable TXs Tunable RXs

Control 15.4. Sketch a block schematic for a programmable FDL, where an optical packet needs to be delayed with a variable delay from zero to 0.7 μs in steps of 0.1 μs. Design the FDL. Solution: The FDL schematic is shown below with four 2 × 2 optical switches and three fiber segments having the delay values of 0.1 μs, 0.2 μs, and 0.4 μs. While using fiber segments with the velocity of light in -6 the core as ≈ 2 × 108 m/s, a delay of 0.1 μs would require a fiber length of (2 × 108) × (0.1 ×10 ) = 20 m. When all of the four switches are in the bar state, the FDL would delay the input signal by 0.7 μs, and with the input/output switches in the cross state and the rest configured in the bar state, the input signal will bypass all the delay elements with zero delay.

15.2

0.1 μs

0.2 μs

0.4 μs

Input

Output 2×2 switch

2×2 switch

2×2 switch

2×2 switch

15.5. Consider that an optical burst has to reach from its ingress node to the egress node through 5 intermediate nodes by using JET protocol. The total length of the chosen route through the intermediate nodes across the network is 200 km along the optical fiber links, and each node in the network needs 100 ns to process the incoming control packet for the subsequent burst transmission, while the network transmits over each wavelength at a bit rate of 10 Gbps. a) Calculate the necessary offset time for the control packet with respect to the burst for which it is being sent. b) Assume that the average burst size in the network is 10,000 bytes, and the first packet in a burst in the ingress node has to wait with an average waiting time of 10 μs before the burst carrying this packet is transmitted. Estimate the average delay incurred by the first packet in a burst from the given pair of ingress and egress nodes.

Solution: a) There are 5 intermediate nodes between the ingress and egress nodes, and the processing time τp needed therein should be included in the offset time Δ. Thus, we obtain Δ as ∆= 5𝜏𝜏𝑝𝑝 = 5 × 100 ns = 500 ns = 0.5 μs.

b) Average burst duration is obtained as 𝜏𝜏𝐵𝐵 = 10,000 × 8 × 10−10 s = 8 μs, and the propagation delay along the path is 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 200 × 5 μs = 1000 μs. Furthermore, the first packet of the burst has to wait in the queue on the average for 𝜏𝜏𝐵𝐵𝐵𝐵 = 10 μs. Hence, the overall delay DP for the first packet of the burst can be obtained as 𝐷𝐷𝑃𝑃 = 𝜏𝜏𝐵𝐵 + 𝜏𝜏𝐵𝐵𝐵𝐵 + Δ + 𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 8 + 10 + 0.5 + 1000 = 1018.5 μs = 1.0185 ms.

15.6. An OBS network is realized with its nodes using TWCs, however without any FDL. The network operates with five wavelengths, each transmitting at 10 Gbps. Consider a core node in this network, where optical bursts arrive at a rate of 50,000 bursts/second with an average burst size of 125,000 bytes. Estimate the burst-loss probability at the node using the M/M/c/c queuing model.

Solution: The number of wavelengths M = 5, burst arrival rate λ = 50,000 bursts/s, average burst size B = 12,500 byte = 100,000 bit. Hence, the average burst duration τB = 100,000 × 10-10 s = 10-5 s, implying that the service rate μ is μ = 1/ τB = 105 bursts/s.

Using the values of λ and μ, we obtain the traffic as 𝜌𝜌 =

𝜆𝜆 50,000 = = 0.5.

105 𝜇𝜇 15.3

Hence, following Erlang B formula, we obtain the probability of burst loss PBL as 𝑃𝑃𝐵𝐵𝐵𝐵 =

𝜌𝜌𝑀𝑀 ⁄𝑀𝑀!

0.55 ⁄5! = = 1.58 × 10−4 .

5 𝑖𝑖 ⁄𝑖𝑖! 𝑖𝑖 ⁄𝑖𝑖! ∑𝑀𝑀 𝜌𝜌 ∑ 0.5 𝑖𝑖=0 𝑖𝑖=0

15.7. An OBS network operates with the DR-based JET signaling by using FDLs. Assume that the two control packets CP1 and CP2 have arrived one after another with a difference of 0.8 μs (CP1 has arrived earlier) (earlier it was 0.1µs in the book). The offset times for CP1 and CP2 are 0.2 μs and 0.25 μs, respectively. The durations of the corresponding data bursts, i.e., B1 (for CP1) and B2 (for CP2), are 1μs and 1.2 μs, respectively. Determine the minimum length of the FDL needed to prevent contention between the two data bursts.

Solution: With reference to Fig. 15.9, we express the required minimum FDL delay as 𝑑𝑑𝑚𝑚𝑚𝑚𝑚𝑚 = (𝑡𝑡01 + Δ1 + 𝜏𝜏𝐵𝐵1 ) − (𝑡𝑡02 + Δ2 ) = 𝜏𝜏𝐵𝐵1 − (𝑡𝑡02 − 𝑡𝑡01 ) − (Δ2 − Δ1 ).

In the present case, 𝑡𝑡02 − 𝑡𝑡01 = 0.8 𝜇𝜇𝜇𝜇, Δ2 − Δ1 = 0.05 𝜇𝜇𝜇𝜇, and 𝜏𝜏𝐵𝐵1 = 1 𝜇𝜇𝜇𝜇. Hence, we obtain dmin as 𝑑𝑑𝑚𝑚𝑚𝑚𝑚𝑚 = 1 − 0.8 − 0.05 = 0.15 𝜇𝜇𝜇𝜇,

-6 which will need an FDL length Lfdl = v × dmin = (2 × 108 m/s) × (0.15 × 10 s) = 30 m.

15.4