Two Numbers To Focus on For Dividends

Dividends are a source of regular income for investors. Because of that, it should be no surprise that the dividend yield of a share is an important part of the investor’s decision-making process as it clues an investor in to the current income he or she can receive.

But, it can be dangerous to focus on a share’s dividend yield alone.

Instead, here are two metrics that investors can include in their tool-kit for a better analysis of a company’s dividends: Free Cash Flow; and Net Cash Balance.

What’s a Free Cash Flow?

In the daily course of every business, it either generates or consumes cash. When a business is able to generate cash, it then uses the cash to touch-up existing income-producing assets. After which, any left-overs can then be allocated by management to pay out dividends; make acquisitions; expand the business’s asset base; store it in the proverbial “vault”; or buy-back shares, among others.

That left-over cash is known as Free Cash Flow. Mathematically, the numbers can be gleaned from a company’s Cash Flow Statement and the equation’s given as:

Free Cash Flow = Cash Flow from Operations – Purchase of Property, Plant & Equipment (otherwise known as Capital Expenditures)

… And What’s a Net Cash Balance?

The Net Cash Balance tells us how much cash a company has stored up for future use, net of all debt. Companies with a large Net Cash Balance have an additional safety net to rely on to pay creditors, bills, suppliers, wages, loans, even dividends etc. when its business runs into any temporary slowdown in which it is unable to generate sufficient cash.

The required figures are found in the company’s Balance Sheet and its equation is given as:

Net Cash Balance = Total Cash & Equivalents – Short-Term Debts & Long Term Debts

Why the cash is important

While dividend investors often like to focus on the pay-out ratio, which measures dividends as a percentage of net income, we have to bear in mind that ultimately, dividends are paid out in cash.

Companies can report great profits but without booking the actual cash into its coffers, it is going to find it difficult to pay out dividends in the future. So, even though it’s important for companies to have profits, we should also keep an eye on the cash.

A Tale of Two Dividends

Let’s use Japanese ramen-restaurant operator Japan Foods Holding (SGX: 5OI) and industrial fishing company China Fishery Group (SGX: B0Z) as examples on the importance of the two metrics: Free Cash Flow and Net Cash Balance.

Company	Price per share	Dividend Yield*
Japan Foods	S$0.79	3.2%
China Fishery	S$0.375	5.1%
*Based on last completed financial year’s full-year payout

Both companies have dividend yields higher than the Straits Times Index’s (SGX: ^STI) yield of around 2.6% (using the index tracker SPDR STI ETF’s (SGX: E3B) data as a proxy) so they might both be considered attractive dividend shares on that basis alone.

China Fishery might even be considered the more lucrative of the two based on its higher yield. But, a different picture emerges when we consider the companies’ Free Cash Flow and Net Cash Balance, as seen from the charts below.

We can see from the charts that Japan Foods has seen both its Free Cash Flow and Net Cash Balance grow steadily over its last five completed financial years (FYs). Meanwhile, China Fishery has had cumulative Free Cash Flow of negative US$15m over its last five completed FYs and consistently carried more debt than cash.

Under such a backdrop, it’s not that surprising to see China Fishery’s dividend fall to 1.9 Singapore cents per share for FY 2012 from a bonus-share-adjusted 5.48 Singapore cents per share for FY 2008.

Investors who tracked profits alone would likely be left scratching their heads on the big drop in dividends as the industrial fishing company’s earnings per share dropped by ‘only’ 30% from US$0.11 to US$0.076 in the same time period.

But for those who had been keeping an eye on China Fishery’s Free Cash Flow and Net Cash Balance, it would have been clearer to them that the company was just not able to bring in sufficient cash.

Meanwhile, Japan Foods – with growing Free Cash Flows and a balance sheet that becomes healthier with each passing year – has been able to grow its dividends from a bonus-share-adjusted 0.167 Singapore cents in FY 2009 to 2.5 Singapore cents in FY 2013 as the company is able to generate ample cash from its daily job of dishing out ramen and more to diners.

The Japanese ramen-restaurant operator, with its historical ability to generate Free Cash Flow and to maintain a healthy balance sheet, could be seen as a better choice than China Fishery for future dividends, ceteris paribus.

That’s especially so in light of the higher perceived risk of the latter dropping its dividends still further if it continues to struggle in producing meaningful Free Cash Flows from its business in the future.

Foolish Bottom Line

Every publicly listed company has to prepare three financial statements: the Income Statement; Cash Flow Statement; and Balance Sheet. All three have to be studied together to paint a holistic picture of a company’s financials and business condition.

That’s important for every investor – not just for dividend investors – to keep in mind.

EIGRP - Auto Summarization

I dont like auto-summarization, especially since you can always manual summarize networks as you please in EIGRP. But it seems to be enabled by default in all the IOS versions I use and it's probably a good idea to know how it works for the lab.

EIGRP auto-summarizes at classful network boundaries. This means that if one interface is in 1.1.0.0/16 and another interface is in 2.2.2.0/24 then EIGRP will summarize the networks before advertising them out of the opposite interface. This is because each subnet belongs to a separate major classful network: 1.0.0.0/8 and 2.0.0.0/8 in this case.

However if you have one interface in 2.2.2.0/24 and another interface in 2.3.0.0 then auto-summarization will not take affect because both interfaces belong to the major network 2.0.0.0/8. EIGRP may auto-summarize for some networks and not others in the same EIGRP process. Here we take a look:

[R1]----[R2]

R1 and R2 are connected to the same LAN 12.0.0.0/16

R1 has two loopbacks:

R1#show run | section interface Loopback
interface Loopback0
ip address 1.1.1.1 255.255.255.255
interface Loopback1
ip address 12.1.1.1 255.255.255.0

R2 also has two loopbacks:

R2#show run | section interface Loopback
interface Loopback0
ip address 2.2.2.2 255.255.255.255
interface Loopback1
ip address 12.2.2.2 255.255.255.0

All interface have EIGRP enabled with auto-summary on (default).

R1's route table:

R1#show ip route | begin Gate
Gateway of last resort is not set
1.0.0.0/8 is variably subnetted, 2 subnets, 2 masks
C 1.1.1.1/32 is directly connected, Loopback0
D 1.0.0.0/8 is a summary, 00:09:40, Null0
D 2.0.0.0/8 [90/156160] via 12.0.0.2, 00:09:34, FastEthernet0/0
12.0.0.0/8 is variably subnetted, 4 subnets, 3 masks
D 12.2.2.0/24 [90/156160] via 12.0.0.2, 00:09:43, FastEthernet0/0
C 12.1.1.0/24 is directly connected, Loopback1
C 12.0.0.0/16 is directly connected, FastEthernet0/0
D 12.0.0.0/8 is a summary, 00:09:40, Null0

R2's route table:
R2#show ip route | begin Gate
Gateway of last resort is not set

D 1.0.0.0/8 [90/156160] via 12.0.0.1, 00:19:01, FastEthernet0/0
2.0.0.0/8 is variably subnetted, 2 subnets, 2 masks
C 2.2.2.2/32 is directly connected, Loopback0
D 2.0.0.0/8 is a summary, 00:18:55, Null0
12.0.0.0/8 is variably subnetted, 4 subnets, 3 masks
C 12.2.2.0/24 is directly connected, Loopback1
D 12.1.1.0/24 [90/156160] via 12.0.0.1, 00:00:02, FastEthernet0/0
C 12.0.0.0/16 is directly connected, FastEthernet0/0
D 12.0.0.0/8 is a summary, 00:18:55, Null0
R2#

Analysis:

Looking at R1's route table you can see that R2 has summarized the route 2.2.2.2/32 into 2.0.0.0/8. This is because R2 sits in the boundary between two classful networks 12.0.0.0 and 2.0.0.0. It does not summarize loopback 1 which is 12.2.2.0/24. This shows up on R1 with the 24 bit mask.

On R2 the same can be said for routes from R1. R1 summarizes it's loopback0 interface into 1.0.0.0/8 because it sits on the boundary between 1.0.0.0 and 12.0.0.0. It does not summarize 12.1.1.1/24.

RIP version 2 auto-summarization works pretty much the same way as far as I can tell. If you know any differences please let me know.

EIGRP Variance & Traffic Distribution Ratios

If you have been given two links, one with 2 Mbps & another is 512Kbps. How would you load balance in the below ratio with EIGRP?

• 5:2
• 4:3

What does the above mean is, for 5 packets sent on 2 Mbps link, 2 packets should be sent over 512Kbps. Same with the next ratio, 4 packets sent over 2 Mbps, 3 packets should be on 512Kbps

Look at the below diagram

By looking at the diagram, we notice there are two paths to reach R4's Loopback 0 interface 1.1.1.1/32. One is via the Fastethernet interface & another via serial interface. So, what are the technical stuffs coming into your end w.r.t EIGRP. Assume only "Delay" parameter is used to perform the composite metric calculation.

1) Path via R2 is the preferred path currently as the Delay parameters are least. 
2) Path via R3 is the feasible successor because the Relative distance advertised by R3 to R1 is the same as advertised by R2 to R1. (i.e 100 + 5000 = 5100)

What is that we can do in order for path via R3 to be used for load balancing? To achieve this, we need to use EIGRP variance feature. Lets calculate the metric to reach R4's loopback from R1 manually. Below is the formula

Metric = 256 * (Sum of Delay/10) 

Note : Delay is in tens of microseconds

Metric via R2 is = 256 * (100+100+5000)/10 = 133120
Metric via R3 is = 256 * (20000 + 100 + 5000) = 642560

So, we know that Metric via R2 will be the preferred path of the least Feasible distance. So, lets use our R3 path as well. 

Remember, Variance = FD of the Feasible successor / FD of the successor
So, Variance = 642560/133120 = 4.82
Lets round that up to a whole number. So, Variance = 5

Now, by setting variance 5 under the EIGRP routing process, you would have the load balancing between two paths up. 
Note : For the variance command to take effect, the path must be a feasible successor

But how to achieve 5:2 or 4:3 ratio load balancing? For this to happen, you need to consider the below things

1) How much delay should be set on the path via R3(the serial interface path) in order to get 5:2 ratio?
2) What would be the calculation to obtain the delay value?

To calculate the delay, below is the formula

Lowest Metric * = 256 * [delay/10]
* Lowest Metric is the path with the lowest feasible distance (in our case path via R2) 

133120 * (5/2) = 256 * [delay]/10
332800 = 256 * [delay/10]
332800/256 = delay/10
1300=delay/10
13000=delay

Maths :) 

The above delay value we calculated is the delay of the entire path (end-to-end). Now, we need to determine the delay value of the local interface. To find that, you need to subtract the advertised delay received from R3. 

So, it would be 13000-5100 = 7900

The actual delay to be set on the serial interface of R1 is 790. Yes, re-read 790. Delay is read in tens of microseconds. So, by setting up 790, you would receive a ratio of 5:2 now.

I haven't added any screenshots of the configs as it's left to you to perform it :-D Oh ya, even 4:3 calculation is left to you for practice !!!!

EIGRP Unequal Cost Load Balancing

In This article I would like to discuss the unequal cost load balancing feature of EIGRP and how to calculate the ratio of traffic that will be sent over each connection.

EIGRP supports load sharing traffic over multiple unequal cost paths based on the composite metric of each path.  In working on the CCIE exam it may be a requirement for you to configure both unequal cost load balancing and define the amount of traffic that should be load shared on each path.  For instance a question may state ”send 66% of traffic thru the primary path and 33% on the secondary path”.  Let’s work thru the process of calculating these ratios.

First and foremost you must know how to calculate the metric of EIGRP.  Metric by default (As only K1 and K3 are not 0 by default) is calculated using the following formula:

Using this same formula, if we already know the metric we want to achieve, we can simply change this formula to obtain either the delay or bandwidth needed to achieve a required ratio.  To calculate delay and bandwidth we simply change the formula as follows:

Delay:

Or for Bandwidth:

For memories sake you only need to remember the first formula, as the last two are just derivatives of the first.  To test this concept let’s work thru a few examples and see what we can do to achieve unequal cost load balancing.  We will be using the following topology:

As shown on this topology we have a network on R2, 200.0.0.2/32, that R5 and R6 are learning.  We will first work on a requirement on R5.

We want to load share traffic to 200.0.0.2/32 from R5 to R2 using Serial0/1/0 and Serial0/2/0.  The requirement is that for every 5 packets sent over Serial0/2/0 1 packet be sent over Serial0/1/0 (or achieve a 5 to 1 ratio).  We are given a bandwidth of 100 MB and a delay of 100 microseconds for Serial0/2/0. We are not allowed to change the delay of Serial0/1/0 so we must determine the bandwidth to achieve a ratio of 5 to 1.

With this problem we will be using the bandwidth formula shown above.  First we need to find out what the feasible distance is.  Let’s look at the EIGRP topology for this route.

R5#show ip eigrp topology 200.0.0.2/32
IP-EIGRP (AS 1001): Topology entry for 200.0.0.2/32
  State is Passive, Query origin flag is 1, 1 Successor(s), FD is 156160
  Routing Descriptor Blocks:
  150.100.25.2 (Serial0/2/0), from 150.100.25.2, Send flag is 0x0
      Composite metric is (156160/128256), Route is Internal
      Vector metric:
        Minimum bandwidth is 100000 Kbit
        Total delay is 5100 microseconds
        Reliability is 255/255
        Load is 1/255
        Minimum MTU is 1500
        Hop count is 1
  150.100.100.2 (Serial0/1/0), from 150.100.100.2, Send flag is 0x0
      Composite metric is (2297856/128256), Route is Internal
      Vector metric:
        Minimum bandwidth is 1544 Kbit
        Total delay is 25000 microseconds
        Reliability is 255/255
        Load is 1/255
        Minimum MTU is 1500
        Hop count is 1
R5#

When the route is inserted into the routing table on R2 it automatically penalized with a 5 millisecond delay value.  This is why we show a 5100 microsecond delay on the Serial0/2/0 interface.  We also know the path thru Serial0/1/0 has a delay of 20000 microseconds, or an accumulitive delay of 25000 microseconds.  The feasible distance (or lowest metric) for this route is 156160.  To get a five to one ratio on this route the higher metric, (less preferred) needs to be five times the feasible distance (5×156160 = 780800).

To calculate the bandwidth Serial0/1/0 must have on R5 we need to plug these values into the formula:

10^7 / ( (780800/256) – (25000/10) ) = 18181.8 Kb

On R5 go into EIGRP 1001 and add the variance as 5 (5 to 1) and put 18182 Kb as the bandwidth on Serial0/1/0.  By doing this I should achieve my 5 to 1 ratio.

R5#conf t
Enter configuration commands, one per line.  End with CNTL/Z.
R5(config)#router eigrp 1001
R5(config-router)#variance 5
R5(config-router)#end
R5#
May  3 19:38:17.423: %SYS-5-CONFIG_I: Configured from console by console
R5#show ip route 200.0.0.2
Routing entry for 200.0.0.2/32
  Known via "eigrp 1001", distance 90, metric 156160, type internal
  Redistributing via eigrp 1001
  Last update from 150.100.100.2 on Serial0/1/0, 00:00:01 ago
  Routing Descriptor Blocks:
    150.100.100.2, from 150.100.100.2, 00:00:01 ago, via Serial0/1/0
      Route metric is 780544, traffic share count is 1
      Total delay is 25000 microseconds, minimum bandwidth is 18214 Kbit
      Reliability 255/255, minimum MTU 1500 bytes
      Loading 1/255, Hops 1
  * 150.100.25.2, from 150.100.25.2, 00:00:01 ago, via Serial0/2/0
      Route metric is 156160, traffic share count is 5
      Total delay is 5100 microseconds, minimum bandwidth is 100000 Kbit
      Reliability 255/255, minimum MTU 1500 bytes
      Loading 1/255, Hops 1

R5#

Notice the traffic share count is exactly 5 to 1 over the two interfaces.

Now let’s move down the line to R6 and work on the next requirement.  On R6 we have a requirement to load balance traffic between Fa0/0, Fa0/1, and Serial0/1/0.  We should send traffic based on a ratio of 8/4/1; 8 packets on Fa0/0, 4 packets on Fa0/1, and 1 packet on Serial0/1/0.  Again we need to start by determining our feasible distance and to begin the process of working thru our calculations.

R6#show ip eigrp topology 1001 200.0.0.2/32
IP-EIGRP (AS 1001): Topology entry for 200.0.0.2/32
  State is Passive, Query origin flag is 1, 2 Successor(s), FD is 158720
  Routing Descriptor Blocks:
  150.100.220.5 (FastEthernet0/0), from 150.100.220.5, Send flag is 0x0
      Composite metric is (158720/156160), Route is Internal
      Vector metric:
        Minimum bandwidth is 100000 Kbit
        Total delay is 5200 microseconds
        Reliability is 255/255
        Load is 1/255
        Minimum MTU is 1500
        Hop count is 2
  150.100.221.5 (FastEthernet0/1), from 150.100.221.5, Send flag is 0x0
      Composite metric is (158720/156160), Route is Internal
      Vector metric:
        Minimum bandwidth is 100000 Kbit
        Total delay is 5200 microseconds
        Reliability is 255/255
        Load is 1/255
        Minimum MTU is 1500
        Hop count is 2
  150.100.100.2 (Serial0/1/0), from 150.100.100.2, Send flag is 0x0
      Composite metric is (2297856/128256), Route is Internal
      Vector metric:
        Minimum bandwidth is 1544 Kbit
        Total delay is 25000 microseconds
        Reliability is 255/255
        Load is 1/255
        Minimum MTU is 1500
        Hop count is 1
R6#

First let’s get the easier ratio.  8 to 4 is also the same as 1 to 2. Looking at our feasible distance of 158720 the metric for the path thru Fa0/1 would need to be two times this to achieve a 1 to 2 ratio: (2×158720 = 317440).  We then need to plug this into our formula to get the value:  (We have a delay of 5200 on both Fa0/0 and Fa0/1 and we want to figure out the bandwidth).

10^7 / ( (317440/256) – (5200/10) ) = 13888.8 Kb

OK so that gives us our bandwidth for Fa0/1.  In the end we want to have a 8/4/1 ratio between these three paths so on R6 enter o variance of 8 for the routing process.  And on Fa0/1 configure the bandwidth as 13889 Kb.

R6#conf t
Enter configuration commands, one per line.  End with CNTL/Z.
R6(config)#router eigrp 1001
R6(config-router)#variance 8
R6(config)#int f0/1
R6(config-if)#band 13889
R6(config-if)#end
R6#
May  3 19:57:18.079: %SYS-5-CONFIG_I: Configured from console by console
R6#show ip route 200.0.0.2
Routing entry for 200.0.0.2/32
  Known via "eigrp 1001", distance 90, metric 158720, type internal
  Redistributing via eigrp 1001, eigrp 6009
  Advertised by eigrp 6009
  Last update from 150.100.221.5 on FastEthernet0/1, 00:00:06 ago
  Routing Descriptor Blocks:
  * 150.100.221.5, from 150.100.221.5, 00:00:06 ago, via FastEthernet0/1
      Route metric is 317184, traffic share count is 1
      Total delay is 5200 microseconds, minimum bandwidth is 13908 Kbit
      Reliability 255/255, minimum MTU 1500 bytes
      Loading 1/255, Hops 2
    150.100.220.5, from 150.100.220.5, 00:00:06 ago, via FastEthernet0/0
      Route metric is 158720, traffic share count is 2
      Total delay is 5200 microseconds, minimum bandwidth is 100000 Kbit
      Reliability 255/255, minimum MTU 1500 bytes
      Loading 1/255, Hops 2

R6#

We now have a 1 to 2 ratio for these two paths.  Let’s now add the third in.  The delay is 25000 and the metric needs to be: (8×158720=1269760).  Plug this into the formula Again.

10^7 / ( (1269760/256) – (25000/10) ) = 4065 Kb

R6#conf t
Enter configuration commands, one per line.  End with CNTL/Z.
R6(config)#int s0/1/0
R6(config-if)#band 4065
R6(config-if)#end
R6#
May  3 20:01:39.759: %SYS-5-CONFIG_I: Configured from console by console
R6#clear ip route *
R6#show ip route 200.0.0.2
Routing entry for 200.0.0.2/32
  Known via "eigrp 1001", distance 90, metric 158720, type internal
  Redistributing via eigrp 1001, eigrp 6009
  Advertised by eigrp 6009
  Last update from 150.100.100.2 on Serial0/1/0, 00:00:01 ago
  Routing Descriptor Blocks:
    150.100.221.5, from 150.100.221.5, 00:00:01 ago, via FastEthernet0/1
      Route metric is 317184, traffic share count is 4
      Total delay is 5200 microseconds, minimum bandwidth is 13908 Kbit
      Reliability 255/255, minimum MTU 1500 bytes
      Loading 1/255, Hops 2
  * 150.100.220.5, from 150.100.220.5, 00:00:01 ago, via FastEthernet0/0
      Route metric is 158720, traffic share count is 8
      Total delay is 5200 microseconds, minimum bandwidth is 100000 Kbit
      Reliability 255/255, minimum MTU 1500 bytes
      Loading 1/255, Hops 2
    150.100.100.2, from 150.100.100.2, 00:00:01 ago, via Serial0/1/0
      Route metric is 1269760, traffic share count is 1
      Total delay is 25000 microseconds, minimum bandwidth is 4065 Kbit
      Reliability 255/255, minimum MTU 1500 bytes
      Loading 1/255, Hops 1

R6#

And there we have it.  We have achieved an 8/4/1 ratio for load balancing traffic over these three paths.

Remembering the initial formula and how to convert it I believe is the hardest part of this task.  The rest is pretty logical.  We just need to have some basic algebra skills.

Regards,

Tyson Scott – CCIE # 13513 (R&S/Security/SP)
Senior Technical Instructor – IPexpert, Inc.