Yet More Gift Wrapping

Quote of the Day

Intentionally adding value to others today will bring you fulfillment every day.

— John C. Maxwell. I think about this all the time – am I making the people around me better?


My sister and I talk about these practical math problems all the time. Here is another good video that covers gift wrapping with a mathematical slant. Yes, I have discussed this before.

Posted in General Mathematics | Comments Off on Yet More Gift Wrapping

A Couple of Cost Versus Run Rate Examples

Quote of the Day

As a child, you want something. As a young adult, you beg for relief from something. When you become old, you pray to give thanks for something.

— Kevin Kling, my characterization of his three stages of prayer. His observation very much reminds me of the Benedictine philosophy I heard in my youth, which taught that if you want to be happy think of something to be grateful for.


Figure 1: Engineering Balanced Time, Technology, and Cost.Figure 1: Engineering Balanced Time, Technology, and Cost.

Figure 1: Engineering
Balances Time, Cost,
and Technology.

I spend a fair amount of time estimating the cost of electronic components with different purchase quantities. In earlier posts, I showed how component cost quotes often reduce by ~7 % per doubling of quantity purchased. In this post, I will present quotes from two different vendors for the same part (normalized to preserve confidentiality). The component quotes from each vendor reduced by 3.6% per doubling of quantity purchased, which differs significantly from my 7% rule. While each vendor had different costs, their rates of variation with purchase quantity were identical. As I thought about these quotes, their deviation from what I usually see made sense.

Normally, I am asking a vendor to deliver parts for which they have already built thousands or even millions of units. In this case, I am asking a vendor to deliver a part that they have never built before. Because of the risk associated with building a component they have never seen, the vendors are conservative and their cost quotes show only a  3.6% per doubling of component purchased. Note that the rules I am working with here are for purchases within a single year. A related rule (Wright's law) holds for year-over-year pricing (blog post).

Figure 2 shows the normalized quotes and my 3.6% cost reduction per doubling curve. I am documenting this information here in the hope it may be useful for those of you who must estimate cost reductions based on purchase quantities.

Figure 2(a): Quote from Vendor 1. Figure 2(b): Quote from Vendor 2.

Addendum

I will add quote charts here as I get them. Figure 3 shows a more typical example of how the cost quotes look. The device in Figure 3 includes significant amounts of passive optics and active electronics. I consider this component a mainstream device – no development risk.

FIgure 3: More Typical Quotation Plot Showing 7.8% Cost Reduction Per Doubling of Volume.

Figure 3: More Typical Quotation Plot Showing 7.8% Cost Reduction Per Doubling of Volume.

Figure 4 shows the quote for a new passive optical component with rather aggressive specifications. Observe that the vendor did not price as aggressively for a new design.

Figure 4: Totally Passive Optical Device with 3.8% Cost Reduction Per Doubling of Volume.

Figure 4: Totally Passive Optical Device with 3.8% Cost Reduction Per Doubling of Volume.

 

Posted in Electronics, Financial | Comments Off on A Couple of Cost Versus Run Rate Examples

Not The Way To Take A Chimney Down

Quote of the Day

Compound interest is the eighth wonder of the world. He who understands it, earns it… he who doesn’t… pays it.

— Albert Einstein


I have worked on construction sites and farms, and the lack of imagination that people show with respect to safety never ceases to amaze me. This video shows what happens when someone does not think about what could happen.

In my own neighborhood, I saw one neighbor cut a tree down that fell onto another neighbor's house. The same neighbor who dropped a tree on the house next door also built a second-floor deck that dropped on top of him during construction – all the beams miraculously missed him.

Posted in Construction, Humor | Comments Off on Not The Way To Take A Chimney Down

Network Availability

Quote of the Day

A theory has to be simpler than the data it explains.

— Leibnitz


Introduction

Figure 1: I am sure even Godzilla has tough days.

Figure 1: I am sure even Godzilla has tough days (Source).

I was in an interminable meeting the other day where we were discussing the MTBF and availability of a system. My issue with this discussion is that each person in the room preferred to think about these terms in different ways. In this post, I will show that the four people in the meeting were actually in violent agreement and simply did not understand that their arguments were mathematically equivalent.

I wish I could say that this was the first time in my career that this had happened, but that would not be true. It happens all the time.

Background

The Argument

I will try to summarize the argument as simply as I can:

Person 1 The system must conform to GR-909 – a telecommunications specification that specifies system availability.
Person 2 The system must have an availability of at least 99.999%.
Person 3 The system must have a downtime (i.e. unavailability) of less than 5 minutes per year.
Person 4 The system must have a Mean-Time-Between Failure (MTBF) of 68.4 years.

Definitions

Availability
The ratio of (a) the total time a functional unit is capable of being used during a given interval to (b) the length of the interval. For example, a unit that is capable of being used 100 hours per week (168 hours) would have an availability of 100/168. In high availability applications, a metric known as "nines", corresponding to the number of nines following the decimal point, is used. With this convention, "five nines" equals 0.99999 (or 99.999%) availability (Source).
Mean Time Between Failures (MTBF)
MTBF describes the expected time between two failures for a repairable system (Source).
Mean Time To Repair (MTTR)
MTTR represents the average time required to repair a failed component or device (Source).
Mean Time to Failure (MTTF)
MTTF denotes the expected time to failure for a system that requires a repair with an MTTR of a given value. For our purposes here, MTTF=\frac{MTBF}{MTBF+MTTR}.
Failure Rate (FR)
Failure rate is the frequency with which an engineered system or component fails, expressed in failures per unit of time (e.g. 1E9 hours).

Analysis

Figure 2 summarizes my demonstration of the equality of each person's argument.

Figure 1: Equality of the Reliabilty Arguments.

Figure 2: Equality of the Reliability Arguments.

Conclusion

It took about 30 minutes to get everyone in the meeting to understand that they all were stating the same requirement. The problem originates in that different departments work in terms of different units. System engineers and industry specifications speak in terms of availability. Hardware engineers speak in terms of MTBF. Customer Service people speak in terms of downtime per year.

The "elephant in the room" was that fact that most systems fail because of software bugs and these reliability calculations ignore software bugs.

Posted in Electronics, General Science | 4 Comments

A Little Sandpaper Kirigami

Quote of the Day

Leaders who do not listen will eventually be surrounded by people have nothing to say.

— Andy Stanley


I just ran across this video that does a nice job of illustrating the sandpaper folding technique that I have been using for a number of years. It works well.

Posted in Construction | Comments Off on A Little Sandpaper Kirigami

Modeling Component Costs Over Time

Quote of the Day

There are two ways to do great mathematics. The first is to be smarter than everybody else. The second way is to be stupider than everybody else -- but persistent.

— Raoul Bott, Hungarian-American mathematician. His first way is not an option for me, so I will need to focus on his second way.


Introduction

Figure 1: Experience Curve for the Ford Model T.

Figure 1: Experience Curve for the Ford Model T (Source).

I frequently am asked by marketing and finance people about how component costs will vary with time. Their motivations are clear – most market segments are strongly driven by unit cost, and the marketing folks need to determine when costs will drop enough to enlarge the Total Addressable Market (TAM) for their products.

I use experience curves, which are sometimes called learning curves. These curves are described using Wright's law, which states that product cost decreases as a power law of cumulative production. Like of all of these empirical "laws", it really is more of a "rule of thumb" than a law. T.P. Wright first published his law in an article on airplane production back in 1936, but it has been found to apply in many industries (Figures 1 and 2).

Figure 2: PV Experience Curve.

Figure 2: Photovoltaic  Experience Curve (Source).

Recently, I saw a magazine article that compared five commonly used costing models, one of which was Wright's law. The article stated that Wright's law was somewhat better than its closest competitor, Moore's law, which states that product costs decrease as a power law of time.

In this post, I will:

  • discuss how I use Wright's law to model product cost.
  • present some normalized cost data from actual fiber optics equipment.

Background

Wright's Law

Wright stated that the average time needed to produce a component will reduce as you build more of the components. Specially, the average build time need for a component after building N units will decrease by a fixed fraction after you have built 2· N units. We can state his law as shown in Equation 1.

Eq. 1 \displaystyle {{t}_{n}}={{t}_{1}}\cdot {{n}^{{{{{\log }}_{2}}\left( K \right)}}}

where

  • n is the cumulative number of units built.
  • t1 is time need to manufacture the first unit of the product.
  • tn is time need to manufacture the nth unit of the product.
  • K is the percentage of the time required to build the 2·ith product relative to the ith product. This fraction is assumed to be constant regardless of the value of i.

While Wright was focused on the time needed to build nth component relative to the first one, it has been found that his rule also holds true for product cost, i.e. {{c}_{n}}={{c}_{1}}\cdot {{n}^{{{{{\log }}_{2}}\left( K \right)}}}, where cn is the cost of the nth unit and c1 is the cost of the first unit. In the discussion to follow, we will be focused on the variation in unit product cost versus cumulative production and time.

Analysis

Objective

The objective of this analysis it to develop a standard approach for predicting product cost versus time.

Logistic Curve

Applying Wright's law means that we need a model for the cumulative production of a product. The logistic curve (Figure 3) is the most commonly used analytic model of the cumulative production of a product. Figure 2 can be thought of as representing the fraction of cumulative production completed at time t assuming time is referenced to the 50% level.

Figure 3: Standard Logistic Curve (Source).

Figure 3: Standard Logistic Curve (Source).

For practical applications, it is useful to scale the time access to represent common units of time (e.g. months, quarters, years) and different "ramp rates" – the slope of the center portion of the curve. Equation 2 shows the functional form scaled and slope-adjusted logistic curve.

Eq. 2 \displaystyle n(t)=\frac{1}{{1+{{\text{e}}^{{-k(t-{{t}_{0}})}}}}}

where

  • k is a scaling parameter that is a function of the "ramp" time of the component .
  • t is the specific time (usually quarter) that the component is purchased.
  • τ is time-shift needed so that t=0 corresponds to the time when we begin purchasing the component.
  • n(t) is the percentage of cumulative production completed at time = t. I usually referred to n(t) as the normalized production.

Cost Modeling

Wright modeled the cost of an item using Equation 3, which uses the cost of the first item (c0) as its reference value.

Eq. 3 \displaystyle c(n)={{c}_{1}}\cdot {{\left( n \right)}^{{{{{\log }}_{2}}\left( K \right)}}}

where

  • c(n) is the cost of the nth item produced.
  • c1 is cost of the 1st item.

Because I would only rarely know value of c1, I will modify Equation 3 so that I use the cost at any level of production to determine my curve. In Equation 4, I assume that I can compute solve for c1 assuming that I know the unit cost after n(t=0) have been built. I refer to this cost as c(n(0)). This allows me to restate Equation 3 as a follows.

Eq. 4 \displaystyle c\left[ {n\left( 0 \right)} \right]={{c}_{0}}\cdot {{\left[ {n\left( 0 \right)} \right]}^{{{{{\log }}_{2}}\left( K \right)}}}
\displaystyle c(n)=\frac{{c\left[ {n\left( 0 \right)} \right]}}{{{{{\left[ {n\left( 0 \right)} \right]}}^{{{{{\log }}_{2}}\left( K \right)}}}}}\cdot {{\left[ {n\left( t \right)} \right]}^{{{{{\log }}_{2}}\left( K \right)}}}=c\left[ {n\left( 0 \right)} \right]\cdot {{\left( {\frac{{n\left( t \right)}}{{n\left( 0 \right)}}} \right)}^{{{{{\log }}_{2}}\left( K \right)}}}

Combined Law

Shifted and Scaled Logistic Curve

To apply Wright's law, I need to determine the specific logistic curve for our component. This means that I must determine:

  • τ, the time-shift parameter that will make t=0 the time we begin buying the component.
  • k, the ramp rate parameter of the logistic curve.

We can compute k and τ as functions of (1) the 10%-to-90% rise time of the cumulative production curve (∆T), and (2) the normalized production rate at t=0, n(0). I often have to estimate ∆T based on my personal experience because I do not have direct access to the total volumes a manufacturer is producing. However, I do usually know the total size and annual shipments into my market, so I roughly know both ∆T and n(t). Figure 4 shows how to derive values for both τ and k given ∆T and n(t).

Figure 4: Derivation of Scaled and Time-Shifted Logistic Curve.

Figure 4: Derivation of Scaled and Time-Shifted Logistic Curve.

The complete set of formulas for the scaled and shifted logistic curve are shown in Equation 4.

Eq. 4 \displaystyle k=\frac{{2\cdot \ln (9)}}{{\Delta T}}\;\;\;\;\tau =\frac{{\ln \left( {\frac{1}{{n\left( 0 \right)}}} \right)}}{k}
\displaystyle n\left( t \right)=\frac{1}{{1+{{e}^{{k\cdot \left( {t-\tau } \right)}}}}}

Integrating the Logistic Curve with Cost Function

Equation 5 shows the result of integrating Equations 4 and Equation 3.

Eq. 5 \displaystyle p(t,n\left( 0 \right),\Delta T,K)={{\left( {\frac{{n\left( {t,n\left( 0 \right),\Delta T} \right)}}{{n\left( 0 \right)}}} \right)}^{{{{{\log }}_{2}}\left( K \right)}}}

To illustrate the use of Equation 5, I show the agreement between Equation 5 and a real fiber-optic transceiver in Figure 5, where I show the normalized price (i.e. cost versus time relative to the initial cost). The agreement is reasonable considering that Wright's law is really a rule of thumb. I originally estimate the parameters of the model (K, n(0)ΔT ) based on my experience with a similar part that we first purchased years ago.

Figure 5: Normalized Fiber Optic Transceiver Cost Versus Time and Overall Production.

Figure 5: Normalized Fiber Optic Transceiver Cost Versus Time and Overall Production.

Conclusion

I am in the middle of some cost projection work right now. Hopefully, this post will help some of you with your cost projection tasks.

Posted in Electronics, Financial, Management | 2 Comments

I am Battling Ladybugs This Year

Quote of the Day

Exhilaration is that feeling you get just after a great idea hits you, and just before you realize what's wrong with it.

— Signature on a forum post


We have had a warm winter in Minnesota so far this year. As with every transition from fall to winter, bugs seek warmth by taking up residence in our fiber-optic enclosures. Here is my latest example, a ladybug on a returned circuit board.

Figure 1: Ladybug in the Electronics.

Figure 1: Ladybug in the Electronics.

Posted in Electronics, Fiber Optics, Humor | 4 Comments

Modeling Coal Energy Output

Quote of the Day

With hard work, difficult material can be grasped. Step by step, incrementally, the novice can become the master.

— Joshua Waitzkin, World Tai Chi champion and subject of the book 'Searching for Bobby Fischer'. He is working with Khan Academy to promote learning through hard work.


Introduction

Figure 1: Lump of Bituminous Coal (Source).

Figure 1: Lump of Bituminous Coal (Source).

I had never seen coal until my first trip to China when I saw people on bicycles transporting coal to their homes for heat.  I started to wonder just how much heat a given amount of coal would generate. I have seen numerous values for the heat content of the various types of coal. I recalled from primary school that there were three types of coal: anthracite, bituminous, and lignite. So I would have expected three values for the heat output of coal. When I actually looked, I found dozens of grades of three primary types of coal. Each of the different grades would generate different amounts of heat per kilogram. I thought I would take a closer look at how the heat output from coal could be modeled using regression and a simplified model based on chemical heats of formation.

I am currently taking an online statistics class and this exercise provided me some good R practice.

Background

Approach

My approach to this problem is simple:

  • Scrape some coal data from the web (source).
  • Perform a regression on the coal data.
  • Compare the results of the regression to a simple model based on chemical heats of formation.

Definitions

Standard Enthalpy of Formation (aka Heat of Formation)
The standard enthalpy of formation of a compound is the change of enthalpy during the formation of 1 mole of the compound from its constituent elements, with all substances in their standard states at 1 atmosphere (1 atm or 101.3 kPa). Its symbol is ΔHfO or ΔfHO.
Coal Ash
The waste left over after coal is burned to generate power, contains concentrated amounts of heavy metals, such as lead, mercury, arsenic, chromium, and selenium, which are hazardous to human health, and to wildlife.
Proximate Analysis
Proximate analysis entails first weighting a sample, then heating the sample of coal to a high enough temperature drive off the water from the sample. and then reweighing the sample. The weight loss divided by the initial weight gives the coal moisture content, M. The remaining material is then heated at a much higher temperature, in the absence of oxygen, for along enough time to drive off gases. The resulting weight-loss fraction gives the volatile matter content, VM, of the coal. The remainder of the sample is then burned in air until only noncombustibles remain. The weight loss gives the fixed carbon, FC, and the remaining material is identified as non-combustible mineral matter or ash, A. The proximate analysis may be reported as percentages (or fractions) of the four quantities: moisture, ash, volatile matter, and fixed carbon. This analysis has the advantage of being simple to perform, but it does not related the coal sample's heat of formation with the specific chemical contents. I will not be using proximate analysis for my modeling, but I mention it here because it is reports include it in discussions of coal grades.
Ultimate Analysis
A more sophisticated and useful analysis is the ultimate analysis, a chemical analysis that provides the elemental mass fractions of carbon (C), hydrogen (H2), nitrogen (N2), oxygen (O2), and sulfur (S), usually on a dry, ash-free basis. The ash content of the coal and heating value are sometimes provided also. I will be focusing on the results of ultimate analysis for my modeling.

Coal Constituents

For the discussion here, I will assume that coal is composed of the following elements. Different types of coal are are characterized by the constituent percentages of these elements. These elements are measured as part of the ultimate analysis of a coal product.

  • Carbon (C)

    Carbon reacts with atmospheric oxygen to produce the bulk of the heat generated by coal. The primary reaction produces CO2, but there is a secondary reaction that produces CO. I will ignore the heat generated by the CO reaction.

  • Hydrogen (H2)

    When hydrogen burns with oxygen it forms water (H2O). There are two heats of formation listed for H2 because the water formed is in the form of steam. The amount of heat realized depends on whether you let the water condense or you just let the steam escape into the atmosphere.

  • Sulfur (S)

    A source of heat and SO2, a major source of atmospheric pollution.

  • Nitrogen (N2)

    There a number of nitrogen reactions that produce some heat, but they are not dominant.

  • Oxygen (O2)

    The heat of formation of O2 is not a significant source of heat from coal.

  • Ash

    Ash is left over from burning coal and presence major disposal challenges because it contains unsafe compounds.

Analysis

Examination of the Data

Figure 2 shows a pair-wise plot of the coal data to help me determine which factors are significant (click on the Figure 2 to see the full-size plot). As I look at the plot, I see that C and H2 look the most significant. I should note that many papers include the energy from burning sulfur, but the data here does not show it to be significant.

Figure 2: Pair-Wise Analysis of Ultimate Analysis Variables.

Figure 2: Pair-Wise Analysis of Ultimate Analysis Variables.

Figure 3 shows my plot of the coal data from Appendix B. We can see that the data is fairly linear with respect to the C and H2 percentages. I have included my regression line, the coefficients for which I will generate in the following section.

Figure M: Plot of Coal Data Versus Coal and Hydrogen Percentages.

Figure 3: Plot of Coal Data Versus Coal and Hydrogen Percentages.
The red line is my regression model, which is documented below.

Generate a Linear Model

I started my analysis of the data by performing linear regression assuming all the ultimate parametrics C and  H2.

 Figure M: Regression Analysis of Coal Data.

The regression of Figure 3 produces the following linear model (Equation 1) for the energy output in BTU for a pound-mass of coal.

Eq. 1 \displaystyle {{H}_{{Total}}}=159.4\cdot {{k}_{C}}+406.2\cdot {{k}_{{H2}}}-457.2

As you can see in Figure 3, the regression line fits the data well.

Chemical Models

I now want to compare my linear regression (Equation 1) with an approximate chemical model. The chemical model is approximate because I only deal with the dominant chemical reactions in burning coal. Equation 2 shows my simplified model. My chemical model has no intercept term.

Eq. 2 \displaystyle {{H}_{{Total}}}={{\Delta }_{f}}H{{~}^{o}}\left( {C{{O}_{2}}} \right)\cdot \frac{{{{k}_{C}}}}{{M{{W}_{C}}}}+{{\Delta }_{f}}H{{~}^{o}}\left( {{{H}_{2}}O} \right)\cdot \frac{{{{k}_{{H2}}}}}{{M{{W}_{{H2}}}}}

where

  • \Delta_{f}H^{o}\left(H_2O\right) is the heat of formation for H2O at STP.
  • \Delta_{f}H^{o}\left(CO_2\right) is the heat of formation for CO2 at STP.
  • MWC is the molecular weight of carbon.
  • MWH2 is the molecular weight of diatomic hydrogen.
  • kC is the percentage of carbon in the coal grade in question.
  • kH2 is the percentage of diatomic hydrogen in the coal grade in question.

We can estimate the heat output based on the heats of formation (Appendix A) for the dominant C, H2, and S chemical reactions, which I show in Figure 3.

Figure 4: Derivation of Coefficients Using Heats of Formation.

Figure 3: Derivation of Coefficients Using Heats of Formation.

We can substitute the values from Figure 3 into Equation 3 to see my estimate using the key heats of formation.

Eq. 3 \displaystyle {{H}_{{Total}}}=141.0\cdot {{k}_{C}}+567.1\cdot {{k}_{{H2}}}

You can see that the coefficients of Equation 3 is similar to those of Equation 2, so the models are consistent.

Conclusion

I was looking for a good multivariate R exercise and this coal data provided that exercise.

Appendix A: Heats of Formation for Key Coal Reactions.

Table 1 shows the assumed heats of formation for the various compounds associated with coal.

Table 1: Heat of Formation for 1-Bar Pressure and 298.15K  Temperature.
Substance State Molecular
Weight
Hf (kJ/mole) Hf (kJ/kg) Hf (Btu/lbm)
H2 Gas 2 0 0 0
H20 Liquid 18 -285.83 -15,866 6,835
H20 Gas 18 -241.826 -13,423 5,783
O2 Gas 32 0 0 0
N2 Gas 28 0 0 0
NO Gas 30 90.291 3,009 1,296
NO2 Gas 46 33.1 719 310
NO3 Gas 62 71.13 1,147 494
C Solid 12 0 0 0
CO Gas 28 -110.53 -3,946 1,700
CO2 Gas 44 -393.522 -8,942 3,852
CH4 Gas 16 -74.873 -4,667 2,011
S Sol 32 0 0 0
SO2 Gas 64 -296.842 -4,634 -1,997

Appendix B: Examples of Coal Grades.

While Table 2 is a bit of an eye chart, it does provide some useful examples of the different grades of coal. There energy output per lb varies dramatically – from a maximum of 14,820 BTU per pound down to 4,918 BTU per pound. This is a dynamic range slightly over 3 to 1.

Table 2: Coal Constituents By Grade.
County and State Geological Name Type Ash VM FC C H2 O2 N2 S Heating Value (BTU/lb)
Schuylkill, Pa. Anthracite culm Raw 59.1 8.3 32.6 33.5 1.2 5.1 0.6 0.5 4918
Schuylkill, Pa. Anthracite culm Clean 1 10.6 11.1 78.3 79.9 2.7 5 0.9 0.9 12733
Schuylkill, Pa. Anthracite culm Clean 2 27.7 16.2 56 61.6 2.6 5.8 0.9 1.4 10135
Freestone, Tex. Big brown lignite Raw 18.4 45.5 36.1 63.2 4.8 11.3 1.2 1.1 9954
Freestone, Tex. Big brown lignite Clean 1 9.1 47.1 43.8 64.8 4.7 19.1 1.2 1 11369
Freestone, Tex. Big brown lignite Clean 2 9.6 45.3 45.1 63.8 4.7 20.1 1 0.9 11222
Freestone, Tex. Big brown lignite Clean 3 14 44.3 41.7 66.3 4.9 12.7 1.2 0.9 10687
Freestone, Tex. Big brown lignite Clean 4 10.2 45.8 44 61.9 4.3 21.05 1.1 0.9 11159
Indiana, Pa. Freeport (upper) Raw 31.7 22.3 46 57.1 3.5 4 1.7 1.9 10239
Indiana, Pa. Freeport (upper) Clean 1 16.4 26.2 57.3 71.7 4.3 4.2 1.9 1.4 12846
Indiana, Pa. Freeport (upper) Clean 2 20.4 26.4 53.2 67.6 4.1 5.2 0.8 1.8 12193
Clearfield, Pa. Freeport (upper) Raw 12.3 24.8 62.9 76.3 4.5 3.4 1.8 1.7 13559
Clearfield, Pa. Freeport (lower) Clean 1 6.5 27.6 65.9 82.2 4.9 4 1.6 0.9 14541
Clearfield, Pa. 50% of each Clean 2 7 26.3 66.7 81.4 4.7 4.8 1.4 0.7 14476
Clearfield, Pa. 50% of each Clean 3 9.2 26.5 64.3 78.9 4.7 4.8 1.7 0.7 14187
Clearfield, Pa. 50% of each Clean 4 8.9 26.3 64.7 79.2 4.7 5.1 1.3 0.7 14168
Indiana, Pa. Freeport (upper) Raw 33.6 19.4 47 54.3 3.3 3.8 1.2 3.8 9780
Indiana, Pa. Kittanning (lower) Clean 1 12.6 22.1 65.3 75.5 4.4 4 1.6 1.9 13563
Indiana, Pa. 20% F, 80% K Clean 2 17 22 61 71.8 4.2 3.2 1.1 2.7 12814
British Columbia, Canada Hat Creek A zone Raw 40.8 31.9 27.3 32.7 2.8 22.3 0.6 0.8 6642
British Columbia, Canada Hat Creek A zone Clean 1 20.3 39 40.8 53.2 3.9 20.8 1 0.9 9496
Nova Scotia, Canada Hub seam Raw 23.4 30.8 45.8 61.1 4.1 6.1 1.3 4 10994
Nova Scotia, Canada Hub seam Clean 1 2.1 40.5 57.4 79.3 5.4 9.9 1.5 1.7 14564
Nova Scotia, Canada Hub seam Clean 2 6.5 37.7 55.7 74.7 5 8.5 1.5 3.8 13770
Perry, III. Illinois no. 6 Raw 18.5 36.7 44.7 61.6 4.2 9.7 1.3 4.6 11345
Perry, III. Illinois no. 6 Clean 1 11.1 40.8 48.1 70.5 4.8 9.1 1.4 3.1 12627
Perry, III. Illinois no. 6 Clean 2 8.6 41.9 49.5 72.4 5.1 9.8 1.2 2.9 13023
Perry, III. Illinois no. 6 Clean 3 10.5 39.5 50 71.1 4.8 9.1 1.4 3.1 12818
Perry, III. Illinois no. 6 Clean 4 9.5 39.7 50.8 70.6 6.2 9.5 1.3 3 12865
Perry, III. Illinois no. 6 Clean 5 8.8 40.4 50.8 72.3 5 9.5 1.4 3 13025
Muhlenberg, Ky. Kentucky no. 9 Raw 16.1 36.3 47.5 64.8 4.5 9.2 1.5 3.9 11952
Muhlenberg, Ky. Kentucky no. 9 Clean 1 6.3 43.2 50.6 73.5 5.1 10.5 1.8 2.8 13536
Muhlenberg, Ky. Kentucky no. 9 Clean 2 7.3 42.6 50.1 72.4 5 10.4 1.8 3 13446
Muhlenberg, Ky. Kentucky no. 9 Clean 3 8.7 39.3 52 71 5 10.5 1.6 3.2 13107
Muhlenberg, Ky. Kentucky no. 9 Clean 4 7.9 40.7 51.3 72.4 5 10 1.6 2.9 13260
Union, Ky. Kentucky no. 11 Raw 39.3 29.9 30.8 45.9 3.2 6.3 1.3 4 8117
Union, Ky. Kentucky no. 11 Clean 1 6.8 41.7 51.5 72.7 5 10.4 1.9 3.2 13254
Union, Ky. Kentucky no. 11 Clean 2 9 40.8 50.2 70.8 4.9 10 1.9 3.4 12856
Union, Ky. Kentucky no. 11 Clean 3 7.9 42 50.1 71.4 4.9 10.8 1.9 3.1 13085
Cambria, Pa. Kittanning (lower) Raw 24.8 17.2 58 65.9 3.6 3.1 1.2 1.4 11510
Cambria, Pa. Kittanning (lower) Clean 1 9.7 19.1 71.2 80.5 4.3 2.7 1.8 1 14238
Cambria, Pa. Kittanning (lower) Clean 2 10 18.8 71.2 79.9 4.2 3.5 1.5 0.9 14069
Cambria, Pa. Kittanning (lower) Clean 3 5.8 19.9 74.3 84.7 4.5 2.5 1.7 0.8 14820
Cambria, Pa. Kittanning (lower) Clean 4 7.8 19.4 72.8 81.9 4.3 3.6 1.6 0.8 14467
Nicholas, W.Va. Kittanning (upper) Raw 15.2 32.3 52.5 70.2 4.6 7.6 1.4 1.2 12704
Nicholas, W.Va. Kittanning (upper) Clean 1 6.3 36.5 57.2 79.1 5 6.5 2 1.2 14240
Nicholas, W.Va. Kittanning (upper) Clean 2 5.4 37.2 57.4 80 5.1 6.2 2.1 1.2 14401
Belmont, Ohio Pittsburgh Raw 33 32.4 34.6 51.5 3.7 6.3 1 4.5 9493
Belmont, Ohio Pittsburgh Clean 1 6.2 43.2 50.6 75.5 5.3 8.3 1.3 3.4 13922
Belmont, Ohio Pittsburgh Clean 2 6.7 43.4 49.9 74.7 5.3 8.2 1.2 3.9 13820
Big Horn, Mon. Robinson Raw 10.6 37.4 52 64.5 4 19 1 0.9 11280
Big Horn, Mon. Robinson Clean 1 6.3 37.7 56 65.9 4.2 22.3 0.9 0.4 11981
Big Horn, Mon. Robinson Clean 2 6.7 37.6 55.7 65.7 4.3 22.5 0.4 0.4 11833
Big Horn, Mon. Robinson Clean 3 8.2 37.8 54 65.8 4.3 20.3 1 0.4 11868
Greene, Pa. Sewickley Raw 33.6 29.1 37.2 54.5 3.8 5.6 1.1 1.3 9786
Greene, Pa. Sewickley Clean 1 7.6 36.3 56.1 78.7 5.2 5.9 1.7 0.9 14021
Greene, Pa. Sewickley Clean 2 8.7 37 54.3 77.3 5.1 6.3 1.7 0.9 13805
Greene, Pa. Sewickley Clean 3 10.4 35.8 53.8 74.8 4.9 7.2 1.6 1 13531
Greene, Pa. Sewickley Clean 4 10.9 35.9 53.2 74.8 5 6.6 1.7 1.1 13442
Kanawha, W. Va. Stockton-Lewiston Raw 40.4 24.2 35.4 48 3.3 6.7 1 0.6 8294
Kanawha, W. Va. Stockton-Lewiston Clean 1 17.5 31.9 50.6 68 4.5 8 1.3 0.7 12174
Belmont, Ohio Washington (lower) Raw 23.8 34.1 42.1 59.1 4.1 7.9 1.3 3.8 11001
Belmont, Ohio Washington (lower) Clean 1 12.1 39.2 48.7 69.5 4.8 9.5 1.2 2.9 12807
Belmont, Ohio Washington (upper) Raw 27.1 31.8 41.1 57.3 4 9 1 1.6 10423
Belmont, Ohio Washington (upper) Clean 1 22.3 37.3 40.4 63.4 4.3 8.2 1.3 1.4 11276
Belmont, Ohio 20% Washington (I) Raw 25.6 32.6 41.8 58.6 4.1 7.9 1.2 2.6 10606
Belmont, Ohio 38% Washington (u) Clean 1 17.6 36.7 45.7 65.6 4.6 8.5 1.5 2.3 11881
Belmont, Ohio 42% Waynesburg Clean 2 17.5 36.1 46.4 66 4.6 8.6 1.1 2.1 11910
Belmont, Ohio Waynesburg Raw 27.7 32.7 39.6 56 3.9 7 1.2 4.1 10315
Posted in General Science, Statistics | 1 Comment

Gross Margin Versus Labor and Material Markups

Quote of the Day

Most corporate planning is like a ritual rain dance: it has no effect on the weather that follows, but it makes those who engage in it feel that they are in control. Most discussions of the role of models in planning are directed at improving the dancing, not the weather.

— Russell L. Ackoff


Introduction

Figure 1: Bid and Proposal Process.

Figure 1: Bid and Proposal Process.

I have spent a fair portion of my career working as an engineering contractor. In fact, I spent five years working on new business development for a large contract firm. During this time, I was involved in writing dozens of proposals on large projects. A proposal occurs at the end of the bid and proposal process (Figure 1) and constitutes a contract firm's attempt to win the project development contract. It is a key part of new business development for many companies.

In general, a proposal contains a technical portion and a cost portion (sometimes called volumes). While both portions are important, the cost portion is usually the most controversial. The controversy occurs because most proposals are won or lost on cost.

All the companies that I have worked for require projects to make a minimum level of profit, which is usually represented as a percentage of total project price. This profit percentage is known as the gross margin percentage. These companies also have standard markups on their labor and material costs. I assume that the labor markup is always higher than the material markup, which is the most common case.

In this post, I will show that a project's total price must have a minimum labor percentage in order to meet the company's gross margin requirement. Equation 1 calculates the minimum labor percentage required for a project to meet a gross margin requirement.

Eq. 1 \displaystyle {{k}_{{Labor}}}=\frac{1}{{1-\frac{{1+{{k}_{{LMU}}}\cdot \left( {1-\frac{1}{{GM\%}}} \right)}}{{1+{{k}_{{MMU}}}\cdot \left( {1-\frac{1}{{GM\%}}} \right)}}}}

where

  • kLabor is the percentage of project cost in labor.
  • kLMU is the percentage of labor markup.
  • kMMU is the percentage of material markup.
  • GM% is the gross margin percentage.

Background

Definitions

I use these definitions all the time. The key thing to remember is that markups are always expressed relative to costs, and margins are always expressed relative to price.

cost
For this discussion, cost is the sum of all the component costs, assembly/construction costs, and shipping costs.
price
Price is what you charge a customer for a project. Price (P) equals total cost (CT) plus profit (p), which we would symbolically express as P=C_T+p.
markup
Markup is the percentage increase in the raw cost of a component that you charge a customer. Think of it in terms of this formula, P_{Component}=\left(1+k_{MMU}\right)\cdot C_{Component}, where PComponent is the price you charge the customer for the component, and CComponent is your price for the component.
gross margin %
Gross Margin % (GM%) is the profit percentage in your total project price. Think of it in terms of this formula, GM\%=\frac{p}{P}.

Critical Questions

During the bidding process, I always ask questions like:

  • How much money does the customer have?

    Many customers have unrealistic expectations of how much a project will cost. Companies have limited amounts of bid and proposal money and they do not want to squander it on projects that can never happen.

  • How much money does the customer expect to pay?

    A key proposal function is to set expectations – both for the customer and the management of the contract firm. The proposal must educate the project's stakeholders on what performance can be achieved for a given level of investment and risk.

  • What fraction of the project will be material versus labor?

    Most companies have different markups for material and labor. In the case of government contracts, these rates are negotiated. On commercial contracts, these rates may or may not be known to customers. However, all large companies have some form of rate structure.

  • What gross margin do I need to make on the project?

    Companies must generate returns on investment. If the returns are not commensurate with the risks involved, the investment money will go elsewhere and the business will fold.

Analysis

Derivation

Figure 2 shows how to derive Equation 1.

Figure 2: Derivation of Equation 1.

Figure 2: Derivation of Equation 1.

Solution Implications

Maximum Possible Gross Margin

Let's assume that the markup on labor is higher than the markup on material (i.e.k_{LMU}>k_{MMU} ), which is usually the case. This means that I will have the greatest GM% when I have a job that is 100% labor.

In the hardware development business, labor is never 100% of a project's cost. I typically see labor taking up 70% to 85% of a project's cost. However, it is useful to look at the limiting cases because some businesses have projects that are 100% labor (e.g. study contracts). Equation 2 shows how to compute the maximum possible gross margin assuming a given labor markup that is greater than your material markup.

Eq. 2 \displaystyle GM\%_{Max}\triangleq \frac{{P-{{C}_{T}}}}{P}=\frac{{\left( {1+{{k}_{{LMU}}}} \right)\cdot {{C}_{T}}-{{C}_{T}}}}{{\left( {1+{{k}_{{LMU}}}} \right)\cdot {{C}_{T}}}}=1-\frac{1}{{1+{{k}_{{LMU}}}}}=\frac{{{{k}_{{LMU}}}}}{{1+{{k}_{{LMU}}}}}

For example, if my kLMU =25%, my maximum possible gross margin is 50%.

As an aside, the minimum gross margin is set by projects that are all material – a situation that also never occurs in real-life. Equation 3 gives the minimum possible gross margin.

Eq. 3 \displaystyle GM\%_{Min}=\frac{{{{k}_{{MMU}}}}}{{1+{{k}_{{MMU}}}}}

For example, if my kMMU =25%, my minimum possible gross margin is 20%.

Required Labor Percentage Versus Gross Margin

Figure 3 shows the required labor percentage to hit a given gross margin – this chart assumes kMMU =25%. You can see that with a low labor markup, your range of possible gross margin's is limited.

Figure 3: Labor Percentage Versus Gross Margin.

Figure 3: Labor Percentage Versus Gross Margin.

Since all three cases have the same kMMU, you can see that the all have the same minimum gross margin.

Conclusion

I have a number of new grad engineers on my team and this material is part of some training material I put together as part of their onboarding process. It is important that all engineers understand how product pricing affects the overall success of a business. This post provides the basic background these engineers will need to understand how the products they design are priced and why it is so important for them to meet their assigned cost targets.

Posted in Financial | 1 Comment

Old Way Of Specifying Phone Wire Diameter

Quote of the Day

Hard work spotlights the character of people: Some turn up their sleeves, some turn up their noses, and some don't turn up at all.

— Sam Ewing, baseball player


Introduction

FIgure 1: We have had phones for a very long time.

FIgure 1: We have had phones for a very long time.

I still work on old copper phone networks, and today I encountered wire specified as "300 pound". I had never seen a specification like this for phone wire before. As I thought about it, this specification seemed very similar to how the diameter of thread is still specified, which is by the weight in grams of a 9000 meters of fiber – a unit of measure called the denier.

In this post, I derive a formula for the diameter of wire specified in terms of weight of annealed copper per mile of distance. This is not a problem, but it does illustrate how our common units of measure evolve over time. In the case of wire, it used to be difficult to measure the diameter of fine wire, but it was easy to measure the weight of given length of material. This simple fact drove the use of this specification.

Analysis

Figure 2 shows my derivation and some worked examples. The derivation assumes the length of copper wire is a long, thin cylinder. I was able to confirm that the actual wire diameter agrees with my calculations.

Figure 2: Wire Diameter Formula with Examples.

Figure 2: Wire Diameter Formula with Examples.

Conclusion

Landline phone service had its beginning back in 1876. It is amazing that even today I encounter vestiges of the old system that date back to the early part of the 20th century.

Posted in Electronics, General Mathematics, History of Science and Technology | Comments Off on Old Way Of Specifying Phone Wire Diameter