span dir rtl span

					   Managing capital and lines of business
Capital Allocation :
Bank assign their capital to absorb the risk in each of their
activities .
The bank can allocate its capital across these line of
business on a risk adjusted basis .
To measure the productivity , the bank calculate return on
equity for each line of business .( table 9.7 p. 333)
A complete line of business reporting system requires
specialized accounting system to determine factors
underlying the report .
For each line the system must account for :
1- Capital required.
2- Revenues generation .
3- Fund used .
4- Overhead consumed .
                Capital Required
An accounting system must determine how to allocate
capital to each division .
There are guidelines that facilitate it.
Capital is assigned to a business line in two part . First , to
account for losses that are expected , second , to
account for losses that are unexpected .
Banks already are accustomed to dealing with expected
losses , they simply assign provision for loan losses
based on historical experience and known
characteristics of the assets in the particular business .
Dealing with unexpected losses is more abstract because
the institution must draw upon expectation of risk and
associated losses they may never experienced
- Revenue generation and fund used
 banks depend on well develop funds transfer pricing
system to estimate the line of business revenues
and cost of raising funds .
The analysis of revenues credited to line of business
and uses of funds by the line is based on financial
market tests of profitability of an institution return on
assets , cost of funds , and mismatch of assets and
liabilities maturities .
Overhead Consumed
The line of business are charged for their use of bank
overhead included , personnel , plant , furnishing , ….
System must consist of well-developed cost accounting
methods for allocating the expenses of
organizationally lower bank support units to business
line units .
                    Capital management
Its important to compare the capital allocation •
decision bank management makes with the
capital adequacy rules set forth in
administrative regulation . (Table 9.8 p. 335 )
If bank have a capital shortage or excess capital   •
There are several types of management action •
for this situation ( table 9.9 p.336)
                  Dividend policy
The bank dividend policy should be to maximize the
value of the stockholders returns , which in turn ,
should benefit the bank .
-The bank management should make assurances that
past dividend action will be continued in the future ,
this is an important factor in evaluating the worth of
bank stock .
-Cash dividends should be at a point where the bank is
capable of maintaining the various type of business
and economic conditions for reasonable period of time
until condition improve .
-Cutting dividends is a negative sign to investors and
reflects a pessimistic view of future by management .
-Consistency in dividends policy is desirable .
- Fluctuation in the amount of the cash dividend paid is
undesirable
             Dividend policy
In deciding on the level and pattern of
dividends . Management must consider
several factors , these factors include :
1- determine whether the bank has a shortage
or excess of capital .
2- the rate of return a bank can earn on its
capital .
3- the bank earning stability .
4- the bank plans for future growth .
         Stock repurchase strategy
When banks has excess capital it always implement
stock repurchase strategy .
The repurchase of stock increased return on equity ,
earning per share and book value per share
- The repurchase strategy enhanced the value of the
remaining common share , because earning per
share are growing faster .
Show table 9.10 p. 338
 •
             Sustainable internal growth
A general rule is that if a bank can finance all of its capital
needs internally without hurting its owners or stock
price , it should do.
Retained earning are not free sources of capital (the cost
of retained earning include the different between cash
value of money today and cash received in future
years.)
There are three variable that combine to determine how
much of a bank growth can b e sustained through the
retention of earning :
1- the amount of capital the bank and its regulators
determine to be adequate .
 2- The earning the bank is able to generate .
 3- the proportion of these earning that is retained in the
bank.

   •
        Formula for calculating sustainable growth
(a) Sustainable growth SG= (PM)×(AY) ×(1-D)
                              EC/TA- (PM) ×(AY) ×(1-D)
 (b) SG = (PM) ×(AY) × (LM) ×(1-D)
          1- (PM) ×(AY) ×(LM) × (1-D)
 (c) SG = (ROA) × (1-D)
           EC/TA – (ROA) ×(1-D)
 (d) SG = (ROE) ×(1-D)
           1 – (ROE) ×(1-D)
Where
SG= sustainable growth , or the annual rate of increases in average total assets that can be
supported by internally generated equity capital .
PM= profit margin or net income after tax divided by total operating income .
AY = assets yield , or total operating income divided by total average assets .
D= percentage of after tax net income paid in cash dividends .
EC = average equity capital .
 TA= average total assets .
 LM = leverage multiplier , or average total assets divided by divided by average equity
capital .
 ROA = return on average total assets
ROE = return on equity
Notice: Equation (c) is the most common formula for calculating sustainable growth
Example p.341
The Effect of Capital Requirements
   on Bank Operating Policies
 Limiting Asset Growth
 The change in total bank assets is restricted by the
 amount of bank equity


           ROA  (1  DR)  ΔEC/TA
  ΔTA/TA 
 Where
            EQ/TA  ROA  (1  DR)
 TA = Total Assets
 EQ = Equity Capital
 ROA = Return on Assets
 DR = Dividend Payout Ratio
 EC = New External Capital
Maintaining Capital Ratios With
  Asset Growth: Application
 Case 1: 8% asset growth, dividend payout = 40%, and capital ratio = 8%.
          What is ROA?
                         ROA(1  0.40)  0
                 0.08 
                        0.08  ROA(1  0.40)
       Solve for ROA  0.99%

 Case 2: 12% asset growth, dividend payout = 40%, and capital ratio = 8%.
          What is required ROA to support the 12% asset growth?
                         ROA(1  0.40)  0
                 0.12 
                        0.08  ROA(1  0.40)
       Solve for ROA  1.43%

 Case 3: ROA = 0.99%, 12% asset growth, and capital ratio = 8%.
          What is the required  dividend payout to support the 12% asset growth?
                       0.99(1 DR)  0
               0.12 
                      0.08  0.99(1 DR)
       Solve for DR  13.42%

 Case 4: ROA = 0.99%, 12% asset growth, capital ratio = 8%, and dividend payout = 40%.
          What is the required  external capital to support the 12% asset growth?
                           0.99(1 0.40)  ΔEC/TA
                    0.12 
                             0.08  0.99(1 0.40)
       Solve for EC/TA  0.29%
                   ΔEC  $294,720
Maintaining Capital Ratios With
  Asset Growth: Application
                                                                  Case 1 Case 2 Case 3   Case 4
                                                       Intitial Initial 8% 12%    12% 12% Growth:
                                                      Position Asset Growth: Growth:  External
Ratio                                                            Growth  ROA  ROA      Capital
Asset growth rate (percent)                                         8.00% 12.00% 12.00%    12.00%
Asset size (millions of $)                              100.00      108.00 112.00 112.00    112.00
ROA (percent)a                                                   0.99%     1.43%     0.99%     0.99%
Dividend payout rate (percent)                                  40.00%    40.00%    13.42%    40.00%
Undivided Profits (millions of $)                        4.00      4.64      4.96      4.96     4.665
Total capital less undivide profits (millions of $)      4.00      4.00      4.00      4.00     4.295
Total capital / total assets (percent)                 8.00%     8.00%     8.00%     8.00%     8.00%
e modified the
     classic form. Sharp, angular bends in the mainstream
* Tributaries control by rock features.
* The direction of angulations indicate the rock type
     - sandstone develops
          jointing patterns
     - limestone cleavage
          joints that intercept at acute angles
Rectangular:
    patterns are variation of a dendritic system
    - tributaries join the main stream at right angle and form
    rectangular shapes
    - controlled by bed rock jointing, foliations or fracturing
    - stronger pattern, the thinner the soil cover
    - forms in slate, schist
    - form in gneiss, in resistive sandstone in arid climates
Parallel
Systems develop on homogeneous, gentle, uniformly sloping
     surface and main stream may indicate a fault or fracture
- Tributaries join stream at right angle
- Landform of young coastal plains
- Large basalt flows

Karst
With surface and subsurface drainage network
- Result from solution weathering of limestone
- Few streams ends in sinkholes
- Scattered sinkholes depressions
- Some gullies lead into sinkholes
Trellis
Is modified dendritic forms with parallel tributaries and short
      parallel gullies occurring at right angles.
- Indicate structure not rock type
- tilted, interbedded, sedimentary rocks in which the main
      parallel channels follow the strike of the beds
Radial:
Is a circular network of almost parallel channel flowing a way
     from a central high
         - A major stream is found in point the bottom of
     topographic features
         - volcanoes
         - isolated hills
         - domelike landforms

Annular:
Develops on topography forms similar to those associated with
   radial patters, but joints control the parallel tributary
       - Sedimentary domes
       - Granite domes
Braided
is found in alluvial plains in arid regions
     - coarse soil
Centripetal
is variation of radial
     - drainage directed downward toward a central point
     - In basin or sinkhole
     - eroded anticline or syncline
Pinnate
Is modified dendritic patterns
     - indicate a high silt content soil
     - found in loess
     - fine tex. Flood plains
Subdendritic
Is complex combination of rectangular and parallel dendritic
     - Indicated different condition, rock type
Internal
Is not have of an integrated drainage
     * granular material, high permeability
     * porous rock materials
     * in alluvium areas
     * beach ridges
     * sand dunes
Deranged
In nonintegrated drainage
     * flat or undulating surface
     * high water table
     * swamps depressions, or lakes
     * in flood plains, till plains
Intrusive rocks
Landforms
- Batholith
      * 40mi2(60km2)area
      * dome shaped
      * Irregular massave domed roof
      * enlarges downward
- Stocks
      * loss then 40mi2
      * similar to batholiths
- Laccolith
      * leans shaped
      * concordant with surrounding rock structure & cause uplift
- Sills
      * similar to the laccolith except no uplift
      * Dikes enter a crack between rock
                           Intrusive rocks
Tone: light gray in acidic rocks dark gray in basic rocks
Drainage: Dendritic patterns
Soil: thin layer of residual soil (2feet thick) SM with ML
Trenching: Excavation of material for pipelines is expensive to
     dig 6 feet deep trench
                         Intrusive rock
Construction material:
       sand is mixed with silt and clay “not suitable”
       Aggregate is not recommended for use in concrete
       Building stone is suitable, “excellent building stone”
Land slides:
    may occur due to fracturing and water
Ground water supply:
    not sufficient as water supply
Dam construction:
    Good areas for dam
Foundation: high load bearing capacities
                       Sedimentary rock
The sediments originates from
    * Remnants of decomposes or disintegrated igneous
    metamorphic or sedimentary rocks (clastic rocks such as
    conglomerate, sandstone, shale)
    * Derived from chemical reactions (Limestone and gypsum)
    * Derived from organic sources like coral
                    Characteristics of Sed. Rocks
* Diversity of their physical and engineering properties
     (strength, porosity, permeability)
* Numerous type of structures
     - Bedding
     - interbeding of different rocks flat, tilted
     - Solution cavities
     - Anticline / syncline
     - Uplifting: cause drying and further jointing perpendicular
     to the bedding
                             Sandstone
Sandstone consolidated sand grains (silica or orthoclase)
* form: flat table rock of equal elevation due to its bedding
      : Rugged topography due to its relative resistance to
     weathering

Interpretation of pattern elements
Tone: light, banded
Drainage: angular or rectangular
Soil: GM (silty gravel)
      GC (clayey gravel)
      SM (silty sand)
      SC (clayey sand)
                           Sandstone
Trenching: may require blasting seepage may occur
Construction material:
    Sand good source from sand dunes
    Agg. From fair to excellent
    Building stone strong sandstone is good and important
       source of stone
Ground water: very good aquifer
Dam: seepage problem
Foundation: Mostly high load bearing capacities
                            Limestone
Limestone is formed by processes as:
1- organic deposition (diatoms, coral)
2- chemical precipitation ‫ ترسيب‬contains impurities such as sand,
     silt or clay
3- chemical reactions (Dolomite, replacement of Ca++ by Mg++)

                      Weathering properties
1- weather by chemical process as carbonation
2- water dissolves limestone through joints and forms channels
     and solution cavities
3- form depression
4-form table rocks with vertical faces ‫نجد سهل واسع مرتفع‬
5-coral form jagged cliff ‫حرف ما وخشن ومسنن‬
                             Limestone
Tone: light gray uniform banded
Drainage: Angler
sail: GM , ml
      GC , CL
Trenching: may require blasting coral       power equipment
Construction:
Material: sand not suitable
          Agg. Good (limestone)
                poor (coral)
          Building good
           stone
           facing stone very good-excellent
Ground Water: cam be found but very hard
Dam: seepage         due to channel-cavities
                     channel-cavities
Slate: foliated meta. rock formed by heat and pressure on shale
    * has parallel foliation planes (slaty cleavage)
    * low grade of met am orphism
    * weathers very quickly by mechanical means
    * develops rugged topography with sharp ridges
                           Slate
Tone: uniform light gray
Drainage: Rectangular
Soil: CM , silty clays
Trenching: Easy with equipment
Sand: not suitable
Aggregate: not suitable
Facing stone: not suitable
Ground water: maybe available
Dam construction: not suitable
Foundation: shallow fourdatim
                               Schist
Tone: light faint parallel banding
Drainage: rectangular – angular
Soil: ML,CL
Trenching: easy with power equipment
Sand: not suitable
Aggregate: not suitable
Building stone: not suitable
Facing stone: good
Ground water: suitable in weathered and fractured
Dam seepage along fracture
Foundation: should take care due to clay and schistosity
                            Gneiss
Tone: Uniform light gray
Drainage: rectangular – angular
Soil: SM silty sand
      clayey sand (SC),ML
      MH,CL and CH
Trenching: heavy equipment and blasting
Sand: not suitable
Aggregate: good to fair
Facing stone: good to poor
Ground water: may occur in fractured
Dam: suitable
Foundation: excellent foundation material
- Terrain evaluation is study of large area and used in
     - planning
     - site investigation
- Objective
     - classification of the area into terrain class
     - Engineering terrain class can be studied
     - similar eng. Properties can be assumed for similar class
- Type of classification
PUCE Pattern Unit Component Evaluation
Basis of the PUCE
    - topography
    - underlying rock and structure
    - soil and vegetation cover
PUCE operates of 4 levels as:
    1- Provinces
    2- Terrain Pattern
    3- Terrain unit
    4- Terrain component
Terrain pattern

Terrain unit (1:25000 or larger)

Terrain component (1:2500 or larger)
Terrain provinces
                      1:250000
Terrain pattern

Patterns and units are described using,
      * Aerial photography Interpretation
      * Field validation
                     Process of terrain analysis
1- To classify terrain on the basis of similarity or homogeneity of
     certais properties, attributes
2- To assess (qualitative) or evaluate (qualitative) like area for
     the properties of the terrain that are significant for the
     desired purpose
PUCE class is composed repetitive association ‫ إتحاد تكراري‬of
    members of the next class in the hierarchy ‫تسلسل هرمي‬
a) aprovince consist of association of terrain pattern
b) a terrain pattern consist of association of terrain units
c) a terrain units consists of a repetitive association of terrain
    component
The pattern and unites are describe using the following criteria:
a) Slope categories
Flat, gently undulating area
b) Soil categories
Shallow soil, sand, uniform soil, organic soil
c) Vegetation categories
Grassland, open wood land, forest area
d) Land use categories
Forestry, unused area, recreation, urban development
                         Terrain classes
A province
     - rock with uniform age
     - determined from geology map of scale 1:250000
     - Association of terrain patterns
A terrain pattern
     - uniform landscape
     - Recurring ‫ متكرر‬topography soil associations, natural veg
     amplitude
     - characteristics drainage pattern
     - uniform drainage density
A terrain unit (area)
     - consist of a single land form
     - characteristic soil and vegetation formation
A terrain component
     - A part from a topography
     - Uniform underlying litho logy and a uniform structure
     - a consistent association of are class in the USC system
     - a characteristic vegetation association
               Class ification nomenclature ‫مصطلح رمز‬
The nomenclature used in PUCE system is numerical
a) Terrain components are allocated ‫ مقسمه‬eigh digits
b) Terrain units are allocated four digits
c) Terrain pattern are allocated three digits
d) Province are five digits
Province (35.003)
    35 carboniferous system
    .003 third recognized ‫ تعرف‬province of carb. age
Terrain pattern 25/2
    2 relicf amplitude to 75 m
    5 Drainage density
    /2 second recogrize ‫ تعرف‬pattern in the province
Terrain unit 1.4.36
    1.4 strongly undulating surface
    .3 clay soil
    6 forest
Terrain component 44203101
    4 slope major axis cor
    4 slope major axis to 10º
    2 slope major axis to 2º
    03 soil profile
    1 land use – forestry
    01 vegetation

				
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