--

Systems Techniques in.Water Resources 177

Inflow:
Storage:

Release R; R; R~
Fig. 2.14 I Trace Back of Optimal Solution for the Reservoir Operation Problem
The following numerical example illustrates the solution procedure for the
reservoir operation problem.
Inflows during four seasons to a reservoir with storage ca-
pacity of 4 units are, respectively, 2,1,3, and 2 units. Only discrete values, 0,
1, 2, ... , are considered for storage and release. Overflows from the reservoir
are also included in the release. Reservoir storage at the beginning of the year
is 0 units. Release from the reservoir during a season results in the following
benefits which are same for all the four seasons.
Release Benefits
0 -100
1 250
2 320
3 480
4 520
5 520
6 410
7 120

To obtain the release policy we use backward recursive equation, starting

with the last stage.

With t = 4 at n = 1, we seek a solution for all possible values of S4 , the

storage at the beginning of period (season) 4, These possible values are 0, 1,
2, 3, and 4, thie storage capacity being 4 units. Inflow during the period being
2 units, some values of release will be infeasible if the end-of-period storage
resulting from such values is more than 4 units. For example, for S4 = 3,
release. R = 0 is in.feasible~:1)ecal.ise;S4+(24~R-(-=-3-+-2--0 =§~is more than-----
the capacity, 4. The following tables give the calculations for each stage.
Stage 1 : t = 4; n = 1
Q4 = 2 f ~ (S4 ) = max [B4(R4)]
o s; R4 s; s4 + Q4
s4 + Q4 - R4 s; 4
--
Systems Techniques if! Water Resources I as
n
L xj :::;; Q (water availability constraint), and
j=l
n
I. x /wj :::;; A (land availability constraint)
j=l .
Note that x/wj is the land allocated to crop j. With non-negativity of vari-
ables included, this problem may be solved as a linear programming problem.
To formulate this problem as a DP problem, we need to define two state
variables: Sj, the water available at stage j, and Lj, the land available at stage j.
The backward recursion may be written as
:h(Sj, L) =max [ R/x) + fj + 1 (Sj- xj, Lj- x/w)]
o:::;;xjs;sj
X/Wj:::;; Li
This recursive equation should be solved for all values of Sj and Lj satisfying
0 :::;; Sj :::;; Q and 0 :::;; Lj :::;; A.
While solving DP problems on computers, with two state variables, there-
fore, we need to store two-dimensional vectors for all the stages, as they are
required for tracing back the solution after computations for all stages are
completed. In general, for a k-state variable problem, we need to store a k
dimensional vector in the computer memory, and as the number of state vari-
ables increases, the computer memory requirements increase rapidly. This is
called the Curse of Dimensionality of DP. Till recently, most common DP
applications .dealt with only a few (3 or 4) state variables. With rapid progress
in the computer speed and memory availability in recent years; however, prob-
lems with more than 6 state variables have also been solved using DP (e.g.
Mujumdar and Ramesh, 1997).

'I
...

Chapter 13
Econontic
Considerations in
Water Resources
Systents

3.1 BASICS OF ENGINEERING ECONOMICS

3.1.1 General Principles
In this section, we discuss the value of money at different points of time and,
based on it, a methodology of comparing alternative plans on an economic
basis.
The first and foremost thing to realize is that money has time v~lue. The
same amount of money is worth more today than tomorrow. Money can be put
to productive use by way of investment. Comparison of amounts of money
must be made based on a common time reference.
Interest may be thought of as the price one has to pay for money borrowed
from someone else, or as the return derived from capital invested; or as the
reward for making available money for someone who needs it. Because of the,
interest rate, a given amount of money today has a higher value when produc-
tively invested than the same amount in future.
If one unit of money were invested in a bank at a nominal interest rate, i
(expfesseaas% j:feryear by default), if wol.lTd accumulate iii one y'earTo\1 ~· i).
However, if the money is compounded m times in a year (i.e. if there are m
compounding periods in a year), the amount accumulated is calculated as fol-
lows:
Let there be m periods in a year and let the money be compounded at the
end of each period. The interest rate per period is ilm. If Re 1 is invested in a
i
bank, it would accumulate in a year to (1 + ilm)m. The effective annual interest
rate then is {(1 + i/m)m - 1}, or Effective rate of interest, ie = (1 + i/m)m - 1.
~.,, '

92 1 Basics of Systems Techniques

Example 3.1.3 The nominal rate of interest is 10%. Determine the effective
rate of interest when money is compounded
1. .yearly
2. half yearly
3. quarterly
4. daily
1. yearly: ie is the same as i = 10%
2. No. of corresponding periods in a year= 2

Interest rate per 6 months = i/2 = .:!Q

= .05
2
:. One rupee becomes (1.05 2) at the end of the year= 1.1025
:. Effective rate of interest= 1.1025- 1 = .1025 or 10.25%
3. No. of compounding periods= 4

Interest rate per period = i/4 = lQ

= .025
I
I
4
One rupee becomes (1.025) 4 = 1.10381
:. Effective interest rate= 1.10381 - 1 = .10381 or 10.381%
4. No. of compounding periods= m = 365

Interest rate per day = z"/m = .10

365
.10 )365
:. Effective rate of interest= ( 1 + - 1 = 1.10516- 1
365
= .10516
= 10.516%

Discount rate is the interest rate used in discounting future cash flows. It is
an expression of the time value of capital used in equivalence calculations
comparing alternatives, and is essentially a value judgment based on a compro-
mise between present consumption and capital formation from the viewpoint of
the decision maker (James and Lee, 1971).
Depreciation is the consumption of investment in property, or is the reduc-
tion in the value of property due to its decreased ability to perform present and
future service.
__________ SinJdng fund is a fund created by_makit:J_g periodic (usually_~_gg~--c!.~J29~it§_p
at compound interest in order to accumulate a given sum at a given future time r
for some specific purpose. If we buy .a pump for Rs 10,000, which has a
service life of say 10 years with no salvage value (salvage value of a property
is the net worth of money that can be ·realized at the end of its service life),
then to compensate for the depreciation of the pump with. time, we raise a
sinking fund by paying equal amounts of money at regular intervals so that the
total sum (with compound interest) accumulates to Rs 10,000 at the end of 10
years at the specified discount rate.
 Economic Considerations in Water Resources Systems I 93

. If the salvage value (market value at the end of the service life) of the pump
at the end of 10 years is L, then the sinking fund to be raised is the cost of the
pump, P, minus the salvage value, L, i.e. (P- L) ignoring inflation. We can
buy another new pump at the end of 10 years with this sum, (P - L). We ignore
the effect of inflation throughout our discussion on economic analysis.
Sometimes sinking fund is simply computed based on the straight-line de-
preciation method. The annual depreciation, in this method, is calculated as the
ratio of the total depreciable value, (P - L) to the service life, n in years. But
the most accurate method of computing the sinking fund is through the use of
the sinking fund factor, defined later in this section under discount factors,
which precisely accounts the time value of money.
3.1.2 Discount Factors
Discounting refers to translating future money to its present value, and com-
pounding refers to translating present money to a future period in time.
For equivalence calculation oLmoney at different points of time and for
comparison of engineering alternatives, we need to know the commonly used
discounting factors:
The following notation is used in this section.
i = interest rate per year expressed in per cent
P = present sum of money
F = future sum of money
A = annual payment, or payment per period
Usually, three values out of these four would be known and it will be
required to compute the fourth.
A cash flow diagram is usually drawn in each case, depicting the payments
on one side and receipts on the other side of a horizontal time line, in order to ,
facilitate computations of the value of money at different times.
1. Single Payment Factors
(a) Compound Amount Factor (FIP, i%, n) This factor multiplied by the
sum, P, invested initially at interest rate, i, for n years, gives .the accumulated
future sum, F. Unless otherwise mentioned, the interest is assumed to be com-
pounded annually.
F = P (1 + it or FIP = (1 + it.
(b) Present Worth Factor (PIF, i%, n) This factor gives the present worth,
- P, of a future sum, F,aiscounted at interesnate,-1, over n years.
P = F/(1 + i)n or PIF = 1/(1 + it
2. Uniform-Annual Series
(a) Sinking Fund Factor (A'IF, i%, n) Assume we buy a machine for
Rs 10,000 with a service life of 10 years with no salvage value. After 10 years,
to utilize the benefit of service from the equipment, we raise a sinking fund by
putting equal amount of money at the end of each year for the service life of
the machine. Let the discount rate be i. The annual amount A', which when
 94 I Basics of Systems Techniques

invested in a bank at i interest rate for n years will accumulate to the cost of
replacement at the end of the service life of the machine, is the annual payment
for the sinking fund.
The sinking fund factor (when multiplied by F) gives the amount of money,
A', to be invested at the end of each year for n years to yield an amount F, at
the end of n years. A' is the annual equivalent of sinking fund.

It can be seen from the cash flow diagram that

F =A'+ A'(l + i) + A'(1 + ii + ... + A'(1 + i)n-l
=A' [(1 + il- 1]/i
The sinking fund factor, denoted by the notation (A'IF, i%, n) is equal to
A'IF= i/[(1 + il- 1]
In this case, a uniform series of equal annual payments is considered. In all
the following formulae, all payments are assumed to be made at the end of the
period, unless otherwise mentioned.
(b) Compound Amount Factor (FIA', i%, n) This factor gives the amount,
F, that will accumulate at the end of n years, if an amount A' is paid annually at
the end of each year for n years.
This is the reciprocal of the sinking fund factor (A'IF, i%, n). The compound
amount factor (FIA', i%, n) is given by
FIA' = [(1 + i)"- 1]/i
(c) Capital Recovery Factor -(AlP, i%, n) If an amount P is invested in a
bank at interest rate i for n years, the Capital Recovery Factor (CRF) gives the
annual amount of money, A, that can be withdrawn at the end of each year for
n years, such that the entire amount of the initial investment is recovered with
interest at the end of the nth year.
Put in another way, if a loan of Pis taken from a bank at i interest rate, the
, 1 Capital Recovery Factor computes the annual amount of money, A, to be paid
at the end of each year for n years, such that the initial loan of P is repaid with
interest at the end of the nth year.

-~-- ~_A-~ ,4___,4 _______ - - -----~ ___c.:A -

t t 3t2
t
n
p

Referring to the cash flow diagram, we can write

A A A A
") +
p = -(1 2 + 3 + ... +
+l (1 + i) (1 + i) (1 + i) n
 Economic Considerations in Water Resources Systems I 95

A[(l + i)n '--1]

= i(l + i)n
Therefore,
AlP = [i (1 + it]l[(1 + it - 1]
Note that both A and A' refer to annual values. The notation A is used with
reference to the present amount P, and A' with reference to the future money F,
to express the fact that A and A' are different in value.
:I
(d) Present Worth Factor (PIA, i%, n) This factor gives thepresent worth of
equal annual payments made at the end of each year for n years if the interest
rate is i.
This is the reciprocal of the Capital Recovery Factor and is equal to
PIA = [(1 + i)n ..:. 1]1[i (1 + it]
It may be noted that the Capital Recovery Factor, (AlP, i%, n), accounts for 'I
I
both the interest and the sinking fund due to depreciation and is equal to the
sum of the interest rate arid the sinking fund factor, as- -
(AlP, i%, n) = [i (1 + i)n]l[(l + it- 1]
= [i { (1+ it- 1} + i ]1[(1 + it - 1]
= i + i/[(1 + it - 1]
=interest rate, i + sinking fund factor, (A'IF)
This means that the capital recovery is equal to the interest plus depreciation
on an annual basis. The equivalence may be seen from the following example.
Note: As n ~ =,AlP~. i, meaning that over a large repayment period, the
capital recovery boils down to· the interest with negligible amount to be raised
as the sinking fund.
Lt (AlP, i%, n) = i
n~=

If A is the constant annual payment to be made on a perpetual basis, the

equivalent present cost indicated by the value, Ali, is termed as the capitalized
cost.
In the examples illustrated in this section, the numerical values of the dis-
count factors, where necessary, were directly used as obtained from the appro-
priate expressions, for the required values of the interest rate, i, and the number
of years, n.
o • Suppose we buy a pump for Rs 10,000 today. It has a service
life of 10 years with no salvage value at the end of its service life. We take a
loan of Rs 10,000 from the bank at 6% interest rate.
The amount required to be paid at the end of each year to the bank to repay
the loan completely in 10 years with interest is
A= 10,000 x Capital Recovery Factor, [AlP, 6%, 10], with i = 6 and n = 10.
= 10,000 x 0.13587 = Rs 1,358.7
 96 I Basics of Systems Techniques

The loan of Rs 10,000 will be repaid with interest if Rs 1,358.7 is paid to

the bank at the end of each year for 10 years.
On the other hand, we bought the pump for Rs 10,000 (bon·owed from the
bank) and it is in our hands today (but we have no other money). Every year,
we raise a fund and, at the end of the year, deposit in the bank to have enough
money at the end of 10 years to buy another new pump. Assuming that the
cost of the pump remains the same, and that money grows at the same interest
rate i (as when we borrowed the money to buy the pump), the annual sinking
fund is
A'= 10,000 x sinking fund factor [A'IF, 6%, 10] with i = 6 and n = 10
= 10,000 x 0.07587 = Rs 758.7 (This is the monetary equivalent of the
annual depreciation of the pump).
With this annual payment, we will have Rs 10,000 at the end of 10 years in
the bank to buy a new pump. Only then, at the end of 10 years we will be in
the same position as we are now. But remember, we should also pay the bank
interest annually on the borrowed money of Rs 10,000 at 6% per year. This
annual interest amount is
Annual interest = Pi
= 10,000 X .06 = Rs 600
which is the exact difference between the amounts A and A',
(A - A' = 600),
i.e. Amount of capital recovery = Interest + Sinking fund (annual)

3.1.3 Amortization
Amortization is the term generally used to indicate payment of a debt in equal
instalments at uniform intervals of time. Part of each payment is credited as
interest and the remaining towards repayment of loan. For example, if we
borrow Rs 100,000 from the bank to be repaid in 10 years at 6 per cent interest
rate, then the equal instalment of money to be paid at the end of each year is
A = Annual instalment = Principal amount x Capital recovery factor
= P (AlP, i%, n) = 100,000 x (AlP, 6%, 10) = 100,000 (0.13587) = 13,587
This is the exact amount to be paid at the end of each year. For any given
.year, this has two components, one is the interest on the outstanding balance of
the loan at the beginning of the year and the other is the amount credited
towards repayment of the loan itself.
Each year, though we pay to the bank the same amount of Rs 13,587
· annua:Uy;-the~amount cre-diled fowaras--friferesf aiid___fowards debt reduction
(repayment) will be different in different years, because the interest in a year is
computed based on the outstanding balance (of debt) at the beginning of the
year. For example, for the first year, the interest component will be 6% of
100,000, which is 6,000. At the end of first year, then, out of Rs 13,587 paid to
the bank, the bank would credit Rs 6,000 as interest, and the remaining amount
of Rs 7,587 (= Rs 13,587- Rs 6,000) towards repayment of the loan. Thus the
outstanding loan at the beginning of the second year is Rs 100,000- Rs 7,587
= Rs 92,413, and so on.
-
Economic Considerations in Water Resources Systems I 97

We can determine the components of interest and repayment (of loan) out of
the equal annual instalment paici, at the end of any year, x, as follows:
We first find out the repayment component in the year x. ·
Repayment I

I
Repayment is made by building a sinking fund through equal annual payments,
A', at the end .of each year. The total fund in n years at i% interest rate should
accumulate to P, when an amount A' is deposited in the bank at the end of each
year for n years.
Thus A'= (Principal borrowed) x (Sinking fund factor)= P (A'!F, i%, n)
The total amount paid towards debt or loan at any point of time is equal in
value to the sinking fund raised till that point of time.

A'f A' f A' f

3 ...

The total amount paid towards repaying the loan till the end of year x
Dx =A' +A' (1 + i) +A' (1 + ii + ... +A' (1 + i)x-l.
=A' [(1 + iY - 1]/i
Similarly, the total amount paid towards repaying the loan till the end of year
x- 1 is
Dx- I = A' [(1 + i)x-1 - 1]/i

Therefore, the amount paid towards repayment of loan in the year x,

fll>x = Dx- Dx-1 =A' {[(1 + i/-- 1]/i- A' [(1 + i)x-l- 1]/i}
= A' (1 + i)x-1
We know that A', the annual sinking fund, is equal to
A'= P (A,'!F, i%, n) = P {i/[(1 + iY- 1]}
1
Therefore, fll>x = Pi(1 + iY- 1[(1 + it- 1],
which is the repayment component in the year x. Since A = Interest + Repay-
ment, the interest component for year x = A - fll> x·
For the problem mentioned earlier, where P = 100,000,
i = 6% and n = 10 yrs, the yearly components of repayment and interest are
shown in the following table .
.. ~

P = 100,000; i = 6%, n = 10 yrs

A = 13,586.80 A' = 7,586.80
Year Repayment, !:JJx Balance Interest
coli col2 col3 col4
1 7,586.80 92,413.20 6,000.00
2 8,042.00 84,371.20 5,544.79
3 8,524.52 75,846.68 5,062.27
4 9,036.00 66,810.68 4,550.80
(Contd.)
'1"''j\'
.I

98 I Basics of Systems Techniques

(Contd.)
P = 100,000; i = 6%, n = 10 yrs
A = 13,586.80 A' = 7,586.80
Year Repayment, Wx Balance Interest
coli col2 col3 col4
5 9,578.15 57,232.53 4,008.64
6 10,152.84 47,079.68 3,433.95
7 10,762.01 36,317.67 2,824.78
8 11,407.74 24,909.93 2,179.06
9 12,092.20 '12,817.73 1,494.60
10 12,817.73 0 769.06

col 1 : year number,

I
col 2: repayment in the year, f:ll)x (end of year value),
I
I,
col 3 : outstanding balance of loan at the beginning of the year in col 1
=P-Dx,
col 4: Interest for the year credited at the end of the year= A- f:ll)x · ,- --
Note that the interest component will be relatively high and the repayment
low in the initial years, while the trend reverses in the latter years.

A person bonows a sum of Rs 50,000 @ 6 per cent interest

for 10 years. Determine the equal annual sum to be paid at the end of each
year to repay the loan, and the amounts credited towards interest and repay-
ment in the 5th year.
1. The amount of money to be paid at the end of each year
= 50,000 [PIA, 6%, 10] = 50,000 (.1359) = 6795
2. Year x = 5; A= 6795
Annual sum paid at the end of 5th year, A = Rs 6795
The sinking fund, A' = 50,000 (A'!F, 6%, 10)
= 50,000 (.0759) = 3795
The repayment component (reduction in the pri1:1cipal bonowed)
in the 5th year = A' (1 + i)x-I
Payment, Wx = 3795(1 + 0.06) 4 = 4791
. . Interest component =A- Wx
= 6795= 4791 = 2004
. . At the en~ of 5th year,
interest compol!~llt.==_:fS~_~QQ4 _ u

repayment component= Rs 4791

out of the annual amount of Rs 6795 paid
Annual Costs The annual cost (AC) of a project consists of the following:
1. Interest on the bonowed capital
2. Depreciation or amortization cost
3. Operation and maintenance cost (O&M cost), and
4. Other costs, specified, if any
The first two costs sum up to the capital t:eCO"®ry on an annual basis.
-
Economic Considerations in Water Resources Systems I 99

3.1.4 Comparison of Alternative Plans

The background infonnation dealt with so far will be useful 1n the economic
evaluation of alternative proposals for specified needs. The comparison may be
based on any of several available- methods, but a proper economic evaluation
(based on any method) should result in an identical selection. Two of the most
common methods used are illustrated in the following pages: present worth
method and the equivalent annual cost method. The use of these two methods
is illustrated next with examples.
Note that the present worth is a function of the duration of analysis and,
therefore, any comparison of alternatives should be based on the same period
of analysis for each alternative. In this respect, comparison on the basis of
equivalent annual cost is relatively easier.
Which of the following plans is more economical at 6%
interest rate?
Plan A Plan B
Cost of Equipment 50,000 35,000
Annual 0 & M Costs 2,000 2,500
Salvage value 7,000 6,000
Service life 30 years 15 years
Let us compare the two alternate plans by different methods
1. Equivalent Annual Cost (AC)
PlanA
Annual Cost = Interest + Depreciation + 0 & M costs
Interest on borrowed capital = 50,000 x .06 = 3000
Depreciation = (P - L) (A'IF, 6%, 30)
= (50,000- 7,000) (.01265) = 543.95
0 & M Costs = 2000
Total AC = 3000 + 543.95 + 2000 = 5543.95

0 & M Cost = 2500

Total AC = 2100 + 1245.84 + 2500
= 5845.84
Plan A is preferable because of lower annual cost.
2. Present Worth (PW) Comparison
As mentioned earlier, PW should be compared for equal periods of
analysis in both plans. Let us compare the plans on a 30-year basis. For
I
this purpose, the life of plan B is extended by another 15 years with II
'!i
, '
.

1oo I Basics of Systems Techniques

identical expenditure pattern as in the first 15 years. A compar~son of

present worth of all costs can be made then, based on an analysis period
of 30 years.
Plan A (30 years)
PW of initial cost 50,000 = Rs 50,000
PW of salvage value= -7000 (PIF, 6%, 30)
(negative cost) = -7000 (0.17411) = -1218.77
I
PW of 0 & M Costs = 2000 (PIA, 6%, 30)
I I

i. = 2000 (13.7648) = 27,529.60

PW of total cost= 50,000 - 1218.77 + 27,529.60
= 76,310.83
i.
Plan B (for 30 years)
I
PW of initial cost of 35,000 = 35,000
PW of 35,000 at 15 years= 35,000 (PIF, 6%, 15)
.. = 35,000 (.41727)= 14604.45
PW of 6000 (negative cost at 15 years)
= -6000 (P/F, 6%, 15)
= -6000 (.41727) = -2503.62
PW of 6000 (negative cost at 30 years)
=-6000 (PIF, 6%, 30)
= -6000(.17411) = -1044.66
PW of 0 & M Costs = 2500 (PIA, 6%, 30)
= 2500 (13.7648) = 34,412
PW of Total Costs
= 35,000 + 14,604.45- 2503.62- 1044.66 + 34,4i2
= 80467.72
Because of lower costs Plan A is preferred to Plan B.
Note: The annual costs worked out in method 1 may be obtained from the
present worth calculated above by applying the capital recovery factor, (AlP,
6%, 30) = 0.07265
AC of Plan A = (PW)A .(AlP, 6%, 30)
= (76,310.83) (.07265) = 5543.98
··-·-------- -Ab'-Gf-fl1an-B-={f!.W-)B--.--(AtF,6%, 30)
= (80,465.72) (.07265) = 5845.98
which are same as before except for rounding off of errors.

There are two alternatives to water supply in an irrigation

district. Plan A is to construct an open channel at an initial cost of
Rs 50,00,000 with 0 & M cost of Rs 400,000 per year. The alternative, Plan
B, is to go for a piping system at an initial cost of Rs 90,00,000 and an 0 & M
 Economic Considerations in Water Resources Sy_st_em_s_ ____.IL-:.1...:;'0~1-

cost of Rs 50,000/year. Money is available for 6% interest rate, and sinking

fund will improve at 4% interest. Useful life of the project in both cases is 20
years. Select the more economical of the two alternatives.
Let us work out the annual costs in both cases.
· Plan A Plan B
Interest at 6% 300,000 540,000
Depreciation @ 4% 167,500 · 301,500
(A'IF, 4%, 20)=.0335 on P
0 & M Costs 400,000 50,000
Total 867,500 · 891,000
. . Plan A is preferable to Plan B.
Note: In this example, the interest rate on borrowed money and the discount
rate for depreciation are different.

Both pipe sizes have a life

rate is 6o/o.
(i) Which is the most economical pipe size, if the total number of
hours of pumping per year is 5000. Compare equivalent annual
costs. ''
(ii) How many hours of pumping per year are. required to make the
two pipe sizes equally economical. ··
(Ans: (i) pipe size A (ii) 4120 hrs/yr)
 102 1 Basics of Systems Techniques .

3;2 ECONOMIC ANALYSIS

A discussion of the market supply and demand is necessary before we go into
the benefit cost analysis of project planning ..
Types of Goods in the Market In general, goods can be classified into two
categories:
1. Normal goods, .or personal goods, or private goods, and
2. Collective goods, or public goods.
Normal goods have the property .of being consumed by individuals, such
that, given a fixed quantity, more units for one means fewer for others and vice
versa. Most of the goods for which a real market exists come U)lder this
category. For example, for a fixed number of automobiles produced, the more
one section of the people buy them, the less they are avai~able for others (to
buy), and vice versa. However, there are some other goods for which this is not
true. National defence and flood control are examples of these. Here each one
"consumes" the same amount of national dt;fence, a,nd, although each persQn
added to the population consumes national defence, his/her consumption does
not necessarily make it less available for the rest of the population. These
goods are called collective goods, or public goods, in contrast to the normal,
or personal or private goods.
3.2.1 Market Demand and Supply
Demand is the quantity per unit time that people will be willing to buy, ex-
pressed as a function of the price, all other factors remaining constant. The
demand curve has the property that it slopes downward to the right because
lower prices increase sales (more number of buyers), and higher prices de- .
crease them (less number of buyers).
Price Elasticity of Demand
Let us consider a demand curve, Fig. 3.1, which depicts a linear relation be-
tween price and the quantity demanded.

A
Price, P

E=O

Quantity, Q B

Fig. 3.1 I Hypothe~caf_pemand Cu!ve

A demand curve indicates the change in the demand or sales for a given
change in the price. This relationship is given by the price elasticity of demand,
-
:'1
I '!
Economic Considerations in Water Resources Systems I 1o3

E, which is defined as the ratio of the relative change in the quantity, for a
given relative change in the price. If Q is the demand at price P, then the price
elasticity of demand E is given by
E = -(!!.QIQ) + (MIP) = -(!!.Q/M)(P/Q) (12.1)
The negative sign indicates that Q increases as P decreases (slope of the
demand curve is negative). In the case of the linear demand curve shown in
Fig. 3.1, though the slope remains constant at all points on it, the elasticity of
demand changes from point to point on the curve. It varies from infinity along
the vertical axis (E = oo) to zero along the horizontal axis (E = 0). -
A value for E of infinity (point A on the linear demand curve) indicates a
perfectly elastic product, which no one will buy if the price is raised. At this
price, goods become perfectly elastic and are completely priced out of the ·, I

market. A value for E of zero (point B on the linear demand curve) indicates a
perfectly inelastic product, i.e. one for which the price has no effect on demand.
In between these two extremes, a given product may be elastic or inelastic at a
given price, depending on its relation to the demand as defined by the demand
curve, in other words, on the price elasticity of demand, E.
As the price is reduced frorri point A, elasticity decreases, until it reaches
unity at some point C, when the product is no longer said to be elastic. For all
points on the line above C, the price elasticity of demand, or simply the elastic-
ity as it is often called, is E > 1. The total revenue, which is the product of the
price and the quantity of goods sold (P.Q) increases up to this point, as the
increase in· sales (demand) offsets the reduction in price resulting in an increase
in the gross revenue. The point C (point of unit elasticity, E = 1) provides the
supplier the largest revenue. If the price is reduced further down the point C,
though demand continues to increase, the increase will not be fast enough to
offset the decreasing price resulting in a decline in the gross revenue from that
at point C. For all points below C, the price elasticity will be less than unity,
(E < 1). In this region, the product is said to be inelastic. It is to be noted that
the same product is inelastic at low prices and elastic at high prices.
Supply Curve
The supply curve shows the relationship between the quantity produced and the
price at which the producers are willing to sell, other things remaining
. sQn~t:mt.Iy£i~~!Y,__ ::l_supply_cury~ slopes upward to th.e right, meaning th(L~ .
more goods will be produced, and more sellers will enter the market as the
price increases.
Market Price Determination .
A market consists of sellers and buyers. The behavior of each group is, how-
ever, different. The buyers' behavior is reflected by the demand curve and that
of the sellers by the supply curve. The buyers would like to minimize the
expenditure and the sellers would like to maximize tl1eir revenue from sales.
 104 1 Basics of Systems Techniques

The demand curve and the supply curve. therefore combine to establish the
equilibrium market price. The equilibrium price is the minimum under conditions
of pure competition that each individual buyer must pay per unit quantity
purchased and the maximum that each seller can receive for each unit quantity
sold (James and Lee, 1971).
That this equilibrium point corresponds to a J?Oint of unit elasticity on the
demand curve can be shown as follows:
The total revenue from sales or the total expenditure, G, for goods pur-
chased is given by
G = price x quantity = P · Q
For this to be optimum, dG/dP = 0
dG/dP = P dQ/dP + Q = 0
or dQ/dP = -QIP
or - (PIQ) · dQ/dP = E = 1
--
The left-hand side is the price elasticity of demand by definition. Therefore,
the market equilibrium prevails at the price corresponding to E = 1 on the
demand curve. This analysis presumes that free market conditions exist with no
restrictions and there is free entry for an unlimited number of buyers and
sellers. At this equilibrium price, the sellers would have maximized their re-
turns and the buyers would have minimized their total expenditure.
We will not discuss in this book the consequences of a shift in demand or
the effect of subsidies on market behavior.
3.2.2 Aggregation of Demand
The procedure for aggregating demand curves depends on the type of product.
In the case of a normal product (market good), where consumption is mutually
exclusive (what is consumed by one is not available to the others-e.g. drink-
ing water), the combined demand curve is generated by adding the demands of
each individual at a given price. Thus, the combined demand is computed at
each price and plotted to get the aggregate demand curve. This horizontal
addition of demands is characteristic of normal or market goods.
In the case of a collective good, however, where the consumption is not
mutually exclusive (example-flood control), the prices are added up for a
-----~~gi~en demand an<L_fur_all d_emandleyels._Thistypec oLverticaLaddition of--pl"ice
is characteristic of collective or public goods.
Figures 3.2 and 3.3 show examples of how this aggregation is done graphi-
cally for normal goods (such as irrigation water) and for collective goods (such
as flood control), respectively.
The aggregate demand curves can be arrived at analytically as well, if the
equations of individual demand curves are given.

I
i.l,
 Economic Considerations in Water Resources Systems I 1os

Price
Individual Demand Curve
K

A D

Q
L N Irrigation water

KL : demand curve 1 AD= AB + AC

MN : demand curve 2
MPQ :aggregate demand curve

Fig. 3.2 I Aggregate Demand Curve for Personal Good

p--

Aggregate Demand Curve

Price

Individual Demand Curve

N
L Flood ·control' •
KL : demand curve 1 AD= AB + AC
MN : demand curve 2
PQN : aggregate demand curve

Fig. 3.3 I Aggregate Demand Curve for Collective Good

-.
Water is supplied from a project for two types of uses: rural
and urban. The demand curve for rural use is given by P + 3y = 30, and the
demand curve for urban use is 4P + y = 40, where y is the demand and Pis
price in appropriate units. Determine the combined demand curve.
Graphical Method
The rural and urban demand curves are first plotted individually. The com-
bined demand curve is obtained by adding the individual demands at a given
price as, in this case, water is a normal good.
i I

Basics of Systems Techniques

1o6 1
30 A

F
50
AB : rural demand curve
CD : urban demand curve
AEF : combined demand curve

Analytical method
Let rural demand =y r at price P
urban demand= Yu at price P
y = Yr + Yu at the same price P for a normal good.
Yr = (30- P)/3
Yu = 40- 4P
For P:::; 10 (or y ~ 20/3),
y = Yr + Yu = (30 - P)/3 + 40 - 4P = 50 - (13/3)P, or 13P + 3y = 150,
where y is the combined demand.
For P ~ 10, (or y:::; 20/3),
y = Yr = (30- P)/3 or P + 3y = 30
Combined demand curve is given by
P + 3y = 30 for P ~ 10
and 13P + 3y = 150 for P :::; 10.

Example 3.2.2 Two different groups of people are affected by a flood con-
trol project. The demand for flood control from the two groups are as follows:
Group 1 P + 3y = 30
Group 2 4P + y = 40
where P is price and y is the level of protection provided.
Determine the aggregate demand curve.
Graphical Method In this case, the 'consumption' is not mutually exclusive
as flood control is a 'collective good'. Therefore, the combined demand is

40 F

D
40
y

AB =group 1 demand curve

CD = group 2 demand curve
FED= combined demand curve
--
Economic Considerations in Water Resources Systems 107

obtained by adding the prices the groups are willing to pay for the same level
of protection, vertically.
Let P 1 =price for group 1 at demand level y.
P2 =price for group 2 at demand level y.
PI= 30- 3y
P 2 = (40- y)/4.
For y::::; 10, P = P 1 + P2 = (30- 3y) + (40- y)/4, or 4P + 13y = 160.
For y ;::: 10, P = P 1 = 30 - 3y or P + 3y = 30.

Problems I
3.2.1 A water project is proposed to supply water for municipal and irrigation
uses. Municipal demand is given by P + 2Y = 10, and irrigation demand
is given by 2P + Y = 20, where P is the price and Y is the demand.
(i) Determine the aggregate demand curve.
(ii) Assuming the total cost curve is given by C = }'4: Y 2 + Y, determine
the optimal level of Y.
(iii) Determine the share of municipal and irrigation supplies at optimal
level of Y.
(Ans: (ii) Y = 10, (iii) Y (municipal) = 2, Y (irrigation) = 8)
3.2.2 Cons~der the combined demand curve for rural and urban users obtained
in Example 3.2.1. If the combined demand with and witho:ut the water
proj~ct is 5 and 20 respectively, estimate the benefits from the project.
(Ans: 133.65) ....

3.3 CONDITIONS OF PROJECT OPTIMALITY

3.3.1 Water Resources as a Production Process
Water resources development is a production process. The basic purpose of
production is to convert a set of inputs to a set of outputs. Examples· of output
for a water resources project are inigation, hydropower generation and. flood
damage alleviation. Examples of inputs are natural streamflow, cement, con-
crete, steel and turbines.
Let the elements of the input vector, X, consist of individual inputs, x 1, x 2 ,
... , xm. Similarly, the output vector, Y, consists of individual outputs, y 1, Yz, y 3 ,
... , Yn'
The Production Function
The production function defines the relationship between the set of outputs that
can be produced from a set of inputs, reflecting, in a way, the efficiency of the
production process itself. The following is an example of a production function
for two outputs, say irrigation water and hydropower generation from a reser-
voir, in which power is produced from the water released downstream of it (in
a riverbed powerhouse).
 1os 1 Basics of Systems Techniques

Production Function

Hydropower, Y2

.Y1
Irrigation Water Supply, y 1

Fig. 3.41 A Two-output Production Function

The region below the curve containing the origin is the technologically
feasible region. Any combination of outputs above the curve (on the side of the
curve away from the origin) is infeasible. A combination. of outpuJ.s~ reJ;2r~­
sented by points within the feasible region will be inefficient. For example, for
a given amount of irrigation water supply, y 1, there is a maximum of hydro-
power, y 2 , that can be produced from the reservoir, represented by the point A
on the curve. Anything below A is feasible, but ineffident, and anything above
A is infeasible. Thus A repre Jents the maximum possible y2 coiTesponding to
y 1. Thus, the curve represents the l()cus of efficient points in the production
process. This is also called the efficiency frontier curve or production possibil-
ity frontier. The production function is a mathematic3.I representation of this
line. It· is related to the input and output vectors and, putting all the terms on
the left-hand side, it is expressed as
f(X, Y) = 0 (3.1)
While every point on the production function represents an efficient point
(effident operation of the system), the selection of the best point, however,
requires ·value judgment. Each point on. the production function (also termed a
pareto-adillissible solution)· represents a rel'ative value assodated with .each
variable. The objective fuhc:tion is a furiction of both the input.and the output
vectO:rs. The net benefit objective function, u = u (X, Y), is expre~sed a~ ·
U,=.}:.l?;Y;- 2.,c1x1 (3.2)
. j

where b; refers to the unit benefit associated with the ith output element, and c1
- refers-to-the-unit-cosrassuciateliwiththejth: inpurele0-enC ~- .-~··· · ·· ·· ~ · ·
Objective Function
The objective is to maximize net benefits such .that the solution point lies on
the production frontier line, i.e. to max:imiz;e the objective function, u = u(X,
Y), subject to the constraint, f(X, Y) = ,0, where X and Yare input and output
vectors containing m and n elements, respectively.
The Lagrangean formed from the objective and the constraint is written as
L = u(X, Y)- A-j(X, Y) (3.3)
-
Economic Considerations in Water Resources Systems 1 1o9
where A is the Lagrangean multiplier and L is the function to be maximized.
The necessary conditions for optimality are obtained by differentiating L par-
tially with respect to each X;, each Yj and A, and setting each partial differential
to zero. Thus, there will be m + n + 1 equations with m + n + 1 unknowns (m
inputs, n outputs, and A). Solving this set of simultaneous equations results in
the optimal levels of inputs and outputs.
du(X, Y)ldx; = Adj(X, Y)ldx; for i = 1, 2, ... , m (3.4)
du(X, Y)ldyj = Adf(X, Y)ldyj for j = 1, 2, ... , n (3.5)
du(X, Y)ldA = f(X, Y)
Setting this last equation to zero means that the constraint, Eq. (3.1) must be
satisfied.
By dividing Eq. (3.4) for two inputs x 1 and x 2 , Eq. (3.5) for two outputs y 1
and y 2 , and pairs of equations for one input, X;, ar1d one output, yj, one obtains
the following equations:
[du(X, Y)ldx 1]/[du(X, Y)ldx 2 ] = [df(X, Y)ldxdl[df(X, Y)ldx 2 ] (3.6)
[du(X, Y)ldy 1]/[du(X, Y)ldy 2 ] = [df(X, Y)ldydl[df(X, Y)ldJ2] (3.7)
[du(X, Y)ldx;]l[du(X, Y)ldyj] = [df(X, Y)ldx;]l[df(X, Y)ldy) (3.8)
Since f(X, Y) must equal zero, an increase in one element must be offset by
a decrease in another. Thus:
[df(X, Y)ldx 1]/[df(X, Y)ldx 2] = -[dx21dx1] (3.9)
[df(X, Y)ldydl[df(X, Y)ldy 2 ] = -[dy21dyj] (3..10)
[ df(X, Y)l dx;]l[ df(X, f)/ dy) = - [dy/ dx;] (3 .11)
Combining Eqs (3.6) arid (3.9), (3.7) and (3.10), and (3.8) and (3.11), one
finally gets
[du(X, Y)ldx 1]/[du(X, Y)ldx 2 ] = -[dx2 /Jxd (3.12)
[du(X, Y)ldydl[du(X, Y)ldy 2 ] = -[dy2/dy 1] (3.13)
[du(X, Y)ldx;]l[du(X, Y)ldy) = -[dy/dx;] (3.14)
3.3.2 Conditions of Optimality
The following can be interpreted easily.
du(X, Y)ldx; =rate of char1ge of the net benefit function with respect
tb input i, or the marginal cost of input X;= MC;
du(X, Y)/ dyj = rate of change of benefit function with respect to output
j, or the marginal benefit of output Yj =
MBj
It is to be emphasized that, whenever the change of one element with respect -
to another is considered, all other el~ments are held at constar1t levels.
The right-hand side of Eq. (3.12) is termed as the marginal rate of
substitution, MRS 21 , which is defined as the marginal rate at which the second
input needs to be substituted for the first input, holding the level of production
constant. Thus, Eq. (3.12) gives
MCrfMC2 = MRS21 (3.15)

·-·~·
~
 110 1 Basics of Systems Techniques

The right-hand side of Eq. (3.13) is termed as the marginal rate of transfor-
mation, MRT21 , Thus Eq. (3.13) gives ·
MB 1/MB 2 = MRTz 1 (3.16)
The right-hand side of Eq (3.14) is termed as the marginal physical produc-
tivity of the ith input when devoted to the jth output, or the marginal physical
product, MPPij Thus Eq. (3.14) gives
MC/MB1 = MPPiJ (3.17)
Equations (3.15), (3.16), and (3.17) are the three conditions of optimality
that are necessary to achieve maximum net benefit.
Note: It must be noted that the conditions of optimality are necessary but not
sufficient conditions. Sufficiency requires that there be no higher value of the
objective function than is indicated by the solution. Even then, the solution can
only be a local maximum and need not be a global maximum. For a more
practical interpretation of those theoretically derived expressions, see Maass et
al. (1968) and James and Lee (1971). ·
Combination of Inputs (Marginal Rate of Substitution)
A water resources project has the option of using two inputs, reservoir water
and groundwater, to provide irrigation to a given area. All other things being
equal, the need is satisfied by the combination of inputs, x 1 and x 2, as shown
in Fig. 3.5. To determine the best combination of these inputs, the first
optimality condition says that the levels of x 1 and x 2 should be in the ratio of
MC2 to MC 1.

90,-----------------------------~

80
70
60 x1 =50; x2 = 45 (opt)
50
X1 40
30
20
10
0+-~~~~--~~~~~~-.~~~
0 100 300 400

Fig. 3.5 I Combination of Inputs

. ·- ----------

MC1 = 10; MC2 = 4; (MRS2l)opt = MC/MC2 = 2.5

SINo. XI x2 1&-11 1&-21 MRS 21 Total cost
80 10 840
2 65 25 15 15 1 750
3 55 40 10 15 1.5 710
4 45 65 10 25 2.5 710
5 20 190 25 125 5 960
6 5 340
I
J!
..
Economic Considerations in Water Resources Systems I 111

For the data shown in the table, .the optimal marginal rate of substitution
is MC1/MC2 = 2.5. This would correspond to x 1 = 50 and x2 = 45, as shown
in the figure. At this point, it is seen that a line with a slope of

-Jxz =2.5(= MC1 ) is tangential to the curve as desired. The total cost would
Jx1 MC2
be a minimum (= 680, in the present example) at this point.

Example ,3.3.2 Combination of Inputs (Marginal Rate of Substitution)

Two inputs x 1 and x2 should be combined to satisfy the relationship x 1x2 = 16,
in order to produce constant results (output), all other things being the same.
x1 and x2 , for example, may represent the quantum of surface water and
groundwater to provide a given level of irrigation to a given area. If the
marginal costs of x 1 and x2 are 4 and 1, respectively, determine the optimum
values of x 1 and x2 .
The desired point (xi> x 2) should lie on the curve x 1x2 = 16. From the first
condition of optimality for two inputs,
Jx2 MC1 4
- 04_ = MRS21 = MCz =T = 4
Jx2 Jx2 x2
where x 1x2 = 16; x1 : ; - + Xz = 0 or :;- =--
ox1 ox1 x1

~ ~~ = :: = ~~~ =4 :. x2 = 4xl
As (x1, x2) should lie on the curve, x 1 (4x 1) = 16
or 4xf = 16
.. x1 =2
and x 2 = 4x 1 = 8
The minimum cost at this level of input combination= x 1 (MC1) + x 2 (MC2)
= 2(4) + 8(1) = 16
Alternate Method
C =Total cost of production= x 1(MC 1) + x2(MC2 )
= 4x 1 + 1x2

For C to be a minimum, ~C
ox1
= 0, or 4 + ~Xz = 0
ox1

or Jx~=-4
Jxl
The minimum point also lies on the curve x 1x2 = 16, which by differentiation
gives
dxz
x1 dx- + x2 = 0 or
1
 112 1 Basics of Systems Techniques

.. x2 = 4x1
As x 1x 2 = 16, x 1 = 2 and x2 = 8, as before.
Note: The results are the same in both methods as both tend to maximize the
objective of net benefits. But in this case, since the benefits are constant,
optimization boils down to minimizing the cost.

Combination of Outputs (Marginal Rate of Transformation)

Two outputs can be produced from a production process such that· the levels
of the two outputs, y 1 and J2, satisfy the relation, Yf + 4 YI = 4, keeping all
other things constant. If the marginal benefits of y 1 and Y2· are respectively
equal to 10 and 5, determine the optimum levels of y 1 and h
y 1 and y 2 can denote, for example, irrigated area and hydropower generated
respectively. It is assumed that the water used for these two is mutually
exclusive. What is used for one is not available fot the other. This can- result
if water is withdrawn from reservoir storage for diversion into canals for
irrigation, and hydropower is produced only from the water released
downstream of the reservoir.
Optimum value of y 1 and Y2 should lie on the curve (production function)
Yf+4YI = 4
dJ2
Diff. with respect to y 1, 2yl + 8Y2 dyi =0
dy2 _1_11_
or . dyj = 4 Y2
From the second condition of optimality;
dy 2 MB .·
- dyi = MRT21 = MB 1 ; and MB 1 = 10, MB 2 = 5
2
1 Y1 10 ·
! I
I +4. Y2 = S = 2, or YI = 8y2

As Yf + 4yi = 4, 64yi + 4yi = 4 or y2 = ll.fU and y 1 = 8/.ffi

··'·-·-··-----1-'1-'ho-l-.-o-n-,,,f; ..-.,t--th-;·c~h.:.-tol--,,,,;n be maximum = YI(Mlf~Y +y~(MB2)
= 8/.fU (10) + u-[0 (5)
= 85/JU
Alternatively, the problem may be solved directly using the Lagrangean
multiplier method.
Maximize u = 10y 1 + 5J2

LJj
..
in_ffi_a_t_er_R_e_s_ou_r_ce_s_S~ys_re_m_s______~l. 113
___________E_c_o_no_m_i_c_C_o_ns_w_e_ro_t_io_ns__

subject to
Necessary Condition:
Lagrangean L = 10y1 + 5Yz- A( Yf + 4yi - 4)
dL
--:\
ayl
= 10- 2A-yl = 0

~L = 5 - 8AYz =0
OYz

a:. =- (yf+4y~- 4) = 0.

Solving y1 = y..J0
- Yz =-Ym-
and B = sy..J0
[The student may check this for the sufficiency condition.]

Combination of an Input and an Output (Marginal

Physical Product, MPP)
Consider a reservoir project operated for irrigation and hydropower in
which the input element is the water supply, x, and the output element is the
hydropower generated, y. Let the marginal cost of water supply be MCx = a,
and the marginal benefit of hydropower generation is MBy = [3, where a and f3
are constants. Let x and y be related as y = Fx
within the feasible ranges of x
and y. The question is how much of x should be used to produce y under
optimal conditions, all other quantities being the same. The rule says:

- dy = MPP = MCX
dx xy MBy
i.e., increase the value of x until the point where the ratio of marginal output
to marginal input equals the marginal physical product, which is the ratio of
the ·marginal input cost to the marginal output benefit.
-Nore: -tmnn:a:rginal-cosr,-as-deterrtrine·d -by-partial-differentiation-of the-net --
benefits, (f3y- ax), is -a.

When v- 1 1
= 0 - dy = - - x = _!_ (f3la) 2
- -..j X ' dX 2.[;- = -a:
f3 ' · 4 . .

For maximum net benefits 7; x = 114([3/cxi andy= t ([3/a)

Alternatively, using the Lagrangean Multiplier method directly,
~
!i

'
I
II.'~

' 114 1 Basics of Systems Techniques

I ,i
Maximize u = f3y- ax
I
I,
subject to y- -rx = 0.

L = f3y- ax- A,(y- --.lx)

dL=-a+ A-,=O.
dx 2....; x
dL
dy = f3- A,= 0

dL =-(y- --./x)=O
()A,

Solving x = 41 (f3!a) 2, y = 21 (j3/a)

This is a necessary condition for local m~i!llull!-
[The student may check this for the sufficiency condition.]

Problems 1.··
3.3.1

3.4 BENEFIT COST ANALYSIS

The primary purpose of a water resources project is its utility to the public.
Obviously, a given project will not be beneficial to everyone who may be
affected by it. Some will be positively benefited and some negatively. Benefi-
 Economic Considerations in Water Resources Systems 1 11s
ciaries can be individuals, communities, groups of people, or entities such as
administrative districts. For example, a reservoir may benefit all those down-
stream by providing the benefits of irrigation, low-flow augmentation, hydro-
power generation, flood-damag~ alleviation etc; whereas, those in the upstream
suffer from submergence, inconvenience in rehabilitation and resettlement, etc.
In this context, a plan will be considered economically feasible only if the
positive benefits outweigh the total costs involved in implementing it or, put-
ting it in monetary terms, the benefit cost ratio should be more than one, or the
net benefits (gross benefits minus the total costs) should be positive. It is
important to note that the benefit cost analysis be made with and without the
project rather than before and after. This is because some of the after effects
may occur in course of time, even without the project and therefore cannot be
counted as a justification for the project. Benefit cost analysis provides an
objective assessment of the economic feasibility of each project and provides a
means of selecting the best among those short-listed.
3.4.1 Benefits-an-d Costs
Benefit cost analysis requires the estimate of both benefits and costs associated
with a plan and should take into account all the parties affected one way or
another by it. Benefits accrue through the use of a commodity or service. The
value of the use, expressed in monetary terms, is the benefit associated with the
plan. There may not exist a market related to the commodity or service and,
therefore, estimation of benefits is not always straightforward and easy. Benefit
cost analysis requires estimation of both benefits and costs in monetary terms.
Benefits are to be estimated by aggregating them to whomsoever they may
accrue from the plan. Cost estimates should ideally reflect opportunity costs
rather than just the market prices for a true comparison .. This is rarely done in
practice, however.
3.4.2 Cost and Benefit Curves
The total cost comprises fixed cost and variable cost. Fixed costs, as the term
indicates, are fixed and do not depend on the level of project output. Variable
costs are marginal costs and vary with level of output. While fixed costs are not
marginal, they do affect the computations of the total benefit cost ratio and the
decision whether the project should be built at all. For example, in a reservoir
sizing problem using LP, if the objective is to minimize the total cost of
reservoir construction (sum-of fixeo ahd-vai"iable costs, -vru:i:a15le costs depend'=
ing on the storage), and if the optimal required storage, works out to zero then,
obviously, no reservoir need be built and no costs incurred. · ·
If TC = f(Y) is the total cost at output Y, (Fig. 3.6), for a project of single
output, then the average cost, AC, is given by
AC = TCIY =f(Y)IY
Average cost curves are usually U-shaped. They decrease initially because
of economies of scale, and increase again, as production becomes very large.
 --
1
I

I,,
1.

I!
116 I Basics of Systems Techniques
'

Cost TC, AC or MC

Total Cost
I TC = f(Y)
•i

Marginal Cost
MC = f'(Y)
AC = f(Y)/Y

Output Y

Fig. 3.6 I Representative Total, Average, and Marginal _Cost Curves

The marginal cost, MC, is given by
MC = dj(Y)IdY =f' (Y)
and is represented by the slope of the total cost curve. The slope represents the
change in total cost with a unit change in the output. Unless the market price
exceeds the marginal cost, no firm would produce an extra unit, and therefore,
the rising limb of the marginal cost curve is a supply curve. It indicates the
price necessary to induce an extra unit of production.
The total cost of production of an output, Y, is the area of the marginal cost
curve to the left of Y. It may be noted that the marginal cost curve intersects the
average cost curve at the latter's minimum, as
AC = j(Y)IY is minimum when d{j(Y)IY}IdYis zero, or
{Yf'(Y) -f(Y)}IY 2 = 0
or f'(Y) = f(Y)!Y
i.e. MC = AC
Similarly, the marginal benefit is the change in the total benefit per unit of
,,
output. The marginal benefit curve is the demand curve because no buyer will
purchase an extra unit unless it. provides him a value that exceeds the cost

r:---
,I

incurred. It indicates the maximum price that a firm can afford to spend to
;I 1
acquire an extra unit. Marginal curves, like average curves, are generally
:.:~~::::::;~:::::(James >ndtee; 1971)····-------
As mentioned earlier, benefits and costs need to be estimated in monetary
units. Depending on the ease with which such estimates may be arrived at,
there could possibly be four different situations in relation to the availability of
a market, and its prices representing marginal social values.
1. Market prices exist and reflect true marginal social values, such as in the
case of pure competition (ideal condition).
...
Economic Considerations in Water Resources Systems . 1 117
2. Market prices exist but, for various reasons, do not reflect marginal social
values. Example: Subsidized agriculturaf commodities.
3. Market prices are essentially non-existent, but it is possible to simulate a
market-like process to estimate what users (or consumers) would pay if a
market existed. Example:. Outdoor recreation.
4. No real or simulated market-like process is conceivable. Example: Historic
monuments, temples, scenic amenities, etc.
For the first three categories (1, 2, 3) mentioned, benefits and costs can be
measured as the aggregate net willingness to pay of those affected by the
project.
Willingness-to-Pay Criterion The concept of willingness-to-pay is based on
the common experience that a person will not buy a commodity at a cost if, in
his view, the value that he accrues by using the commodity is any less than the
price that he has to pay in acquiring it. In other words, the net benefit that
accrues to him (over and above the cost he pays for it) is what induces him to
purchase the commodity. in the market. Assume (Loucks et al., 1981), all
people that are benefited by the pl<:m X are willing to
pay B(X) rather than
forego the project. This represents the aggregate value of the project to the
beneficiaries. Let D~X) equal the amount. that all the nonbeneficiaries of plan
X are willing to pay to prevent it from being implemented. The aggregate net
willingness to pay, W(X), for plan X, is equal to the difference between B(X)
and D(X), i.e. W(X) = B(X)- D(X).
The rationale implied in the willingness-to-pay criterion is that if B (X) >
D (X), the beneficiaries could compensate the nonbeneficiaries and everyone
would benefit from the project. This, however, has some implications such as
the marginal social value of income to all affected parties is the same. If the
beneficiaries are essentially rich and the nonbeneficiaries are poor, then B (X)
may be larger than D (X) simply because the beneficiaries can afford to pay
more than the nonbeneficiaries.
Market Prices Equal Social Values Let us assume that the price demanded is
p(x), when the quantity available in the market is x (demand curve given). Let
there be an increase in the level of the output from x 1 without the project to x 2
with the project. Then the willingness-to-pay (which is a proxy for the accrued
benefits) is given by the area under the demand curve for the increased portion
of the output, Fig. 3.7.

Price

x1 x2
Output

Fig. 3.71/ncrease in Output with the Project

 11s 1 Basics of Systems Techniques

Xz
Willingness to pay = f p(x)dx (3.2)

which is the area under the demand curve between x = x 1 and x = x2 • This is an
estimate of the benefit attributable to the project under contemplation.
The demand curve for recreation at a reservoir site is
determined as 4P + Y = 30, where Y is the annual demand curve and P is
price in appropriate units. The annual demand was 10 without the reservoir
and is expected to increase to 20 with the construction of the reservoir.
Estimate the benefits of recreation arising from the construction of the reservoir.
The demand curve plots as a straight line, EF, as shown in the following
figure:

E
7.5
A
p5.0
8
2.5

0 F
0 10 30 y
Demand Curve for Recreation

The area under the demand curve between y = 10 andy= 20 gives the total
willingness to pay for recreation or, in other words, the benefits due to recre-
ation at the reservoir. The area ACDB = (5.0 + 2.5)/2. x 10 = 37.50. The
benefits are thus estimated to be 37.50.

Market Prices do not Equal Social Values There are several procedures that
can be used depending on the situation. A rather common method is to estimate
the cost of the least expensive alternative and project it as the benefit. However,
the· alternative cost approach is meaningless 'in the absence of proof that the
second best alternative would be built if the best were not~ As an example, let
the benefits from generating hydroelectricity be estimated using this approach.
Let the alternatives considered be geothermal or nuclear sources. both of which,
let us assume, are more expensive t.lJ.an the hydroelectric scheme. Then the cost
1
of t:l:}e least expensive alternative of the two sources will be higher than the cost
1

,I
of .
the proposed hydroelectric
. ..
scheme itself,. . and hence
.
the proposal
. .
will
.
have----a
~-~-----::::oenef!t cost ratiofiiglier than-one--:llowever~ ili.fsapproacl:lio henefiTestimat:lon
I is valid only when there is a commitment to actually build the less expensive of
the geothermal and the nuclear schemes, in case the hydroelectric scheme is
not built. Therefore, this method is rather tricky in that it is possible to propose
and justify a plan by choosing expensive alternatives (which cost higher thfu'1
the proposed plan), and therefore, should be used with extreme caution.
The following example shows how a demand curve can be derived to estimate
the value of outdoor recreation. This is a case where market prices do not exist
 I' i
I

Economic Considerations in Water Resources Systems 1 119

but a market like process can be simulated. There are many underlying assump-
tions and· limitations of the approach used herein, but the example illustrates
the conceptual approach that may be used in cases where an imputed demand
curve is needed to estimate recreation benefits.
A recreation area is proposed to be developed near a reservoir,
which will serve the population from two towns, Town A and Town B. Town
A has a population of 50,000 and Town B of 150,000. It is estimated from
surveys that 150,000 visits per year will be made at a travel cost of Rs 10 per
visit from town A, and 300,000 visits per year will be made at a travel cost of
Rs 20 from Town B. We have to construct a demand curve for recreation and
estimate the annual benefits due to recreation.
The data are tabulated as follows:
Population Visitslyr Visits/capita Cost/visit (Rs)
Town A 50,000 150,000 3 10
TownE 150,000 300,000 2 20

We see that three visits per capita will be made at a cost of Rs 10 per visit,
and two visits per capita will be made. at Rs 20. Thus we have two points that
can be plotted, as in the following figure.

£
::1'J', . .
1:1 ',,',,~',
0 +------,· '
--,-----,-----'~-----1
0 2 3 4 5
Visits per capita
Demand Curve for Recreation

We shall assume that the visitation rate of the population from both the
towns depends only on the total cost incurred per visit, .and estimate the user
response for increasea~ Ieveis~of1otai cosf(travercosr + additional cost such- ~ ~
as admission cost).
Firct cons~der an added cost ofRs lQ over and above the travel cost making
it Rs 20 per visit from Town A, and Rs.30 per visit from Town B.
It is assumed that the relationship between the cost/visit and expected
number of visits per capita will be the same as indicated in the figure.··
Now we shall estimate user response based on the total cost per visit
(travel cost+ additional cost). At an additional cost of Rs 10 per visit the total
 120 I Basics of Systems Techniques

cost per visit will be Rs 20 from town A .and Rs 30 from Town B. It is seen
that only two visits/capita will be made at Rs 20, and only one visit/capita at
Rs 30, from the previous figure. It is also :>een that at a total cost of Rs 40per
visit, no visits will be made from either town. Also, where there is no addition
to travel cost, the total number of visits per year will be 150,000 from Town A
+ 300,000 from Town B making it a total of 450,000 visits per year.
The following table shows the number of visits per year from each town
'I
and the total number of visits per year (recreation demand) that can be expected
to be made at different levels of added cost.
Total Visits/capita No. of I Total no. of
I cost I I visits/yr visits/yr
Added Cost = 0
Town A I 10 I 3 I 1so,ooo 1
450,000
TownE I 20 I 2 I 3oo,ooo 1

Added Cost = 10
Town A I 20 I 2 I 1oo,ooo 1
250,000
Town B I 30 I 1 I 1so,ooo 1

Added Cost = 20
Town A I 30 1 1 I 50,000 _I
50,000
TownE I 40 I 0 I 0 I
Added Cost = 30
Town A I 40 1 0 I 0 _I
0
TownE I 50 I 0 I 0 I
The following figure shows the additional cost vs. total number of visits/yr
due to recreation, which is an estimate of the demand for recreation at the
reservoir, or the demand curve.
40
~ 30
~
g_ 20
t5
8 10

--- 0 .
--~ ----~--~--o~so--10o-lso-2oo-2so·soo
. . .
-sso-4oo- 450
Recreation Demand (1 000 visits per year)

The total benefits due to recreation at the reservoir is the area under the
demand curve, which is Rs 5,250,000/year.
No Market Process In the absence of any market-like process, real or
simulated, it is extremely difficult to quantify benefits in monetary terms. The
benefits associated with aesthetics, for example, are considered intangible.
.....

Economic Considerations in Vl(ater Resources Systems 1 121

Negative benefits due to submergence of temples or other places of worship
and environmental impacts of a long-lasting nature belong to this category.
Project Size A practical goal in planning a water resource project is to plan
the size of the project, such that it yields the maximum net benefits, Fig. 3.8. At
the optimum output level, the slopes of the benefit and cost curves should be
the same, i.e. marginal benefit should equal marginal cost.

Total Benefit I
Cost

Max Net Benefit

Output

Fig. 3.~0aximum Net Benefits

Water is supplied from a project for two types of users: rural

and urban. The benefits to rural community are given by B, = 30 Yr- 3/2 Y?:
and those to the urban community are given by Bu = 10 Yu-
2
y;
18. If the total
cost of the project is C = Y !2 + 2Y, where Y is the aggregate demand,
determine the optimum level of total water supply. Also determine the
corresponding components of rural and urban water supply levels.
Rural B, = 30 Yr- 3/2 Y?:
The demand curve is given by the marginal benefit curve.
dBr
P = -dy = 30 - 3yr or P + 3yr = 30

Urban
The demand curve for urban use is given by
dBu
... - ._ ·---- ------- _ _f_= __dy- =_lQ -:_yj4o_r4P .+ Yu-=:4.0: ··-···
For these two types of demand, the aggregate demand curve is given by the
line AEF in the figure in Example 3.2.1. This line is the marginal benefit
curve.
Cost Curve
The Total Cost C = Y212 + 2Y
dC
Marginal Cost, MC=- = Y+2
dy
 122 1 Basics of Systems Techniques ·

. At optimum level of water supply Y, · the marginal benefit and marginal

cost should be equal. In other words, determine the intersection of the curve P
= y + 2, and the aggregate demand curve (the line AEF in the figure in
Example 3.2.1).
150 -3Y
Assuming Y2: 20/3, the combined demand curve is given by P = -...,--::--
13
Equating this to the marginal cost gives

P = 1501; 3
= Y + 2; Y = 31/4 (2: 20/3).
y
Optimal Price P (corresponding to Y = 31/4) = 39/4.
Since the marginal price P should be the same for both types of users, the
individual demands can be obtained from the individual demand curves at
p = 39/4.
Rural: P = 39/4 = (30- 3yr) or Yr = 81112.
Urban: P = 39/4 = (40- Yu)/4 or Yu = 1
Check: Optimum Total Demand = Yr + Yu = (81/12) + 1 = 93112 = 31/4, as
before.

1. James, Douglas, L. and Lee, Robert R., Economics of Water Resources

Planning, McGraw-Hill, Inc., New York, 1971.
2. Loucks, D. P., J. R. Stedinger and D. A. Haith, Water Resource Systems
Planning and Analysis, Prentice-Hall, Englewood Cliffs, N.J., 1981.
3. Maass, A., M.M. Hufschmidt, R. Dorfman, Jr. H.A. Thomas, S.A. Marglin,
and G.M. Fair, Design of Water Resource Systems, Macmillan, 1968.

1. Aguilar, R.I., Systems Analysis and Design in Engineering, Architecture,

Construction and Planning, Prentice-Hall, Englewood Cliffs, N. J., 1973.
2. Kuiper, E., Water Resources Development Planning, Engineering and
Economics, Butterworths, London, 1965.
3. Ossenbruggen, P.J., Systems Analysis for Civil Engineers, John Wiley and
Sons, New York, 1984:. ·--~-~------------- ____________________ _
4. Young, R.A., (1996) Water Economics, Ch. 3 in Lary W. Mays (Ed.)
Water Resources Handbook, McGraw-Hill, Inc., New York, 1996.
p

Reservoir Systems-Deterministic Inflow 1 141

The required reservoir capacity, when evaporation losses are neglected, is
588.11 Mm3 ; whereas, with evaporation losses taken into account, the re-
quired capacity works out to 617.986 Mm 3.

1. Compute the active storage capacity of a reservoir to supply a constant

maximum annual yield, given the following sequence of annual flows:
{8, 4, 6, 2,4, 6}
2. Assume a year has two distinct seasons, wet and dry. Eighty per cent of
annual flow occurs in the wet season each year, and eighty per cent of
annual yield is demanded in the dry season each year. Determine the
percentage increase in the required storage capacity compared to 1 above.
Use the sequent peak method, neglecting evaporation and seepage
loss.
Solution:
Assuming the annual inflow sequence repeats, the maximum constant yield
possible (without losses)= average annual inflow= (8 + 4 + 6 + 2 + 4 + 6)/6
= 5.
1. The following table shows the computations of K 1 in each period.
K0 = 0
Period, t R 1 =release Q1 =inflow K1 =max {0, K1_ 1 + R1 - Q1 }
1 5 8 0
2 5 4 1
3 5 6 0
4 5 2 3
5 5 4 4 (maximum)
6 5 6 3
1 5 8 0

In this example, computations can stop after period 1 in the second

cycle as the value of K 1 for period 1 is the same (= 0) as that for the
period 1 in the earlier cycle. The entries for the subsequent periods
simply repeat (as those for the corresponding periods in the first cycle).
Required active storage capacity= max {K 1 } = 4.
With this capacity, the reservoir will be full at the end of the first
- ·and·third --periods--and-will-be- empty-af'the- end-of- the -fifth' .period.
(verify). Hint: Examine S1 + Q1 -R 1 = St+! over a few cycles (sequences)
starting from any initial storage, S1, in any period, t, till steady state is
reached, i.e., st for given t is the same in different cycles.
2. Inflow in the wet season = 0.8 (annual flow) each year, and inflow in
the dry season = 0.2 (annual flow) each year.
Demand in the wet season= 0.2 (5) = 1 each year, and
demand in the dry season= 0.8 (5) = 4 each year.
 148 )1 Model Development

a constant rate per season to

peat every year. Con-

y season to be 5% of
other losses. Formulate
constant maximum rate of
water supply per season, if the reservoir capacity is 450 units.
(Ans. 938.75)
5.1.4 A reservoir is to be constructed to supply water at a maximum con-
stant rate per season for a city. The infJows in the six seasons of the
year are 3, 12, 7, 3, 2, and 3, respectively. Determine the minimum
required reservoir capacity using sequent peak method. Neglect all
losses. (Ans. 9)
5.1.5 In general, if the storage continuity constraints are used as expressed
below, in the LP formulation to determine reservoir capacity,
(1 - a 3 St + Qt - L, - D, 5 (1 + a,) for all t,
analyze the nature of the solution for feasibility.
(Hint: First analyze !the case without evaporation losses).

5.2 RESERVOIR OPERATION

A reservoir operating policy is a sequence of release decisions in operational
periods (such as months), specified as a function of the state of the system. The
state of the system in a period is generally defined by the reservoir storage at
the beginning of a period and the inflow to the reservoir during the period.
Once the operating policy is known, the reservoir operation can be simulated in
-time with a given inflow sequence. A number of optimization algorithms have
been developed for deriving reservoir operating policies. However, the most
common policy implemented in practice is the so-called standard operating
policy, which is discussed first in this section. This policy by itself is not based
on or derived from any optimization algorithm.
5.2.1 Standard Operating Policy
The standard operating policy (SOP) aims to best meet the demand in each
period based on the water availability in that period. It thus uses no foresight
g -
tn tho
- Reservoir Systems-Deterministic Inflow 1 149

and R represent, respectively, the demand and the release in a period. Let the
of the reservoir be K. Then the standard operating policy for the
period is represented as illustrated in Fig. 5.3. The available water in any
period is the sum of the storage, S, at the beginning of the period, and the
inflow & during the period. The release is made as per the line OABC in the
figure.
Release, R

Spills
Filling Phase
*-----* /

0 D D + K Water ~vailable,S + Ci
Fig. 5.3 ) Standard Operating Policy
Figure 5.3 implies the following:
Along OA: Release = water available; reservoir will be empty after'release.
Along AB: Release = demand; excess water is stored in the reservoir (filling
phase).
At A: Reservoir is empty after release.
At B: Reservoir is full after release.
Along BC: Release = demand + excess of availability over the capacity (spill)
h other words, the release in any time period is equal to the availability, S +
Q, or demand, D, whichever is less, as long as the availability does not exceed
the sum of the demand and the capacity. Once the availability exceeds the sum
of the demand and the capacity, the release is equal to demand plus excess
available over the capacity. It is to be noted that the releases made as per the
standard operating policy are not necessarily optimum as no optimization crite-
rion is used in the release decisions. For highly stressed systems (systems in
which water availability is less than the demand in most periods), the standard
operating policy performs poorly in terms of distributing the deficits across the
periods in a year.

. - --

R,= D , i f S , + Q , - E t 2 D,
= St + Q, - Et, otherwise
0, = (St + Q, - E, - D,) - K if positive
= 0 otherwise
S t + , = St + Q, - E, - R, - 0, with R, and 0,determined as above
 150 1 Model Development

St is the storage at the beginning of the period t, Q, is the inflow during the
period t, D,is the demand during the period t, E, is the evaporation loss during
the period t, R , is the release during the period t, and 0,is the spill (overflow)
during the period t. Note that = K, if 0,> 0.
An illustrative example oi^ the standard operating policy is
5.3, for a reservoir with capacity K = 350 units and an initial
storage of 200 units. Inflow Q,, demand D ,i d evaporation E, are known
values, as given in the table.

Table 5.3 ) Reservoir Operation with Standard Operating Policy (SOP)

K = 350

5.2.2 Optimal Operating Policy Using LP

One of the classical problems in water resources systems modelling is the
derivation of an optimal operating policy for a reservoir to meet a long-term
objective. Modelling techniques to be used depend on whether the reservoir
inflows are treated deterministic or stochastic In this chapter only the case of
deterministic inflows is dealt with. Operating polities considering stochastic
inflows are discussed in Chapter 6.
Given a reservoir of known capacity K, and a sequence of inflows, the
reservoir operation problem involves determining the sequence of releases R,,
that optimizes an objective function. In general, the objective function may be
a function of the storage volume and/or the release. A single, simplified reser-
voir system
- --
-
is represented
-. -
-- -----
as shown in Fig. 5.4. - - -- ----- -

Evaporation, Et
Reservoir

Release
Rt

Fig. 5.46 Singie Reservoir Operation

 Reservoir Sysrems-Deterministic Inflow 1 151

LP Formulation
Consider the simplest objective of meeting the demand to the best extent pos-
sible (the same objective as considered in the standard operating policy), such
that the sum of the demands met over a year is maxir;..km. This may be formu-
lated as a LP problem as follows:
Max C R t (5.2.1)
L

Subject to
St+i St + Qt- R,- Et - 0, 'dt (5.2.2)
R, I D, Vt (5.2.3)
St 5 K Vt (5.2.4)
Rt2 0 Vt (5.2.5)
$2 0 Vt (5.2.6)
ST+I= s1 (5.2.7)
where T is the last period in the year, and all other terns are as defined in
Section 5.2.1.
The constraint (5.2.3) restricts the release during a period to the correspond-
ing demand, while the objective function (5.2.1) maximizes the sum of the
releases. Thus the model aims to make the release as close to the demand as
~ possible over the year. To ensure that the overflows 0,assume a nonzero value
in the solution only when the storage at the end of the period is equal to the
reservoir capacity, K, integer variables may be used as discussed in Section
5.1.4. Constraint (5.2.7) makes the end of the year storage equal to the begin-
ning of the (next) year's storage, so that a steady state solution is achieved.
When the initial storage at the beginning of the first period is known, an
additional constraint of the form, S1 = So, may be included, where So is the
known initial storage, in which case the sequence of releases obtained would
be optimal only with respect to the particular initial storage. The end of the
year storage, ST+1,may be set equal to the known initial storage (5.2.7), if a
steady state solution is desired, or may be left free (i.e. constraint 5.2.7 is
excluded) if only the release sequence for one year, with known initial storage,
So, is of interest.
Table 5.4 below is constructed with results obtained by solv-
el (5.2.1) to (5.2.7) with inflow Q,, demand D,, and evapora-
 (Contd)
t st Qt Dt Rt E, &+I 0, -

6 350.00 45.00 203.99 39.00 6 350.00 0.00

7 350.00 19.06 203.99 108.87 5 255.19 0.00
8 255.19 14.27 179.47 179.47 5 84.99 0.00
9 84.99 10.77 89.76 89.76 6 0.00 0.00
10 0.00 8.69 0.00 0.00 8 0.69 0.00
11 0.69 9.48 0.00 0.00 8 2.17 0.00
12 2.17 18.19 0.00 0.00 10 10.36 0.00
Note that the values of R, in Tables 5.3 and 5.4 are not expected to be identical, as the two
policies arc not strictly identical.
I
Rule Cuwes for Reservoir Operation Reservoir operation rules are guides
for those responsible for reservoir operation. They apply to reservoirs which
are .to be operated in a steady state condition. A rule curve indicates the
desired reservoir release or storage volume at a given time of the year. Some
rules identify storage volume targets that the operator is to maintain, as far as
possible, and others identify storage zones, each associated with a particular
release policy. The rule curves are best derived through simulation for a
specified objective, although for some simple cases, it may be possible to
derive them using optimization.
I
I In Example 5.2.2 given earlier, the end of period storages, define the
rule curve. They are the desired end-of-period storages (target storages) to be
1 maintained in operating the reservoir in different time periods.
I
The resulting rule curve is as shown in Fig. 5.5. Note that if the objective
function is changed, in general, a different rule curve is obtained.
I

1 Fig. 5.5 1 Reservoir Rule Curve for Example 5.2.2

Multireservoir Operation
A multireservoir system operation problem may be similarly formulated as an
LP problem. Consider, for example, the three-reservoir system shown in
I

I Fig. 5.6. The system serves the purposes of water supply, flood control and
hydropower generation. Release for water supply is passed through the
 Reservoir Systems-Deterministic Inflow 1 153

Fig. 5.6 f Multiresen/oir Operation Problem

powerhouse generating power, and losses in the powerhouse are negligible.

Benefits from hydropower are expressed as a function of storage alone. B ; ~Bit,
and B;, , are respectively the net benefits associated with unit release, unit
available flood freeboard (= reservoir capacity - available storage), and unit
storage for reservoir i in period t. A portion al and g of the release made at
reservoirs 1 and 2 respectively add to the natural inflow of reservoir 3. Further,
a minimum flood storage F- needs to be ensured during flood season at the
reservoir i. Maximum release that may be made at reservoir i is R;,. It is
assumed, in the following illustration, that
B;', = B: 'di
2Iit = B; Vi
and = B: 'di
The LP problem may be stated as

max z3

z=lt=l
T
[ B ~ R+; B: ( K , - $1 + B:s:]
subject to

s , ' : + , = s , ' : + Q , ' : + ~ , R : + ~ , R ~ - R ~ - E'dt;

~-o~i =. 3

K, - S: 2 FA, 'di;'dt E Flood Season

 154 1 Model Development

~:20;R:20.
In the model, the index i refers to a reservoir (i = 1, 2, 3), and t to time period
(t = 1, 2,..., T). Q is the natural inflow, K is reservoir capacity, S is storage, R
is release, E is evaporation, and 0 is overflow (spill). The storage S, is the
storage at the beginning of period t.

11 1 Fig. 5.6.
Consider the data given in the table below for a three-period,
system. The reservoir system configuration is as shown in

Reservoir Inflow K Fmirz B,* B; B,*

t=l t=2 t=3 t=l t=2 t=3
1 25 10 15 10 3 2 7 50 10 25
2 20 30 15 15 2 3 4 60 10 30
3 20 12 15 20 2 3 5 70 10 35
*The benefit coefficients B,, Bz, and B3 are assumed constant for all three periods.

f. I a, = 0.2 and or, = 0.3

The solution of the LP problem is as follows:
Reservoir I Reservoir 2 Reservoir 3 I
t=I t=2 t=3 t=I t=2 t=3 t=I t=2 t=3
st 0.0 8.0 3.0 2.0 12.0 11.0 0.0 17.0 15.0
R, 17.0 15.0 18.0 0.0 31.0 24.0 6.4 26.3 40.8
(K-s,)" 10.0 2.0 7.0 13.0 3.0 4.0 20.0 3.0 5.0
*(K-SJ is the available flood storage in period t.

Note that the spill from a reservoir is included in the release values shown

~ for that reservoir. Also note that the mass balance for reservoir 3 includes the
contributions from the two upstream reservoirs. In this hypothetical example,
the benefit coefficients for the release are much higher than those for the
flood storage. The solution will be quite sensitive to variations in these coeffi-
cients. It would be instructive for the student to generate and analyze a num-
ber of solutions, varying the benefit coefficients relative to each other, for the
Isame lnflow data provided in this example.
-- - - - -

5.2.3 Stationary Policy Using DP

A general reservoir operation problem is to derive a stationary policy for a
repeating sequence of inflows {Q,) every year, where t is a 'within-year' period,
i.e. less than one year. The stationary policy, which may be derived using DP,
specifies the release as a function of the storage in a period. The objective is to
i ~
111
derive an operating policy which results in the maximized annual net benefit in
the long run (steady state).
 Applications of Linear Programming 1 215

Allocations obtained from the LP model account for competition among

crops for water through the crop sensitivity factors Ky,f. The optimal alloca-
tions in a period depend on a number of factors such as water available, time
of season, soil moistures of individual crops in the period, potential
evapotranspiration of the crops, rainfall in the period, and crop areas. Note
that in this example, the water available in intraseasonal periods across the
crop season is assumed to be known. In planning reservoir releases for
irrigation, however, the water available for irrigation itself becomes a deci-
with cropwater allocations, as shown in Secs. 7.3 and 7.4.

7.2 MULTIRESERVOIR SYSTEM FOR IRRIGATION PLANNING

A river basin is considered for expanding its irrigation operations in its rela-
tively less developed upper basin, without detriment to the irrigation needs of
the lower basin, which is fully developed.
A four-reservoir system in the upper basin is considered for development
with given reservoir capacities. The crops proposed to be grown and the land
area available for cultivation in the command area of each reservoir are given.
Irrigation requirements of each crop per unit area are worked out at each site
and are known.
The four-reservoir system comprising the upper basin is modelled with
monthly time periods by using LP to determine the crops to be irrigated and the
extent of irrigation at each reservoir subject to land, water, and downstream
release constraints. Under conditions of limited water supply, there are two
options of irrigation considered: one is to provide intensive irrigation to those
crops that yield higher economic returns (these are also the crops that need
relatively more water) and leave the others unimgated; the other is to allocate
the available water to crops over a wider area to provide extensive irrigation.
Accordingly, two objectives are studied: one of maximizing the net economic
benefits (from both irrigated as well as unirrigated cropped areas at all reservoirs)
and the other of maximizing the total irrigated cropped area at all the reservoirs
in the upper basin. These two conflicting objectives are studied for analyzing
the tradeoff between them from the point of view of multiobjective planning,
using the constraint method (Section 4.2.2).
The model decides the crops to be irrigated and thg e-xtent of irrigation. If a
crop is considered at all for irrigation, under either option mentioned above, it
receives its full water requirements in all periods of the crop season.
The four-reservoir system in the upper basin is shown in Fig. 7.9. There are
two reservoirs B and D on the main river, and the other two, A and C, on the
tributaries. There is existing irrigation at reservoirs B and D. Expansion of
irrigation at B and new cropped areas at A and C are proposed.
 Basin

Fig. 7.9 1Schematic Diagram of the River Basin under Study

-

The reservoir capacities and land area proposed for irrigation are as shown
in Table 7.3.

*This is in addition to the existing irrigation at Reservoir B.

The Model
A linear programming model is formulated with monthly time periods to deter-
mine the crops and crop areas to be irrigated at each of the sites A, B, and C,
subject to minimum downstream releases from reservoir D to meet the existing
irrigation requirement in the lower basin. The model considers the option of
irrigating or not irrigating any or all of the cropped area under each crop, for
each objective.
Objective 1 Maximize the total net economic benefits from all crops in the
upper basin.
The objective function is written as
-- ;--Mi..; -
?-MY--=
max iia,],+ x z&U,J
1=1 ]=I '31
-

aij net benefits at site j per unit area of irrigated crop i (j= 1, reservoir A;
j = 2, reservoir B; and j = 3, reservoir C);
Pij net benefits at site j per unit area of unirrigated crop i;
I, irrigated area at site j under crop i;
- Uii unirrigated area at site j under crop i; and
. .
M, total number of crops c o n s i d e l . e d a t :
 Applications of Linear Programming 1 217
Objective 2 Maximize total irrigated area from all crops in the upper basin.

Constraints
Land Allocation Constraints The same land should be made available for a
given crop throughout the growing season.
For all the growing months m of crop i at site j
Iqn7= Iq for all nz,
where I,,, is the irrigated area at site j under crop i in pcriod m.
For all the growing months nz of crop i at site j,
Uij, =: U,. for all m,
where Uij, is the unirrigated area at site j under crop i in period m.
Storage-Continuity Equations The equationsfor the four sites j = 1, 2. 3, 4
are written as follows:
These constraints are valid for all time periods, m = 1, 2, ..., 12.
Forj= 1

1=1
For j = 2

where
Si, storage at site j at the beginning of month m;
p,,,, water diversion requirement of crop i (in depth units) at j during its
-- - -
growingmonth m, - i d equ-d-to zerCifor3lie nongiiowiiig riZonths;
R d/s release from the reservoir at site j during the month m;
Ej,, Reservoir evaporation at site j during the month rn;
F,,, mean inflow (unregulated) at site j during month rn; and
Wjm water diversion requirement at site j for all existing crops during month
m (this term will appear only for j = 2, and for j = 4, as per existing
irrigation requirements. Additional irrigation to ten proposed crops, i, at
site j = 2 is considered) - .,
 218 1 Applications

Land Area Constraints The total cropped area (irrigated andlor unirrigated)
in each month is less than the possible maximum at each site.

where L, = land available under the command of the reservoir at site j.

For a given crop calendar, the total number of governing constraints at each
site may be less than 12 because some constraints may be repetitive or may
form a subset of other constraints. In a month where one crop ends and another
begins, the area of the crop ending in that month is not included in the con-
straint to permit the possible use of the land for both crops in that month.
Downstream Release Constraints The release from the reservoir at site 4,
R4,, should be greater than the minimum downstream release specified by
(DR),,, which is arrived at on the basis of the requirements for existing irriga-
tion below reservoir D. Then
R4,2 (DR),, b' m
Capacity Constraints The storage in each reservoir is limited to its capacity
(live storage) at all times.
Sjnz5 Sjmax j, m
where Sjm, is the live storage capacity of the reservoir at site j.
Cropping Data
The first month (m = 1) corresponds to June as per the water year in India
(June 1 to May 31). Twelve principal crops are proposed to be grown in the
command area of reservoir A, 10 crops in the command area of reservoir B, and
14 in the command area of reservoir C. Crop water requirements are estimated
as per standard procedures. Estimated net benefit from irrigated and unirrigated
crops per unit area (Table 7.4), and release requirements downstream of
Reservoir D (based on the existing cropping in the lower basin, as given in
Table 7.5) are the other inputs used in the model.

Groundnut
Potato
Wheat

"Obtained by deducting estimated cost of fertilizer from produce value at wholesale prices.
 Applications of Linear Programming [ 219

Solution
Objective 1 Maximization of net economic benefits (annual): The optimal
cropping for this case is shown in Table 7.6. The table shows what crops are to
be grown, whether they should be irrigated or not, to what extent they should
be grown in each of the areas, and the growing seasons. The maximized net
benefits and the corresponding values of the total diversion and the irrigated
cropped area in the upper basin are also given in the table. -

Reservoir B

Reservoir C

Maximum net benefits = Rs. 2084.709 million; total irrigated cropped area = 527,570 ha.
(exclusive of fixed mulberry crop area of 20,700 ha.)
*I = irrigated, U = unirrigated.

Objective 2 Maximization of irrigated cropped area (ICA): The solution

obtained for this case is given in Table 7.7.
- The maximum- net- benefits (Objective 1)-are- Rs-2084~7-million and--the- -

corresponding irrigated area is 527,520 ha. The maximum irrigated cropped

area (Objective 2) is 755,600 ha, an increase of 4396, and the corresponding
net benefits are Rs 1659.6 million, a reduction of 20% when compared to the
results of maximizing Objective 1. In this case, increase in the cropped area
was made possible by replacing the rice and soyabean (November-February)
crops with less water-requiring crops such as ragi, jowar, wheat, vegetables,
and pulses. When the irrigated cropped area is maximized, the model chooses
 220 ) Applications

Reservoir B

PulsesNegetables Nov-Feb

Reservoir C

Maximum irrigated cropped area = 755,600 ha; total net benefits = Rs. 1659.623 million
(excluding mulberry over a fixed area of 20,700 ha).
*I = irrigated, U = unirrigated,

crops needing relatively lower water requirement, as the total amount of water
available is limited. These are also the crops that yield lower economic
benefits.
Summary of Results
Variable maximized Net Benefits Irrigated Cropped Area (ICA)
Net benefits (lo6 Rs 1970-72) 2084.709" 5275.700
Irrigated cropped area (100 ha) 1659.623 7556.00 *
("maximum value)

A Multiobjective Approach As the two objectives are conflicting with each

other, multiobjective analysis deals with finding the tradeoff between net in-
come and inrigated cropped area at different levels of development. Figure 7.10
shows the efficiency frontier or the transformation curve for these outputs.
Each of the plotted points in the fi~rureis obtained by maximizing &t net
benefits for a fixed value of the irrigated cropped area, using constraint ap- I
proach (Section 4.2.2), and corresponds to a plan of cropping such as the one
given in Table 7.6. Any point on the curve defined by these points is Pareto
I
admissible. I
I

Implicit Tradeoff A total irrigated cropped area of 576,800 ha. (at reservoirs
A, B, and C) is projected for development by the concerned planning agency,
based on rainfall distribution and local conditions. It is interesting to see that
 Applications of Linear Programming 1 221

1850 1' I I 1
4800 5800 6800 7800
Irrigated Cropped Area (100 ha)
Fig. 7.1 0 %j Transformation Curve for the Four-reservoir System

the proposed area lies within the range of values obtained for the two objec-
tives (between 527,570 and 755,600 ha.). A chance constrained form of the
model may be used to determine the results for inflows at a specified reliability
level, though the study reported herein does not attempt it.
A tradeoff between the net benefits and the irrigated cropped area can be
found from the curve above (Fig. 7.10) by determining the slope of the curve at
a given value of the irrigated area. For the projected level of 576,800 ha., the
tradeoff works out to Rs 360 per ha. This means that the amount of net benefits
foregone for an increase in the irrigated cropped area (at the level of projection)
is Rs. 360 per hectare, subject to the assumptions and limitations of the model.
For details of the study, refer to Vedula and Rogers (1981).

7.3 RELIABILITY CAPACITY TRADEOFF FOR

MULTICROP IRRIGATION
In this application, the formulation runs along similar lines to the problem
discussed in Section 6.3.1, except that in the present case we discuss how the
reservoir capacity tradeoff with reliability can be arrived at in the case of
multiple crops, taking into account the soil moisture balance in the root zone of
individual crops (as against lumped demands considered earlier).
A single reservoir is considered for multicrop irrigation. Data on crops,
soils, and potential evapotraspiration of each crop are known, A reliability-
based reservoir-irrigation model is formulated to determine (i) minimum size-of _- - _-
the reservoir needed to meet crop irrigation requirements at a specified level of
reliability, and (ii) maximum reliability of meeting irrigation requirements as-
sociated with a given reservoir capacity, using chance constrained linear pro-
gramming. Reservoir inflow is considered random, and rainfall in the com-
mand area is considered deterministic. The model takes into account water
allocation to multiple crops, soil moisture balance in the root zone of the
cropped area, heterogeneous soil types, and crop root growth with time, and