# CH 5170 Optimization Problem set 1

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1. Water distribution networks, as the name suggests transport water from a source (or multiple
sources) and deliver them to demand points. Generally speaking, pressure drives flow, i.e., the
upstream pressure is greater than the downstream pressure. Water can also flow aided by gravity,
i.e., water flows from a higher level to a lower level.

Thus, a difference in elevation and/or pressure provides the driving force for flow. The two are often combined into a single quantity called the
”head”, (expressed in meters) If water flows from point A to point B, the head at point A, HA is greater
than the head at point B, HB.

If the flow rate of water Q is known, the head loss ∆H = HA − HB is
related to the flow rate, diameter and length of the pipe through the following correlation:
HA − HB = ∆H = 4.457 × 108LQ1.85
D4.87 , (1)
where Q is the flow rate in m3/min, HA, HB, ∆H are the heads in meters, L is the length of pipe in
meter, D is the inner diameter of the pipe in mm.

A networks consists of a source node (or multiple sources) and demand points or nodes. The
topography of the network (i.e., where the source is situated, where the demand points are located
etc.), the interconnections between the nodes is usually decided beforehand. The water demand
at the individual demand points is also known.

The task at hand is to decide the diameters of the
pipes used to transport the water. There are two competing factors that have to be traded off when
deciding the optimum pipe diameters: the capital cost of the pipe and pump operating costs. In the
following problems we consider flow by gravity and hence, pump operating costs are zero. A large
diameter pipe would result in a high capital cost, but low head loss. Likewise, a smaller diameter
pipe would result in small capital cost, high head loss. As expected, one aims to minimize the total
cost.

Constraints to be enforced are the demand flow rates at the demand points. In addition, it
is common to specify a minimum head at the demand nodes. The task at hand is to decide the
individual pipe diameters that minimize the total cost while satisfying the constraints.

(a) Consider the network shown in Figure 1. (Refer Fig. 1)
The water distribution network consists of a source node 0 and three demand nodes 1,2 and 3.
Node 0 and node 1 are connected by link I. Likewise, link II connects nodes 1 and 2. Link III
connects nodes 1 and 3. The diameters of the pipes are D1, D2, D3 respectively (in mm). Flow
is by gravity and it is possible to achieve a head of 100 m at source node 0. Minimum head
values at nodes 1, 2 and 3 are 80, 85 and 80 m respectively. Lengths of links I, II and III
1
Figure 1
0
3
2
0 1
I
II

III Figure 1: Water distribution network
are 300, 500 and 400 m respectively. Flow rates in links I, II and III are 9,3 and 2 m3/min
respectively. The cost of the pipe per unit length (in Rs.) can be estimated by the correlation:
c = 1.2654D1.327
, (2)
where D is the diameter of the pipe. Since flow is by gravity, there are no operating costs
associated with this network, i.e., total cost is simply the sum of capital costs of the pipes. The
task at hand is to decide the diameters of the pipes, D1, D2, D3 respectively.

(b) Write down expressions for the heads at nodes 1, 2 and 3 in terms of D1, D2, D3. Hint: Head
loss across pipes connected in series is equal to sum of individual head losses.

(c) Write down expression for the total cost, i.e., sum of costs of links I, II and III.
(d) Formulate the appropriate optimization problem to minimize the costs while ensuring that the
heads at nodes 1, 2 and 3 are greater than the minimum values.

(e) At the optimum, what do you expect the values of the head at nodes 1, 2 and 3 to be? Give
reasons.
(f) Unlike previously, we will consider the heads at nodes 1,2 and 3, viz., H1, H2, H3 to be the
decision variables.

(g) Reformulate the problem to minimize pipe cost subject to the minimum head requirements at
1, 2 and 3 in terms of H1, H2 and H3.

(h) Reduce this formulation to an equality constrained problem.

(i) The optimization formulation that you derived in Question has two issues: The first is that it is
a nonlinear constrained optimization problem. We have not yet discussed solution techniques
for the same in class. Fortunately for you, I will provide the optimal solution: D1 = 303.5,
D2 = 209.3 and D3 = 155.8, all in mm.

The second issue is that these are odd sizes, i.e.,
commercially available pipes are available only in certain diameters, e.g., 80 mm, 100mm, 125
mm etc. Table 1 contains a list of the commercially available pipes and their unit costs. Clearly,
pipes with diameters 303.5 mm, 209.3 mm and 155.8 mm are not available commercially. One
solution to avoid this is to restrict the solution space to pipes from the discrete set. However,
this can lead to a combinatorially difficult problem.

In this example, this would mean examining
potentially 143 combinations. Another solution is to simply ”round” off the pipe diameter to the
next commercially available pipe size. A more sophisticated approach is to use 2 closest available pipe sizes in series, i.e., we replace the odd size by two pipes having adjacent diameters
and connect them in series. One of the pipes has diameter larger and the other smaller than

Table 1: Commercially available pipe diameters
Sr no Pipe dia (mm) Unit costs (Rs./m)
1 80 424
2 100 570
3 125 767
4 150 977
5 200 1431
6 250 1924
7 300 2451
8 350 3008
9 400 3591
10 450 4198
11 500 4828
12 600 6149
13 700 7545
14 750 8269
the odd diameter. E.g., the pipe of diameter 303.5 mm is replaced by two pipes of diameter
300 mm and 350 mm, such that the total length is still 300 m. If L
I
1 and L
I
2 are the lengths
of the pipe having 300 and 350 mm diameter respectively, we have that L
I
1 + L
I
2 = 300.

The
advantage of this approach is that it results in lower cost than going in for a single pipe of a
commercially available size. Similarly replace link II with pipes of lengths L
II
1
(200 mm dia)
and L
II
1
(250 mm dia) respectively. Repeat for link III with pipes of diameter 150 and 200 mm
diameters.

(j) Rewrite the expression for the cost in terms of the new variables L
I
1
, L
I
2
,L
II
1
, L
II
2
, L
III
1
, L
III
2
.
(k) Neglecting the head loss in the joints, rewrite the expression for the heads in terms of the new
variables.

(l) Reformulate the optimization problem to minimize the total cost in terms of the new decision
variables, while meeting the minimum head requirements.
(m) Show that it is a Linear Program and express it in the form:
min c
0x,s.t. Ax ≤ b, Aeqx = beq, x ≥ 0.
Write out the variable x and the matrices and vectors clearly.

2. We will now consider a larger pipe network shown in Figure 2.
The network consists of source node 0 and 5 demand nodes 1,2,3,4,5. The flow in links I, II,
III, IV and V are 8.5, 5.8, 1.2, 1.6 and 1.3 m3/min respectively. The minimum heads required at
nodes 1,2,3,4 and 5 are 90,85,80,80 and 80 m respectively. The lengths of links I, II, III, IV and
V are 1000, 600, 400, 300 and 300 m respectively. The head available at node 0 is 115 m.

(a) Repeat parts (1d), (1g) making appropriate changes.

(b) Solution of the nonlinear optimization results in pipes of 304.9, 267.5, 151.9, 159.8 and 119.8
mm diameter respectively. As before, replace the odd pipe by using adjacent pipe diameters,
formulate the optimization problem as in (1m).
3
0
1 2 3
5 4
I II III
IV V
Figure 2: Water distribution network 2

3. In both networks, we agreed to replace an odd size pipe by a combination of 2 adjacent pipe
sizes. In this exercise, you will allow any link to be composed of a series combination of all of the
14 commercially available pipe sizes, i.e., if there is a link of length 300 m, we will allow it to be
replaced by L
I
1 of diameter 80 mm, L
II
1 of diameter 100 mm, . . . LI
14 of diameter 750 mm. This
problem continues to a be a LP, but with a much higher dimensionality, viz., 14 × 3 = 42 variables
and 14×5 respectively. Will you get the same solution as what you got by replacing the odd diameter
pipe by adjacent diameter pipes? Discuss.

4. I cannot claim that I have all the answers, so I am looking to see how you approach the problem. If you
can help me answer them, all the better! Discussion encouraged In general, how many segments do you
expect the final solution to consist of? 1,2,3 all 14? Can you prove it? Will they be adjacent? If
so, what additional conditions must be imposed?

I can extend this idea to a pipe of continuously
varying diameters by considering allowing pipes having incrementally increasing diameters, i.e.,
given a diameter range [d1, d2, divide into N equi-spaced intervals of size ∆. So, I have pipes
of diameters d1, d1 + ∆, d1 + 2∆, . . . , dN = d2. Let l1, l2, . . . , lN be corresponding lengths of pipe.
If in the earlier question, you were able to prove that the final solution will consist of only k pipe
sizes, you can extend this idea here to show that even allowing for continuously varying diameters,
the final solution consists of only finite set of diameters.

You may find the following materials and
references therein useful:
Optimal design of water distribution networks by P R Bhave. (One copy supposedly available in
library.)

“Two Adjacent Pipe Diameters at the Optimal Solution in the Water Distribution Network Models”,
Fujiwara and Dey, Water Resources Research, 23(8), 1457 − 1460, 1987.

5. Formulate the coin changing problem as an optimization problem.
6. Given an infinite supply of coins of denomination 2
0
, 2
1
, 2
2
, 2
3
, . . ., prove that the greedy algorithm
will solve the coin changing problem exactly.

7. Determine a set of available coin denominations and an amount such that the greedy algorithm
does not work.

8. You are given n different solid items denoted by x1, x2, . . . , xn each of weight wi and value vi
. You
are a door-to-door salesman and have to pack some or all of these items in a bag. You can carry
a maximum weight W. Obviously, you want to maximize the value. Formulate an optimization
problem to achieve this. What is the nature of this optimization problem?

9. This is similar to the previous problem except that you have multiple copies of each item, C1, C2, . . . , Cn.
i.e., the number of copies of item x1 is C1, and so on. Formulate an optimization problem to maximize the value subject to the maximum weight capacity and the availability of each item. What is
the nature of this optimization problem?

10. This is similar to the previous problem except that you have large number of copies (practically infinite) of each item, Formulate an optimization problem to maximize the value subject to the maximum
weight capacity. What is the nature of this optimization problem?

11. Again, similar to the previous problem, except that the items are liquids and you can carry fractional
amounts. The value of each item is obviously proportional to the volume of each item and this
proportionality constant is the values per unit volume, i.e., ρ1, ρ2, . . . , ρn. Formulate an optimization
problem that maximizes the value of the liquids carried subject to overall volume being smaller than
V . Is there an efficient algorithm to solve this?

12. Consider the following inequality constrained quadratic program
min f(x) = (x1 − 4)2 + (x2 − 5)2
,
subject to
2×2 − 4 − x1 ≤ 0, (3)
x1 + x2 − 5 ≤ 0, (4)
x1 ≤ 3, (5)
−x2 ≤ 0, (6)
−x1 ≤ 0. (7)
Interpret the problem geometrically and determine the solution based on geometry. Sketch the
feasible set.