In this section, we will extend the range of application of LP to include problems that can be modeled as those of optimizing a convex piecewise linear objective function subject to linear constraints. These problems can be transformed easily into LPs in terms of additional variables. This material is from Murty (under preparation).
Letθ(λ) be a real valued function of a single variableλ∈R1.θ(λ) is said to be apiecewise linear (PL) function if it is continuous and if there exists a partition of R1 into intervals
TABLE 1.6 The PL Functionθ(λ)
Interval forλ Slope Value ofθ(λ)
λ≤λ1 c1 c1λ
λ1≤λ≤λ2 c2 θ(λ1) +c2(λ−λ1) λ2≤λ≤λ3 c3 θ(λ2) +c3(λ−λ2)
.. .
.. .
.. .
λ≥λr cr+1 θ(λr) +cr+1(λ−λr)
of the form [−∞, λ1] ={λ≤λ1}, [λ1, λ2], . . .,[λr−1, λr], [λr,∞] (where λ1< λ2<ã ã ã< λr are the breakpoints in this partition) such that inside each interval the slope of θ(λ) is a constant. If these slopes in the various intervals are c1, c2, . . ., cr+1, the values of this function at various values ofλare tabulated in Table 1.6.
This PL function is said to beconvexif its slope is monotonic increasing withλ, that is, if c1< c2ã ã ã< cr+1. If this condition is not satisfied it is nonconvex. Here are some numerical examples of PL functions of the single variableλ(Tables 1.7 and 1.8).
Example 1.3: PL Functionθ(λ)
TABLE 1.7 PL Convex Functionθ(λ)
Interval forλ Slope Value ofθ(λ)
λ≤10 3 3λ
10≤λ≤25 5 30 + 5(λ−10)
λ≥25 9 105 + 9(λ−25)
Example 1.4: PL Functiong(λ)
TABLE 1.8 Nonconvex PL Functionθ(λ)
Interval forλ Slope Value ofθ(λ)
λ≤100 10 10λ
100≤λ≤300 5 1000 + 5(λ−100) 300≤λ≤1000 11 2000 + 11(λ−300)
λ≥1000 20 9700 + 20(λ−1000)
Both functions θ(λ), g(λ) are continuous functions and PL functions. θ(λ) is convex because its slope is monotonic increasing, butg(λ) is not convex as its slope is not monotonic increasing withλ.
A PL function h(λ) of the single variable λ∈R1 is said to be a PL concave function iff −h(λ) is a PL convex function; that is, iff the slope ofh(λ) is monotonic decreasing as λincreases.
PL Functions of Many Variables
Letf(x) be a real valued function ofx= (x1, . . ., xn)T.f(x) is said to be a PL (piecewise lin- ear) function ofxif there exists a partition ofRninto convex polyhedral regionsK1, . . ., Kr such that f(x) is linear within each Kt, for t= 1 to r; and a PL convex function if it is also convex. It can be proved mathematically that f(x) is a PL convex function iff there exists a finite number,rsay, of linear (more precisely affine) functionsct0+ctx, (wherect0,
andct∈Rnare the given coefficient vectors for thet-th linear function)t= 1 torsuch that for each x∈Rn
f(x) = Maximum{ct0+ctx:t= 1, . . ., r} (1.3) A function f(x) defined by Equation 1.3 is called the pointwise maximum(or supremum) function of the linear functionsct0+ctx,t= 1 tor. PL convex functions of many variables that do not satisfy the additivity hypothesis always appear in this form (Equation 1.3) in real world applications.
Similarly, a PL function h(x) ofx= (x1, . . ., xn)T is said to be aPL concave functionif there exist a finite numbersof affine functionsdt0+dtx,t= 1 tos, such thath(x) is their pointwise infimum, that is, for eachx∈Rn
h(x) = Minimum{dt0+dtx:t= 1, . . ., s}
Now we show how to transform various types of problems of minimizing a PL convex function subject to linear constraints into LPs, and applications of these transformations.
Minimizing a Separable PL Convex Function Subject to Linear Constraints
A real valued functionz(x) of variablesx= (x1, . . ., xn)T is said to beseparableif it satisfies the additivity hypothesis, that is, if it can be written as the sum of n functions, each one involving only one variable as in: z(x) =z1(x1) +z2(x2) +ã ã ã +zn(xn). Consider the following general problem of this type:
Minimize z(x) =z1(x1) +ã ã ã+zn(xn)
subject to Ax=b (1.4)
x≥0
where eachzj(xj) is a PL convex function with slopes in intervals as in Table 1.9,rj+ 1 is the number of different slopescj1< cj2<ã ã ã< cj,rj+1 ofzj(xj), andj1,j2, . . ., j,rj+1 are the lengths of the various intervals in which these slopes apply.
As the objective function to be minimized does not satisfy the proportionality assumption, this is not an LP. However, the convexity property can be used to transform this into an LP by introducing additional variables. This transformation expresses each variablexj as a sum ofrj+ 1 variables, one associated with each interval in which its slope is constant. Denot- ing these variables byxj1, xj2, . . ., xj,rj+1, the variablexj becomes =xj1+ã ã ã+xj,rj+1 and zj(xj) becomes the linear functioncj1xj1+ã ã ã+cj,rj+1xj,rj+1in terms of the new variables.
The reason for this is that as the slopes are monotonic (i.e., cj1< cj2<ã ã ã< cj,rj+1), for any value of ¯xj≥0, if (¯xj1,x¯j2, . . .,x¯j,rj+1) is an optimum
Minimize cj1xj1+ã ã ã+cj,rj+1xj,rj+1
Subject to xj1+ã ã ã+xj,rj+1= ¯xj
0≤xjt< jt, t= 1, . . ., rj+ 1 TABLE 1.9 The PL FunctionZj(xj)
Interval Slope in Interval Value ofzj(xj) Length of Interval
0≤xj≤kj1 cj1 cj1xj j1=kj1
kj1≤xj≤kj2 cj2 zj(kj1) +cj2(xj−kj1) j2=kj2−kj1
.. .
.. .
.. .
.. . kj,rj≤xj cj,rj+1 zj(kj,rj) +cj,rj+1(xj−kj,rj) j,rj+1=∞
x¯j,t+1 will not be positive unless ¯xjk=jk for k= 1 to t, for each t= 1 to rj. Hence the optimum objective value in this problem will be equal tozj(xj). This shows that our original problem (Equation 1.4) is equivalent to the following transformed problem which is an LP.
Minimize
j=n
j=1
t=rj+1 t=1
cjtxjt
subject to xj=
t=rj+1 t=1
xjt, j= 1, . . ., n Ax=b
0≤xjt≤jt, t= 1, . . ., rj+ 1; j= 1, . . . , n x≥0.
The same type of transformation can be used to transform a problem involving the maximization of a separable PL concave function subject to linear constraints into an LP.
Example 1.5
A company makes productsP1, P2, P3 using limestone (LI), electricity (EP), water (W), fuel (F), and labor (L) as inputs. Labor is measured in man hours, and other inputs in
suitable units.
Each input is available from one or more sources. The company has its own quarry for LI, which can supply up to 250 units/day at a cost of $20/unit. Beyond that, LI can be purchased in any amounts from an outside supplier at $50/unit. EP is only available from the local utility. Their charges for EP are: $30/unit for the first 1000 units/day, $45/unit for up to an additional 500 units/day beyond the initial 1000 units/day, $75/unit for amounts beyond 1500 units/day. Up to 800 units/day of water is available from the local utility at
$6/unit; beyond that they charge $7/unit of water/day. There is a single supplier for F who can supply at most 3000 units/day at $40/unit; beyond that there is currently no supplier for F. From their regular workforce they have up to 640 man hours of labor/day at $10/man hour; beyond that they can get up to 160 man hours/day at $17/man hour from a pool of workers.
They can sell up to 50 units ofP1 at $3000/unit/day in an upscale market; beyond that they can sell up to 50 more units/day ofP1 to a wholesaler at $250/unit. They can sell up to 100 units/day ofP2 at $3500/unit. They can sell any quantity ofP3 produced at a constant rate of $4500/unit.
Data on the inputs needed to make the various products are given in Table 1.10. Formulate the product mix problem to maximize the net profit/day at this company.
Maximizing the net profit is the same as minimizing its negative, which is = (the costs of all the inputs used/day)−(sales revenue/day). We verify that each term in this sum
TABLE 1.10 I/O Data
Input Units/Unit Made
Product LI EP W F L
P1 1
2 3 1 1 2
P2 1 2 1
4 1 1
P3 3
2 5 2 3 1
is a PL convex function. So, we can model this problem as an LP in terms of variables corresponding to each interval of constant slope of each of the input and output quantities.
LetLI, EP,W,F, L denote the quantities of the respective inputs used/day; andP1, P2,P3denote the quantities of the respective products made and sold/day. LetLI1andLI2
denote the units of limestone used daily from own quarry and outside supplier. Let EP1, EP2, andEP3denote the units of electricity used/day at $30, 45, 75/unit, respectively. Let W1 andW2 denote the units of water used/day at rates of $6, 7/unit, respectively. LetL1
andL2 denote the man hours of labor used/day from regular workforce, pool, respectively.
LetP11 and P12 denote the units ofP1 sold at the upscale market and to the wholesaler, respectively.
Then the LP model for the problem is:
Minimize z= 20LI1+ 50LI2+ 30EP1+ 45EP2+ 75EP3+ 6W1+ 7W2+ 40F+ 10L1
+ 17L2−3000P11−250P12−3500P2−4500P3
subject to
(1/2)P1+P2+ (3/2)P3=LI 3P1+ 2P2+ 5P3=EP P1+ (1/4)P2+ 2P3=W
P1+P2+ 3P3=F 2P1+P2+P3=L
LI1+LI2=LI, W1+W2=W EP1+EP2+EP3=EP
L1+L2=L, P11+P12=P1, all variables≥0 (LI1, EP1, EP2, W1)≤(250,1000,500,800)
(F, L1, L2)≤(3000,640,160) (P11, P12, P2)≤(50,50,100).
1.3.1 Min–Max, Max–Min Problems
As discussed above, a PL convex function in variables x= (x1, . . ., xn)T can be expressed as the pointwise maximum of a finite set of linear functions. Minimizing a function like that is appropriately known as a min–max problem. Similarly, a PL concave function inx can be expressed as the pointwise minimum of a finite set of linear functions. Maximizing a function like that is appropriately known as a max–min problem. Both min–max and max–min problems can be expressed as LPs in terms of just one additional variable.
If the PL convex function f(x) = min{ct0+ctx:t= 1, . . ., r}, then −f(x) = max{−ct0− ctx:t= 1, . . ., r} is PL concave and conversely. Using this, any min–max problem can be posed as a max–min problem and vice versa. So, it is sufficient to discuss max–min problems.
Consider the max–min problem
maxz(x) = min{c10+c1x, . . ., cr0+crx}
subject to Ax=b x≥0
To transform this problem into an LP, introduce the new variable xn+1 to denote the value of the objective functionz(x) to be maximized. Then the equivalent LP with additional
linear constraints is:
max xn+1
subject to xn+1≤c10+c1x xn+1≤c20+c2x
...
xn+1≤cr0+crx Ax=b
x≥0
The fact that xn+1 is being maximized and the additional constraints together imply that if (¯x,x¯n+1) is an optimum solution of this LP model, then ¯xn+1= min{c10+c1x, . . ., c¯ r0+ crx}¯ =z(¯x), and that ¯xn+1is the maximum value ofz(x) in the original max–min problem.
Example 1.6: Application in Worst Case Analysis
Consider the fertilizer maker’s product mix problem with decision variablesx1andx2(Hi- ph, Lo-ph fertilizers to be made daily in the next period) discussed in Example 1.2, Sec- tion 1.2. There we discussed the case where the net profit coefficients c1 and c2 of these variables are estimated to be $15 and $10, respectively. In reality, the prices of fertilizers are random variables that fluctuate daily. Because of unstable conditions, and new agri- cultural research announcements, suppose market analysts have only been able to estimate that the expected net profit coefficient vector (c1, c2) is likely to be one of {(15, 10), (10, 15), (12, 12)} without giving a single point estimate. So, here we have three possible scenarios. In scenario 1, (c1, c2) = (15, 10), expected net profit = 15x1+ 10x2; in scenario 2, (c1, c2) = (10, 15), expected net profit = 10x1+ 15x2; in scenario 3 (c1, c2) = (12, 12), expected net profit = 12x1+ 12x2. Suppose the raw material availability data in the prob- lem is expected to remain unchanged. The important question is: which objective function to optimize for determining the production plan for the next period.
Irrespective of which of the three possible scenarios materializes, at the worst the minimum expected net profit of the company will be p(x) = min{15x1+ 10x2,10x1+ 15x2,12x1+ 12x2} under the production plan x= (x1, x2)T. Worst case analysis is an approach that advocates determining the production plan to optimize this worst case net profitp(x) in this situation. This leads to the max–min model: maximizep(x) = min{15x1+ 10x2,10x1+ 15x2,12x1+ 12x2}subject to the constraints in Equation 1.2. The equivalent LP model cor- responding to this is:
max p
subject to p≤15x1+ 10x2
p≤10x1+ 15x2
p≤12x1+ 12x2
2x1+x2≤1500 x1+x2≤1200 x1≤500, x1, x2≥0
1.3.2 Minimizing Positive Linear Combinations of Absolute Values of Affine Functions
Consider the problem
min z(x) =w1|c10+c1x|+ã ã ã+wr|cr0+crx|
subject to Ax≥b
where the weights w1, . . ., wr are all strictly positive. In this problem the objective func- tion to be minimized, z(x), is a PL convex function; hence this problem can be trans- formed into an LP. To transform, define for eacht= 1 to rtwo new nonnegative variables u+t = max{0, ct0+ctx}, u−t =−min{0, ct0+ctx}. u+t is called the positive partof ct0+ctx, andu−t itsnegative part. It can be verified that (u+t)(u−t) is zero by definition, and because of this we have:ct0+ctx= (u+t)−(u−t) and|ct0+ctx|= (u+t) + (u−t). Using this, we can trans- form the above problem into the following LP:
min w1[(u+1) + (u−1)] +ã ã ã+wr[(u+r) + (u−r)]
subject to c10+c1x = (u+t)−(u−t) ... ...
cr0+crx = (u+r)−(u−r) Ax ≥ b
(u+t),(u−t) ≥ 0, t= 1, . . ., r
Using the special structure of this problem it can be shown that the condition (u+t)(u−t) = 0 for allt= 1 to rwill hold at all optimum solutions of this LP. This shows that this trans- formation is correct.
An application of this transformation is discussed in the next section. This is an important model that finds many applications.