Operations Research : Applications and Algorithms
Operations Research : Applications and Algorithms
4th Edition
ISBN: 9780534380588
Author: Wayne L. Winston
Publisher: Brooks Cole
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Chapter 4.16, Problem 11P

Explanation of Solution

Using preemptive goal programming to achieve the goal:

Consider the case of building recreational facilities at Gotham city. Goal course, Swimming pool, Gymnasiums and Tennis Courts are the recreational facilities under consideration. The facilities can be built at any of the 6 locations.

Lets,

xij={1  If facility, i is built in location j2  otherwise

Since, Golf course can only be built at site 1 or 6, so

x11+x16=1

Since each facility can be built at only one site, add the following constraints.

x22+x23+x24=1x32+x33+x34=1x42+x43+x44=1x52+x53+x54=1

The land available at site 2,3,4,5 is 70,80,95, and 120 respectively. Land required, building annual maintenance cost is tabulated below,

SiteConstruction costMaintenance costLand required
Golf34080Not relevant
Swimming3003629
Gymnasium8405038
Tennis curt851745
Site123456
Golf31NANANANA27
SwimmingNA25213232NA
GymnasiumNA37292838NA
Tennis courtNA20232220NA

a.

Priority set are given below,

Priority 1: Limit land use at each site to the land available. Thus the following constraints are added.

29x22+38x32+45x427029x23+38x33+45x438029x24+38x34+45x449529x25+38x35+45x45120

Priority 2: Construction costs should not exceed $1.2 million. Thus the following constraints are added.

340x11+340x16+300x22+300x23+300x24+300x25840x32+840x33+840x44+840x3585x42+85x43+85x44+85x451200

Priority 3: User days should exceed 200,000. Thus the following constraints are added.

31x11+27x16+25x22+21x23+32x24+32x2537x32+29x33+28x44+38x3520x42+23x43+22x44+20x45200

Priority 4: Annual maintenance cost should not exceed $200,000. Thus the following constraints are added.

80x11+80x16+36x22+36x23+36x24+36x2550x32+50x33+50x3517x42+17x43+17x44+17x45200

From the above equations it is found that these set of constraints there is no feasible region. That is all constraints cannot be met. So assign a cost value incurred if any of the priorities or goal is not met.

So, introduce the following deviational variables.

Si= Amount by which numerically under the ith goal

Si+= Amount by which numerically exceed the ith goal

Therefore, the constraints become,

29x22+38x32+45x42+S1+S1=7029x23+38x33+45x43+S2+S2=8029x24+38x34+45x44+S3+S3=9529x25+38x35+45x45+S4+S4=120

340x11+340x16+300x22+300x23+300x24+300x25840x32+840x33+840x44+840x3585x42+85x43+85x44+85x45+S5+S5=1200

31x11+27x16+25x22+21x23+32x24+32x2537x32+29x33+28x44+38x3520x42+23x43+22x44+20x45+S6+S6=200

80x11+80x16+36x22+36x23+36x24+36x2550x32+50x33+50x3517x42+17x43+17x44+17x45+S7+S7=200

Now, the goal is to minimize the deviation from each goal. Hence if the left side of constraint was less than right hand side than Si is included in minimization equation. If left hand side is more than right hand side, then Si+ is included in minimizations equation.

Hence the minimization equation is,

Minimize, z=P1S1+P1S2+P1S3+P1S4+P2S5+P3S6++P4S7

For each Priority assign a value of P1=4,P2=3,P3=2,P1=1 for priority 1,2,3,4 respectively

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Chapter 4 Solutions

Operations Research : Applications and Algorithms

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Operations Research : Applications and Algorithms
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ISBN:9780534380588
Author:Wayne L. Winston
Publisher:Brooks Cole