Operations Research
“Research on operations” is applied to problems that concern as to how to conduct and
coordinate the operations or activities within an organization.
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Builds mathematical model of system, deterministic or probabilistic.
Establishes objectives that are consistent with those of the overall organization.
Attempts to find the best or optimal solution to the problem under consideration with a
team effort.
In summary, the study involves:
1. Formulating the problem – study and develop a well defined statement of the problem
that include objectives, constraints, interrelationships between the area of study and
other areas of the organization, possible alternative courses of action.
2. Model – reformulate the problem into a form that is convenient for analysis.
3. Solution – optimal or best solution
4. Testing – validity is very important.
5. Control – model is valid as long as the conditions do not change.
6. Implementation – develop procedures for how to implement. Needs support of top
management and operations management. Benefits are reaped here.
Formulation of Problem
Example: Company X manufactures two types of wooden toys: soldiers and trains. A soldier
sells for $27 and uses $10 worth of raw materials. Each soldier that is manufactured increases
Co. X’s variable labour and overhead costs by $14. A train sells for $21 and uses $9 worth of
raw materials. Each train built increases variable labour and overhead costs by $10. The
manufacture of wooden soldiersand trains requires two types of skilled labour: carpentry and
finishing. A soldier requires 2 hours of finishing labour and 1 hour of carpentry labour. A
train requires 1 hour of finishing and 1 hour of carpentry labour. Each week, Co. X can obtain
all the needed raw material but only 100 finishing hours and 80 carpentry hours. Demand for
trains is unlimited, but at most 40 soldiers are bought each week. Co. X wants to maximize
weekly profit (revenues – costs). Formulate a mathematical model of Co. X’s situation that
can be used to maximize weekly profit.
Decision variables – we begin by defining the relevant decision variables, i.e. how many
soldiers and trains should be manufactured each week. We define,
X1 = number of soldiers produced each week
X2 = number of trains produced each week.
Objective function – normally, if we are dealing with profits, we want to maximize the
profits. When dealing with costs only, we wish to minimize the costs. The function to be
maximized or minimized is called the objective function. Then, in Co.X’s case,
Weekly revenue = revenue from soldiers + revenue from trains = 27.X1 + 21.X2
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Also, weekly raw materials costs = 10.X1 + 9.X2
Other weekly variable costs = 14.X1 + 10.X2
Profit = (27.X1 + 21.X2) – (10.X1 + 9.X2) – (14.X1 + 10.X2) = 3.X1 + 2.X2
Then, objective function is,
Maximize Z = 3.X1 + 2.X2
(The coefficient of a variable in the objective function is called the objective function
coefficient of the variable.
Constraints – without these, there would be no limit to how much we would produce as the
profits inrease with the numbers produced. However, there are limited resources available.
Constraint 1 – each week, no more than 100 hours of finishingtime may be used.
Constraint 2 – Each week, no more than 80 hours of carpentry time may be used.
Constraint 3 – due to limited demand, at most 40 soldiers should be produced each week.
Hence, formulating the constraints,
2.X1 + 1.X2 ≤ 100
1.X1 + 1.X2 ≤ 80
X1
≤ 40
Constraint 1
Constraint 2
Constraint3
(The coefficients of the decision variables in the constraints are called technological
coefficients).
Sign restriction – if decision variable can only assume non-negative values, we add sign
restriction Xi ≥ 0. If both +ve and –ve values can be assumed, Xi is unrestricted in sign (urs).
In the example considered, X1 ≥ 0 and X2 ≥ 0.
Combining all of the above,
Max Z = 3.X1 + 2.X2
2.X1 + 1.X2 ≤ 100
1.X1 + 1.X2 ≤ 80
X1
≤ 40
X1 ≥ 0
X2 ≥ 0
Constraint 1
Constraint 2
Constraint3
Sign restriction.
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Other Formulation Problems
1. Agricultural production is being planned in three regions under the umbrella of a
confederation. Agricultural output is limited by both available land and water allocated.
Available land and water allocation are given below for the three regions
Region
Usable land, acres
Water allocation, acre feet
1
400
600
2
600
800
3
300
375
The crops to be produced in the confederation (in the three regions), the total acreage for each
crop, the water consumption for each crop, and the net return from each crop are given in the
following Table.
Crop
Max. land quota
Water consumption Net return
Acres
acres feet / acre
dollars /acre
Sugar beets
600
3
400
Cotton
500
2
300
Sorghum
325
1
100
It is agreed that each region will plant the same proportion of its available irrigable land.
However, any combination of the crops may be grown at any of the regions. The job of the
farmer’s coordinating office is to plan how many acres to devote to each crop at the respective
region while satisfying the given restrictions. The objective is to maximize the total net return
for the confederation as a whole.
Let Xj (j = 1,2,....9) be the decision variables where Xj gives the number of acres to devote to
each of the three crops at each region.
Region
Crop
1
2
3
Sugar beets
X1
X2
X3
Cotton
X4
X5
X6
Sorghum
X7
X8
X9
Objective function:
Maximize Z = 400.(X1 + X2 + X3) + 300.(X4 + X5 + X6) + 100.(X7 + X8 + X9)
Subject to the following constraints:
1. Land.
X1 + X4 + X7 ≤ 400
X2 + X5 + X8 ≤ 600
X3 + X6 + X9 ≤ 300
2. Water
3.X1 + 2.X4 + X7 ≤ 600
3.X2 + 2.X5 + X8 ≤ 800
3.X3 + 2.X6 + X9 ≤ 375
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3. Crop
X1 + X2 + X3 ≤ 600
X4 + X5 + X6 ≤ 500
X7 + X8 + X9 ≤ 325
4. Social
(X1 + X4 + X7)/400 = (X2 + X5 + X8)/600
(X2 + X5 + X8)/600 = (X3 + X6 + X9)/300
(X3 + X6 + X9)/300 = (X1 + X4 + X7)/400
Or, these may be written as,
3.(X1 + X4 + X7) – 2.(X2 + X5 + X8) = 0
(X2 + X5 + X8) – 2.(X3 + X6 + X9) = 0
4.(X3 + X6 + X9) – 3.(X1 + X4 + X7) = 0
5. Non-negativity
Xj ≥ 0, for j = 1, 2, ......9.
Another way of representing the decision variables would be: Xij where i = 1, 2, 3 represent
the regions and j = 1, 2, 3 represent the crops. Then, for example, X11 would be eqivalent to
X1, X12 to X4, X13 to X7, X23 to X8, X32 to X6, etc.
2. A farmer is raising sheep for market, and wishes to determine the quantities of the available
types of feed that should be given to each sheep to meet certain nutritional requirements at a
minimum cost. The number of units of each type of basic nutritional ingredient contained
within a kilogram of each feed type is given in the following table, along with the daily
nutritional requirements and feed costs:
Nutritional
Ingredient
Kilogram
of Corn
Kilogram
of Tankage
Kilogram
of Alfalfa
Minimum Daily
Requirement
Carbohydrates
Protein
Vitamins
90
30
10
20
80
20
40
60
60
200
180
150
Costs (cents)
42
36
30
Formulate the linear programming model for this problem.
Let X1 : corn
s.t.
X2: tankage
X3: alfalfa
Min Z = 42.X1 + 36.X2 + 30.X3
90.X1 + 20.X2 + 40.X3 ≥ 200
30.X1 + 80.X2 + 60.X3 ≥ 180
10.X1 + 20.X2 + 60.X3 ≥ 150
X1, X2, X3 ≥ 0
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