Chapter 7. Quantitative Business Analysis-Beneficial Tools for Business
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Figure 7.1
Microsoft Excel 2003 with Solver.
Figure 7.2
Solver add-in process for Microsoft Excel 2007.
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7.1 Linear Programming
Let us use a simple production schedule problem to explain the solving procedure in detail.
MMI Example—Part I
Micro Motors Inc. (MMI) produces multilayer actuators (MLAs) and ultrasonic motors (USMs)
every month. Table 7.1 summarizes the necessary data for producing MLAs and USMs. Though
USMs generate a larger gross proﬁt per unit with a smaller amount of raw materials per unit than
MLAs, USMs require a working period ﬁve times longer than MLAs. MMI has two major constraints: (1) labor—total 500 h/month maximum (= 3 engineers × 40 h/week × 4.17 weeks/month),
and (2) raw materials available—total 300 g/month maximum. MMI would like to determine the
optimum production quantities (unit number) X for MLAs and Y for USMs in order to maximize
the gross proﬁt.
Part I seeks the solutions for nonnegative continuous numbers of X and Y .
7.1.1 Mathematical Modeling
The production quantities (unit number) X for MLAs and Y for USMs are called decision variables .
The total gross proﬁt is provided by the sum of MLA proﬁt $25X and USM proﬁt $100Y :
25X + 100Y
which MMI would like to m aximize. This is called objective function. Two major physical constraints at MMI are expressed with functional constraints, ≤, ≥, or =:
Labor: 5X + 25Y ≤ 500
Raw materials: 5X + 2Y ≤ 300
Now we have the following constrained mathematical model for this problem:
MAXIMIZE:
SU
25X + 100Y (T otal proﬁt)
5X + 25Y ≤ 500
BJECT TO:
(Labor)
5X + 2Y ≤ 300 (Raw materials)
X, Y ≥ 0
(Nonnegativity)
(The condition that “X and Y are integers” is not added in Part I.)
Table 7.1 Production Data for Multilayer Actuators (MLAs) and Ultrasonic Motors
(USMs)
Gross
Proﬁt/Unit
Required
Labor/Unit
Gross
Proﬁt/Labor
Required
Material/Unit
MLA
$25
5h
$5/h
5g
USM
$100
25 h
$4/h
2g
Product
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7.1.2 Graphical Solution
Two constraints can be transformed as
⎛ 1⎞
Y ≤ − ⎜ ⎟ X + 20
⎝ 5⎠
and
Y ≤ −2.5X + 150
Taking into account X and Y nonnegative, the constraints exhibit a quadrilateral region shown in
gray in Figure 7.3. The coordinates for four corner points are obtained as (0,0), (0,20), (60,0), and
M .The red-star point M can be obtained from the equation
⎛ 1⎞
Y = − ⎜ ⎟ X + 20 = −2.5X + 150
⎝ 5⎠
as (56.52, 8.696). This point of the feasible region is called the extreme point, which is the intersection
of the limit on labor time and the limit for materials quantity.
Equating the total proﬁt to K (where 25X + 100Y = K ), we obtain
⎛ 1⎞
K
Y = −⎜ ⎟ X +
100
⎝ 4⎠
160
140
USM Quantity Y
120
100
Labor
80
Materials
60
40
K > 2283
20
0
Figure 7.3
K = 2283
K < 2283
0
20
M
40
60
80
MLA Quantity X
100
Graphical solution of the production schedule problem.
120
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By scanning this line upward and downward, we obtain the maximum K value. When K > 2283,
this line will not overlap with the quadrilateral region, while for K < 2283, the line overlaps with the
quadrilateral area. As you can clearly see, the K value becomes the maximum when this line crosses
the star point M (56.52, 8.696), that is, the extreme point. Now we can obtain the maximum proﬁt
$2283 for the MLA production of 56.52 units and the USM quantity of 8.696 units.
7.1.3 Excel Spreadsheet Solver
We will now trace the same solution by using the Solver function of the Excel spreadsheet. Solver
is an option found in the Tools menu. If you do not see Solver listed, you must check the Solver
Add-In box of the Add-Ins option under the Tools menu.
We will start from Figure 7.4, which shows how to assign the cells. To use Solver, I recommend
you designate cells to contain the following:
◾ Values of the decision variables (changing cells)—cells C5 and D5
◾ Value of the objective function (target cell)—cell E7
◾ Total value of the left-hand side of the constraints—cells E9 and E10
Also familiarize yourself with Excel’s SUM and SUMPRODUCT functions for this setting. Cell
E7 is total proﬁt, which is provided by the sum of MLA and USM proﬁts. It should be equal to
SUMPRODUCT(C5:D5,C7:D7), which means that this cell is computed by
(C5 × C7 + D5 × D7)
As you may recognize, if this cell E7 is copied to E9, the row numbers are automatically shifted to
(C7 × C9 + D7 × D9)
Figure 7.4
Microsoft Excel 2007 with Solver for an MMI production schedule.
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Because we sh ould n ot sh ift t he ro w number 5, w hich i s t he va riable u nit number ( X a nd Y )
row, we must ﬁ x the row number. For this purpose, we will use the “$” symbol. Thus, instead of
SUMPRODUCT(C5:D5,C7:D7), we type here:
SUMPRODUCT($C$5:$D$5,C7:D7)
The F4 f unction key at t he top of the keyboard is easy to u se for this exchange. Highlight only
the ﬁrst array (C5:D5) of the formula in the formula bar and press the F4 key; then you can insert
dollar signs.
Next, c opy E7 i nto E 9 a nd E10, so t hat yo u o btain t he n umbers 3 0 a nd 7 i n t hese c ells,
because the initial unit numbers in C5 and D5 are tentatively 1 for both.
We now click Solver in the Tools menu. This gives the dialog box shown in Figure 7.1. Note
that the dialog box is the same for both Excel 2003 and 2007.
Step 1: Set Target Cell. This is t he cell for t he objective f unction. With t he cursor in t he Set
Target Cell box, click on cell E7.
Step 2: E qual To. Problem t ype (maximization or minimization) is set. L eave t he button for
Max highlighted. (When you need to compute the minimum, choose “Min.”)
Step 3: Changing Cells. Changing cells are the cells that contain the decision variables. With the
cursor in the By Changing Cells box, highlight cells C5 and D5.
Step 4: Subject to the Constraints. To introduce the constraints, click the Add button in Subject
to t he Constraints. You will get t he Add Constraint dialog box shown in Figure 7.5. The
constraint options include ≤, =, ≥, i nt, a nd bin. The last t wo options re strict a c ell to b e
integer-valued or binary (0 or 1), respectively. With t he cursor in t he Cell Reference box,
highlight cells in E9 and E10. Click the direction as ≤ in this case. With the cursor in the
Constraint box, highlight cells G9 and G10. Click Add, if we have more constraints. If
ﬁnished, click OK.
Step 5: Options. Click Options to open the dialog box shown in Figure 7.6. It is important to
check both Assume Linear Model and Assume Non-Negative. Click OK.
Step 6: Solve. Figure 7.7 shows the completed Solver dialog box for the MMI product schedule.
To solve for the optimal solution, click Solve.
Step 7: Rep orts. O nce a n optimal solution i s found, t he S olver R esults d ialog b ox shows u p
(Figure 7.8). At the same time, Figure 7.4 is automatically changed to Figure 7.9. You can
ﬁnd that the solutions (quantity X = 56.52, quantity Y = 8.696, and the maximum proﬁt
$2283) are exactly the same in the calculation error range.
Two further reports are available: Answer Report and Sensitivity Report. Highlight one of them
and click OK. Note that these t wo reports are created on separate spreadsheets. No new dialog
boxes show up.
Answer Report: Click on the Answer Report tab at the bottom of the spreadsheet to access the
window shown in Figure 7.10. The A nswer Report includes three parts: Target Cell, Adjustable
Cells, and Constraints.
Target Cell provides the optimal value of the objective function in the Final Value column,
with the initial input ($125 in our trial). Similarly, the optimum values for the decision variables
are shown in the Final Value column of the Adjustable Cells section. In the Constraints section,
the Cell Value gives the total values of the left side of the constraints. The information entered in
the Constraint dialog box of Solver appears in the Formula column. The Slack column shows the
amount of slack for each constraint. Note that if the slack is 0, the word “Binding” is printed in
the Status column; “Not Binding” is printed when the slack is positive.
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Figure 7.5
Adding the Constraint dialog box in Solver.
Figure 7.6
Solver Options dialog box.
Figure 7.7
Solver Parameters dialog box.
Figure 7.8
Solver Results dialog box.
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Figure 7.9
Figure 7.10
Computed results for the optimal production schedule.
Answer report.
Sensitivity R eport: The S ensitivity R eport ( Figure 7.11) e xhibits i mportant i nformation c oncerning the eﬀects of changes to either an objective function coeﬃcient or a constraint value.
In the A djustable Cells, sensitivity analysis of an objectiv e function coeﬃ cient answers the
question, “K eeping all other factors the same, ho w much can an objectiv e function coeﬃ cient
change without changing the optimal solution (that is, without changing the production quantities
of X and Y )?” For example, the O bjective Coeﬃ cient for ML A (proﬁt per unit) is initially $25.
Without changing the extr eme point position, ho w much can w e change this number? I n other
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Figure 7.11
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Sensitivity report.
words, how much can we swing the slope of the objective function curve in Figure 7.3? The answer
is provided by Allowable Increase and Allowable Decrease. The MLA proﬁt can be changed fr om
$20 (= 25 − 5) to $250 ( = 25 + 225). This range is called range of optimality . Similarly, the gross
proﬁt for USM can be changed from $10 (= 100 − 90) to $125 (100 + 25). Note that the value of
the objective function (total proﬁt) will, of course, change, according to the unit proﬁt change.
Shadow price is a useful number fr om a practical operation vie wpoint. The shadow for a constraint is the change to the objectiv e function v alue per unit incr ease to its constraint v alue. For
example, MMI set the maximum labor (wor king time) as 500 h. I f MMI asks the r esearcher to
work one more hour (up to 501 h), ho w much will total proﬁt increase (a sort of “what-if ” question)? The answer can be found in the S hadow Price column as $3.91. I f this amount (additional
proﬁt) is higher than the additionally required cost (such as an extra portion of the overwork payment, after subtracting the regular salary rate, utility, etc.), MMI should move in this direction.
The range of feasibility is the range of v alues for a constraint v alue in which the shado w
prices for the constraints r emain unchanged. The range is pr ovided by Allowable Increase and
Allowable Decrease. In the labor constraint in MMI, the range will be from 300 h (= 500 − 200)
to 3750 h (= 500 + 3250), while in the materials constraint, fr om 40 g (= 300 − 260) to 500 g
(= 300 + 200).
7.1.4 Integer Model
MMI Example—Part II
As described in Part I, MMI produces MLAs and USMs every month. Table 7.1 summarized the
necessary data for producing MLAs and USMs. Though USMs generate a larger gross proﬁt per
unit, with a smaller amount of raw materials per unit than MLAs, USMs require a working period
ﬁve times longer than MLAs. MMI has two major constraints: (1) labor—total 500 h/month maximum (= 3 engineers × 40 h/week × 4.17 weeks/month), and (2) raw materials available—total
300 g/month maximum. MMI would like to determine the optimum production quantities (unit
number) X for MLAs and Y for USMs in order to maximize the gross proﬁt.
Part II seeks the solutions for nonnegative integer numbers of X and Y .
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Figure 7.12
Solver Options dialog box (integer analysis).
Solution—Part II
In Part I, we found that one solution (quantity X = 56.52, quantity Y = 8.696, and the maximum
proﬁt $2283) is the optimal solution. However, the product unit should be an integer, since we
cannot sell 0.696 unit of USM, in practice.
If we initially consider the graphical solution ﬁrst, the integer optimum solution should
exist around the MLA production of 56.52 units and the USM quantity of 8.696 units. Let us take
into account the following positions: (X,Y) = (56,8), (57,8), (58,8), (59,8); (53,9), (54,9), (55,9), and
(56,9). Only (56,8), (53,9), (54,9), and (55,9) pass the criteria: labor 5 X + 25 Y ≤ 500, and raw materials 5X + 2Y ≤ 300. Thus, it is obvious the maximum proﬁt can be obtained for the position (55,9).
Let us now solve the same question on Excel Solver by adding the integer constraint to the decision
variables (quantities X and Y). We will add one more condition to “Subject to the Constraints”:
$C$5:$D$5 = integer
as shown in Figure 7.12 (ﬁnd the difference from Figure 7.7). The solution can easily be found
(Figure 7.13). The Answer Report is shown in Figure 7.14, where the optimum production
schedule—55 units MLA, 9 units USM, and total proﬁt $2275—is the same as we analyzed
graphically. Note that there is Slack in Materials Constraint (7 g material remained), leading to a
“Not Binding” note in the Status column. Also note that there is a slight decrease in the total proﬁt
($2275) in comparison with the proﬁt obtained from the continuous parameter model.
It is important to know that the Excel Solver does not generate sensitivity analysis for the integer
analysis. Thus, I strongly recommend you use the continuous parameter analysis in parallel to the
integer analysis in order to obtain the sensitivity analysis.
7.1.5 Binary Model
MMI Example—Part III
MMI produces MLAs and USMs every month. Table 7.1 summarizes the necessary data for
producing MLAs and USMs. MMI has two major constraints: (1) labor—total 500 h/month
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Figure 7.13
Computed results for the optimal production schedule (integer analysis).
Figure 7.14
Answer report (integer analysis).
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maximum (= 3 engineers × 40 h/week × 4.17 weeks/month), and (2) raw materials available—
total 300 g/month maximum. MMI would like to determine the optimum production quantities
(unit number) X for MLAs and Y for USMs in order to maximize the gross proﬁt.
Part III seeks the solution for choosing only one product, either MLAs or USMs.
Appropriate u se of binary va riables c an help u s i n e xpressing c omparative relationships. To
illustrate, su ppose Z1 , Z2, a nd Z3 a re b inary va riables rep resenting w hether e ach o f t hree p roduction lines should be set ( Zi = 1) or not set ( Zi = 0). The following relationships can then be
expressed by these variables:
◾ At least two production lines must be set: Z1 + Z2 + Z3 ≥ 2
◾ If line 1 is set, line 2 must not be set: Z1 + Z2 ≤ 1
◾ If line 1 is set, line 2 must be set: Z1 − Z2 ≤ 0
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◾ One but not both of lines 1 and 2 must be set: Z1 + Z2 = 1
◾ Both or neither of lines 1 and 2 must be set: Z1 − Z2 = 0
◾ Total line construction budget cannot exceed $K, and the setting costs for lines 1, 2, and 3
are $A ,$B, and $C, respectively: AZ1 + BZ2 + CZ3 ≤ K
Binary variables can also be used to indicate restrictions in certain conditional situations. Suppose
X denotes the production amount in MLA line ( X ≥ 0). If the MLA line is set, there is no other
restriction on the value of X, but if it is not set, X must be 0. This relation can be expressed by
X ≤ M Z1
In this equation, M denotes an extremely large number such as 1020 (or 1E + 20) that does not
restrict the value of X if Z1 = 1. Because M is large, X ≤ M will not restrict the number of MLA
products. But if the MLA line is not set (Z1 = 0), the constraint becomes X ≤ 0, leading to X = 0;
that is, no product is made in the MLA line. In summary, the binary parameters can be used to
set On/Oﬀ or used for the selection of projects.
Solution—Part III
Now, we will apply the binary variables to the MMI production schedule. Because MMI production line space is limited, Barb Shay decided to set only one production line; that is, either
MLAs or USMs, not both. The problem is which to set—MLA or USM lines? We introduce
here Z1 and Z 2 for On/Off parameters for the MLA and USM production lines (C5 and D5,
respectively; see Figure 7.15). Then, introducing an extremely large number M (1E+20) in G5,
the products MZ1 and MZ 2 are prepared in H5 and I5 cells. Subject to the Constraints in the
Solver Parameter dialog box is set as in Figure 7.16. Three different points from Figure 7.12 are
as follows: (1) $C$5 and $D$5 are binary parameters; (2) $C$5 and $D$5 are equal to or smaller
than M × X and M × Y, respectively; and, most importantly, (3) Z1 + Z 2 = 1; that is, one but not
both production lines must be set. The results are shown in Figure 7.15 and its Answer Report
is in Figure 7.17. The computation “chose” the USM production (rather than the MLA production), if MMI needs to set only one production line. Twenty units of USMs generate the total
gross proﬁt of $2000.
Figure 7.15
Computed results for the optimal production schedule (binary model).