Supplement to Chapter 06 - Linear Programming
SUPPLEMENT TO CHAPTER 6:
LINEAR PROGRAMMING
Answers to Discussion and Review Questions
1.
2.
3.
4.
5.
6.
Linear programming is well-suited to an environment of certainty.
The “area of feasibility,” or feasible solution space is the set of all combinations of values of
the decision variables which satisfy the constraints. Hence, this area is determined by the
constraints.
Redundant constraints do not affect the feasible region for a linear programming problem.
Therefore, they can be dropped from a linear programming problem without affecting the
optimal solution.
An iso-cost line represents the set of all possible combinations of two inputs that will result in
a given cost. Likewise, an iso-profit line represents all of the possible combinations of two
outputs which will yield a given profit.
Sliding an objective function line towards the origin represents a decrease in its value (i.e.,
lower cost, profit, etc.). Sliding an objective function line away from the origin represents an
increase in its value.
a. Basic variable: In a linear programming solution, it is a variable not required to equal
zero.
b. Shadow price: It is the change in the value of the objective function per unit increase in a
constraint right hand side.
c. Range of feasibility: The range of values over which a constraint’s right hand side value
may vary without changing the optimal basic feasible solution.
d. Range of optimality: The range of values over which a variable’s objective function value
may vary without changing the current optimal basic feasible solution.
6S-1
Supplement to Chapter 06 - Linear Programming
Solution to Problems
1.
a.
1. The optimal values of the decision variables are x1 = 2, x2 = 9 and the optimal
objective function value = z = 35.
2. None of the constraints have any slack. Both constraints are binding.
3. Neither of the constraints have any surplus (there are no greater than or equal to
constraints).
4. No, there are no redundant constraints.
X2 18
16
14
12
Optimum
10
8
6
ProfitLine
4
Material
Labor
2
2
4
6
8
10
12
Simultaneous solution:
*2 (6x1 + 4 x2 = 48)
*1 (4 x1 + 8 x2 = 80)

(12x1 + 8x2 = 96) * –1
(4x1 + 8 x2 = 80)
6S-2
14
16
18
20
X1
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
–12x1
– 8x2
= –96
4x1
+ 8x2
= 80
4x1
+ 8x2
= 80
4(2)
+ 8x2
= 80
8x2
= 72
x2
= 9
–8x1
= –16
x1
=
2
z = 4x1 + 3x2 = 4(2) + 3(9) = 35
b. 1. The optimal values of the decision variables are x1 = 1.5, x2 = 6.25 and the optimal
objective function value z = 65.5.
2. None of the constraints have any slack.
3. The second constraint has a surplus of 15. (There are 15 lbs. of excess material).
4. No, there are no redundant constraints.
*R & T (Optimum)
10x1 + 4x2
x1 + 2x2
10x1 + 4x2
–2x1 – 4x2
8x1
R & S**
10x1 +
x1 +
10x1 +
–10x1 –
–
= 40
= 14
= 40
= –28
= 12
x1 = 1.5
4x2
6x2
4x2
60x2
56x2
= 40
= 24
= 40
= –240
= –200
x2 = 3.57
1.5 + 2x2 = 14
2x2 = 12.5
x1 + 21.42 = 24
x2 = 6.25
x1 = 2.58
x1 = (1.5) (2), x2 = (6.25) (10)
x1 = 3,
x2 = 62.5
x1 + x2 = 65.5
S & T***
x1 + 6x2 = 24
x1 + 2x2 = 14
x1 + 6x2 = 24
–x1 – 2x2 = –14
4x2 = 10
x2 =
2.5
x1 + 5 = 14
x1 =
9
x1 = (9) (2) = x2 = (2.5) (10)
x1 = 18
x2 = 25
* R = durability
** S = Strength
*** T = Time
6S-3
x1 = (2.58) (2), x2 = (3.57) (10)
x1 = 5.16,
x2 = 35.7
x1 + x2 = 40.86
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
X2
24
22
20
18
16
14
12
10
8
Optimum
6
4
2
R
S
T
0
0
2
4
6
8
10
12
14
16
18
20
22
24
X1
6S-4
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
c. 1. The optimal values of the decision variables are A = 24, B = 20 and the optimal
objective function value = z = 204.
2. The third constraint has a slack of 120. In other words, there are 120 hours of
unutilized labor hours.
3. There are no greater than or equal to constraints, therefore no surplus is possible.
4. No, there are no redundant constraints.
B
100
90
80
70
Material
60
50
40
Optimal Solution
30
20
Profit
Line
Labor
10
Machinery
10
20
30
40
50
60
70
80
90
100
A
At the intersection of Machinery and Material constraints:
*5
(20A + 6B = 600)
*–4 (25A + 20B = 1000)
100A
–100A
+ 30B =
(Material)
(Machinery)
3000
– 80B = –4000
–50B = –1000
B =
20
6S-5
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
From the material constraint
2.
20A
+ 6B
= 600
substituting B = 20
20A
+ 6(20)
= 600
20A = 480
A = 24
a. 1. The optimal values of the decision variables are: S = 8 and T = 20. The optimal
objective function value = z = 58.4.
2. Since all of the constraints are greater than or equal to type, none of the constraints
have any slack.
3. The third and fourth constraints have surpluses of 92 grams and 10 grams
respectively.
4. Yes, the third constraint is redundant. It does not affect the feasible region.
b. 1. The optimal values of the decision variables are: x1 = 4.2 and x2 = 1.6. The optimal
value of the objective function = z = 13.2.
2. Yes, the constraint F has a slack of 4.6.
3. No, there is no surplus.
4. No, there are no redundant constraints.
X2
12
11
10
9
8
D
7
6
5
4
E
3
2
F
Optimum
1
00
1
2
3
4
5
6S-6
6
7
8
9
10
11
12
X1
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
D&E
D&F
4x1 + 2x2
= 20
4x1 +
2x2 = 20
2x1 + 6x2
= 18
x1 +
2x2 = 12
4x1 + 2x2
= 20
4x1 +
2x2 = 20
–4x1 – 12x2
= –36
–x1 –
2x2 = –12
–10x2
= –16
3x1
x2 =
1.6
x1 =
4x1 + 3.2 = 20
8
2.67
2.67 + 2x2 = 12
4x1 = 16.8
3.
=
2x2 =
9.33
x1 =
4.2
x2 =
4.67
x1 = (2) (4.2) =
8.4
x1 = (2) (2.67) =
5.34
x2 = (3) (1.6) =
4.8
x2 = (3) (4.67) = 14.01
Total = 13.2
Total = 19.35
Maximize: $40H + $30W
Subject to:
fabrication 4H + 2W  600 hours
300
250
assembly 2H + 6W  480 hours
a. Optimum:
H = 132
W = 36
Z = $6,360
W
fabrication
200
150
b. 0,80: Z=$2,400
150,0: Z = $6,000
132,36: Z = $6,360 (optimum)
100
optimum
50
0

100
assembly
150
200
250
H
6S-7
Supplement to Chapter 06 - Linear Programming
D = 110
Solutions (continued)
4.
Peanuts cost $.60/lb.
Deluxe revenue is $2.90/lb.
Raisins cost $1.50/lb.
Standard revenue is $2.55 /lb.
Deluxe mix has 1/3 lb. peanuts, 2/3 lb. raisins. Hence, deluxe mix cost is
1
2
($.60) +
($1.50) = $1.20/lb.
3
3
The standard mix has ½ lb. peanuts and ½ lb. raisins. Hence, the standard mix cost is ½ ($.60)
+ ½ ($1.50) = $1.05/lb.
Profits are $2.90 – $1.20 = $1.70/lb. for deluxe and $2.55 – $1.05 = $1.50/lb. for the standard
Standard
mix. Thus, the objective function is:
220
Maximize: Z = $1.70D + $1.50S
200
Subject to:
180
2
1
raisins
D + S  90 lb.
160
3
2
raisins
140
1
1
peanuts
D + S  60 lb.
120
3
2
S = 110
D
 110 lb.
100
S  110 lb.
80
Optimum:

60
40
D = 90 lb.
peanuts
20
S = 60 lb.
0
Profit = $243
20
40 60 80 100 120 140 160 180 200 220 240
5.
Maximize: $1.50A + $1.20G
Deluxe
Grape
Subject to:
1200
sugar:
1.5A + 2.0G  1,200 cups
flour:
3.0A + 3.0G  2,100 cups
1000
time:
6.0A + 3.0G  3,600 min.
Optimum:
800
A = 500 pieces
600
G = 200 pieces
Revenue = $990
400

200
0
200
400
sugar
600
flour
time
800
1000
Apple
6S-8
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
Unused supplies
sugar:
1.5(500) + 2(200) = 1,150 cups used. Hence, 1,200 – 1,150 = 50 cups
remaining.
flour:
3.0(500) + 3.0(200) = 2,100 cups used. No flour remains.
time:
6.0(500) + 3.0(200) = 3,000 minutes. No time remains.
6.
a. The optimal value of the decision variables are: x1 = 4, x2 = 0, x3 = 18. The optimal value
of the objective function value = z = 106.
b. The optimal value of the decision variables are: x1 = 15, x2 = 10, x3 = 0. The optimal
value of the objective function value = z = 210.
7.
a. For problem 6a, the range of feasibility for the three constraints are as follows:
Constraint 1:
22 to infinity ()
Constraint 2:
10 to 47.5
Constraint 3:
20 to 45
b. For problem 6a, the range of optimality for the three coefficients of the objective function
are:
Variable 1 (x1):
2.5 to 15
Variable 2 (x2):
– to 10.6
Variable 3 (x3):
8.
1.333 to 8
a. For problem 6b, the range of feasibility for the three constraints are as follows:
Constraint 1:
20 to 26.6667
Constraint 2:
35 to 50
Constraint 3:
35 to 
b. For problem 6b, the range of optimality for the three coefficients of the objective function
are:
9.
Variable 1 (X1):
6 to 12
Variable 2 (X2):
5 to 10
Variable 3 (X3):
– to 20
The optimal value of the decision variables are: x1 = 0, x2 = 80, x3 = 50. The optimal value of
the objective function is z = 350. The range of optimality for the profit coefficient of each
variable is as follows:
x1:
– to 3.042
x2:
1.95 to 3.75
x3:
2 to 5
6S-9
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
10.
x1 = number of containers of orange juice
x2 = number of containers of grapefruit juice
x3 = number of containers of pineapple juice
x4 = number of containers of All-in-One
Orange Juice Grapefruit Juice
Revenue per qt.
$1.00
$.90
Pineapple Juice
$.80
All-in-One
$1.10
Cost per qt.
.50
.40
.35
.417
Profit per qt.
$.50
$.50
$.45
$.683
Maximize .50x1 + .50x2 + .45x3 + .683x4
s.t.
Orange juice:
Grapefr. juice:
Ratio:
+.33x4  1200 qt.
1x2
Pineapple juice:
Grapefr. cont.:
+.33x4  1600 qt.
1x1
1x3
+.33x4 
–.30x1 +.70x2 –.30x3 –.30x4 
5x1
800 qt.
0 cont.

0
x1, x2, x3, x4 
0
–7x3
The optimal values of the decision variables are: x1 = 800, x2 = 400, x4 = 2,424.24. The
optimal value of the objective function coefficient: z = 2,255.78.
6S-10
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
11.
x1 = qty. of chopping boards
x2 = qty. of knife holders
maximize
2x1
+ 6x2
s.t.
Cutting
1.4x1 + .8x2
Gluing
5x1
Finishing
12x1 + 3x2
 56 minutes
+ 13x2  650
x1, x2
 360
 0
x2
Opt. is: x1 = 0
x2 = 50
z = 300
120
Slack:
finishing
Cutting 16 minutes
Gluing
0
Finishing 210 minutes
70
50
Opt.
gluing
cutting
0
12.
30 40
130
x1
x1 = qty. ham spread
x2 = qty. deli spread
maximize
2x1
+ 4x2
(profit) or minimize 3x1 + 3x2 (cost)
s.t.
x2
mayo
1.4x1 + 1.0x2  70 lb.
mayo
1.4x1 + 1.0x2  112 lb.
ham
deli
x1
112
 20 pans
x2  18 pans
x1, x2  0
70
a. x1 = 37.14, x2 = 18, Cost = $165.42
b. x1 = 20, x2 = 84, Profit = $376.
mayo = 112
mayo = 70
18
0
6S-11
x1
20
50
80
Supplement to Chapter 06 - Linear Programming
Solutions (Continued)
13.
A = quantity of product A
B = quantity of product B
C = quantity of product C
A
Revenue
$80
Cost
B
$90
C
$70
Mat’l #1
2
x $ 5 = $10 1
x $ 5 = $ 5 6 x $ 5 = $30
Mat’l #2
3
x $ 4 = $12 5
x $ 4 = $20
Labor
3.2 x $10 = $32 1.5 x $10 = $15 2 x $10 = $20
Total
$54
$40
$50
Profit
$26
$50
$20
Maximize 26A + 50B + 20C (profit)
s.t.
Mat’l #1
2A + 1B + 6C  200 lb.
Mat’l #2
3A + 5B
 300 lb.
Labor
3.2A + 1.5B + 2.0C  150 hr.
A output
2/3A – 1/3 B – 1/3C  0
2A –
A/B
A
A
3B
=0
5
A, B, C  0
Solution: Optimal values of the decision variables are:
A = 18.75
B = 12.50
C = 25.00
Optimal value of the objective function is:
z = $1,612.50
14.
x1 = boxes of regular mix
x2 =
“mix” deluxe
x3 =
“mix” cashews
x4 =
“mix” raisins
x5 =
“mix” caramels
x6 =
“mix” chocolates
6S-12
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
maximize
.80x1 + .90x2 + .70x3 + .60x4 + .50x5 + .75x6
s.t.
cashews
.25x1 + .50x2 +
raisins
.25x1
caramels
.25x1
chocolate
.25x1 + .50x2
 120 lb./day
x3
+
 200 lb./day
x4
+
 100 lb./day
x5
+
x6  160 lb./day
boxes:
regular
 20 boxes
x1
deluxe
 20 boxes
x2
cashews
 20 boxes
x3
raisins
 20 boxes
x4
caramels
x5
 20 boxes
x6  20 boxes
chocolates
x1, x2, x3, x4, x5, x6  0
Solution:
15.
x1 = 320
x4 = 120
x2 = 40
x5 = 20
x3 = 20
x6 = 60
Z = 433
a. The first constraint (machine) and the third constraint (material) are binding because S1
and S3 are not in the solution (are not basic variables). Therefore as nonbasic variables,
they each have a value of zero. In other words, there are no excess machine hours or
materials.
b. The range of optimality for the objective function coefficient of product 3 is from 13.5 to
36. Therefore an increase from 15 to 22 would not change the value of the decision
variables. However, the objective function value would increase from 792 to 792 + 48 (22
– 15). Therefore the new value of z = 1128.
c. The range of optimality for the objective function coefficient of product 1 is from – to
22.2. Since 22 is within the range, the change would not affect the value of decision
variables. Since x1 is not a basic variable, the objective function value will not be affected
(we are not producing any units of product 1).
d. We have a slack of 56 hours (S2 = 56), and the range of feasibility lower limit for the
second constraint is 232. Therefore, reducing the available labor hours by 10 (288 – 10 =
278) will not affect the value of the decision variables. The objective function value will
not change either. However, there will be 10 hours less slack. Thus, the new value of S2 =
56 – 10 = 46.
6S-13
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
e. If no additional machine hours and materials are obtained, there would not be any change
in the profit (z). No change is allowed in the objective function value because all machine
hours and all materials are used (constraint 1 and constraint 3 are binding).
f. To determine if the changes are within the range for multiple changes, we first compute
the ratio of the amount of each change to the end of the range in the same direction.
1
1
For product 1, it is
=
= 9.8%
22.2 – 12
10.2
1
1
For product 2, it is
=
= 50%
20 – 18
2
1
1
For product 3, it is
=
= 4.76%
36 – 15
21
The sum of the ratios = .098 + .50 + .0476 = .646
Since .646  1.0, we conclude that these values are within the range. Therefore, the
optimal values of the decision variables will not change (x1 = 0, x2 = 4, x3 = 48). However,
the objective function value will change. The new objective function value = z = (19 x 4)
+ (16 x 48) = 844 or 792 + 4 + 48 = 844.
16.
Instructor Note: The RHS of the machine constraint should be 660 minutes.
a. The marginal value (shadow price) of a pound of bark is $1.50. This marginal value is in
effect from 510 lbs. to 750 lbs. of bark (range of feasibility for the first constraint right
hand side).
b. 1.50 per pound.
c. The marginal value (shadow price) of 1 labor hour is zero because we currently have 105
excess labor hours remaining. This marginal value is in effect from 375 hours to infinity.
d. We can not use any additional machine hours because we currently have 135 minutes of
excess machine time.
e. Maximum possible increase for pine bark constraint is 150 lbs. (750 – 600). Maximum
possible increase for storage constraint is 14.21 bags.
(1.50) (150) = $225 (expected increase in profit for pine bark)
(1.50) (14.21) = $21.32 (expected increase in profit for storage)
Therefore, add 150 pounds of pine bark.
f. The range of optimality for the objective function coefficient of chips (x3) is from 5.4 to 9.
Therefore an increase from $6 to $7 would not change the value of the decision variables.
However, the optimal objective function value (z) would increase from 1125 to 1125 +
1(75 units) = $1200.
g. To determine if the changes are within the range for multiple changes, we first compute
the ratio of the amount of each change to the end of the range in the same direction.
1
For chips (x3), it is
= .333
9–6
6S-14
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
.6
= .600
9–8
The sum of the ratios = .333 + .600 = .933
Since .933  1.0, we conclude that these values are within the range. Therefore, the
optimal values of the decision variables will not change (x1 = 75, x2 = 0, x3 = 75).
However, the optimal value of the objective function will change. The new z = (8.4 x 75)
+ (7 x 75) = $1,155.
h. To determine if the changes are within the range for multiple changes, we first compute
the ratio of the amount of each change to the end of the range in the same direction.
15
For pine barks (first constraint), it is
= .10
750 – 600
27
For machine time (second constraint), it is
= .36
600 – 525
5
For storage capacity (fourth constraint), it is
= .352
164.21 – 150
The sum of the ratios = .10 + .36 + .352 = 1.112
Since 1.112> 1.0, we conclude that these values are not within the range. Therefore, the
optimal values of the decision variables will not change (x1 = 75, x2 = 0, x3 = 75). The
optimal value of the objective function (z) will change.
For nuggets (x1), it is
Solution to Son, Ltd. Case
1.
Q = quantity of Product Q
L = quantity of labor
R = quantity of Product R
W = quantity of Product W
A = quantity of Material A
B = quantity of Material B
Maximize 122Q + 115R + 76W – 8L – 4A – 4B
subject to
Labor
5Q + 4R + 2W – L  0 hr.
Material A 2Q
+ 2R + 1/2W – A  0 lb.
Material B 1Q
+ 2W – B  0 lb.
Product R
Budget
 85 units
R
8L
+ 4A +
 $11,980
4B
All variables
Optimal:
R = 85
0
Labor =
1,000 hr.
W = 330 Mat A =
335 lb.
Mat B =
660 lb.
6S-15
Contribution = $22,875
Supplement to Chapter 06 - Linear Programming
2.
E = equal quantities of Q, R, and W
[E contribution is 122 + 115 + 76 = 313]
[An alternate approach would be T = total amount, with an average contribution of 313/3 =
104.333]
Maximize 313E – 8L – 4A – 4B
subject to
Labor
11E
–L
0
Mat’l A
4.5E
–A
0
Mat’l B
3E
–B
0
Product R
E
 85
Budget
8L + 4A + 4B
 $11,980
All variables
0
Optimal: E = 101.53 [i.e., Q = 101.53, R = 101.53, and W = 101.53.]
Labor = 1,116.78 hr.
Material A = 456.86 lb.
Material B = 304.58 lb.
Contribution = $19,798.89
The contribution is less by: $22,875 – $19,798.89 = $3,076.11
3.
5% waste on A:
4.5E – .95A  0
6S-16
Supplement to Chapter 06 - Linear Programming
Case: Custom Cabinets, Inc.
Problem Formulation:
Semi-custom Cabinets
A = quantity of Type A
B = quantity of Type B
C = quantity of Type C
D = quantity of Type D
Standard Cabinets
S10 = quantity of Type S10
S20 = quantity of Type S20
S30 = quantity of Type S30
S40 = quantity of Type S40
Max Z = 325A + 575B + 257C + 275D +175S10 + 210S20 + 260S30 + 230S40
s.t.
Wood:
125A + 160B + 140C + 200D + 60S10 + 110S20 + 200S30 + 180S40 < 400,000
Trim:
27A + 42B + 35C + 52D + 21S10 + 28S20 + 50S30 + 43S40 < 140,000
Granite:
175A + 243B
< 45,000
Solid Surface:
160C + 140D + 112S10
< 150,000
Laminate:
+ 135S20 + 254S30 + 176S40 < 400,000
Assembly:
37A + 57B + 30C + 35D + 21S10 + 25S20 + 30S30 + 27S40 < 100,000
Finish:
7A + 12B + 5C + 7D + 3S10 + 5S20 + 7S30 + 5S40 < 25,000
A > 117
B > 92
C > 130
D > 150
S10 > 475
S10 < 875
S20 > 363
S20 < 713
S30 > 510
S30 < 960
S40 > 412
S40 < 887
All variables > 0
Optimal Values
A = 117
B = 101
C = 193
D = 150
S10 = 875
S20 = 713
S30 = 535
S40 = 412
Z = $723,831.60
6S-17
Supplement to Chapter 06 - Linear Programming
Sensitivity Analysis
Both Assembly and Finishing have shadow prices equal to 0, so don’t work overtime.
Laminate also has a shadow price of 0, so don’t purchase additional laminate.
Wood has a shadow price of $1.30, and an allowable increase of 462,136.8 board feet. Purchase that
amount (500,000 board feet is available) at a cost per board foot of $.50, for a net increase in profit of
$.80 per board foot.
Using the additional wood, all decision variables values remain the same except S30, which increases
by 250 units to 849.6821. The revised profit increases by $260(250) = $65,000 to $776,617.36
Enrichment Module: The Simplex Method
The simplex method is a general-purpose linear-programming algorithm widely used to solve largescale problems. Although it lacks the intuitive appeal of the graphical approach, its ability to handle
problems with more than two decision variables makes it extremely valuable for solving problems
often encountered in operations management.
When teaching the simplex method, please consider the following points:
1.
A computer package for simplex is highly desirable because it permits assigning a range of
problems and concentrating on interpretation of solutions rather than on technique.
2.
Students should solve a few problems manually to gain some knowledge of what is actually
taking place during computations, and gain some insight as to why.
3.
Insight receives a boost when simplex and graphical solutions are compared for the same
problem.
4.
Computations are best done without calculators; students should keep numbers in fractional
form.
5.
Minimization, artificial variables and ranging can be skipped without seriously impairing
appreciation and understanding of the simplex method.
The simplex technique involves a series of iterations; successive improvements are made until an
optimal solution is achieved. The technique requires simple mathematical operations (addition,
subtraction, multiplication, and division), but the computations are lengthy and tedious, and the
slightest error can lead to a good deal of frustration. For these reasons, most users of the technique rely
on computers to handle the computations while they concentrate on the solutions. Still, some
familiarity with manual computations is helpful in understanding the simplex process. You will
discover that it is better not to use your calculator in working through these problems because
rounding can easily distort the results. Instead, it is better to work with numbers in fractional form.
Even though simplex can readily handle three or more decision variables, you will gain considerable
insight on the technique if we use a two-variable problem to illustrate it because you can compare
what is happening in the simplex calculations with a graphical solution to the problem.
Let’s consider the simplex solution to the following problem:
6S-18
Supplement to Chapter 06 - Linear Programming
Maximize
Subject to
Z=
4x1
+ 5x2
x1
+ 3x2
 12
4x1
+ 3x2
 24
x1, x2
0
The solution is shown graphically in Figure 1. Now let’s see how the simplex technique can be used to
obtain the solution.
Figure 1. Graphical Solution
X2
10
8
6
4X1 + 3X2 = 24
Objective
function
Optimum
4
2
X1 + 3X2 = 12
4X1 + 5X2 = 20
0
2
4
6
8
10
12
X1
The simplex technique involves generating a series of solutions in tabular form, called tableaus. By
inspecting the bottom row of each tableau, one can immediately tell if it represents the optimal
solution. Each tableau corresponds to a corner point of the feasible solution space. The first tableau
corresponds to the origin. Subsequent tableaus are developed by shifting to an adjacent corner point in
the direction that yields the highest rate of profit. This process continues as long as a positive rate of
profit exists. Thus, the process involves the following steps:
1.
Set up the initial tableau.
2.
Develop a revised tableau using the information contained in the first tableau.
3.
Inspect to see if it is optimum.
4.
Repeat steps 2 and 3 until no further improvement is possible.
Setting Up the Initial Tableau
Obtaining the initial tableau is a two-step process. First, we must rewrite the constraints to make them
equalities and modify the objective function slightly. Then we put this information into a table and
supply a few computations that are needed to complete the table.
Rewriting the objective function and constraints involves the addition of slack variables, one for each
constraint. Slack variables represent the amount of each resource that will not be used if the solution is
implemented. In the initial solution, with each of the real variables equal to zero, the solution consists
solely of slack. The constraints with slack added become equalities:
6S-19
Supplement to Chapter 06 - Linear Programming
1)
x1 + 3x2
2)
4x1 + 3x2
+ 1s1
= 12
+ 1s2
= 24
It is useful in setting up the table to represent each slack variable in every equation. Hence, we can
write these equations in an equivalent form:
1)
x1 + 3x2
+ 1s1
+ 0s2
= 12
2)
4x1 + 3x2
+ 0s1
+ 1s2
= 24
The objective function can be written in similar form:
Z = 4x1 + 5x2 + 0s1 + 0s2
The slack variables are given coefficients of zero in the objective function because they do not
produce any contributions to profits. Thus, the information above can be summarized as:
Maximize Z = 4x1 + 5x2 + 0s1 + 0s2
Subject to
1)
x1 + 3x2
+ 1s1
+ 0s2
= 12
2)
4x1 + 3x2
+ 0s1
+ 1s2
= 24
This forms the basis of our initial tableau, which is shown in Table 5S–1.
To complete the first tableau, we will need two additional rows, a Z row and a C – Z row. The Z row
values indicate the reduction in profit that would occur if one unit of the variable in that column were
added to the solution. The C – Z row shows the potential for increasing profit if one unit of the
variable in that column were added to the solution.
To compute the Z values, multiply the coefficients in each column by their respective row profit per
unit amounts, and sum within columns. To begin with, all values are zero:
C
0
x1
(1)0
x2
3(0)
s1
1(0)
s2
(0)0
Quantity
(12)0
(1)0
(24)0
0
0
0
4(0)
3(0)
0(0)
Z
0
0
0
The last value in the Z row indicates the total profit associated with a given solution (tableau). Since
the initial solution always consists of the slack variables, it is not surprising that profit is 0.
Values in the C – Z row are computed by subtracting the Z value in each column from the value of the
objective row for that column. Thus,
Variable row
x1
x2
s1
s2
Objective row (C)
4
5
0
0
Z
0
0
0
0
C–Z
4
6S-20
5
0
0
Supplement to Chapter 06 - Linear Programming
Table 1 Partial Initial Tableau
Profit per unit
for variables
in solution
C
Variables
in solution
Decision
Variables
4
5
0
0
Objective
row
x1
x2
s1
s2
Solution
quantity
0
s1
1
3
1
0
12
0
s2
4
3
0
1
24
The completed tableau is shown in Table 2.
The Test for Optimality
If all the values in the C – Z row of any tableau are zero or negative, the optimal solution has been
obtained. In this case, the C – Z row contains two positive values, 4 and 5, indicating that
improvement is possible.
Developing the Second Tableau
Values in the C – Z row reflect the profit potential for each unit of the variable in a given column. For
instance, the 4 indicates that each unit of variable x1 added to the solution will increase profits by $4.
Similarly, the 5 indicates that each unit of x2 will contribute $5 to profits. Given a choice between $4
per unit and $5 per unit, we select the larger and focus on that column, which means that x2 will come
into the solution. Now we must determine which variable will leave the solution. (At each tableau, one
variable will come into the solution, and one will go out of solution, keeping the number of variables
in the solution constant. Note that the number of variables in the solution must always equal the
number of constraints. Thus, since this problem has two constraints, all solutions will have two
variables.)
To determine which variable will leave the solution, we use the numbers in the body of the table in the
column of the entering variable (i.e., 3 and 3). These are called row pivot values. Divide each one into
the corresponding solution quantity amount, as shown in Table 3. The smaller of these two ratios
indicates the variable that will leave the solution. Thus, variable s1 will leave and be replaced with x2.
In graphical terms, we have moved up the x2 axis to the next corner point. By determining the smallest
ratio, we have found which constraint is the most limiting. In Figure 1, note that the two constraints
intersect the x2 axis at 4 and 8, the two row ratios we have just computed. The second tableau will
describe the corner point where x2 = 4 and x1 = 0; it will indicate the profits and quantities associated
with that corner point. It will also reveal if the corner point is an optimum, or if we must develop
another tableau.
6S-21
Supplement to Chapter 06 - Linear Programming
Table 2 Completed Initial Tableau.
C
4
5
0
0
0
Variables
in solution
s1
x1
1
x2
3
s1
1
s2
0
0
s2
4
3
0
1
24
Z
0
0
0
0
0
C–Z
4
5
0
0

C
Solution
quantity
12
4
5
0
0
0
Variables
in solution
s1
x1
1
x2
3
s1
1
s2
0
Solution
quantity
12/3 = 4
0
s2
4
3
0
1
24/3 = 8
Z
0
0
0
0
0
C–Z
4
5
0
0
 Smallest positive
ratio

Largest
positive
At this point we can begin to develop the second tableau. The row of the leaving variable will be
transformed into the new pivot row of the second tableau. This will serve as a foundation on which to
develop the other rows. To obtain this new pivot row, we simply divide each element in the s1 row by
the row pivot value (intersection of the entering column and leaving row), which is 3. The resulting
numbers are:
Pivot-row value
x1
x2
s1
s2
Solution
quantity
1/3
1
1/3
0
4
These numbers become the new x2 row of the second tableau.
The pivot-row numbers are used to compute the values for the other constraint rows (in this instance,
the only other constraint row is the s2 row). The procedure is:
1.
Find the value that is at the intersection of the constraint row (i.e., the s2 row) and the entering
variable column. It is 3.
2.
Multiply each value in the new pivot row by this value.
3.
Subtract the resulting values, column by column, from the current row values.
6S-22
Supplement to Chapter 06 - Linear Programming
x1
4
Current value:
–3 x (pivot row)
x2
3
–3(1/3)
New row value
s1
0
–3(1)
3
–3(1/3)
–1
0
s2
1
Quantity
24
–3(0)
–3(4)
1
12
The two new rows are shown in Table 4. The new Z row can now be computed. Multiply the row unit
profits and the coefficients in each column for each row. Sum the results within each column. Thus,
Row
x2
Profit
5
s1
0
New Z row
x1
5(1/3)
x2
5(1)
s1
5(1/3)
s2
5(0)
Quantity
5(4)
0(3)
0(0)
0(–1)
0(1)
0(12)
5/3
5
5/3
0
20
Next, we compute the C – Z row:
C
x1
4
x2
5
s1
0
s2
0
Z
5/3
5
5/3
0
C–Z
7/3
0
–5/3
0
Table 4 partially completed second tableau
C
5
Variables
in solution
x2
0
s2
4
5
0
0
x1
1/3
x2
3
s1
1/3
s2
0
3
0
–1
1
4
5
0
0
x1
1/3
x2
1
s1
1/3
s2
0
Solution
quantity
4
Solution
quantity
4
12
Table 5 completed second tableau
C
5
Variables
in solution
x2
0
s2
3
0
–1
1
12
Z
5/3
5
5/3
0
20
C–Z
7/3
0
–5/3
0
6S-23
Supplement to Chapter 06 - Linear Programming
The completed second tableau is shown in Table 5. It tells us that at this point 4 units of variable x2 are
the most we can make (see column Solution quantity, row x2) and that the profit associated with x2 =
4, x1= 0 is $20 (see row Z, column Solution quantity).
The fact that there is a positive value in the C – Z row tells us that this is not the optimal solution.
Consequently, we must develop another tableau.
Developing the Third Tableau
The third tableau will be developed in the same manner as the previous one.
1.
Determine the entering variable: Find the column with the largest positive value in the C – Z
row (7/3, in the x1 column).
2.
Determine the leaving variable: Divide the solution quantity in each row by the row pivot.
Hence,
4
= 12
12/3 = 4
1/3
The smaller ratio indicates the leaving variable, s2. See Table 5S–6.
3.
Divide each value in the row of the leaving variable by the row pivot value (3) to obtain the
new pivot-row values:
Current value
x1
3
x2
0
s1
–1
s2
1
Quantity
12
New pivot-row value
1
0/3
–1/3
1/3
12/3 = 4
5
0
Table 6 Leaving/Entering Variables
C
4
Variables
in solution
0
x1
x2
s1
s2
Solution
quantity
5
x2
1/3
1
1/3
0
4
1/3
0
s2
3
0
–1
1
12/3 = 4
Z
5/3
5
5/3
0
20
C–Z
7/3
0
–5/3
0

Entering
variable
is x1
6S-24
= 12
Leaving
 variable
is s2
Supplement to Chapter 06 - Linear Programming
4.
Compute values for the x2 row: Multiply each new pivot-row value by the x2 row pivot value
(i.e., 1/3) and subtract the product from corresponding current values. Thus,
x1
1/3
Current value:
–1/3 x (pivot row)
x2
1
–1/3(1)
New row value
s1
Quantity
4
–1/3(1/3)
–1/3(4)
1/3
–1/3(0)
0
s2
0
–1/3(–1/3)
1
–1/9
4/9
8/3
At this point, it will be useful to consider the tableaus in relation to a graph of the feasible solution
space. This is shown in Figure 2.
5. Compute new Z row values. Note that now variable x1 has been added to the solution mix; that
row’s unit profit is $4.
Row
x2
Profit
$5
x1
5(0)
x2
5(1)
s1
5(4/9)
s2
5(–1/9)
x1
$4
4(1)
4(0)
4(–1/3)
4(1/3)
4(4)
4
5
8/9
7/9
88/3
New Z row
6.
Compute the C – Z row values:
C
x1
4
x2
5
s1
0
s2
0
Z
4
5
8/9
7/9
C–Z
0
0
–8/9
–7/9
Figure 2 Graphical Solution and Simplex Tableaus
X2
10
8
6
3rd tableau
4
2nd tableau
2
1st tableau
0
2
4
6
6S-25
8
10
12
X1
Quantity
5(8/3)
Supplement to Chapter 06 - Linear Programming
Table 7. Optimal Solution
C
4
5
0
0
5
Variables
in solution
x2
x1
0
x2
1
s1
s2
4/9 –1/9
Solution
quantity
8/3
4
x1
1
0
–1/3 1/3
4
Z
4
5
8/9 7/9
88/3
C–Z
0
0
–8/9 –7/9
The resulting values of the third tableau are shown in Table 7. Note that each of the C – Z
values is either 0 or negative, indicating that this is the final solution. The optimal values of x1
and x2 are indicated in the quantity column: x2 = 8/3, or 2 2/3, and x1 = 4. (The x2 quantity is
in the x2 row and the x1 quantity in the x1 row.) Total profit is 88/3, or 29.33 (quantity
column, Z row).
Handling  and  Constraints
Up to this point, we have worked with  constraints. Constraints that involve equalities and 
constraints are handled in a slightly different way.
When an equality constraint is present, use of the simplex method requires addition of an artificial
variable. The purpose of such variables is merely to permit development of an initial solution. For
example, the equalities
(1) 7x1 + 4x2 = 65
(2) 5x1 + 3x2 = 40
would be rewritten in the following manner using artificial variables a1 and a2:
(1) 7x1 + 4x2 + 1a1 + 0a2 = 65
(2) 5x1 + 3x2 + 0a1 + 1a2 = 40
Slack variables would not be added. The objective function, say Z = 2x1 + 3x2, would be rewritten as:
Z = 2x1 + 3x2 + Ma1 + Ma2
where
M = A large number (e.g., 999)
Since the artificial variables are not desired in the final solution, selecting a large value of M (much
larger than the other objective coefficients) will insure their deletion during the solution process.
For  constraints, surplus variables must be subtracted instead of added to each constraint. For
example, the constraints
(1) 3x1 + 2x2 + 4x3  80
(2) 5x1 + 4x2 + x3  70
(3) 2x1 + 8x2 + 2x3  68
6S-26
Supplement to Chapter 06 - Linear Programming
would be rewritten as equalities:
(1) 3x1 + 2x2 + 4x3 – 1s1 – 0s2 – 0s3  80
(2) 5x1 + 4x2 + x3 – 0s1 – 1s2 – 0s3  70
(3) 2x1 + 8x2 + 2x3 – 0s1 – 0s2 – 1s3  68
As equalities, each constraint must then be adjusted by inclusion of an artificial variable. The final
result looks like this:
(1) 3x1 + 2x2 + 4x3 – 1s1 – 0s2 – 0s3 + 1a1 + 0a2 + 0a3  80
(2) 5x1 + 4x2 + x3 – 0s1 – 1s2 – 0s3 + 0a1 + 1a2 + 0a3  70
(3) 2x1 + 8x2 + 2x3 – 0s1 – 0s2 – 1s3 + 0a1 + 0a2 + 1a3  68
If the objective function happened to be
5x1 + 2x2 + 7x3
it would become
5x1 + 2x2 + 7x3 + 0s1 + 0s2 + 0s3 + Ma1 + Ma2 + Ma3
Summary of Maximization Procedure
The main steps in solving a maximization problem with only  constraints using the simplex algorithm
are these:
1. Set up the initial tableau.
a. Rewrite the constraints so that they become equalities; add a slack variable to each constraint.
b. Rewrite the objective function to include the slack variables. Give slack variables coefficients
of 0.
c. Put the objective coefficients and constraint coefficients into tableau form.
d. Compute values for the Z row; multiply the values in each constraint row by the row’s C value.
Add the results within each column.
e. Compute values for the C – Z row.
2. Set up subsequent tableaus.
a. Determine the entering variable (the largest positive value in the C– Z row). If a tie exists,
choose one column arbitrarily.
b. Determine the leaving variable: Divide each constraint row’s solution quantity by the row’s
pivot value; the smallest positive ratio indicates the leaving variable. If a tie occurs, divide the
values in each row by the row pivot value, beginning with slack columns and then other
columns, moving from left to right. The leaving variable is indicated by the lowest ratio in the
first column with unequal ratios.
c. Form the new pivot row of the next tableau: Divide each number in the leaving row by the
row’s pivot value. Enter these values in the next tableau in the same row positions.
d. Compute new values for remaining constraint rows: For each row, multiply the values in the
new pivot row by the constraint row’s pivot value, and subtract the resulting values, column by
column, from the original row values. Enter these in the new tableau in the same positions as
the original row.
e. Compute values for Z and C – Z rows.
f. Check to see if any values in the C – Z row are positive; if they are, repeat 2a–2f. Otherwise,
the optimal solution has been obtained.
6S-27
Supplement to Chapter 06 - Linear Programming
Minimization Problems
The simplex method handles minimization problems in essentially the same way it handles
maximization problems. However, there are a few differences. One is the need to adjust for 
constraints, which requires both artificial variables and surplus variables. This tends to make manual
solution more involved. A second major difference is the test for the optimum: A solution is optimal if
there are no negative values in the C – Z row.
Example
Solve the following problem for the quantities of x1 and x2 that will minimize cost.
Minimize
Z
= 12x1 + 10x2
x1 + 4x2
8
3x1 + 2x2
6
x1, x2
0
Subject to
Solution to example
1. Rewrite the constraints so that they are in the proper form:
x1 + 4x2  8 becomes x1 + 4x2 – 1s1 – 0s2 + 1a1 + 0a2 = 8
3x1 + 2x2  6 becomes 3x1 + 2x2 – 0s1 – 1s2 + 0a1 + 1a2 = 6
2. Rewrite the objective function (coefficients of C row):
12x1 + 10x2 + 0s1 + 0s2 + 999a1 + 999a2
3. Compute values for rows Z and C – Z:
C
999
x1
1(999)
x2
4(999)
s1
–1(999)
s2
0(999)
a1
1(999)
a2
0(999)
Quantity
8(999)
999
3(999)
2(999)
0(999)
–1(999)
0(999)
1(999)
6(999)
Z
3,996
5,994
–999
–999
999
999
13,986
C–Z
–3,984
–5,984
999
999
0
0
4. Set up the initial tableau. (Note that the initial solution has all artificial variables.)
C
12
10
0
0
999
999
999
Variables
in solution
a1
x1
1
x2
4
s1
–1
s2
0
a1
1
a2
0
Solution
Quantity
8
999
a2
3
2
0
–1
0
1
6
Z
3,996
5,994
999
999
13,986
0
0
C–Z
–3,984 –5,984
–999 –999
999
999
5. Find the entering variable (largest negative C – Z value: x2 column) and leaving variable
(smaller of 8/4 = 2 and 6/2 =3; hence, row a1).
6S-28
Supplement to Chapter 06 - Linear Programming
6. Divide each number in the leaving row by the pivot value (4, in this case) to obtain values for
the new pivot row of the second tableau:
1/4
–1/4
4/4 = 1
0/4
1/4
0/4
8/4 = 2
7. Compute values for other rows; a2 is:
x1
3
x2
2
–2 x (new pivot row)
–2/4
New row
10/4
Current value
s1
0
s2
–1
a1
0
a2
1
Quantity
6
–2
2/4
–0/4
–2/4
–0/4
–4
0
+2/4
–1
–2/4
1
2
8. Compute a new Z row:
Row
x2
Cost
10
x1
10(1/4)
x2
10(1)
s1
10(–1/4)
s2
10(0)
a1
10(1/4)
a2
10(0)
Quantity
10(2)
a2
999
999(10/4)
999(0)
999(2/4)
999(–1)
999(–2/4)
999(1)
999(2)
2,500
10
497
–999
–497
999
2,018
Z
9. Compute the C – Z row:
C
x1
12
x2
10
Z
2,500
C–Z
–2,488
s1
0
s2
0
a1
999
a2
999
10
497
–999
–497
999
0
–497
999
1,496
0
10. Set up the second tableau:
C
10
12
10
0
0
999
999
Variables
in solution
a1
x1
1/4
x2
1
s1
–1/4
s2
0
a1
1/4
a2
0
Solution
Quantity
2
a2
10/4
0
2/4
–1
–2/4
1
2
Z
2,500
10
497
–999 –497
999
2,018
C–Z
–2,488
0
–497
999
999
1,496
0
11. Repeat the process.
a. Check for optimality: It is not optimum because of negatives in C – Z row.
b. Determine the entering variable: The largest negative is in column x1.
c. Determine the leaving variable: 2/(1/4) = 8, 2/(10/4) = 0.8. Therefore, it is row a2.
d. Find new pivot-row value using the pivot value of 10/4:
1
0
0.2
–0.4
–0.2
0.4
6S-29
0.8
Supplement to Chapter 06 - Linear Programming
e. Determine values for new x2 row:
0
f.
–0.3
1
0.1
–0.1
0.3
1.8
Determine new values for row Z:
Row
x2
Cost
10
x1
10(0)
x2
10(1)
s1
10(–0.3)
s2
10(0.1)
a1
10(0.3)
a2
10(–0.1)
Quantity
10(1.8)
x1
12
12(1)
12(0)
12(0.2)
12(–0.4)
12(–0.2)
12(0.4)
12(0.8)
12
10
–0.6
–3.8
0.6
3.8
27.6
Z
g. Determine values for the C – Z row:
C
x1
12
x2
10
s1
0
s2
0
a1
999
a2
999
Z
12
10
–0.6
–3.8
0.6
3.8
C–Z
0
0
0.6
3.8
998.4
995.2
h. Set up the next tableau. Since no C – Z values are negative, the solution is optimal. Hence,
x1 = 0.8, x2 = 1.8, and minimum cost is 27.60.
C
12
10
0
0
999
999
10
Variables
in solution
a1
x1
0
x2
1
s1
–0.3
s2
0.1
a1
0.3
a2
–0.1
Quantity
1.8
12
a2
1
0
0.2
–0.4
–0.2
0.4
0.8
Z
12
10
–0.6
–3.8
0.6
3.8
27.6
C–Z
0
0
0.6
3.8
6S-30
998.4 995.2
Supplement to Chapter 06 - Linear Programming
Problems for the enrichment module (simplex)
1.
Given this information:
Maximize
Z = 10.50x + 11.75y + 10.80z
Subject to
Cutting
5x + 12y + 8z  1,400 minutes
Stapling
7x + 9y + 9z  1,250 minutes
Wrapping
4x + 3y + 6z  720 minutes
x, y, z  0
Solve for the quantities of products x, y, and z that will maximize revenue.
2.
Use the simplex method to solve these problems:
a.
Minimize
Z = 21x1 + 18x2
Subject to
(1) 5x1 + 10x2  100
(2) 2x1 + 1x2  10
x1, x2  0
b.
Minimize
Subject to
Z = 2x + 5y + 3z
(1) 16x + 10y + 18z  340
(2) 11x + 12y + 13z  300
(3) 2x + 6y + 5z  120
x, y, z  0
3.
Use the simplex method to solve the following problem.
Minimize
Z = 3x1 + 4x2 + 8x3
Subject to
2x1 + x2  6
x2 + 2x3  4
x1, x2, x3  0
4.
Use the simplex method to solve the following problem.
Maximize
Z = 8x1 + 2x2
Subject to
4x1 + 5x2  20
2x1 + 6x2  18
x1, x2  0
Note: Row operations in problems 3 and 4 are computationally easy.
6S-31
Supplement to Chapter 06 - Linear Programming
Solutions-Enrichment Module (SIMPLEX)
1.
C
10.5
11.75
10.80
0
0
0
Var
x
y
z
S1
S2
S3
0
S1
5
12
8
1
0
0
1,400
116.67
0
S2
7
9
9
0
1
0
1,250
138.89
4
S3
4
3
6
0
0
1
720
0
10.80
0
0
0
0
0
0
0
Z
C–Z
C
Var
0
10.5
0
11.75
x
y
z
S1
S2
S3
bi
bi
11.45 y
0
S2
5/12
13/4
1
0
2/3
3
1/12
–3/4
0
1
0
0
0
11/4
0
4
–1/4
0
1
0
0
0
0
1,370.83
S2
S3
bi
S3
240
ratio
1,400/12 280
200
61.54
370
134.54
Z
C–Z
4.896 11.75
5.604 0
Var
x
y
z
11.75 y
0
1
11/39
7/39
–5/39
0
91.026
10.5
x
1
0
12/13
–3/13
4/13
0
61.54
0
S3
0
0
19/13
5/13
–11/13
1
2610/13 522
11.75
0
13.01 –0.314 1.724
–2.206 0.314 –1.724
0
0
1,715.73
C
7.833 0.979
2.967 –0.979
ratio
S1
Z
C–Z
10.5
0
C
Var
x
y
S1
S2
S3
bi
0
10.5
S1
x
0
1
39/7
9/7
11/7
117/91
1
0
–5/7
–1/7
0
0
507.14
178.57
0
S3
0
–15/7
78/91
0
–4/7
1
5.72
10.5
0
13.5
–1.75
13.5
–2.7
0
0
1.5
–1.5
0
0
1,874.99
Z
C–Z
z
Optimal solution is x = 178.57, y = 0, z = 0, and optimal solution = 1874.9
6S-32
ratio
507.1
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
2.
a. Minimize Z =
21x1 + 18x2
5x1 + 10x2 + A1 – S1
s.t.
2x1 +
C
I.
II.
= 100
+ A2 – S2 = 10
1x2
21
18
M
0
M
0
C Var
x1
x2
A1
S1
A2
S2
bi
M A1
5
10
1
–1
0
0
100
10
M A2
2
1
0
0
1
–1
10
10
M
–0
–M
M
M
–0
–M
M
110M
M
Z
C–Z
7M
11M
[21–7M] [18–11M]
C
ratio
21
18
M
0
M
0
C Var
x1
x2
A1
S1
A2
S2
bi
ratio
18 x2
0.5
1
0.1
–0.1
0
0
10
20
M A2
1.5
0
–0.1
0.1
1 –1
0
0
M –M
–0 M
180
Z
C–Z
[1.5M+9]
[12–1.5M]
18
0
C
21
18
0
III. C Var
[1.8–0.1M] [0.1M–1.8]
[1.1M–1.8] [1.8–.1M]
x1
x2
18 x2
0
1
–0.1333
0.333
10
21 x1
1
0
+0.0667
–0.667
0
21
0
18
0
–0.99999
0.99999
–8.000
8.000
180
Z
C–Z
S1
0
S2
The optimal solution: x1 = 0; x2 = 10; Z = 180
6S-33
bi
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
2. b.
I.
C
2
5
3
M
0
M
0
M
0
Var
x
y
z
A1
S1
A2
S2
A3
S3
bi
M
M
A1
A2
16
11
10
12
18
13
1
0
–1
0
0
1
0
–1
0
0
0
0
340
300
M
A3
2
6
5
0
0
0
0
+1
–1
120
M
0
–M
M
M
0
–M
M
M
0
–M
M
Z
29M
28M
36M
C–Z [–29M+2] [–28M+5] [–36M+3]
II.
C
Var
x
y
z
3
M
Z
A2
.8889
–.5556
.5556
4.778
1
0
.0556
–.722
M
A3
3.222*
0
–.2778
[8M+1.7]
[–8M+3.3]
3
0
[–M+.17] [M–.17]
[2M–47] [–M+.17]
–2.444
Z
[–3M+2.7]
C–Z [+3M+.7]
III. C
Var
x
y
z
3
M
z
A2
1.31
3.069*
0
0
1
0
5
Y
–0.7586
1
5
0
Z
[3M+.138]
C–Z [–3.1M+1.86]
IV. C
V.
A1
S1
S1
A2
S2
A3
S3
bi
–.0556
.722
0
1
0
–1
0
0
0
0
18.89
54.44
.2778
0
0
1
–1
25.56
M
0
–M
M
M
0
–M
M
A2
S2
–.1034
.3103
0
1
0
–1
.1724
1.483
0
0.08621
0
0
–0.3103
3
0
[.3M+.121]
[–.3M–.121]
0
–M
M
Var
x
y
z
S1
3
2
z
x
0
1
0
0
1
0
–0.236
.1011
5
y
0
1
0
Z
C–Z
2
0
5
0
3
0
S2
C
Var
x
y
z
S1
3
0
z
S1
2.333
9.889
0
0
1
0
0
1
5
y
–1.611
1
0
0
Z
C–Z
–1.06
3.056
5
0
3
0
0
0
.1629
6S-34
S3
[1.5M–1.03]
[–1.5M+1.03]
S3
.427
–.3258
–.2472
–0.6067
0.6067
0.309
–0.309
760M
–0.4607
.4831
.05618
–0.1348
0.1348
S2
S3
–0.3333
–3.222
.6667
4.778
.2778*
+0.3889
–0.3889
–0.7222
–1.6111
1.6111
80M+56.7
bi
14.48
16.55
7.931
[16.55M+83.1]
bi
7.416
5.393
12.02
93.15
bi
20
53.33
3.333
76.67
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
VI. C
Var
x
y
z
S1
S2
S3
bi
3
z
.4
1.2
1
0
0
–0.200
24
0
S1
–8.8
11.6
0
1
0
–3.6
92
0
S2
–5.8
3.6
0
0
1
–2.6
23
1.2
3.6
3
0
0
–0.6
72
.8
1.4
0
0
0
.6
Z
C–Z
Optimal solution is: x = 0; y = 0; z = 24 and Z = 72.0
3.
C
M
M
3
4
8
0
0
M
M
Var
x1
x2
x3
S1
S2
A1
A2
A1
A2
2
0
1
1
0
2
–1
0
0
–1
1
0
0
1
6
4
Zj
Cj–Zj
2M
3–2M
2M
4–2M
2M
8–2M
–M
M
–M
M
M
0
M
0
10M
3
4
8
0
0
Var
x1
x2
x3
S1
S2
x1
A2
1
0
½
1
0
2
–½
0
0
–1
Zj
Cj–Zj
3
0
2M
8–2M
–3/2
3/2
C
3
M
C
3
8
–M
M
3
4
4
8
0
0
Var
x1
x2
x3
S1
S2
x1
x3
1
0
½
½
0
1
–½
0
0
–½
3
2
3
0
11/2
–3/2
8
0
–3/2
3/2
–4
4
25
3
4
8
0
0
Var
x1
x2
x3
S1
S2
x1
x2
1
0
0
1
–1
2
–½
0
½
–1
1
4
3
0
4
0
5
3
–3/2
3/2
–5/2
5/2
19
C
Zj
Cj–Zj
The optimal solution is: x1 = 1; x2 = 4; x3 = 0 and Z = 19.
6S-35
bi/aij
3½=6
41=4
4M+9
3
Zj
Cj–Zj
3
4
M–3/2
5/2 –M
bi
bi
bi
bi
bi/aij
3½=6
2½=4
bi/aij
6/2 = 3
–
Supplement to Chapter 06 - Linear Programming
Solutions (continued)
4.
Cj
0
0
8
2
0
0
Var
x1
x2
S1
S2
S1
S2
4
2
5
6
1
0
0
1
20
18
Zj
Cj–Zj
0
8
0
2
0
0
0
0
0
Var
x1
x2
S1
S2
x1
S2
1
0
Zj
Cj–Zj
8
0
bi
Cj
8
0
5/4
7/2
10
–8
bi
¼
–1/2
0
1
5
8
2
–2
0
0
40
Optimal solution is x1 = 5, x2 = 0, S1 = 0, S2 = 8 and Z = 40.
6S-36