LINEAR PROGRAMMING
1. Two models of color TV A and B are produced by Davao Electronics
Corporation. The company is in the market to make money or to maximize
profit. The profit realized is P3000 from Set A and P2500 from each unit of
Set B. The limitations are as follows:
(1) Availability of only 40 hours of labor each day in the production
department,
(2) A daily availability of only 45 hours of machine time, a machine
constraint,
(3) Inability to sell more than 12 units of Set A each day.
Each set of Model A requires 2 hours of labor whereas each set of B requires
only 1 hour; each set of A requires 1 hour of processing time, while each set
of B, 3 hours. Formulate the problem.
2. In preparing Sungold paint, it is required that the paint have a brilliance
rating of at least 300 degrees and a hue level of at least 250 degrees. The 2
ingredients Alpha and Beta determine brilliance and hew levels. Both Alpha
and Beta contribute equally to the brilliance rating, one ounce of either
producing one degree of brilliance in one case of paint. However, the hue is
controlled entirely by the amount of Alpha, one ounce of it producing three
degrees of hue in one case of paint. The cost of Alpha is P45.00 per ounce,
and the cost of Beta is P12.00 per ounce. Assuming that the objective is to
minimize the cost of the resources, then he problem is to find the quantity of
Alpha and Beta to be used in the preparation of each case of paint.
3. Vitamin Requirement Problem. The minimum daily requirement of
Vitamin D for patients following surgery is 180 units, while the maximum
amount of B12 that they may receive is 150 units. The hospital dietician is
considering liver and steak for the daily diet. Each kilo of liver costs
P150.00, and each kilo of steak costs P170.00. Two grams of liver and 5
grams of steak provide the daily minimum requirement of Vitamin D. A kilo
of steak yields 480 units of Vitamin B12 and an undetermined amount of
Vitamin D, which may be determined. A gram of liver yields 30 units of D
and 20 units of B12. Find the least cost diet per patient following surgery.
4. Painthaus Company advertises its weekly sales in newspapers, television,
and radio. Each peso spent in advertising in newspaper is estimated to reach
an exposure of 12 buying customers, each peso in TV reaches an exposure of
15 buying customers, and each peso in radio reaches an exposure of 10
buying customers. The company has an agreement with all three media
services according to which it will spend not less than 20 percent of its total
budget. Further, it is agreed that the combined newspaper and TV budget will
not be larger than 3 times the radio budget. The company has just decided to
spend more than P17,000.00 on advertising. How much should the company
budget for each medium be if it is interested in reaching as many buying
customers a s possible?
5. The City Council has allocated P500,000.00 for projects designed to
prevent automobile accidents. Four proposals were submitted: (a) TV
advertisement (b) teenage safety education (c) improved airbags (d)
enforcement of driving laws. The projects are expected to result in the
reduction of both fatalities and property damages as follows:
Project
a
b
c
d
Expected Reduction in
fatalities per P1,000.00
0.25
0.32
0.15
0.28
Average Damage Averted
per P1,000.00
P300
500
0
250
The council has decided that no single project will be awarded more than
P250,000.00. They also wish to award at least P50,000.00 for teenage
education. They want to award at least P1 for improved airbags for each peso
awarded to TV advertisement, The average value of a human life has been
assessed as being P400,000.00. Formulate the LP Model.
6. Mindanao Valves , Inc. produces valves. Two alternative lines are
available. The company just received an order for producing 1,500 valves.
Line I can produce 3 times as many as Line II. Line I is available for not
more than 200 hours at P10.00 an hour. Line II is available for at most 170
hours at P7.00 an hour. Determine the best production plan.
7. A company produces two types of hats. Each hat of type A requires twice
as much labor time as that of type B. If all hats produced are of type A
only, the company can produce a total of 500 hats each day. The market
limits daily sales of type A and B to 150 and 250 respectively. Assume
that profit per hat are P8 for type A and P6 for type B. Formulate an LP
model to determine the number of hats of each type to produce in order to
maximize profit.
8. Solve graphically Problem 1
9. Solve graphically Problem 2
10. Determine the optimal solution of Problem 1 using Simplex algorithm.
11. Determine the optimal solution of Problem 2 using Simplex algorithm.
12. Determine the optimal solution given
Max Z = 3X + 4X
Subject To: 5X + 4X  200
3X + 5X  1,500
5X+ 4X  100
8X+ 4X = 240
X, X  0
13. Determine the optimal solution, given
Min Z = 90X + 62X + 40X
Subject to: 2X+ 2X
 30
3X + X + X  20
X, X, X  0
14. Given the Primal LP model in Problem 1, write its dual model, and
discuss thoroughly the implications of the dual variables.
15. Given the following LP model in the Primal:
Max Z = 8X + 6X + 5X
Subject to: X + 2X + X  1500
X
 300
X
 250
X , X , X  0
Discuss the duality and the implications of the dual variables on the LP
problem.
16. Analyze the basic variables, nonbasic variables, and the capacity changes
of the LP models in Problem 14.
17. Analyze the basic variables, nonbasic variables, and the capacity changes
of the LP models in Problem 15.
DISTRIBUTION MODELS
The Transportation Problems
Objective: to assign a schedule where available good from different sources
(row) are transported to different destinations (column) as they are
needed, at minimum transportation cost.
Methods in obtaining Initial Tableau
 NCR (Northwest Corner Rule or Upper Lefthand Corner)
 Least Cost Method – limitations
 VAM (Vogel’s Approximation Method) is the most efficient method
 Trial and Error is not scientific at all
VAM algorithm
1. Determine both row AND column penalties using concept of opportunity
losses
2. Select the largest penalty to indicate the row OR column in which we
assign A stone using least cost criterion
3. After assigning a stone, ALWAYS cross out the satisfied row (column)
and adjust the column (row)
4. Determine both row AND column penalties, not including unit
transportation cost found in a satisfied row or column (we do not need
to compute the penalty of a satisfied row or column)
5. IF
(i)
Only one row or one column is left uncrossed, STOP computing
penalties; then assign stones to unused squares using least cost
criterion in order to satisfy the uncrossed rows AND columns,
until only one row (column) all together is left uncrossed;
(ii)
Only one row (column) is left uncrossed, STOP assigning stones,
obtained schedule makes the initial tableau;
(iii) Otherwise, go to number 2, and continue
Methods of iteration to reach optimal solution
 Stepping Stone Method
 Method of Multipliers or Modified Approach
Terms










Stone square
Unused square
Zero Square
Supply and Demand
Sources and Destinations
Unit Transportation Cost
Entering Variable and Leaving Variable
Profit or Cost Contribution (C for each column)
Objective (Z for each column)
Closed Path
- Starts and ends in the same unused square
- One for every unused square
- Clockwise or counter clockwise
- Use only horizontal and vertical lines
- Shortest path possible (use least number of corners)
- Every corner of the path must coincide on stone squares, avoiding
the zero squares (as a last resort, zero squares must coincide on a +
corner of the path)
 Path may skip over stone, zero, or unused squares
 Evaluation of Improvement Index (C-Z) for each column
Each unused square has its Improvement Index computed (indicator of
whether or not the unused square when assigned as stone, will improve
the value of Z)
- Alternately assign + and – signs on the corner of each closed path,
starting with a + sign at the unused square, following the flow of the
closed path.
- Following the alternately assigned + and – signs, attach the signs to
the unit transportation cost starting with that of the unused square
and that of every corner of the path, thereafter.
- The total value is the value of the improvement index for the specific
unused square.
Stepping Stone Method
1. Draw one closed path for every unused square.
2. To identify the entering variable, evaluate the improvement index of
each closed path. Since we are minimizing transportation cost, entering
variable or best improvement index is the one with most negative value
or the negative value whose absolute value is largest.
3. From among negative corners of the closed path, select the smallest
stone. Move this stone to the unused square, adjusting all stones in the
corners of the path, following the direction of the path and the
alternately assigned + and – signs. This yields a zero square (the leaving
variable).
4. The resulting tableau is the new tableau. This tableau is optimal (best
feasible) if there are no longer negative improvement indices left.
5. Otherwise, go to number 1.
Method of Multiplier or Modified Approach
1. Compute all row multipliers (R) and column multipliers (K), using only
the stone squares, and avoiding the use of zero squares. Initialize at the
leftmost corner.
UTC (stone ij) = R (i) + K (j)
2. For each unused square, compute the improvement index (I).
I (ij) for unused square (ij) = UTC (ij) of unused square (ij) – R (i) – K (j)
3. To identify the entering variable, evaluate the improvement index of
each closed path. Since we are minimizing transportation cost, entering
variable or best improvement index is the one with most negative value
or the negative value whose absolute value is largest.
4. Draw a closed path for the unused square with the best improvement
index.
5. From among negative corners of the closed path, select the smallest
stone. Move this stone to the unused square, adjusting all stones in the
corners of the path, following the direction of the path and the
alternately assigned + and – signs. This yields a zero square (the leaving
variable).
6. The resulting tableau is the new tableau. This tableau is optimal (best
feasible) if there are no longer negative improvement indices left.
7. Otherwise, go to number 1.
Assignment Algorithm
1. Obtain an Opportunity Loss Table
For a max problem:
Select the largest element from each row (column) and subtract every
element in the same row (column) from this largest element.
For a min problem:
Select the smallest element from each row (column) and subtract this
from every element in the same row (column).
2. Cross out all zeroes with minimum number of horizontal and vertical
lines. If the number of lines used to cross out the zeroes is equivalent to
the number of assignment pairs of the problem, STOP iterating. Assign
the assignment pairs where “zeroes” are found.
3. Otherwise, continue the iteration. Go to number 1.
PERT-CPM
PERT - Program Evaluation and Review Technique
- helps managers plan, schedule, monitor, and control complex
projects.
- probabilistic technique. It allows us to find the probability that
the entire project will be completed within a given date.
CPM - Critical Path Method
- deterministic technique. It uses two time estimates: normal and
crash. It allows us to compute for the crash completion time,
which is the shortest time it would take to finish an activity if
additional funds and resources were allotted.
6 Steps common to both PERT and CPM:
1. Define the project and all its significant activities and tasks.
2. Develop relationships among the activities. Decide which activities
must precede or follow the other.
3. Draw the network connecting all the activities.
4. Assign time and / or cost estimates to each activity.
5. Establish the critical path - the longest time path through the
network.
6. Use the network to help plan, schedule, monitor, and control the
project.
PERT - uses the variance of critical path to determine the variance of
the
whole project.
- uses Normal Distribution and Beta Distribution in the
statistical
analysis.
TIME ESTIMATES involved in PERT procedure:
1. optimistic time - only a small probability of this occurring
2. pessimistic time - only a small probability of this occurring
3. most likely time - most realistic time estimate for completion
Using the BETA DISTRIBUTION
Let
o : optimistic time estimate
p : pessimistic time estimate
m : most likely time estimate
Expected Time Estimate, t = o+4m+p
6
2
Dispersion or Variance of Expected Time Estimate = (p-o) / 36
Project Variance = Sum of Variances of Critical Activities
2
Project Variance = Project Standard Deviation
Using NORMAL DISTRIBUTION
Let
z : number of standard deviation the Project Deadline lies
from the mean or expected date or Project
Completion Time
Project Completion Time : shortest time within which the
project would be completed
Project Deadline : time within which project completion is
committed to client
z = Project Deadline - Project Completion Time
Project Standard Deviation
Referring to the Table on Area Under the Normal Curve, the
PROBABILITY THAT THE PROJECT WILL BE DELIVERED
OR COMPLETED WITHIN PROJECT DEADLINE may be
determined.
PROBLEMS IN PERT-CPM
1. Application of PERT analysis
Activities
Preceding
Activities
o
m
p
A
B
C
D
E
F
G
H
A
B
C
C
D, E
F, G
1
2
1
2
1
1
3
1
2
3
2
4
4
2
4
2
3
4
3
6
7
9
11
3
t
Determine the probability that the project will be completed within
16 weeks.
2. Application of CPM
Activity
A
B
C
D
E
F
G
H
Preceding Normal
Activities Time, t
A
B
C
C
D, E
F,G
2
3
2
4
4
3
5
2
Crash
Time
(P000)
Normal
Cost__
(P000)
Crash
Cost
1
2
1
2
3
2
2
1
5
5
3
8
7
2
10
4
8
9
7
12
10
6
16
7
Present a least cost crashing-in schedule in order to expedite the
project.
3. Determine the Project Completion Time, given:
Activity
A
B
C
D
E
F
G
H
Preceding
A
A
B
B
C, E
F
Normal Time, weeks
3
1
3
4
4
5
2
3
4. Determine the Project Completion Time, given:
Activity
A
B
C
D
E
F
G
H
I
Preceding
A
A
C
B
B, D
C
G, F
E, H
Normal Time, weeks
3
5
3
1
3
4
2
3
1
5. Determine the Project Completeion Time, given:
Activity
A
B
C
D
E
F
G
H
Preceding
A
A
B
B
D, E
G, F
Normal Time, weeks
3
5
14
5
4
7
8
5
6. Present a least cost schedule if project were to be expedited.
Activity
Preceding
Activity
A
B
C
D
E
F
G
H
I
NORMAL
Weeks (P000)
A
B
C
C
D, E, F
G
5
5
10
7
6
11
6
5
4
7.
Activity
A
B
C
D
E
F
G
H
I
Preceding
Activity
A, B
B
A
C
E, F
D, F
G, H
4
3
4
4
3
6
3
2
2
p
m
o
Normal
Cost
_(P000)
20
50
30
15
8
50
35
40
50
15
40
20
10
5
35
25
30
40
10
30
10
5
2
20
15
20
30
24
42
34
15
12
24
38
24
38
CRASH
Weeks (P000)
3
1
5
2
2
5
4
1
1
6
5
7
6
5
9
6
4
5
Crash
Cost
Time (P000)
12
35
18
5
5
30
23
24
36
33
52
42
30
12
44
48
36
50
(a) Determine the probability that the project will be completed within
175 weeks.
(b) Present the least cost schedule to fully expedite the project.
(c) Present the least cost schedule to expedite the project if
management only has P260,000 available for necessary
additional expense.