PRE FINAL REVIEW EXERCISE

1. The IT Division of a university processes students’ applications to enter the institution. During peak period, the processing of the applications is done by a large pool of computer operators, some of whom are permanent and some temporary. A permanent operator can process 24 applications per day, whereas a temporary operator can process 16 per day. On average, the IT Division processes at least 600 applications each day. The IT Division has allocated 45 computer workstations to process the applications. A permanent operator generates about 1 application with errors each day, whereas a temporary operator 2 errors per day. The Director of the IT Division wants to limit applications with errors to 60 per day. A permanent operator is paid RM60 per day, and a temporary operator RM35 per day. The Director wants to determine the number of permanent operators and temporary operators to hire in order to minimize costs.

a. Formulate a Linear Programming model for the problem.b. Solve this model by using graphical method.c. If the temporary operator is paid RM40 per day instead of RM35, what will be your

optimal solution?2. A sawmill in Tannassee produces cherry and oak boards for a large furniture manufacturer.

Each month the sawmill must deliver at least 5 tons of wood to the manufacturer. It takes the sawmill 3 days to produce a ton of cherry and 2 days to produce a ton of oak, and the sawmill can allocate 18 days out of a month for this contract. The sawmill can get enough cherry to make 4 tons of wood and enough oak to make 7 tons of wood. The sawmill owner has developed the following linear programming model to determine the number of tons of cherry (x1) and oak (x2) to produce to minimize cost.

Minimize Z = 3x1 + 6x2 (cost, $)Subject to:

3x1 + 2x2 ≤ 18 (production time, days)x1 + x2 ≥ 5 (contract, tons)x1 ≤ 4 (chery, tons)x2 ≤ 7 (oak, tons)x1, x2 ≥ 0 (non-negative condition)

a. Solve this model using the simplex method.b. Suppose, the objective function in this problem is profit ($) maximization [i.e.

Maximize Z = 3x1 + 6x2]; solve this model using the simplex method.

3. Consider the following linear programming problem.

Maximize Z = 5x1 + 6x2 + 4x3

Subject to3x1 + 4x2 + 4x3 ≤ 120 …1x1 + 2x2 + x3 ≤ 50 …2x1 + 2x2 + 3x3 ≥ 30 …3x1, x2, x3 ≥ 0

The optimal simplex tableau is

Basis cj

x1 x2 x3 s1 s2 s3

Soln5 6 4 0 0 0

s3 0 0 4 0 -2 7 1 80

x3 4 0 2 1 1 3 0 30

x1 5 1 0.5 0 0 -2 0 20

zj 5 10.5 4 4 2 0 220

cj - zj 0 -4.5 0 -4 -2 0

a. Find the range of optimality for c1 and c2.

b. Find the range of feasibility for b2.c. Suppose the right-hand side value of the first constraint is decreased from 120 to 110,

what will be the new optimal solution and its value?d. Suppose the right-hand side value of the second constraint is increased from 50 to 60,

what will be the new solution and its value?e. What is the marginal increase (say in $) in the objective function if resource no 1 is

increased by one unit? What is the amount of increase in the objective function if resource no 1 is increased by three units?

f. The cost for a unit of resource no 1 is $2.50. Assume the objective function is profit ($), would you purchase the additional unit in resource no 1. Why?

4. Micro Computers Limited sells microcomputers to universities and colleges on the East Coast and ships them from three distribution warehouses. The firm is able to supply the following numbers of microcomputers to the universities by the beginning of the academic year;

Distribution

Warehouse

Supply

(microcomputers)

1. Richmond 420

2. Atlanta 610

3. Washington, D.C. 340

Total 1370

Four universities have ordered microcomputers that must be delivered and installed by the beginning of the academic year;

University

Demand

(microcomputers)

A. Tech 520

B. A and M 250

C. State 400

D. Central 380

Total 1550

The shipping and installation costs ($) per microcomputer from each distributor to each university are as follows;

From

To

A B C D

1 22 17 30 18

2 15 35 20 25

3 28 21 16 14

a. Find the initial solution using;i. NWCR

ii. LCM, andiii. VAM

b. Using the initial solution with NWCR method in (a) above, proceed to solve the problem for the optimal solution using MODI and Stepping Stone method.

c. Suppose the demand in State is reduced from 400 units to 200 units, find the initial solution using NWCR. All other data remain the same. Proceed to solve the problem for the optimal solution using MODI and Stepping Stone method.

5. MS Consultant has received four consultancy works for the East Coast Development Corridor. The Executive Director is planning to assign the jobs to four of its senior consultants to lead each job so that the total completion time is minimized. The time frame (in weeks) for each of these consultants to complete each job is tabulated in the following table.

ConsultantsJobs

1 2 3 4

A 10 14 16 13

B 12 13 15 12

C 9 12 12 11

D 14 16 18 16

You have been given the task to assist the Executive Director to plan the job assignments. Find the optimal solution using the Hungarian method (Assignment method) so that the total completion time is minimized.

a. State the optimal job assignments to each consultant.b. What is the total completion time for the consultancy works?c. Suppose, the firm has an additional consultant E, who can complete jobs 1, 2, 3, and 4

in 11, 15, 17 and 14 weeks respectively. What will be the optimal solution? The consultant not assigned to any of these jobs will be given a task to do Environmental Impact Assessment (EIA) study for other project. Which consultant will be assigned to this EIA study?

d. Suppose, from the original problem, the firm has secured additional job 5, that can be completed by consultant A, B, C, and D in 15, 14, 13, and 17 weeks respectively. What will be the optimal solution? The job that has no consultant to do will be outsourced to another company. Which job will be outsourced?

6. Doug Casey is in charge of planning and coordinating next spring’s sales management training program for his company. Doug has listed the following activity information for this project.

Activity Description

Immediate

Predecessor

Time (weeks)

Optimistic Most

Probable

Pessimistic

A Plan topic - 1.5 2.0 2.5

B Obtain speakers A 2.0 2.5 6.0

C List meeting locations

- 1.0 2.0 3.0

D Select location C 1.5 2.0 2.5

E Finalize speaker travel plans

B. D 0.5 1.0 1.5

F Make final check with speakers

E 1.0 2.0 3.0

G Prepare and mail brochures

B, D 3.0 3.5 7.0

H Take reservations G 3.0 4.0 5.0

I Handle last-minute details

F, H 1.5 2.0 2.5

a. Draw a project network for this problem.b. Prepare an activity schedule.c. What are the critical activities (i.e. critical path) and what is the expected project

completion time?d. If Doug wants a 0.99 probability of completing the project on time, how far ahead

of the scheduled meeting date should he begin working on the project?e. If Doug wants 90% confidence to complete on time, how long will be the

completion time?f. What is the probability for the project to be completed within 16 weeks?g. What is the likelihood (the probability) that the project is delayed beyond 17

weeks?h. Is it possible to delay the commencement of activities E & F? If yes, by how many

weeks each.

ANSWERS

1. Answera. LPP Formulation

Letx1 = permanent operatorx2 = temporary operator

Min Z = 60x1 + 40x2 minimize costSubject to

24x1 + 16x2 ≥ 600 number of applications processed per dayx1 + x2 ≤ 45 computer workstations availablex1 + 2x2 ≤ 60 applications processed with errors per dayx1, x2 ≥ 0 non-negativity condition

b. Graphical solutionDraw graph, grids 60 x1 and 60 x2

Solution,A(25, 0) ZA = RM1500B(7.5, 26.25) ZB = RM1368.75C(30, 15) ZC = RM1975D(45, 0) ZD = RM2700

10

10

20

20

30

30

40

40

50

50

60

600 x1

x2

D

C

B

A

Z

Therefore, Optimal Solution;ZBmin = RM1368.75x1 = 7.5 (or 8 permanent operators)x2 = 26.25 (or 26 temporary operators)

c. Solution,A(25, 0) ZA = RM1500B(7.5, 26.25) ZB = RM1500C(30, 15) ZC = RM2400D(45, 0) ZD = RM2700

Therefore, alternate optimal solutionZAmin = RM1500x1 = 7.5 (8 permanent operators)x2 = 26.25 (26 temporary operators)

ZBmin = 1500x1 = 25 (25 or all permanent operators)x2 = 0 (0 or without temporary operators)

2.a. Minimization:Standard from:

Minimize Z = 3x1 + 6x2 + 0s1 + 0s2 + 0s3 + 0s4 + Ma2

Subject to:

3x1 + 2x2 + s1 + 0s2 + 0s3 + 0s4 + 0a2= 18x1 + x2 + 0s1 – s2 + 0s3 + 0s4 + a2 = 5x1 + 0s1 + 0s2 + s3 + 0s4 + 0a2 = 4x2 + 0s1 + 0s2 +0s3 + s4 + 0a2 = 7x1, x2, s1, s2, s3, s4, a2 ≥ 0

Convert the minimization problem to maximization in order to proceed with the simplex method

Let Z’ = Z(-1)

Minimize Z = Maximize Z’

Thus: the minimization problem becomes maximization

Therefore the tableau form isMaximize Z’ = -3x1 - 6x2 - 0s1 - 0s2 - 0s3 - 0s4 - Ma2

Subject to:

3x1 + 2x2 + s1 + 0s2 + 0s3 + 0s4 + 0a2 = 18x1 + x2 + 0s1 – s2 + 0s3 + 0s4 + a2 = 5x1 + 0s1 + 0s2 + s3 + 0s4 + 0a2 = 4x2 + 0s1 + 0s2 +0s3 + s4 + 0a2 = 7x1, x2, s1, s2, s3, s4, a2 ≥ 0

Simplex method:

Basis cj

x1 x2 s1 s2 s3 s4 a3

Soln Ratio-3 -6 0 0 0 0 -M

s1 0 3 2 1 0 0 0 0 18 6

a2 -M 1 1 0 -1 0 0 1 5 5

s3 0 1 0 0 0 1 0 0 4 4

s4 0 0 1 0 0 0 1 0 7 -

zj -M -M 0 M 0 0 -M -5M

cj - zj -3+M -6+M 0 -M 0 0 0

s1 0 0 2 1 0 -3 0 0 6 3

a2 -M 0 1 0 -1 -1 0 1 1 1

x1 -3 1 0 0 0 1 0 0 4 -

s4 0 0 1 0 0 0 1 0 7 7

zj -3 -M 0 M -3+M 0 0 -12-M

cj - zj 0 -6+M 0 -M 3-M 0 -M

s1 0 0 0 1 2 -1 0 - 4

x2 -6 0 1 0 -1 -1 0 - 1

x1 -3 1 0 0 0 1 0 - 4

s4 0 0 0 0 1 1 1 - 6

zj -3 -6 0 6 3 0 - -18

cj - zj 0 0 0 -6 -3 0 -

Optimal solution (after three iterations: cj - zj ≥ 0):x1 = 4, x2 = 1, s1 = 4, s4 = 6

Z’max = -18Thus,

Zmin = Z’(-1) = 18x1 = 4, x2 = 1

b. Maximization:Standard from:

Maximize Z = 3x1 + 6x2 + 0s1 + 0s2 + 0s3 + 0s4 - Ma2

Subject to:

3x1 + 2x2 + s1 + 0s2 + 0s3 + 0s4 + 0a2 = 18x1 + x2 + 0s1 – s2 + 0s3 + 0s4 + a2 = 5x1 + 0s1 + 0s2 + s3 + 0s4 + 0a2 = 4x2 + 0s1 + 0s2 +0s3 + s4 + 0a2 = 7x1, x2, s1, s2, s3, s4, a2 ≥ 0

Simplex method:Do it yourself (check your answer by using the simplex tool)

Optimal solution (after five iterations):x1 = 4/3, x2 = 7

Zmax = 46

3. Sensitivity Analysisa. Range of optimality

-4 ≤ c1, c1 ≤ 6;-4 ≤ c1 ≤ 6c2 ≤ 10.5

b. Range of feasibility b2 (original right hand side value, bm = 50)

b’1 a1j

b’2 a2j

. + ∆bi . ≥ 0

. .b’m amj

-10 ≤ ∆b2 ≤ 10

Range of feasibility for bi is (bm + ∆bi) ≤ bi ≤ (bm + ∆bi)

40 ≤ b2 ≤ 60

c. New solution

b’1 a1j b*1

b’2 a2j b*2

. + ∆bi . = . Then find Znew

. . .b’m amj b*m

New optimal solution∆b1 = 110 – 120 = -10

x1 = 20x3 = 20s3 = 100Znew = 180

d. New optimal solution∆b2 = 60 – 50 = 10

x1 = 40x3 = 60s3 = 150Znew = 440

e. Marginal increase in the objective function resulting from an increase by one unit of resource no 1 is $4.00. [see dual price]Amount of increase in the objective function resulting from an increase of resource no 1 by three unit is $12.00

f. Yes, because there is a net increase in the objective function of $1.50 per unit due to the increase in resource no 1 (i.e. Profit $4.00 less Cost $2.50)

4. Check whether the transportation problem is balanced

Supply = 1370; Demand = 1550 Supply ≠ Demand [difference 180 (i.e. 1550 - 1370)]The transportation problem is unbalanced; demand is 180 more than supply.Introduce Dummy Supply of 180 units with transportation cost of $0 per unit in each route.

a. Initial (Hueristic) Solution

NWCR

Warehouses

Universities

SupplyA B C D

1 420 420

2 100 250 260 610

3 140 200 340

Dummy 180 180

Demand 520 250 400 380 1550

Initial Solution:Total Cost: $29,730Routes: 1-A=420; 2-A=100; 2-B=250; 2-C=260; 3-C=140; 3-D=200; Dummy-D=180

2035

1628

25

14

15

21

301722 18

0 0 00

LCM

Warehouses

Universities

SupplyA B C D

1 250 130 40 420

2 340 270 610

3 340 340

Dummy 180 180

Demand 520 250 400 380 1550

Initial solution:Total Cost: $24,130Routes: 1-B=250; 1-C=130; 1-D=40; 2-A=340; 2-C=270; 3-D=340; Dummy-A=180

2035

1628

25

14

15

21

301722 18

0 0 00

VAM

Warehouses

Universities

Supply

row

penaltyA B C D

1 70 350 4201,1,1,1,1,1,1

2 520 90 6105,5,5,-,

-,-,-,

3 310 30 3402,2,2,2,

2,7,-

Dummy 180 1800,-,-,-,-,-,-

Demand 520 250 400 380 1550

Column penalty

15,7,-,-,-,-,-

17,4,4,4,-,-,-

16,4,4,14,-,-,-

14,4,4,4,4,4,7

Initial Solution:Total Cost = $22,470Routes:1-B=70; 1-D=350; 2-A=520; 2-C=90; 3-C=310; 3-D=30; Dummy-B=180

2035

1628

25

14

15

21

301722 18

0 0 00

b. MODI and Stepping Stone Method:

Check whether m + n – 1 = number of basic cellsm + n – 1 = 4 + 4 – 1 = 7number of basic cells = 7

Thus; m + n – 1 = number of basic cells

Therefore we can proceed with MODI and Stepping Stone Method

First iteration; find auxilary indices & net evaluation indices (improvement indices)Check net evaluation indices, whether it is optimal. In this case, it is not, therefore proceed with close path for MODI & Stepping Stone

Warehouses

Universities

Supply uiA B C D

1 420 - (-25) + (3) (-7)420 0

2 100 + 250 - 260 (7)610 -7

3 (17) (-10) 140 200340 -11

Dummy (3) (-17) (-2) 180 180-25

Demand 520 250 400 380 1550

vj 22 42 27 25

2035

1628

25

14

15

21

301722 18

0 0 00

Second iteration; find auxilary indices & net evaluation indices (improvement indices)

Warehouses

Universities

Supply uiA B C D

1 170 - 250 (3) (-7) +420 0

2 350 + (25) 260 - (7)610 -7

3 (17) (15) 140 + 200 -340 -11

Dummy (3) (8) (-2) 180 180-25

Demand 520 250 400 380 1550

vj 22 17 27 25

Third iteration; find auxilary indices & net evaluation indices (improvement indices)

Warehouses

Universities

Supply uiA B C D

1 (7) 250 (10) 170420 0

2 520 (18) 90 (7)610 0

3 (17) (8) 310 - 30 +340 -4

Dummy (3) (1) (-2) + 180 - 180-18

Demand 520 250 400 380 1550

vj 15 17 20 18

2035

1628

25

14

15

21

301722 18

0 0 00

2035

1628

25

14

15

21

301722 18

0 0 00

Fourth iteration; find auxilary indices & net evaluation indices (improvement indices)

Warehouses

Universities

Supply uiA B C D

1 (7) 250 (10) 170420 0

2 520 (18) 90 (7)610 0

3 (17) (8) 130 210340 -4

Dummy (5) (3) 180 (2) 180-20

Demand 520 250 400 380 1550

vj 15 17 20 18

On fouth iteration, the net evaluation indices (or improvement indices) ≥ 0Thus, on fifth iteration, reach optimal solution;

The optimal solution is;Total Cost = $21,930Optimal routes: 1-B=250; 1-D=170; 2-A=520; 2-C=90; 3-C=130; 3-D=210; Dummy-C=180

2035

1628

25

14

15

21

301722 18

0 0 00

c. If demand in State is reduced to 200, total demand = 1350, while total supply = 1370.

Total supply ≠ Total demand; i.e. unbalanced transportation problem: total supply is less than total demand by 20 (1370 – 1350 = 20). Therefore introduce Dummy Demand of 20 units with transportation cost of $0 per unit in each route.

NWCR

Warehouses

Universities

SupplyA B C D Dummy

1 420 420

2 100 250 200 60 610

3 320 20 340

Demand 520 250 200 380 20 1370

Initial Solution:Total Cost: $29,730Routes: 1-A=420; 2-A=100; 2-B=250; 2-C=200; 2-D=60; 3-D=320; 3-Dummy=20

Proceed with MODI & Stepping Stone Method.

m + n – 1 = 7; number of basic cells = 7Therefore: m + n – 1 = number of basic cellsSo you can proceed with MODI and Stepping Stone Method

Do it yourself, using exactly the same procedure: find the values of auxiliary indices and net evaluation indices (improvement indices) and then check for optimality at every iteration until optimal solution is obtained (net evaluation indices ≥ 0). Then state your solution: Total Cost and the Optimal Routes.

2035

1628

0

0

15

21

301722 018

25

14

5.a. To determine job assignments:

Check whether assignment problem is balanced.Number of consultants = 4; number of jobs = 4; thus, assignment problem is balanced [number of consultants = number of jobs].

Therefore, we can proceed with the solution using Hungarian method without adding dummy consultant or dummy job.

[if number of consultant is more, introduce dummy job with completion time 0 for each consultant; if number of job is more, introduce dummy consultant with completion time 0 for each job]

The assignment tableau

ConsultantsJobs

1 2 3 4

A 10 14 16 13

B 12 13 15 12

C 9 12 12 11

D 14 16 18 16

Row operation (compute opportunity costs row-wise)

ConsultantsJobs

1 2 3 4

A 0 4 6 3

B 0 1 3 0

C 0 3 3 2

D 0 2 4 2

Column operation (compute opportunity costs column-wise)

ConsultantsJobs

1 2 3 4

A 0 3 3 3

B 0 0 0 0

C 0 2 0 2

D 0 1 1 2

First iteration: Hungarian Method [draw minimum number of lines covering zeros]

ConsultantsJobs

1 2 3 4

A 0 3 3 3

B 0 0 0 0

C 0 2 0 2

D 0 1 1 2

number of lines ≠ number of rows or columns; Therefore not optimal

Second iteration: Re-computation of opportunity cost [‘adjust’]

ConsultantsJobs

1 2 3 4

A 0 2 3 2

B 1 0 1 0

C 0 1 0 1

D 0 0 1 1

Second iteration: Hungarian method [draw lines]

ConsultantsJobs

1 2 3 4

A 0 2 3 2

B 1 0 1 0

C 0 1 0 1

D 0 0 1 1

no of lines = no of rows or columns; Therefore optimal solution is reached

Assign consultant to jobs; i.e. Optimal assignments

ConsultantsJobs

1 2 3 4

A 0 √

B 0 0 √

C 0 0 √

D 0 0 √

The assignments:Consultant A to Job number 1 with completion time of 10 weeksConsultant B to Job number 4 with completion time of 12 weeks Consultant C to Job number 3 with completion time of 12 weeksConsultant D to Job number 2 with completion time of 16 weeks

b. Total completion time is10 + 12 + 12 + 16 = 50 weeks

c. Consultants are more than Jobs (unbalanced assignment problem)Do it yourself. Introduce dummy job with completion time of 0 for each consultant.Alternative optimal assignment: A-2; B-4; C-3; D-Dummy; E-1: or A-1; B-4; C-3; D-Dummy; E-2: or A-4; B-2; C-3; D-Dummy; A-1. Completion time: 49 weeksConsultant D will be assigned to conduct EIA study.

d. Jobs are more than Consultants (unbalanced assignment problem)Do it yourself. Introduce dummy consultant with completion time of 0 for each job.Optimal assignment: A-1; B-4; C-3; D-2; Dummy-5. Completion time: 50 weeksProject number 5 will be outsourced.

6. a. Network diagram

A B

FinishStart

F

GC

E

D

I

H

e. Activity schedules

Activity Expected Time Variance

A 2 0.03

B 3 0.44

C 2 0.11

D 2 0.03

E 1 0.03

F 2 0.11

G 4 0.44

H 4 0.11

I 2 0.03

Critical activities

Activity

Earliest

Start

Latest

Start

Earliest

Finish

Latest

Finish Slack

Critical

Activity

A 0 0 2 2 0 Yes

B 2 2 5 5 0 Yes

C 0 1 2 3 1

D 2 3 4 5 1

E 5 10 6 11 5

F 6 11 8 13 5

G 5 5 9 9 0 Yes

H 9 9 13 13 0 Yes

I 13 13 15 15 0 Yes

c. Critical Path: Start – A – B – G – H – I – Finish

Expected completion time, E(T) = 2 + 3 + 4 + 4 + 2 = 15 weeks

d. Variance on critical path

σ2 = 0.03 + 0.44 + 0.44 + 0.11 + 0.03 = 1.05

From Probability Distribution Table, we find 0.99 probability occurs at z = +2.33.Thus

z = T - E (T)

= T - 151.05

= 2.33

or

T = 15 + 2.33 1.05 = 17.4 weeks

e. T = 15 + 1.28 √1.05 = 16.3 weeks

f. P(complete ≤ 16 weeks) = (0.5 + 0.3353) = 0.8353 or 83.53%

g. Z = 1.9518, Therefore P(complete ≥ 17 weeks) = (0.5 – 0.4744) = 0.0256 or 2.56%

h. Yes, commencement of both activities can be delayed by 5 weeks each (slack is 5 for each activity)