Linear programming graphing

profiledwsvag
last_week_hw.docx

D3.

Paul Fenster owns and manages a chili-dog and softdrink

stand near the Kean U. campus. While Paul can service 30

customers per hour on the average (m), he gets only 20 customers

per hour (l). Because Paul could wait on 50% more customers

than actually visit his stand, it doesn’t make sense to him that he

should have any waiting lines.

Paul hires you to examine the situation and to determine

some characteristics of his queue. After looking into the problem,

you find it follows the six conditions for a single-server waiting

line (as seen in Model A). What are your findings?

Answer:

Probability of k or More Customers Waiting in Line and/or Being Waited On

 k

 0

0.667

 1

0.444

 2

0.296

 3

0.198

D6.

Calls arrive at Lynn Ann Fish’s hotel switchboard at

a rate of 2 per minute. The average time to handle each is 20 seconds.

There is only one switchboard operator at the current time.

The Poisson and exponential distributions appear to be relevant

in this situation.

a) What is the probability that the operator is busy?

b) What is the average time that a customer must wait before

reaching the operator?

c) What is the average number of calls waiting to be answered?

Answer:

D8.

Virginia’s Ron McPherson Electronics Corporation

retains a service crew to repair machine breakdowns that occur

on average l = 3 per 8-hour workday (approximately Poisson in

nature). The crew can service an average of m = 8 machines per

workday, with a repair time distribution that resembles the exponential

distribution.

a) What is the utilization rate of this service system?

b) What is the average downtime for a broken machine?

c) How many machines are waiting to be serviced at any given

time?

d) What is the probability that more than 1 machine is in the

system? The probability that more than 2 are broken and

waiting to be repaired or being serviced? More than 3? More

than 4?

Answer:

M/M/1 model with = 3, = 8

(a) The utilization rate, , is given by:

(b) The average down time, Ws, is the time the machine waits to be serviced plus the time taken to perform the service.

(c) The number of machines waiting to be served, Lq, is, on average:

(d) Probability that more than one machine is in the system:

Probability that more than 2, 3, or 4 machines are in the system:

B7.

The Attaran Corporation manufactures two electrical

products: portable air conditioners and portable heaters. The

assembly process for each is similar in that both require a certain

amount of wiring and drilling. Each air conditioner takes 3 hours

of wiring and 2 hours of drilling. Each heater must go through

2 hours of wiring and 1 hour of drilling. During the next production

period, 240 hours of wiring time are available and up to

140 hours of drilling time may be used. Each air conditioner sold

yields a profit of $25. Each heater assembled may be sold for a

$15 profit.

Formulate and solve this LP production-mix situation, and

find the best combination of air conditioners and heaters that

yields the highest profit.

ANSWER:

Let x1 number of air conditioners to be produced

x2 number of heaters to be produced

Maximize 25x1  15x2

Subject to 3x1  2x2 240 (wiring)

2x1  1x2 140 (drilling)

x1, x2 0 (non-negativity)

Profit:

@a: (x1  0, x2  0)  Obj  $0

@b: (x1  0, x2  120) Obj  25  0  15  120  $1,800

@c: (x1  40, x2  60) Obj  25  40  15  60  $1,900*

@d: (x1  70, x2  0) Obj  25  70  15  0  $1,750

* The optimal solution is to produce 40 air conditioners and 60 heaters each period. Profit will be $1,900.

B11.

The Sweet Smell Fertilizer Company markets bags

of manure labeled “not less than 60 lb dry weight.” The packaged

manure is a combination of compost and sewage wastes. To

provide good-quality fertilizer, each bag should contain at least

30 lb of compost but no more than 40 lb of sewage. Each pound

of compost costs Sweet Smell 5¢ and each pound of sewage costs

4¢. Use a graphical LP method to determine the least-cost blend

of compost and sewage in each bag.

ANSWER

Let: X1  number of pounds of compost in each bag

X2  number of pounds of sewage waste in each bag

Minimize cost  5X1  4X2 (in cents)

Subject to X1  X2 60 (pounds per bag)

X1 30 (pounds compost per bag)

X2 40 (pounds sewage per bag)

Corner point a:

(X1  30, X2  40) cost  5(30)  (4)(40)  $3.10

Corner point b (which is optimal):

(X1  30, X2  30) cost  5(30)  (4)(30)  $2.70

Corner point c:

(X1  60, X2  0) cost  5(60)  (4)(0)  $3.00

B21.

Par, Inc., produces a standard golf bag and a deluxe

golf bag on a weekly basis. Each golf bag requires time for cutting

and dyeing and time for sewing and finishing, as shown in the following

table:

PRODUCT

CUTTING & DYEING

SEWING & FINISHING

STANDARD BAG

0.50

1.00

DELUXE BAG

1.00

0.67

The profits per bag and weekly hours available for cutting and

dyeing and for sewing and finishing are as follows:

PRODUCT

PROFIT PER UNIT

STANDARD BAG

10.00

DELUXE BAG

8.00

ACTIVITY

WEEKLY HOURS AVAILABLE

CUTTING & DYEING

300.00

SEWING & FINISHING

360.00

Par, Inc., will sell whatever quantities it produces of these two

products.

a) Find the mix of standard and deluxe golf bags to produce per

week that maximizes weekly profit from these activities.

b) What is the value of the profit?

ANSWER

 (a, b) Let S = number of standard bags to produce per week

D = number of deluxe bags to produce per week

Maximize profit = 10S + 8D

Extreme (Corner) Points

Point

Profit

(0, 0)

$0

Macintosh HD:Users:dwsavage254:Library:Caches:TemporaryItems:msoclip:0:clip_image001.png

(0, 300)

$2,400

(360, 0)

$3,600

(240, 180)

$3,840

22

22

(b)

()3(32)3

0.667 time units(minutes)

24

(c) 1.333

()3(32)3

q

q

W

L

l

mml

l

mml

===

--

=

====

--

3

0.375

8

l

r

m

===

(

)

(

)

11

0.2 days, or 1.6 hours

83

s

W

ml

===

--

22

3

0.225 machines waiting

()8(83)

q

L

l

mml

===

--

1

2

1

39

, or 0.141

864

k

nkn

PP

l

m

+

>>

éù

éù

====

êú

êú

ëû

ëû

3

2

4

3

5

4

327

0.053

8512

381

0.020

84096

3243

0.007

832,768

n

n

n

P

P

P

>

>

>

éù

===

êú

ëû

éù

===

êú

ëû

éù

===

êú

ëû

1

+300 ()

2

2

360 ()

3

, 0

SDA

SDB

SD

£

³

22

//1 model: 20, 30

20

2 customers in the system on the average

3020

11

0.1 hour (6 minutes) that the average cu

stomer spends in the total system

3020

20

1.33 cus

()30(3020)

s

s

q

MM

L

W

L

lm

l

ml

ml

l

mml

==

===

--

===

--

===

--

tomers waiting for service in line on th

e average

20

1/15 hour=(4 minutes)=average waiting ti

me of a customer in the queue awaiting s

ervice

()30(3020)

20

=0.67percent of the time that he

30

q

W

l

mml

l

r

m

===

--

===

0

is busy waiting on customers

110.33probability that there are no cust

omers in the system

(being waited on or waiting in the queue

) at any given time

P

l

r

m

=-=-==

l

m

+

>

æö

=

ç÷

èø

1

k

nk

P

180/hour, 120/hour,

//1 model with

or 3/min. and 2/min.

2

(a) 0.667

3

MM

ml

ml

l

r

m

==

==

===