~ 1 ~
Mathematical Model FORMULATION :
بجٕتسٌ جرًَٕ تغاٍص تٍطخنا تجيشبنا : 1 فذٓنا تناد ذٌذحت objective function : تساسذنا ذٍق تهكشًنا فذْ ذٌذحت يا تقٍقدٔ تحضأ ةسٕصب . ٌٕكت ذقٔ حابسا ىٍظؼت maximization profit تناد ٌٕكت اْذُػٔ ىٍظؼتنا عَٕ ٍي فذٓنا ( (Max. Z . فٍناكت ٍَّذت ٔأ ( شئاسخ مٍهقت ) minimization costs(losses) مٍهقتنا عَٕ ٍي فذٓنا تناد ٌٕكت اْذُػٔ ( Min. Z ) . 2 ذٌذت ، تٍنٔلاا دإًنا ثاًٍك ٔا ، مًؼنا ثاػاسك ،تحاتًنا ثاَاكيلإأ دسإًنا واذختسا دٍٕق تٍجاتَلاا آتاقاط ٔا ٍئاكًنا مٍغشت ثاػاس ٔا . ىٍظؼت عَٕ ٍي تناذن تغاٍصنا Max. Z=10X1+15X2 S.T.:- 6X1+4X2 ≤ 30000 work hours ---(1) 2X1+4X2 ≤ 20000 raw material ---(2) X1 ≤ 7000 production rate ---(3) X2 ≤ 8000 production rate ---(4)
X1, X2 ≥0 non – negative constraints تٍَذت عَٕ ٍي تناذن تغاٍصنا Min. Z=10X1+15X2 S.T.:- 6X1+4X2 ≥ 30000 pm hours ---(1) 2X1+4X2 ≥ 20000 production units ---(2) X1 ≤ 7000 failure cost ---(3)
X2 ≤ 8000 spare part cost ---(4)
X1, X2 ≥0 non – negative constraints EX 1 :company manufactures a product consiste of x1 and X2. one unit of product X1 consumes one hour of machine A, 3 hours of machine B and 10 hours of machine C., one unit of X2 consumes one hour of machine A, 8 hours of machine B and 7 hours of machine C. machine A has available 40 hours and machine B has available a capacity of 240 hours and of machine C is 350 hours. the profit of x units Rs.5 and y units are /- and Rs 7. Write the L,P MODEL THAT INCREASED THE PROFIT FOR THE COMPANY ? Maximise Z = 5 + 7 OBJECTIVE FUNCTION + ≤ 40
3 + 8 ≤240 STRUCTURAL CONSTRAINTS 10 + 7 ≤ 350
~ 2 ~
EX2 : A company manufactures two products X and Y, which require,
the following resources. The resources are the capacities machine M1,
M2, and M3. The available capacities are 50,25,and 15 hours respectively
.. Product X requires 1 hour of machine M2 and 1 hour of machine M3. Product Y requires 2 hours of machine M1, 2 hours of machine M2 and 1 hour of machine M3. The profit of products X and Y are Rs.5 and Rs.4 respectively. Write the L,P MODEL THAT INCREASED THE PROFIT FOR THE COMPANY Linear Programming Models (Resource Allocation Models) ?
Maximize Z = 5 + 4 OBJECTIVE FUNCTION.
For Machine M1 0 + 2 ≤50
For Machine M2 + 2 ≤25 STRUCTURAL CONSTRAINTS. For machine M3 + ≤15
, ≥ 0 NON -NEGATIVITY CONSTRAINT
EX3: A company PRODUCE two types of shirts A and B. During a
week the store can sell a maximum of 400 shirts of type A and a maximum of 300 shirts of type B. The storage capacity, is limited to a maximum of 600 of both types. a profit Type A shirt Rs. 2 per shirT. and a profit type B of Rs. 5 per unit. How many of each type the store should stock per week to maximize the total profit? Formulate a mathematical model of the problem.
Maximize Z = 2 + 5
s.t ≤400
≤300
+ ≤600 and X1, ≥0.
EX4: : A company has 3 winches and 3 stores, the capacity limits
are: store A 2000 tons, with volume 100 meters, the store B 3000 tons, with volume 135 meters and the store C 1500 tons, 30 meters. The following the v data about the amount and profit for each units .
Units Amount in tons space Volume for ton Profit
A 6000 60 60 B 4000 50 80 C 2000 25 50
~ 3 ~ Formulate this as linear programming model
The amount of A is 6000 and each ton have 60 meters volume
= 6000/60 = 100 m , And the same for B( 4000/50= 80 ) , and for c (
2000/25= 80 )
Maximise Z = 60 +80 +50 s.t.
6000 + 4000 + 2000 ≤6500 ( 2000+3000+1500) 100 +80 +80 ≤265 ( 100+135+30)
Both and are ≥0.
EX5: A company has 2 machines m1 and m2 . the company produce 2
products A and B .the time process time on each machine is bellow .and the Daily hours requirement per units are at least 40 hurs for M1 and 50 hurs for M2 .Formulate L.P. to minimize the cost of tonics.
A B Daily hours requirement M1 2 4 40
M2 3 2 50
Cost per unit. 5 3 Minimize Z = 5 + 3
s.t 2 + 4 ≥40
3 + 2 ≥50
and ≥0.
EX6: A company produces three products A, B and C by using two raw
materials R and S. 4000 units of R and 6000 units of S are available for production. The requirement of raw materials by each product is given below:
A B C R 2 3 5
~ 4 ~
S 4 2 7
The labour time for each unit of product A is twice that of product B and three times that of product C. The labour force of the company can
produce the equivalent of 2500 units of product A. the minimum demand of products A, B and C are 500, 500 and 375 respectively . However, their ratio of number of units produced must be equal to 3: 2: 5. the profit per units of product A, B and C are 60,, 40, 100 respectively. Formulate the L.P.P. for maximizing the profit.
Solution:
Let the company manufactures a units of A, b units of B and c units of C. The constraints for
raw materials are 2a + 3b + 5c ≤4000 4a +2b + 7c ≤6000
Now let ( t ) be the labour time required for one unit of product A, then the time required for per unit of product B is t/2 and that for product C is
t/3. As 2500 units of A are produced, the total time available is 2500 t.
Hence the constraints for time are:
ta + t/2 b + t/3 c ≤2500t
= a + 1/2 b + 1/3 c ≤2500 the demand constraints are:
a ≥ 500 b ≥ 500 c ≥ 375
As the ratio of production must be 3 : 2 : 5,
= so 2a = 3b = so 5b =2c
Hence the Linear programme is:Maximise Z = 60 a + 40b + 100c St 2a + 3b + 5c ≤4000 4a + 2b + 7c ≤6000 a + 1/2 b + 1/3 c ≤2500
2a-3b =0
5b – 2c =0
a ≥ 500 b ≥ 500 c ≥ 375~ 5 ~ a.b.c ≥ 0
METHODS FOR THE SOLUTION OF A LINEAR ROGRAMMING ---
( 1 ): The Graphical Method: ---
(i) Generally the method is used to solve the problem, when it involves two decision variables
(ii) Always, the solution to the problem lies in first quadrant
EX 7 : A company manufactures two products, X and Y by using tow
machines M1, M2,.Machine M1 has 4 hours of capacity available, the available capacity of machines M2 is 24. One unit of product X requires 1 hour of Machine M1, 3 hours of machine M2 one unit of product Y requires 1 hour, 8 hour M1, M2 respectively. a profit of X Rs. 5 per product and that of Y is Rs. 7 t. Solve the problem by using graphical method to find the optimal product mix.?
Solution: The details given in the problem is given in the table below: Machines Products Available capacity in hours. X Y M1 1 1 4 M2 3 8 24 Profit Rs. 5 7 Maximise Z = 5 + 7 S.T + ≤4 3 1 + 8 ≤24 and ≥ 0.
+ ≤4 let =0 so = 4 , and let =0 so = 4
~ 6 ~ + =4 * 3 3 1 + 8 =24 --- 3 + 3 = 12 -3 – 8 = 24 ( حشطناب ) so ( = 1.6 = 2,4 ) Z( 0.3) = 5 × 0 + 7 × 3 = 21, at point A Z (0,0) = 5 × 0 + 7 × 0 = 0 at point O Z (4,0) = 5 × 4 + 7 × 0 = 20 at point B Z (1.6,2.4) = 5 × 1.6 + 7 × 2.4 = 24.8 at point R = 1.6 = 2.4 Z= 24.8
EX8 :SOLVE L.P USE GRGHICAL METHOD
Max. Z=10 X1 +15X2 ST 6x1+4x2 ≤ 30000 2x1+4x2 ≥ 20000
CAULCALTE THE POINTS FOR DRAWING
6x1+4x2 ≤ 30000 LET X1 =0 , X2 = 7500 AND X2=0 , x1=5000 2x1+4x2 ≥ 20000 LET X1=0 X2 = 5000 , AND X2=0 , X2=10000
~ 7 ~ EX 9 :Min.Z=1000X1+800X2 S.T.:- X1≥30---(1) X2≥20---(2) X1+X2≥60---(3) X1, X2≥0
X1≥30 LET X1 =0 THEN X2= 0 AND X2=0 THEN X1=30 X2≥20 LET X1=0 THEN X2= 20 AND X2=0 THEN X1=0 X1+X2≥60 LET X1=0 THEN X2=60 AND X2=0 THEN x1=60
ةطقن A اينا لحن اهتميق ىلع لىصحلل ثلاثلا و لولاا ديقلا عطاقت نم “ : X1 =30 X1+x2 = 60 so x1=30 , x2= 30 ةطقن B اينا لحن اهتميق ىلع لىصحلل ثلاثلا و يناثلا ديقلا عطاقت نم “ : X2 = 20 X1+x2 = 60 so x2=20 , x1=40 Min z= 1000x1 + 800 x2 (30 , 30 ) z= 1000*30 +800 *30 = 54000 (40 , 20 ) z= 1000*40 + 800 *20 = 56000 Z= 54000 x1= 30 x2= 30
~ 8 ~
EX 10 : Solve graphically the given linear programming problem.
Minimize Z = 3 + 5 –3 + 4 ≤12 2 – ≥– 2 * -1 -2X1 + ≤ 2 2 + 3 ≥12 ≥4 ≥2 and ≥0. SOLUTION :
FOR S.T 1 : let = 0 so =3 , let =0 so X1 =-4 FOR S.T 2 : let = 0 so = 2 , let =0 so X1 = -1 FOR S.T 3 : let = 0 so = 4 , let 0 so X1 =6 FOR S.T 4 : let = 0 so = 0 , let =0 so X1 = 4 FOR S.T 5 : let = 0 so = 2 , let =0 so X1 = 0
~ 9 ~ ≥4 ≥2 so ( 4,2 ) for point A Z= 3*4+5*2 = 22 --- ≥4 –3X1 + 4 ≤12 so ( 4,6 ) for point B Z= Z = 3 + 5 Z = 3*4+5*6=42 THE OPTIMAL X1=4 , = 2 Z= 22
EX 11 : SOLVE THE FOLLOWING L.P USE GRAPHICAL METHOD
. Maximize Z = + 4 + 2 ≤16 + ≤6 + ≤4 and ≥ 0. SOLUTION :
FOR S.T 1 : let = 0 so =8 , let =0 so X1 =4 FOR S.T 2 : let = 0 so =6 , let =0 so X1 =6 FOR S.T 3 : let = 0 so =4 , let =0 so X1 =0
~ 10 ~ X2 ≤4 4 + 2 ≤16 SO X1 = 2 = 4 FOR POINT A Z== X1 + A(=(2,4) Z=(2,4) Z= 2+4=6 ( OPTIMAL ) B (0,4) Z= 0+4=4 C(0,0) Z = 0 +0 = 0 D ( 4,0 ) Z= (4,0) Z= 4+0 =4
H.W : SOLVE THE FOLLOWING L.P USE GRAPHICAL METHOD:
Maximise Z = 8000a + 7000b. S.T 3 a + 1 b ≤66 1 a + 1 b ≤45 1 a + 0 b≤20 0 a + 1 b ≤40 a , b ≥0. Minimise Z = 1.5 + 2.5 S.T + 3 ≥3
~ 11 ~ + 6 ≥2
≥0
(2) :Simplex Method The standerd form :
1- all the constraints are equations
2- the right hand of cositraints is postive 3-all the decision variables are postive 4-the objective function is max or min
5- the inequality change to equality by adding slack variables (+s) if the inequality ≤ type. And subtracting slack variables (-s) , if the inequality ≥ type .
EX 12 : CHANGE THE FOLLOWING L.P FROM THE GENERAL
FORM TO STANDERD FORM
MAX Z= 3 +2 +5 MAX Z = 3 +2 +5
S.T 2 - 3 ≤ 3 2 - 3 + = 3 +2 +3 ≥ 5 +2 +3 -S2= 5 3 +2 ≤ 2 3 +2 +S3 = 2
. , ≥0
*- in Simplex method, the problem may have any number of decision variables
*- the objective function is maximization ( max)
*- in Simplex method, the inequalities are ≤ kind .and converted into equations by:
Adding a SLACK VARIABLE ( s).
EX13: Maximize Z = 10 + 20X2. Maximize Z = 10X1+ 20X2 +
~ 12 ~ 5 + 3 ≤ 30 5 + 3 + = 30 3 + 6 ≤ 36 3 + 6 + S2 = 36 2 + 5 ≤ 20 2 1 + 5 + S3 = 20 , ≥ 0. S2 S3 B cj 10 20 0 0 0 0 5 3 1 0 0 30 0 S2 3 6 0 1 0 36 0 S3 2 5 0 0 1 20 EJ 0 0 0 0 0 CJ-EJ 10 20 0 0 0 INTER Variable = X2 OUT Variable = s3 S2 S3 B cj 10 20 0 0 0 0 3.8 0 1 0 0.6 18 0 S2 0.6 0 0 1 - 1.2 12 20 0.4 1 0 0 0.2 4 EJ 8 20 0 0 4 CJ-EJ 2 0 0 0 - 4 INTER Variable X1 OUT Variable s3 S2 S3 B cj 10 20 0 0 0 10 1 0 5/19 0 -3/19 90/19 0 S2 0 0 -3/19 1 -21/191 174/19 20 0 1 -2/19 0 5/19 40/19 EJ 10 20 0 CJ-EJ 0 0 -10/19 0 - 70.19 = 90/19 = 40/19 Z = 1700/19
~ 13 ~
EX14: A company manufactures three products namely X1, x2 and x3.
Each of the product require processing on three machines, Turning, Milling and Grinding. Product X1 requires 10 hours of turning, 5 hours of milling and 1 hour of grinding. Product x2 requires 5 hours of turning, 10 hours of milling and 1 hour of grinding, and Product x3 requires 2 hours of turning, 4 hours of milling and 2 hours of grinding.the available hoiurs 2700 hours of turning, 2200 hours of milling and 500 hours of grinding. The profit of X1, x2 and x3 are Rs. 10, Rs.15 and Rs. 20 per unit respectively. Find the optimal product mix to maximize the profit.use the simplex method ?
Maximise Z = 10 + 15 + 20 . Maximise Z = 10 + 15 + 20 + 0S1 + 0S2 + 0S3 10 + 5 + 2 ≤ 2,700 10 + 5 + 2 +S1 = 2700 5 + 10 + 4 ≤ 2,200 5 + 10 + 4 + S2 = 2200 + + 2 ≤ 500 + + 2 + S3 = 500 All X1, , ≥ 0 S2 S3 B cj 10 15 20 0 0 0 0 10 5 2 1 0 0 2700 0 S2 5 10 4 0 1 0 2200 0 S3 1 1 2 0 0 1 500 EJ 0 0 0 0 0 0 CJ- EJ 10 15 20 0 0 0 INTER Variable OUT Variable s3 S2 S3 B cj 10 15 20 0 0 0 0 9 4 0 1 0 -1 2200 0 S2 3 8 0 0 1 -2 1200 20 0.5 0. 5 1 0 0 0.5 250 EJ 10 10 20 0 0 10 CJ- EJ 0 5 0 0 0 -10 INTER Variable X2 OUT Variable s2 S2 S3 B
~ 14 ~ cj 10 15 20 0 0 0 0 7.5 0 0 1 -0.5 0 1600 15 3/8 1 0 0 1/8 -0.25 150 20 5/16 0 1 0 -1/16 5/8 175 EJ CJ- EJ -15/8 0 0 0 -5/8 -8.75 =0 X2= 150 = 175 Z= 5750
EX15: A company HAS three products A, B and C. They are processed in
three departments X, Y and Z. Products A require 2 hours of department X, 3 hours of department Y,and product B requires 3 hours,2 hours and 4 hours of department X, Y and Z respectively. Product C requires 2 hours in department Y and 5 hours in department Z respectively. The profit of A,
B and C are Rs. 3,, Rs.5 and Rs. 4 respectively. Find the optimal product
mix for maximising profit. THE available hours 8 hours of department
X, 15 hours of department Y and 10 hours of department Z ,
Maximise Z = 3 +5 + 4 . Maximise Z = 3X1 + 5x2 + 4x3 + 0S1 + 0S2 + 0S3 2 + 3 + 0 ≤ 8 2X1 + 3 + 0 + S1 + 0S2 + 0S3 = 8 3 + 2 + 4 3≤ 15 3X1 + 2 + 4 + 0S1 + S2 + 0S3 = 15 0 1 + 2 + 5 ≤ 10 0X1 + 2 + 5 + 0S1 + 0S2 + S3 = 10 , ,x3 all ≥ 0. S2 S3 B cj 3 5 4 0 0 0 0 2 3 0 1 0 0 8 0 S2 3 2 4 0 1 0 15 0 S3 0 2 5 0 0 1 10 EJ 0 0 0 0 0 0 CJ-EJ 3 5 4 0 0 0 Inter Variable X2 Ratios: { 8/3 7.5 5 } Out Variable out s1
S
2 S3 B
cj 3 5 4 0 0 0
5 2/3 1 0 1/3 0 0 8/3
~ 15 ~ 0 S3 -4/3 0 5 -2/3 0 1 14/3 EJ 10/3 5 0 5/3 0 0 CJ-EJ -1/3 0 4 -5/3 0 0 Inter Variable Ratios{ - 29/12 14/15 } Out Variable s3 S 2 S3 B cj 3 5 4 0 0 0 5 2/3 1 0 1/3 0 0 8/3 0 S2 41/15 0 0 -2/15 1 -0.8 89/15 4 -4/15 0 1 -2/15 0 0.2 14/15 EJ 5 4 0 -0.8 CJ-EJ 11/5 0 0 -17/15 0 0.8 Inter Variable X1 Ratios { 4 89/41 - } Out Variable s2 S1 S2 S3 B cj 3 5 4 0 0 0 5 0 1 0 15/41 -10/41 8/41 50/41 3 1 0 0 -2/41 15/41 -12/41 89/41 4 0 0 1 -6/41 4/41 5/41 62/41 EJ CJ-EJ 0 0 0 -45/41 -11/41 -24/41 = 89/41 , = 50/41 , = 62/41 Z = 765/41 3) : Big M–n Method • دٍٕقنا ٌٕكت ايذُػ تناح ًف تقٌشطنا ِزْ وذختست ” ئاست ٔأ ٍي شبكأ “ دٍٕقنا تناح ًف ٔ تطهتخًنا . • ىهػ تٍساٍقنا تغٍصنا ىنا جرًُٕنا مٌٕحت ذُػ تقٌشطنا ِزْ ةشكف يٕطُت : ذكاس شٍغتي حشط ( Si ) ًػاُطصا شٍغتي تفاضإٔ ( Artificial Variable ) ًتنا دٍٕقنا ىنا ةساشا مًحت ” ئاست ٔأ ٍي شبكأ “ ةساشا مًحت ًتنا دٍٕقهن ًػاُطصا شٍغتي تفاضا ٔ ( = ) ايأٔ ، ةساشا مًحت ًتنا دٍٕقنا ” ئاست ٔأ ٍي شغصأ “ ةذكاس ثاشٍغتي ظقف آن فاضٍف ( Si . ) ٔ ، تٌشفص ثلاياؼًب فذٓنا تناد تنداؼي ىنا ةذكاشنا ثاشٍغتًنا تفاضا كنزك ٔ ( +M ) ٔ مٍهقتنا تناح ًف تٍػاُطصلاا ثاشٍغتًهن ( -M ) ىٍظؼتنا تناح ًف . EX 16: MAX Z= 3 – MAX Z= 3 – +O S1+S2+0S3 – MA1 ST 2 + ≥ 2 2 + –S1 +A1 =2
~ 16 ~
+3 ≤ 3 +3 +S2= 3
≤ 4 +S3 = 4 . ≥ 0
Inter Variable X1 Ratios{ 1, 3 ,α }, out Variable A1
Inter variable s1 . ratio (-2 , 4 , α ) , out variable s2
X1= 3 , x2=0 z= 3x1-x1 z= 3*3 -0 =9
EX 17: solve the following l.p use the big M method . Min z= 2x1+3x2
X1+x2 ≥ 5 X1+2x2 ≥ 6 X1,x2 ≥ 0 Solution :
Min z= 2x1+3x2 max z= -2x1 – 3x2 +0s1+0s2 –MA- MA2
X1 X2 S1 S2 S3 A1 b CJ 3 -1 0 0 0 -m -M A1 (2) 1 -1 0 0 1 2 0 S2 1 3 0 1 0 0 3 0 S3 0 1 0 0 1 0 4 EJ -2m -m m 0 0 -m Cj-Ej 3+2m -1+m -m 0 0 0 X1 X2 S1 S2 S3 b CJ 3 -1 0 0 0 3 X1 1 0.5 - o.5 0 0 1 0 S2 0 2.5 (0.5) 1 0 2 0 S3 0 1 0 0 1 4 EJ 3 1.5 -1.5 0 0 Cj-Ej 0 -2.5 1.5 0 0 X1 X2 S1 S2 S3 b CJ 3 -1 0 0 0 3 X1 1 3 0 1 0 3 0 S1 0 5 1 2 0 4 0 S3 0 1 0 0 1 4 EJ 3 9 0 3 0 Cj-Ej 0 -10 0 -3 0
~ 17 ~
X1+x2 ≥ 5 X1 +X2 –S1 +A1 = 5 X1+2x2 ≥ 6 X1 +2X2 –S2+A2 =6 X1,x2 ≥ 0
Inter variable x2 ratio ( 5 , 3 ) , out variable A2
INTER variable x1 , ratio ( 4 , 6 ) out variable A1
X1= 4 , X2=1 Z= 2*4+3*1= 11
EX 18: SOLVE USING BIG M METHOD
MAX Z= X1+2X2+3X3-X4 MAX Z= X1+2X2+3X3-X4-MA1-MA2-MA3 X1+2X2+3X3=15 X1+2X2+3X3+A1=15 2X1+X2+5X3=20 2X1+X2+5X3+A2=20 X1+2X2_X3+X4=10 X1+2X2_X3+X4+A3=10 X1,X2,X3,X4≥0 X1 X2 S1 S2 A1 A2 b CJ -2 -3 0 0 -M -M -M A1 1 1 -1 0 1 0 5 -M A2 1 (2) 0 -1 0 1 6 EJ -2M -3M M M -M -M Cj-Ej -2+2M -3+3M -M -M 0 0 X1 X2 S1 S2 A1 b CJ -2 -3 0 0 -M -M A1 (1/2) 0 -1 1/2 1 2 -3 X2 1/2 1 0 -1/2 0 3 EJ -(M+3)/2 -3 M 3-M/2 -M Cj-Ej -1+M/2 0 -M -3+M-2 0 X1 X2 S1 S2 b CJ -2 -3 0 0 -2 X1 1 0 -2 1 4 -3 X2 0 1 1 -1 1 EJ -2 -3 1 1 Cj-Ej 0 0 -1 -1
~ 18 ~
INTER variable X3 , RATIO (5,4,10) , OUT variable A2
INTER variable X2 , RATIO (17/7,20,10/3) , OUT variable A1
INTER variable X4 , , OUT variable A3
X1 X2 X3 X4 A1 A2 A3 B CJ 1 2 3 -1 -M -M -M -M A1 1 2 3 0 1 0 0 15 -M A2 2 1 (5) 0 0 1 0 20 -M A3 1 2 1 1 0 0 1 10 EJ -4M -5M -9M -M -M -M -M Cj-Ej 1+4M 2+6M 3+9M -1+M 0 0 0 X1 X2 X3 X4 A1 A3 B CJ 1 2 3 -1 -M -M -M A1 -1/5 (7/5) 0 0 1 0 3 3 X3 2/5 1/5 1 0 0 0 4 -M A3 3/5 9/5 0 1 0 1 6 EJ 6-2M/5 3-16M/5 3 -M -M -M Cj-Ej -1+2M/5 7+16M/5 0 -1+M 0 0 X1 X2 X3 X4 A3 B CJ 1 2 3 -1 -M 2 X2 -1/7 1 0 0 0 15/7 3 X3 3/7 0 1 0 0 25/7 -M A3 6/7 0 0 (1) 1 15/7 EJ 1-6m/7 2 3 -M -M Cj-Ej 6+6m/7 0 0 -1+M 0
~ 19 ~ Inter variable x1 , out variable x4
X1=5/2 , x2= 5/2 x3 = 5/2 x4=0 z=15
EX 19:solve use big M method
Z=3x1+2x2
St 2x1+x2 ≤ 1 3x1+4x2 ≥ 4 X1,x2 ≥ 0 Solution :
Max Z=3x1+2x2 max Z=3x1+2x2+0s1+0s2-MA2 St 2x1+x2 ≤ 1 2x1+x2+S1 = 1 3x1+4x2 ≥ 4 3x1+4x2-S2+A2 = 4 X1,x2 ≥ 0 X1 X2 X3 X4 B CJ 1 2 3 -1 2 X2 -1/7 1 0 0 15/7 3 X3 3/7 0 1 0 25/7 -1 X4 (6/7) 0 0 1 15/7 EJ 1/7 2 3 -1 Cj-Ej 6/7 0 0 0 X1 X2 X3 X4 B CJ 1 2 3 -1 2 X2 0 1 0 1/6 5/2 3 X3 0 0 1 -1/2 5/2 1 X1 1 0 0 7/6 5/2 EJ 1 2 3 0 Cj-Ej 0 0 0 -1 X1 X2 S1 S2 A2 b CJ 3 2 0 0 -M 0 S1 2 (1) 1 0 0 1 -M A2 3 4 0 -1 1 4 EJ -3M -4M 0 M -M
~ 20 ~ INTER variable x2 , out variable s1
X1=0 x2 = 1 z=2 Cj-Ej 3+3M 2+4M 0 -M 0 X1 X2 S1 S2 A2 b CJ 3 2 0 0 -M 2 X2 2 1 1 0 0 1 -M A2 -5 0 -4 -1 1 0 EJ 4+5M 2 2+4M M -M Cj-Ej -1-5M 0 -2-4M -M 0
~ 22 ~ TOW PHASES : 1 -متت دوٌقلل ةٌعانطصلاا و ةٌمهولا تارٌغتملا ةفاضا ددب تاءارجلاا سفن . ةقٌرط ًف رم امك BIG M . 2 نٌتلحرم ًف لحلا متٌ . ةٌمهولا و ةٌلصلاا تارٌغتملا عٌمجل ةٌرفص مٌق ءاطعا متٌ ىلولاا ةلحرملا ًف . ءاطعا متٌو ةمٌق 1 ةٌعانطصلاا تارٌغتملا ىلا . 3 فص مٌق ىلولاا ةلحرملا تناك اذا ( CJ-EJ ) ةبجوم وا ةٌرفص مٌق ًه . ةمٌق ناو ( A ) بناج ًف b رفصلا نم ربكا ةمٌق ًه . لثملاا لحلا ىلا انلصوت دق نوكن . ةٌناثلا ةلحرملا ىلا لاقتنلاا متٌلا و . اذا هسكعبو مٌق تناك ( cj-ej ) ةبجوم وا ةٌرفص مٌق ًه . ةمٌق ناو A بناج ًف b ًه . ةٌناثلا ةلحرملا ىلا لقتنن . 4 فدهلا ةلاد مٌق مادختساب ةٌلصلاا فدهلا ةلاد مادختسا متٌ ةٌناثلا ةلحرملا ًف .
EX 20 : SOLVE THE FOLLOING L.P USE THE TOW
PHASE METHOD .
MAX Z = 5X1 + 3X2 ST 2X1+X2 ≤ 1 X1+4X2 ≥ 6 ≥ 6 X1,X2 ≥ 0
SOLUTION : ( FIRST PHASE )
MAX Z = 5X1 + 3X2 MAX Z W= A1 ST 2X1+X2 ≤ 1 2X1+X2 +S1 = 1 X1+4X2 ≥ 6 ≥ 6 X1+4X2 –S2 +A2 = 6 X1 X2 S1 S2 A2 b CJ 0 0 0 0 1 0 S1 2 (1) 1 0 0 1 1 A1 1 4 0 -1 1 3/2 EJ 1 4 0 -1 1 Cj-Ej -1 -4 0 1 0 X1 X2 S1 S2 A2 b CJ 0 0 0 0 1 0 X2 2 1 1 0 0 1 1 A1 -7 0 -4 -1 1 3/2 EJ -7 0 -4 -1 1 Cj-Ej 7 0 4 1 0
~ 23 ~ تًٍق ( cj-ej ) تبجٕي ٔا تٌشفص تبجٕي . تًٍق ٌأ ( A ) بَاج ًف b شفصنا ٍي شبكا . كنزن مثيلاا محنا ىنا اُهصٕت . تٍَاثنا تهحشًنا ىنا لاقتَلاا ىتٌلا ٔ .
EX 21 : SOLVE THE FOLLOING L.P USE THE TOW PHASE
METHOD .
Max z = 3x1+2x2+2x3 max w = A1 5x1+7x2+4x3 ≤7 5x1+7x2+4x3+S1 = 7 -4x1+7x2+5x3 ≥ -2 4x1-7x2-5x3+S2=- 2 3x1+4x2-6x3≥ 29/7 3x1+4x2-6x3-S3+A3 = 29/7 X1,x2,x3≥ 0 phase 1 : فص ( cj-ej ) تٌشفص ٔا تبجٕي ىٍقنا غًٍج . تًٍق ٌا لاا A ًْ بَاج ًف ( b ) شفص ًْ . تٍَاثنا تهحشًنا ىنا لاقتَلاا ىتٌ كنزن . مثيلاا محنا ىنا مصٕتهن . X1 X2 X3 S1 S2 S3 A2 b CJ 0 0 0 0 0 0 1 0 S1 5 7 4 1 0 0 0 1 0 S2 4 -7 -5 0 1 0 0 2 1 A3 3 4 -6 0 0 -1 1 29/7 EJ 3 4 -6 0 0 -1 1 Cj-Ej -3 -4 6 0 0 1 0 X1 X2 X3 S1 S2 S3 A2 b CJ 0 0 0 0 0 0 1 0 X2 5/7 1 4/7 1/7 0 0 0 1 0 S2 (9) 0 -1 1 1 0 0 9 1 A3 1/7 0 -58/7 -4/7 0 -1 1 1/7 EJ 1/7 0 -58/7 -4/7 0 -1 1 Cj-Ej -1/7 0 58/7 4/7 0 1 0 X1 X2 X3 S1 S2 S3 A2 b CJ 0 0 0 0 0 0 1 0 X2 0 1 41/63 4/63 -5/63 0 0 7/9 0 X1 1 0 -1/9 1/9 1/9 0 0 1 1 A3 0 0 -513/63 -37/63 -1/63 -1 1 0 EJ 0 0 -513/63 -37/63 -1/63 -1 1 Cj-Ej 0 0 513/63 37/63 1/63 1 0~ 24 ~ PHASE 2 : آن تٍهصلاا ثاشٍغتًنا ىٍق غي تٍهصلاا فذٓنا تناد واذختسا ىتٌ تهحشًنا ِزْ ًف . لٔذجنا واذختسابٔ ةشٍخلاا تهحشًنا ٍي شٍخلاا . مثيلاا محنا ىنا اُهصٕت ٌٕكَ تٌشفص ٔا تبناس ىٍقنا غًٍج .
EX 22: : SOLVE THE FOLLOING L.P USE THE TOW PHASE
METHOD
MAX Z = 2X1 +5X2
ST 2X1+4X2≤ 8
X1+
+3X2≥9
X1.X2 ≥0
PHASE 1 :
MAX W = A2
2X1+4X2 +S1 = 8
X1+ +3X2 -S2 +A2 = 9
X 1 X2 X3 S1 S 2 S3 A2 b CJ 3 2 2 0 0 0 -M 0 X2 0 41/63 41/63 4/63 0 -5/63 0 2/7 0 X1 1 -1/9 -1/9 1/9 0 1/9 0 1 -M A3 0 0 -513/63 -37/63 -1 -1/63 1 0 EJ 3 2 61+513m/63 29+37m/63 m 11+m/63 -m Cj-Ej 0 0 -(65+513m)/63 -(29+37m)/63 -m -( 11+m)/63 0~ 25 ~
inter variable x2 out variables1
s1 = 1 3 0 -1 1 9
0.5 1 0.25 0 0 2 *3
---
1 3 0 -1 1 9
-1.5 -3 -0.75 0 0 -6
---
-0.5 0 -0.75 -1 1 3
فص ( cj-ej ) تٌشفص ٔا تبجٕي ىٍقنا غًٍج . تًٍق ٌا لاا A ًْ بَاج ًف ( b ) ٍي شبكا شفصنا . تٍَاثنا تهحشًنا ىنا لاقتَلاا ىتٌ لا كنزن . مثيلاا محنا ىنا اُهصٕت ٌٕكَ ٔ X1 X2 S1 S2 A2 b CJ 0 0 0 0 1 0 S1 2 (4) 1 0 0 8 1 A2 1 3 0 -1 1 9 EJ 1 3 0 -1 1 Cj-Ej -1 -3 0 1 0 X1 X2 S1 S2 A2 b CJ 0 0 0 0 1 0 S1 0.5 1 0.25 0 0 2 1 A2 -0.5 0 -0.75 -1 1 3 EJ -0.5 0 -0.75 -1 1 Cj-Ej 0.5 0 0.75 1 0
~ 26 ~
ADDITION EXAPLES” ---
*-Ex 1 :A company produce 3 products (A,B,C).The profit of each 3,2,4
$ respectively .the processes time as following :
A B C AVAILABALE HOURS MACHIN 1 4 3 5 2000
Machine 2 2 2 4 2500
Accompany must manufacturing at least 100 , 200 , 50 units of each product .but not more 150 for product A .write the L.P MODEL
MAX Z = 3x1+2x2+4x3 s.t 4x1+3x2+5x3 ≤ 2000 2x1+2x2+4x3 ≤ 2500 X1 ≥ 100 X2 ≥ 200 X3 ≥ 50 X1 ≤ 150 X1,x2,x3 ≤ 0
*-EX 2 : A company produce 2 products (A,B).on 2 production lines .the
following represent the raw material required and the out put of each product .. the available raw material of product A IS 200 UNITE . and for product B is 150 unite. A Company must manufacture 100 , 80 unit of product A,B respectively .the profit of A is 3 $ . and for B is 4 $ .write the L.P
~ 27 ~ MAX Z = 3x1 +4x2 s.t 5x1 + 4x2 ≤ 200 3x1 + 5x2 ≤ 150 5x1 +4x2 ≥ 100 8x1 + 4x2 ≥ 80 X1,X2 ≥ 0
*- EX3 : A company produce 4 products (A,B.C,D ). the following Represent the raw material require;WRITE THE L.P THAT MIUMUM TE COST . MAX Z= 41X1 +35X2 +96X3 2X1 + 3X2 +7 X3 ≥ 1250 X1 + X2 ≥ 250 Raw material Prouduct A prouduct B Output Prouduct A prouduct B Line 1 5 3 5 8 Line 2 4 5 4 4
Prouduct Raw material Machine1 m2 m3 Minimum requirement A B C D 2 3 7 1 1 0 5 3 0 6 25 1 1250 250 900 300 Cost of Machine 41 35 96
~ 28 ~ 5X3 + 3X2 ≥ 900 6X1 + 25 X2 + X3 ≥ 300 X1,X2,X3 ≥ 0
*- Ex4 : A company produce 3 products (A,B,C ) .the company produce
not more than 400 units of product B .for every three units of product C , there were one unit of by- product ,which can sell by 3 $ per unit , and only 100 unit of this by- product can sell per a month . the profit is 6,8,4 $ per each product respectively .the following represent the the process time of each product . write the L.P MODEL .
Product A Product B Product C Available Machine 1 2 3 1 900 Machine2 - 1 2 600 Machine3 3 2 2 1200 MAX Z = 6X1 + 8X2 + 4X3 + 3 *( X3 / 3) S.T 2X1 + 3X2 +X3 ≤ 900 X2 + 2X3 ≤ 600 3X1 +2X2 +2X3 ≤ 1200 X2 ≤ 400 X3/3 ≤ 100 X1,X2 ,X3 ≥ 0
*-EX 5 : A company produce3 products (A,B,C).ON 3 MACHINE .the
following represent the process time and the cost and the PRICES for each product .
~ 29 ~
Product A product B product C available hours Machine 1 4 2 3 800 Machine2 2 5 4 1000 Machine 3 3 6 8 1200 Cost 35 20 40 SELLING PRICE 70 30 60 MAX Z = ( 70 – 35 ) x1 + ( 30 – 20 ) x2 + ( 60 – 40 ) x3 4x1 + 2x2 + 3x3 ≤ 800 2x2 + 5x2 + 4x3 ≤ 1000 3x1 + 6x2 + 8x3 ≤ 1200 X1,x2,x3 ≥ 0
*-EX6 : A company produce 2 products (A,B).the company decide to
produce 20000 unite of product A .and 40000 unite of B . BUT there are 45000 bottles are available to use either for product A AND B . product A need 3 hour to fill 1000 bottle of product A . AND 1 hour to fill 1000 bottles of PROUDUCT B . the available hours are 66 hours ,the profit of A IS 8 $ AND FOR B IS $ .WRITE THE L.P .
MAX Z = 8X1 +7X2 S.T X1 ≤ 20000 X2 ≤ 40000 X1 +X2 ≤ 45000 3/1000 X1 + 1/1000 X2 ≤ 66 X1,X2,X3 ≥ 0
~ 30 ~
*- EX7 : A company produce 2 products (A,B).THE muumuu that must
produce 100. 200.150 units respectively . the following represent the raw material required . the profit for each profit is 2,5,4 $ respectively .
Product A product B product C available Machine 1 ½ 1 1 500 Machine 2 2 ½ 1/5 400 Max z = 2x1 + 5x2 + 4x3 ½ x1 + x2 +x3 ≤ 500 2x1 + ½ x2 + 1/5 x3 ≤ 400 X1 ≥ 100 X2 ≥ 200 X3 ≥ 150
*- EX8 : A company produce 3 products (A,B,C ) .THE following
represent the prossecc time on machines >
Product machine1 machine 2 A 0.5 0.6 B 0.7 0.8 C 0.9 1.05
There was 85 hours available for each machine . the operating cost for machine 1 is 5 $.and for machine 2 is 4$. The market orders is at least 90,80,60 units for each product . write the L.p model.
Min z = ( 0.5*5 + 06* 4 ) x1 + ( 0.7*5 + 08* 4 ) x2 + (( 0.9*5 +1. 05* 4 ) x3 s.t 0.5 x1 + 0.7 x2 + 0.9 x3 ≤ 85
0.6 x1 + 0.8 x2 + 1.05 x3 ≤ 85 X1 ≥ 90
~ 31 ~ X2 ≥ 80
X3 ≥ 60 X1.x2.x3 ≥ 0
EX 9: FOR FIRST EXAM :
A company produce two products( A,B ).Write the L.P MODEL .if u know :
1- the following represent production data .
Out put Of product Process time for
product Raw material for
product mach in e B A B A B A 2 3 4 2 3 2 M1 5 2 5 6 5 4 M2
2- The raw material available for product ( A,B) are ( 180 , 160 ) tons respectively .the available process time at least 90 hours on machine 1. And not more 120 hours on machine 2 .
3- the company must produce 110 unite of product (A), but not more than 130 units of product (B) .
4- the operating cost is 3$ per hour for machine 1 . and 4$ per hour for machine 2 .
5- the selling price is 40 $ for each product .
6- the( by- product) from product (A) is not more 15 units which can sell with 5 $ per a unite .
~ 32 ~ Max z = 40 – (2*3+6*4 ) x1 +40- (4*3 + 5*4 )x2+ 15 *5 x1 s.t 2x1 + 4x2 ≤ 180 3x1+5x2 ≤ 160 2x1+4x2 ≥ 90 6x1+5x2 ≤ 120 3x1+2x1 ≥ 110 2x1+ 5x2 ≤ 130 X1 ,x2 ≥ 0
Ex12 : solve the following L.P use the big M method .( for first exam) Min z = x1 +2x2 +x3 min z = x1 + 2x2+ x3 +0s1+0s2+MA2
X1 + 1/2x2 +1/2 x3 ≤1 X1 + 1/2x2 +1/2 x3+S1 =1 3/2x1 + 2x2 + x3 ≥8 3/2x1 + 2x2 + x3 –S2+A2 = 8 بناس شبكا مخاذنا شٍغتًنا فص ( Cj-Ej ) بجٕي -* مخاذنا شٍغتًنا تناح ًف ( MIN TO MAX ) بناس مثيلاا ٔ بجٕي شبكا ْٕ . X1 X2 X3 S1 S2 A2 B CJ 1 2 1 0 0 M 0 S1 1 (1/2) 1/2 1 0 0 1 M A2 3/2 2 1 0 -1 1 8 EJ 3M/2 2M M 0 -M M Cj-Ej 1-3M/2 2-2M 1-M 0 M 0 X1 X2 X3 S1 S2 A2 B CJ 1 2 1 0 0 M 2 X2 2 1 1 2 0 0 2 M A2 -5/2 0 -1 -4 -1 1 4 EJ 4-5M/2 2 2-M 4-4M -M M Cj-Ej -3+5M/2 0 -1+M -4+4M M 0
~ 33 ~
Ex13 : solve the following L.P use the big M method( min to max ) .
For the first exam .
Min z = x1 +2x2 +x3 max z = -x1 - 2x2- x3 +0s1+0s2-MA2 X1 + 1/2x2 +1/2 x3 ≤1 X1 + 1/2x2 +1/2 x3+S1 =1
3/2x1 + 2x2 + x3 ≥8 3/2x1 + 2x2 + x3 –S2+A2 = 8
EX14: SOLVE USE THE BIG M METHOD
MIN Z = X1 +X2 MIN Z = X1 +X2+0S1+0S2 +MA1+MA2 2X1+X2 ≥ 4 2X1+X2 -S1+A1=4 X1 +7X2 ≥ 7 X1 +7X2-S2+A2=7 X2 INTER VERAIBLE X1 X2 X3 S1 S2 A2 B CJ -1 -2 -1 0 0 -M 0 S1 1 (1/2) 1/2 1 0 0 1 -M A2 3/2 2 1 0 -1 1 8 EJ -3M/2 -2M -M 0 +M -M Cj-Ej -1+3M/2 -2+2M -1+M 0 -M 0 X1 X2 X3 S1 S2 A2 B CJ -1 -2 -1 0 0 -M -2 X2 2 1 1 2 0 0 2 -M A2 -5/2 0 -1 -4 -1 1 4 EJ -4+5M/2 -2 -2+M -4+4M +M -M Cj-Ej 3-5M/2 0 1-M 4-4M -M 0 X1 X2 S1 S2 A1 A2 B CJ 1 1 0 0 M M M A1 2 1 -1 0 1 0 4 M A2 1 (7) 0 -1 0 1 7 EJ 3M 8M -M -M M M Cj-Ej 1-3M 1-8M +M 4M 0 0
~ 34 ~
Ex 15: solve by using big M method
min z = 4x1+x1 max z= -4x1+-x1 +0S1+0S2 –MA1- MA2 3X1+X2=3 3X1+X2+A1 =3 4X1+3X2≥ 9 4X1+3X2-S2+A2 =9 X1+2X2≤ 3 X1+2X2+S3 = 3 INTER VARABILE X1 X1 X2 S1 S2 A1 B CJ 1 1 0 0 M M A1 (13/7) 0 -1 1/7 1 3 1 X2 1/7 1 0 -1/7 0 1 EJ 13M+1/7 1 -M M-1/7 M Cj-Ej 6-13M/7 0 +M -M+1/7 0 X1 X2 S1 S2 B CJ 1 1 0 0 1 X1 1 0 -7/13 1/13 3 1 X2 0 1 1/13 -4/21 1 EJ 1 1 -6/13 -13/273 Cj-Ej 0 0 6/13 13/273 X1 X2 S2 S3 A1 A2 B CJ -4 -1 0 0 -M -M -M A1 (3) 1 0 0 1 0 3 -M A2 4 3 -1 0 0 1 6 0 S3 1 2 0 1 0 0 3 EJ -7M -5M M 0 -M -M Cj-Ej -4+7M -1+5M -M 0 0 0
~ 35 ~
EX16:SOLVE THE FOLLOWING L.P USE BIG M METHODE
MAX Z = 3X1+X2+2X3 MAXZ= 3X1+X2+2X3+0S1+0S2-MA3 12X1+3X2 +6X3+3S1 = 9 4X1+X2+2X3+S1=3 8X1+X2-4X3+2S2 = 10 4X1+1/2 X2 +2X3+S2=5 3X1-S3 =0 3X1-S3+A3 =0 X1 X2 S2 S3 A2 B CJ -4 -1 0 0 -M -4 X1 1 1/3 0 0 0 1 -M A2 0 -5/3 -1 0 1 2 0 S3 0 (5/3) 0 1 0 2 EJ -4 (-4-5M)/3 M 0 -M Cj-Ej 0 (1+5M)/3 -M 0 0 X1 X2 S2 S3 A2 B CJ -4 -1 0 0 -M -4 X1 -M A2 -1 X2 0 1 0 3/5 0 6/5 EJ Cj-Ej X1 X2 X3 S1 S2 S3 A3 B CJ 3 1 2 0 0 0 -M 0 S1 4 1 2 1 0 0 0 3 0 S2 4 1/2 -2 0 1 0 0 5 -M A3 (3) 0 0 0 0 -1 1 0 EJ -3M 0 0 0 0 M -M Cj-Ej 3+3 M 1 2 1 0 -M 0 X1 X2 X3 S1 S2 S3 B CJ 3 1 2 0 0 0 0 S1 0 1 (2) 1 0 4/3 3 0 S2 0 1/2 -2 0 1 4/3 5 3 X1 1 0 0 0 0 -1/3 0 EJ 3 0 0 0 0 -1 Cj-Ej 0 1 2 0 0 1
~ 36 ~
Ex 17: solve using tow phase method ( for first exam )
Min z = 2x1+x2 5x1+10x2-s1+A1= 8 X1+X2+S2=1 تظحلاي : تناذنا MIN فاضٌ 1 مثي كنزك MAX . تناذنا ٌلا تٍَاثنا تهحشًنا ًف ٍكنٔ MIN فاضٌ فاضٌ +M . ٌأ A لٔاذجنا ٍي فزحتلا .
(A) value = 0 SO WE GO TO SECOND PAHSE
X1 X2 X3 S1 S2 S3 B CJ 3 1 2 0 0 0 2 X3 0 1/2 1 1/2 0 2/3 3/2 0 S2 0 3/2 0 1 1 8/3 8 3 X1 1 0 0 0 0 -1/3 0 EJ 3 1 2 1 0 1/3 Cj-Ej 0 0 0 -1 0 -1/3 X1 X2 S1 S2 A1 B CJ 0 0 0 0 1 1 A1 5 (10) -1 0 1 8 0 S2 1 1 0 1 0 1 EJ 5 10 -1 0 1 Cj-Ej -5 -10 1 0 0 X1 X2 S1 S2 A1 B CJ 0 0 0 0 1 0 X2 1/2 1 -1/2 0 1/10 8/10 0 S2 1/2 0 1/10 1 -1/10 2/10 EJ 0 0 0 0 0 Cj-Ej 0 0 0 0 0
~ 37 ~
EX18: SOLVE USE TOW PHASE METHOD
MIN Z = 15/2 X1 -3X2 3X1-X2-X3 ≥ 3 3X1-X2-X3-S1+A1= 3 X1- X2 +X3 ≥ 2 X1- X2 +X3-S2+A2=2 تظحلاي MIN اُفضا 1 اُفضا تٍَاثنا تهحشًنا ًفٔ +M ٌأ A ٍي فزحت لا لٔاذجنا . سٕٓظ وذػٔ X3 ئاست آَا ًُؼٌ فذٓنا تناد ًف 0 . X1 X2 S1 S2 A1 B CJ 2 1 0 0 +M 1 X2 1/2 1 -1/2 0 1/10 8/10 0 S2 1/2 0 1/10 1 -1/10 1/5 EJ 1/2 1 -1/2 0 1/10 Cj-Ej 3/2 0 1/2 0 M-1/10 X1 X2 X3 S1 S2 A1 A2 B CJ 0 0 0 0 0 1 1 1 A1 (3) -1 -1 -1 0 1 0 3 1 A2 1 -1 1 0 -1 0 1 2 EJ 4 2 0 -1 -1 1 1 Cj-Ej -4 2 0 1 1 0 0 X1 X2 X3 S1 S2 A1 A2 B CJ 0 0 0 0 0 1 1 0 X1 1 -1/3 -1/3 -1/3 0 1/3 0 1 1 A2 0 -2/3 (4/3) 1/3 -1 -1/3 1 1 EJ 0 -2/3 4/3 1/3 -1 -1/3 1 Cj-Ej 0 2/3 -4/3 -1/3 1 4/3 0 X1 X2 X3 S1 S2 A1 A2 B CJ 0 0 0 0 0 1 1 0 X1 1 -1/2 0 -1/4 -3/12 1/4 3/12 15/12 0 X3 0 -1/2 1 1/4 -3/4 -1/4 3/4 3/9 EJ 0 0 0 0 0 0 0 Cj-Ej 0 0 0 0 0 1 1
~ 38 ~
Ex 19: solve use simplex method .for first exam
Min z = -2x1 – 4x2 max z= 2x1 +4x2 +0s1+0s2
-3x1 + 2x2 ≥ -18 3x1 – 2x2 +s1 = 18
8x1+4x2 ≤24 8x1 +4x2 +s2 = 24
2x2 ≤ 14 2x2 +S3 =14
X1 X2 X3 S1 S2 A1 A2 B CJ 15/2 -3 0 0 0 M M 15/2 X1 1 -1/2 0 -1/4 -3/12 1/4 3/12 15/12 0 X3 0 -1/2 1 1/4 -3/4 -1/4 3/4 3/9 EJ 15/2 -15/4 0 -15/8 -45/24 15/8 45/24 Cj-Ej 0 3/4 0 15/8 45/24 M-15/8 M-45/12 X1 X2 S1 S2 S3 B CJ 2 4 0 0 0 0 S1 3 -2 1 0 0 18 0 S2 8 (4) 0 1 0 24 0 S3 0 2 0 0 1 14 EJ 0 0 0 0 0 Cj-Ej 2 4 0 0 0 X1 X2 S1 S2 S3 B CJ 2 4 0 0 0 0 S1 7 0 1 1/2 0 30 4 X2 2 1 0 1/4 0 6 0 S3 -4 0 0 -1/2 1 2 EJ 8 4 0 1 0 Cj-Ej -6 0 0 -1 0~ 39 ~
Transportation problems --- 1- North west corner method : --- :
1- After u finish yr assignment ,no of cell must be equal to (m+n-1) 2- always the assignment to the cell on the north west of the table . 3-when u do yr assignment if there were 2 same value one must be 0 . 4-the display must be equal to requirement , if not we must add a column or a row with the difference and zero cost .
EX1 :solve the following transportation problem use North west corner method 15 5 5 5 2 6 12 Number of cells = m+n-1 = 4+4-1 = 7
The total cost = 4*15 + 3*5 + 5*6 + 2*5 + 2*5 + 2*6 + 1*12 = 1 1 8 2 4 2 5 3 6 1 7 4 2 5 1 1 requirement 20 5 10 5 8 6 12 15 0 10 5 7 2 18 DISTRIBTION SUPPLY A B PLANT C D require
~ 40 ~
EX2 : FIND THE OPTIMALSOLUTION FOR THE( TP ) USE THE NORTH WEST MOTHED ? 5 5 7 3 10 0 15 No of cells = m+n-1 = 4+4-1 = 7 Cost = 4*5+6*5+2*7+1*3+10*3+0*1+3*15=
Ex3: find the optimal solution use north west method
Not :There was a difrrence between the Display and the requirement 10 , so we added a new column with 10 value and zero cost .
1 5 3 2 1 2 3 3 2 1 5 4 6 5 7 3 10 5 10 3 10 0 15 5 12 7 13 10 15 5 3 2 1 2 3 3 2 1 4 6 5 15 15 15 15 10 25 1 5 5 supply requir ement requir emen t supply
~ 41 ~ 10 5 15 5 10 5 10 No of cell = m+n-1 = 4+4-1 = 7
The total cost = 4*10 +6*5+2*15+2*5+3*10+2*5+0*10 = 2- the minimum cost method :
Here we Distributing to the cell with the minimum cost first , then we repeat with same way >
Ex4 : find the optimal solution use for the transportation problem use minimum cost method .
3 12 2 8 7 18 0
The total cost = 2*3+1*12+2*3+1*8+2*7+6*18+0*8 =
0 5 3 2 1 2 3 3 2 1 0 4 6 5 0 0 15 5 `15 15 10 15 10 25 20 5 1 5 5 10 requir ement supply 9 6 8 2 4 2 5 3 6 1 7 4 2 5 1 8 15 3 10 2 7 18 0 require 20 18 10 3 8 12 supply
~ 42 ~
Ex5: find the optimal solution use minimum cost method
Not :There was a difference between the supply and the requirement 10 , so we added a new column with 10 value and zero cost .
15 0 15 10 5 5 10 3- Vogel method “ ---
*calculate the difference between the smallest and second smallest cost t in each column and row
5 3 2 1 2 3 3 2 1 4 6 5 15 15 15 15 10 25 1 5 5 requir emen t supply 0 5 3 2 1 2 3 3 2 1 0 4 6 5 0 0 15 `15 0 15 5 15 5 supply require 10 25 20 15 10
~ 43 ~ *- chose the big value of the difference ,
*- if the problem is max , the assignment must be to big cost in the column/row , if the problem is min then the assignment is to lowest cost in the column / row .
Ex6 : find the optimal solution for the following transportation problem
that minimize the cost use the Vogel method .
DISTRIBUTION SUPPLY requirement SOLUTION :
* calculate the difference between the smallest cost in column and row , then we chose the big difference , and because the problem is min so we assignment to the lowest cost in the column/ row .
5 8 15 1 0 6 2 3 11 7 1 9 1 PLANTS 2 3 6 1 10 7 5 3 2
~ 44 ~
*- we repeat the same procedure , till we finish all the assignments .
1 5 1 6 3 1
Ex7 : find the optimal solution for the following transportation problem that minimize the cost use the Vogel method .
DISTRIBUTION SUPPLY Retirement 10 5 9 21
* the SUPPLY and requirement not equal so we added row with 15 and zero cost 5 8 15 1 0 6 2 3 11 7 1 9 1 PLANTS 2 3 6/1/0 [1][1][5] 1/0 [1] 10 [3][3][4] 7/6 5 3 2/1 [1] [3] [5] [6] [3] [5] [4] [2] [3] [4[ [2] 5 8 15 1 0 6 2 3 11 7 1 9 1 Plant 2 3 12 8 10
~ 45 ~ 0 5 7 8 10 9 6 Retirement 10 0 5 9 21 15 7 [1] [0] [6] [1] [0] [0] [1] [1] [0] [1] [1] [3] [6] No of cells = m+n-1 = 4+4-1 = 7 Total cost :0*2+3*5+7*7+1*8+5*10+0*9+0*6 =
Ex8 : find the optimal solution for the following transportation problem that minimize the cost use the Vogel method
5 8 15 1 0 6 2 3 11 7 1 9 1 Plant 2 3 dummy 12 [1] [1][1][1] 8 [1][1][1][1] 10 [3][3] 15 6 [0][0][0] 0 0 0 0
~ 46 ~ DISTRIBUTION SUPPLY 8 Solution ---
Not : we must balance the supply with the demand .
DISTRIBUTION SUPPLY 8 25 35 105 20 25 [5] [5] 35 5 [2] [5] [5] 105 85 [2] [2] [2] [2] [3] 20 [0] [0] [0] [0] [0] 15 [0] [0] [0] [0] [0] 14 4 10 13 11 10 4 6 8 13 8 13 1 Plant 2 3 4 50 70 30 50 10 8 13 11 9 14 4 10 13 11 10 4 6 8 13 8 13 Plant 1 2 3 4 50 /25/20 [2][2][2][5] 70 [2][2][2][2][2] 30 [6] 50/30 [1][1][3][5][6] 8 13 11 9 0 0 0 0
~ 47 ~
4- stepping-Stone Method for Testing the optimality of the
solution :
---
We can do that with help of The stepping-Stone Method:
*- calculate the value of( v and u) for assignment cells in the matrix ,not
that v1 always equal to 0 .
*-calculate the cost value for the empty cells by using (c –v-u ).if these
value are positive that mean , the solution is optimal , if not the solution is not optimal .
*- if the solution not optimal , we chose the cell with the most negative value , with sign + ,
*- we draw the path ,and from this path we chose the great value to move it .the path begin with +the – then+ then - ….. etc .
Ex9: test the optimal solution for the following transportation problem . use the stepping-Stone Method:
requireqirment 7 5 3 2
solution :
A- the first stage :
1- calculate the value of( v and u) for assignment cells in the matrix ,not that v1 always equal to 0.
1 5 1 6 3 1 5 8 15 1 0 6 2 3 11 7 1 9 1 Plant 2 3 6 1 10
~ 48 ~ V1=0 V2= 1 V3=10 V4=4 U1=2 U2=-3 U3= 5 U1 + v1 =2 U1+0= 2 U1 =2 u1 + v2 = 3 2+V2 =3 V2 = 1 u2 + v4 = 1 U2+4= 1 U2 = -3 u3 + v1 = 5 U3+0= 5 U3=5 u3 + v3 = 15 5+V3 = 15 V3 = 10 u3 + v4 = 9 5+ V4 = 9 V4= 4
2- calculate the cost value for the empty cells by using (c –v-u ).if these value are positive that mean , the solution is optimal , if not the solution is not optimal V1=0 V2=1 V3= 10 V4=4 U1=2 U2=-3 U3=5 c-v-u = 11-2-10 = -1 , 7-2-4 =1 1+3-0 = 4 , 0+3-1= 2 , 6+3 -10 = -1 , 8- 5-1 = 2 1 4 2 -1 2 5 15 2 3 1 9 8 1 6 11 7 0
~ 49 ~
3- because there was negative value ,so the solution not optimal .here we chose the most negative value -1 .and put + sing in the cell .and draw the path .
requirement 7 5 3 2
B- the second stage : the Evaluation
For the solution from the first stage repeating the points (1,2 ) above . *- calculate( u and v ) value
- 1 5 + 1 + 6 - 3 1 - 5 + 1 + 7 - 2 1 5 1 1 7 2 1 5 15 2 3 1 9 5 15 2 3 1 9 5 8 15 1 0 6 2 3 11 7 1 9 6 1 10
~ 50 ~
V1=0 v2 = 2 v3=1 0 v4 = 4 U1=1
U2= - 3
U3= 5
U1+v2 = 3 1+v2 =3 v2= 2 , U1+v3 = 11 u1+10 = 11 u1 =1 U2+v4 = 1 u2+4 = 1 u2 = -3 . U3+v1=5 u3+0=5 u3=5
U3+v3=15 5+v3= 15 v3=10 ,U3+v4 =9 5+v4= 9 v4= 4 *- cost for empty cells : using [ C-V-U]:
V1=0 v2 = 2 v3=1 0 v4 = 4 U1=1
U2= -3
U3= 5
2-1-0=1 7-1-4 = 2 1+3-0= 4 0+3-2=1 6+3-10=-1 8 -5-2= 1 Since there was one negative value so the solution not optimal , we need to continue .till we get the optimal solution .
Ex10 : SOLVE USE VOGEL MOTHED AND TEST THE SOLUTION
Distribution center supply
Requirement 20 15 25 5 15 9 3 11 1 8 1 0 6 2 7 1 0 6 2 1 2 1 Plant 2 20 40
~ 51 ~
supply
Requirement 20 15 25 5 [1] [1] [4]
*- calculate the (u,v) value for assignments cells > V1=0 v2= -2 v3=4 U2=-2
U2=2
U1+v3 = 2 u1+4=2 u1= -2 , U2+v1 =1 u2+0=1 u2=1 U2+v2=0 2+v2=0 v2=-2 , U2+v3 = 6 2+v3= 6 v3=4 *- calculate the cost for the empty cells using ( C-U-V) :
V1=0 V2= -2 V3= 4
U1=-2
U2=2
(C-U-V) = 2+2-0 =4 1+2+2= 5 ( OPTIMAL ).
EX11 : SOLVE THE FOLLOWING TRANSPORTATION PROBLEM USE THE NORTH WEST CORNER AND TEST THE SOLUTION .
20 20 15 5 20 20 15 5 20 20 15 5 1 0 6 2 1 2 1 Plant 2 20 [1] 40 [1] 1 6 2 0 2 1
~ 52 ~
Distribution center supply
Requirement 20 15 25 supply
Requirement 20 15 25
First stage : *- calculate the (u,v) value for assignments cells
V1=0 V2= -1 V3=5 U1=2
U2=1
U1+V1 = 2 u1+0 =2 u1=2 , U1+V2 = 1 2+v2= 1 v2=- 1 U2+V2 = 0 u2-1=0 u2=1 , U2+V3= 6 1+v3=6 v3=5 *- calculate the cost for the empty cells using ( C-U-V) :
V1=0 V2= -1 V3=5
U1= 2
U2=1
( C-U-V) 2-2-5 = -5 , 1-1-0 = 0 ( not optimal ) 20 0 15 25 20 - 0 + + 15 - 25 1 0 6 2 1 2 1 Plant 2 20 40 1 0 6 2 1 2 1 Plant 2 20 0 40 0 6 2 1 2 1
~ 53 ~
Second stage : the Evaluation
*- calculate the (u,v) value for assignments cells V1=0 v2=-6 v3=0 U1=2
U2= 6
U1+v1 =2 u1+0 = 2 u1=2 , U1+v3 = 2 2+v3 =2 v3=0 U2+v2=0 6+v2 = 0 v2= -6 , U2+v3=6 u2+0=6 u2=6 *- calculate the cost for the empty cells using ( C-U-V) :
V1= 0 v2= -6 v3=0 U1=2
U2=6
( C-U-V) 1-2+6 = 5 , 1-6-0 = -5 ( not optimal) 20 0 15 25 - 20 + 0 + 15 - 25 20 20 15 5 1 1 2 1 2 1 0 6 1 1 0 6 2 2 2 0 6 2
~ 54 ~ *calculate the (u,v) value
V1=0 v2= -2 v3=4 U1=-2
U2=2
* calculate (c-u-v ):
(C-U-V) = 2+2-0 =4 1+2+2= 5 ( OPTIMAL)
5- Transportation simplex method :
---Find the optimal solution for the following transportation problem use the transportation simplex method :
Requirement 5 15 20 20 20 15 5 2 6 0 1 1 2 5 2 2 6 1 4 2 6 3 14 6 20
~ 55 ~
1- check the problem . must be balancing , if not make the balance for it .2- chose any method for the distribution .( for example we use the north -west corner )
5 9
6 1
20
5 15 6 21
3- calculate the( u,v )for the basic variable . use( u1= 0) always . V1= v2= v3= U1=0 U2= U3= 5 9 6 1 20 5 2 2 6 1 4 2 6 3 14 9 7 20 5 2 2 6 1 4 2 6 3
~ 56 ~
Assignment models :
---
Ex1: Solving the assignment problem by enumeration( دادعت)
jobs MACHINE1 X MACHINE2 Y MACHINE 3 Z
JOB1 A 11 16 21 JOB2 B 20 13 17 CJOB3 C 13 15 12 AXBYCZ 11+13+12=26 AXBZCY 11+17+15=43 AYBXCZ 16+20+12=48 AYBZCX 16+17+13=46 AZBYCX 21+13+13=47 AZBXCY 21+20+15
THE OPTIMAL ASSIGNMENT IS ( JOB 1 ON M1,JOB2 ON M2.JOB3 ON M3)THE TOTAL TIME IS 26 .
2- Hungarian Method / Assignment algorithm”
1- make sure that the cost matrix is balanced, if not do that (square matrix WITH ZERO )
2-Deduct the smallest element in each row from the other elements of the row 3- Deduct the smallest element in each COLUMN from the other elements of the row 4- IN THE cost matrix, cover all the zeros by MINIMUM LINES .
5- If the lines are equal to the number of rows or columns), we can make assignment 6-and if the lines not equal to the number of colmumn / rows do the following : (i) Subtract the smallest element from the elements of all other uncovered cells. (ii) Add this element to the elements of the crossed cells.
(iii) Do not alter the elements of covered cells.
7- Once again cover all the zeros by minimum number of horizontal and vertical lines>
~ 57 ~
EX2 : THERE WERE 3 JOBS AND 3 MACHINES . FIND THE OPTIMAL
ASSINGMENT THAT MAKE THE OPTIMAL PROCESS TIME .USE THE Hungarian
Method
2 3 4
5 1 2
3 2 4
FOR THE ROW :
0 1 2
4 0 1
1 0 2
FOR THE CULUMN :
0 1 1 4 0 0 1 0 1 1 1 4 1 0 1 1 1 4 0 1 1 J1 ON M1 = 2 , J2 ON M3 = 2 , J3 ON M2 = 2 . THE TOTAL PROCEES TIME IS 6 MIN
Ji J2 j3 M1 M2 M3 0 0 0 WRONG 0 0 0
~ 58 ~
EX3 : THERE WERE 3 JOBS AND 3 MACHINES . FIND THE OPTIMAL ASSINGMENT
THAT MAKE THE OPTIMAL PROCESS TIME .USE THE Hungarian Method
2 4 5 0 8 6 1 2 3 ةدمعلاا ددع و فوفصلا ددع يواسٌ لا طوطخلا ددع . صٌصختلا متٌلا اذل . ًتلاا متٌ : ةاطغملا رٌغ ماقرلاا نم مقر لقا راٌتخا . ماقرلاا ًقاب نم مقرلا اذه حرطٌ . ىلا فاضٌ و عطاقتلا ماقرا . ًه امك ىقبت ةاطغملا ماقرلاا و . ( WRONG ) J1 ON M2=4 , J2 ON M 2= 0 J3 ON M 3 =3 THE TTAL TIME = 7
0 2 3 0 8 6 0 1 2 0 1 1 0 7 4 0 0 0 0 1 1 0 7 4 0 0 0 0 0 0 0 6 3 1 0 0 0 0 0 6 3 1 0 0 6 3 1 0 Ji J2 j3 M1 M2 M3 0 0 0 0 0 0
~ 59 ~
EX4 : THERE WERE 3 JOBS AND 3 MACHINES . FIND THE OPTIMAL
ASSINGMENT THAT MAKE THE OPTIMAL PROCESS TIME .USE THE Hungarian
Method
EX 5 : THERE WERE 3 JOBS AND 3 MACHINES . FIND THE OPTIMAL
ASSINGMENT THAT MAKE THE OPTIMAL PROCESS TIME .USE THE Hungarian
Method JI=M1 , J2=M2, J3=M3 TOTAL = 11+13+0 = 24 jobs M1 M2 M3 J1 11 16 21 J2 20 13 17 J3 13 15 12 jobs M1 M2 M3 J1 0 5 10 J2 7 0 4 J3 1 3 0 jobs M1 M2 M3 J1 0 5 10 J2 7 0 4 J3 1 3 0 jobs M1 M2 M3 J1 (0) 5 10 J2 7 (0) 4 J3 1 3 (0 jobs M1 M2 J1 11 16 J2 20 13 J3 13 15 jobs M1 M2 M3 J1 11 16 0 J2 20 13 0 J3 13 15 0 jobs M1 M2 M3 J1 0 3 0 J2 9 0 0 J3 2 2 0 jobs M1 M2 M3 J1 11 16 0 J2 20 13 0 J3 13 15 0 jobs M1 M2 M3 J1 (0) 3 0 J2 9 (0) 0 J3 2 2 (0)
~ 60 ~
EX: A company has five jobs J1.J2,J3.L4.J5 and five machines
M1.M2.M3.M4.M5 . The given matrix shows the return. of assigning a
job to a machine. Assign the jobs to machines so as to maximize the total returns. USE THE Hungarian Method
JOBS M1 M2 M3 M4 M5 J1 5 11 10 12 4 J2 2 4 6 3 5 J3 3 12 5 14 6 J4 6 14 4 11 7 J5 7 9 8 12 5 حابرلاا مٌظعت وه فدهلا . لاوا بجٌ كلذل " عٌمج حرط للاخ نم ةٌندت ىلا ةفوفصملا لٌوحت ةفوفصملا ًف مقر ربكا نم ةفوفصملا بف ماقرلاا . اقباس رم امك لمعلا متٌ مت . " JOBS M1 M2 M3 M4 M5 J1 9 3 4 2 10 J2 12 10 8 11 9 J3 11 2 9 0 8 J4 8 0 10 3 7 J5 7 5 6 2 9 JOBS M1 M2 M3 M4 M 5 J1 7 1 2 0 8 J2 4 2 0 3 1 J3 11 2 9 0 8 J4 8 0 10 3 7 J5 5 3 4 0 7 صٌصختلا متٌلا طوطخ ثلاث . ةفوشكملا ماقرلاا نم مقر لقا حرطن . عطاقتلا ىلا فاضٌو . ًه امك ىقبت ةاطغملا ماقرلااو . JOBS M1 M2 M3 M4 M5 J1 3 1 2 0 7 J2 0 2 0 3 0 J3 7 2 9 0 7 J4 4 0 10 3 6 J5 1 3 4 0 6
~ 61 ~ J1=M3=10 , J2=M5 =5 , J3= M4=14 J4=M2=14 . J5=M1=7 TOTAL = 50 JOBS M1 M2 M3 M4 M5 J1 2 0 1 0 6 J2 0 2 0 4 0 J3 6 1 8 0 6 J4 4 0 10 4 6 J5 0 2 3 0 5 JOBS M1 M2 M3 M4 M5 J1 1 0 0 0 5 J2 0 3 0 5 0 J3 5 1 7 0 5 J4 3 0 9 4 5 J5 0 3 3 1 5 JOBS M1 M2 M3 M4 M5 J1 1 0 (0) 0 5 J2 0 3 0 5 (0) J3 5 1 7 (0) 5 J4 3 (0) 9 4 5 J5 (0) 3 3 1 5
~ 62 ~
QUEUING SYSTEM OR PROCESS :
راظتنلاا طوطخ
*- Poisson Arrival / Poisson output / Number of channels / Infinite capacity /
FIFO Model [ M / M / 1 / ( / FIFO) ]
1. Average number of arrivals per unit of time = λ 2. Average number of units served per unit of time = μ 3- 1 / λ : Inter arrival time between two arrivals. 4-1 / μ : Service time between two units or customers. 5. Probability that the system is empty = P0 = (1 - ρ )
6. Probability that there are .n. units in the system = Pn = ρn P0 7. Average number of units in the system = E (n) = or = Lq + λ/ μ
8. Average number of units in the waiting line = EL = = or 9. Average time an arrival spends in the system = E (v) = or 10. System is busy = ρ 11.system Idle time = (1− ρ)
12- ρ = (λ /μ) :System utility or traffic intensity which tells us how much time the system was utilized in a given time. For example given time is 8 hours and if ρ = 3 / 8, it means to say that out of 8 hours the system is used for 3 hours and (8 . 3 = 5) 5 hours the is idle
~ 63 ~
Ex1 : a store has one cashier to serve the customers.. The arrival rate is 9
customers for every 5 minutes and the cashier can serve 10 customers in 5 minutes. Assuming Poisson arrival rate and exponential distribution for service rate,find:
(a) Average number of customers in the system.
(b) Average number of customers in the queue or average queue length. (c) Average time a customer spends in the system.
(d) Average time a customer waits before being served
Solution :
Arrival rate is λ = (9 / 5) = 1.8 customers per minute.
Service rate = μ = (10 / 5) = 2 customers per minute. Hence ρ = (λ /μ) = (1.8 / 2 ) = 0.9
(a) Average number of customers in the system = E (n) = ρ / (1− ρ) = 0.9 / (1 . 0.9) = 0.9 / 0.1
= 9 customers.
(b) Average time a customer spends in the system = E (v) = 1/μ(1− ρ) =1/(μ − λ) = 1 / (2 .
1.8) = 5 minutes.
(c) Average number of customers in the queue = E (L)
= ρ2 /(1− ρ) = λ2 /μ(μ − λ) = (ρ)× λ /(μ − λ) = 0.9 × 1.8 / (2 . 1.8) = 8.1 customers.
(d) Average time a customer spends in the queue = ρ /μ(1− ρ) = λ /μ(μ − λ) = 0.9 / 2 ( 1 . 0.9)
~ 64 ~
Ex1: for a job-shop the time spent on the jobs have an exponential
distribution with mean of 30 minutes, and the arrival is approximately Poisson with an average rate of 10 per 8 hour day, what is the job-shop expected idle time each day? How many jobs are in the job-shop>
Solution:
λ = 10 jobs per 8 hour day = 10 / 8 = 5/4 jobs per hour.
Given 1/μ = 30 minutes, hence μ = (1/30) × 60 = 2 jobs per hour.
Utility ratio = ρ = (λ /μ) = (5/4) / 2 = = 5 / 8. = 0.625. This means out of 8 hours 5 hours the system is busy
Probability that there is no queue = The system is idle = (1− ρ) = 1 - (5 / 8) = 3 / 8 = That is out of 8 hours the job-shop will be idle for 3 hours. Number of jobs in the job-shop = Average number of jobs in system = λ / (μ − λ) = or = ρ / (1− ρ) = 0.625 / (1 . 0.625) = 5 / 3 jobs.
Ex2:The arrivals at a maintenance shop is following Poisson distribution
with an average time of 10 minutes between one arrival and the next. The repair time is distributed exponentially with a mean of 3 minutes.
What is the probability that a job arriving at the shop will have to wait?
Solution :
Time interval between two arrivals = 10 min. = 1/ λ so λ= 1/10 = 0.1 job per min .
~ 65 ~ لمع رما لك ساظتَلاا فص ًف ةدٕجٕي مًػ شيأا ٌٕكت ايذُػ ساظتَلاا ًف ىقبٌ . مًؼنا شيأ ساظتَلاا فص ًف مًػ شيا يأ ذجٌٕلا ايذُػ ساظتَلاا ىنا جاتحٌلا . ساظتَلاا وذػ لاًتحا كنزن ٌٕكٌ : the probability of that an arrival does not wait = P0 = (1− ρ).
So the probability that job has to wait = 1 -The probability job not wait = (1 - P0) = 1 - (1− ρ) = so ρ = 0.3. That means 30% of the time the job
