5 files changed, 127 insertions, 134 deletions
diff --git a/116/CH12/EX12.1/exa12_1.sce b/116/CH12/EX12.1/exa12_1.sce
index 1032ecdda..b604ac666 100755
--- a/116/CH12/EX12.1/exa12_1.sce
+++ b/116/CH12/EX12.1/exa12_1.sce
@@ -1,23 +1,21 @@
-
-//Caption:Program to calculate how often do two calls arrive with less than 0.01 sec between them
- 
-//Example 12.1
- 
-//Page 524
- 
-disp('The average arrival rate is')
-
-lam=(3600/10000)//arrivals per sec
-
-disp('From equation 12.2, the probability on arrival in 0.01-sec interval is')//equation on page 524
-
-P0=(%e^-0.0278)
-
-disp('Thus 2.7% arrivals occur withnin 0.01 sec of the pervious arrival. Since the arrival rate is 2.78 arrivals per second, the rate of occurrence of intervarrival time less than 0.01 sec is')
-
-2.78*0.027
-
-disp("Result")
- 
-disp("0.075 times/sec")
+
+//Example 12.1
+ 
+//Page 524
+ 
+disp('The average arrival rate is')
+
+lam=(3600/10000)//arrivals per sec
+
+disp('From equation 12.2, the probability on arrival in 0.01-sec interval is')//equation on page 524
+
+P0=(%e^-0.0278)
+
+disp('Thus 2.7% arrivals occur withnin 0.01 sec of the pervious arrival. Since the arrival rate is 2.78 arrivals per second, the rate of occurrence of intervarrival time less than 0.01 sec is')
+
+2.78*0.027
+
+disp("Result")
+ 
+disp("0.075 times/sec")
  
 \ No newline at end of file
diff --git a/116/CH12/EX12.2/exa12_2.sce b/116/CH12/EX12.2/exa12_2.sce
index 638ae0ef0..e5e5dd433 100755
--- a/116/CH12/EX12.2/exa12_2.sce
+++ b/116/CH12/EX12.2/exa12_2.sce
@@ -1,35 +1,33 @@
-
-//Caption:Program to calculate the probability that eight or more arrivals occur in an chosen 30 sec
- 
-//Example 12.2
- 
-//Page 526
- 
-disp('The average number of arrivals in a 30 sec interval is,')
-
-lamt=4*(30/60)
-
-disp('The probability of eight or more arrivals is,')
-
-P0=1
-
-P1=[(2^1)/(1)]
-
-P2=[(2^2)/(1*2)]
-
-P3=[(2^3)/(1*2*3)]
-
-P4=[(2^4)/(1*2*3*4)]
-
-P5=[(2^5)/(1*2*3*4*5)]
-
-P6=[(2^6)/(1*2*3*4*5*6)]
-
-P7=[(2^7)/(1*2*3*4*5*6*7)]
-
-i=1-{(%e^-2)*[P0+P1+P2+P3+P4+P5+P6+P7]}
-
-disp("Result")
- 
-disp("P(2) = 0.0011")
+
+//Example 12.2
+ 
+//Page 526
+ 
+disp('The average number of arrivals in a 30 sec interval is,')
+
+lamt=4*(30/60)
+
+disp('The probability of eight or more arrivals is,')
+
+P0=1
+
+P1=[(2^1)/(1)]
+
+P2=[(2^2)/(1*2)]
+
+P3=[(2^3)/(1*2*3)]
+
+P4=[(2^4)/(1*2*3*4)]
+
+P5=[(2^5)/(1*2*3*4*5)]
+
+P6=[(2^6)/(1*2*3*4*5*6)]
+
+P7=[(2^7)/(1*2*3*4*5*6*7)]
+
+i=1-{(%e^-2)*[P0+P1+P2+P3+P4+P5+P6+P7]}
+
+disp("Result")
+ 
+disp("P(2) = 0.0011")
  
 \ No newline at end of file
diff --git a/116/CH12/EX12.3/exa12_3.sce b/116/CH12/EX12.3/exa12_3.sce
index 742b5f27b..2044e15e8 100755
--- a/116/CH12/EX12.3/exa12_3.sce
+++ b/116/CH12/EX12.3/exa12_3.sce
@@ -1,19 +1,18 @@
-  
-//Caption:Program to calculate the probability that a 1000 bit data block experiences exaclty 4 errors while being transmitted over a link having 10^-5 error rate
- 
-//Example 12.3
- 
-//Page 527
- 
-disp('Assuming inpendenterror, we can obtain the probability of exactly 4 errors directly from the Poisson distribution. The average number of errors is,')
-
-lamt=[(10^3)*(10^-5)]
-
-disp('Thus,')
-
-P4={[(0.01^4)/(1*2*3*4)]*%e^-0.01}
-
-disp("Result")
- 
-disp("P(4) = 4.125*10^-10")
+
+ 
+//Example 12.3
+ 
+//Page 527
+ 
+disp('Assuming inpendenterror, we can obtain the probability of exactly 4 errors directly from the Poisson distribution. The average number of errors is,')
+
+lamt=[(10^3)*(10^-5)]
+
+disp('Thus,')
+
+P4={[(0.01^4)/(1*2*3*4)]*%e^-0.01}
+
+disp("Result")
+ 
+disp("P(4) = 4.125*10^-10")
  
 \ No newline at end of file
diff --git a/116/CH12/EX12.4/exa12_4.sce b/116/CH12/EX12.4/exa12_4.sce
index eff220a3f..7ddacf887 100755
--- a/116/CH12/EX12.4/exa12_4.sce
+++ b/116/CH12/EX12.4/exa12_4.sce
@@ -1,34 +1,33 @@
-  
-//Caption:Program to calculate the percentage of total traffic carried by first five ckt and traffic carried by all other remaining
- 
-//Example 12.4
-
-//Page 529
- 
-disp('The Traffic intensity of system is,')
-
-A=1*2
-
-disp('The raffic intensity carried by i active ckt is exactly i erlangs. Hence the traffic carried by 1st 5 ckt is,')
-
-P1=[(1*2^1)/(1)]
- 
-P2=[(2*2^2)/(1*2)]
- 
-P3=[(3*2^3)/(1*2*3)]
-
-P4=[(4*2^4)/(1*2*3*4)]
-
-P5=[(5*2^5)/(1*2*3*4*5)]
-
-A5={(%e^-2)*[P1+P2+P3+P4+P5]}
-
-disp('All of remaining ckts carry,')
-
-Ar=2-1.89
-
-disp("Result")
- 
-disp("A(5) = 1.89 erlangs")
- 
-disp("A(remaining) = 0.11 erlangs")
+  
+
+//Example 12.4
+
+//Page 529
+ 
+disp('The Traffic intensity of system is,')
+
+A=1*2
+
+disp('The raffic intensity carried by i active ckt is exactly i erlangs. Hence the traffic carried by 1st 5 ckt is,')
+
+P1=[(1*2^1)/(1)]
+ 
+P2=[(2*2^2)/(1*2)]
+ 
+P3=[(3*2^3)/(1*2*3)]
+
+P4=[(4*2^4)/(1*2*3*4)]
+
+P5=[(5*2^5)/(1*2*3*4*5)]
+
+A5={(%e^-2)*[P1+P2+P3+P4+P5]}
+
+disp('All of remaining ckts carry,')
+
+Ar=2-1.89
+
+disp("Result")
+ 
+disp("A(5) = 1.89 erlangs")
+ 
+disp("A(remaining) = 0.11 erlangs")
diff --git a/116/CH12/EX12.5/exa12_5.sce b/116/CH12/EX12.5/exa12_5.sce
index 1c77bc6db..26994b43c 100755
--- a/116/CH12/EX12.5/exa12_5.sce
+++ b/116/CH12/EX12.5/exa12_5.sce
@@ -1,27 +1,26 @@
-  
-//Caption:Program to calculate how much traffic can the trunk group carry
- 
-//Example 5.5
- 
-//Page 534
- 
-//Refer figure 12.5 on page 533
- 
-disp('From fig, it can be that the output circuit utilization for B=0.1 and N=24 is 0.8.')
-
-N=24
-
-op=0.8
- 
-disp('Thus the carried traffic intensity is ')
-
-N*op
-
-disp('Since the blocking probability is 0.1, the maximum level of offered traaffic is,')
-
-A=[19.2/(1-0.1)]
-
-disp("Result")
-
-disp("A = 21.3 erlangs")
+
+ 
+//Example 5.5
+ 
+//Page 534
+ 
+//Refer figure 12.5 on page 533
+ 
+disp('From fig, it can be that the output circuit utilization for B=0.1 and N=24 is 0.8.')
+
+N=24
+
+op=0.8
+ 
+disp('Thus the carried traffic intensity is ')
+
+N*op
+
+disp('Since the blocking probability is 0.1, the maximum level of offered traaffic is,')
+
+A=[19.2/(1-0.1)]
+
+disp("Result")
+
+disp("A = 21.3 erlangs")
  
 \ No newline at end of file