2 files changed, 63 insertions, 0 deletions
diff --git a/116/CH10/EX10.1/exa10_1.sce b/116/CH10/EX10.1/exa10_1.sce
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+ 
+//Caption:Program to determine the amount of transmission capacity 
+ 
+//(a)Assume the link-by-link error control (b) Assume end-to-end error control (c)Repeat the calculation for a bit error probability of 10^-5
+ 
+//Example 10.1
+ 
+//Page 472
+ 
+//(a)With link-by-link 
+ 
+frame=1000*10^-8
+
+disp('The expected number of bits of transmission capacity required to retransmit is')
+
+frame*1000
+
+//(b)With end-to-end
+ 
+frames=10*10^-5//corrupted frame
+
+disp('The expected number of bits of transmission capacity required is')
+
+frames*1000 
+ 
+//(c)With bit error 10^-5
+ 
+ans1=1000*10^-5 
+ 
+ans1=1000*10^-5*1000
+
+ans=10*10^-2*1000
+ 
+//Result
+ 
+//(a)0.01 bit/link
+ 
+//(b)0.1 bit/link
+ 
+//(c)1. 10 bits/link
+ 
+//(c)2. 100 bits/link
diff --git a/116/CH10/EX10.3/exa10_3.sce b/116/CH10/EX10.3/exa10_3.sce
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+
+//Caption:Program to determine the probability that the delay of an ATM voice cell 
+ 
+//Example 10.3
+ 
+//Page 488
+ 
+disp('Assuming the access link is 90% utilized on average.')
+ 
+disp('The queuing theory is provided in Chapter 12. It involves determining the probability that the DSI access queue contains enough cells to represent 10msec of transmission time')
+ 
+tm=[(53*8)/(192*8000)]
+
+disp('Therefore, 10 msec delay represents 10/0.276 = 36.2 cell times.')
+
+p=(0.9)*{%e^[-(1-0.9)*36.2]}//Refer to equation 12.25 in chap 12 
+ 
+disp("Result")
+ 
+disp("P(>10msec) = 2.5% delay will be displayed by more than 10 msec ")
+ 
+\ No newline at end of file