1. Invariant Based Methodology
CIS 720
Invariant Based Methodology

2. Producer/consumer problem
do true  do true 
produce pdata
buf = pdata cdata = buf;
consume cdata
od od oc

3. Producer/consumer problem
in1 = out1 = in2 = out2 = 0; co
Producer: Consumer:
do true  {in1=out1} do true  {in2=out2}
produce pdata
in1 = in in2 = in2 + 1
buf = pdata {in1 = out1+1} cdata = buf {in2 = out2+1}
out1 = out out2 = out2 + 1
{in1=out1} consume cdata {in2 = out2}
od od oc
Invariant:

4. Producer/consumer problem
do true  do true 
produce pdata
<await(in1 <=out2)  in1 = in1 + 1> <await(in2 < out1) in2 = in2 + 1>
buf = pdata cdata = buf
<out1 = out1 + 1> <out2 = out2 + 1>
consume cdata
od od oc
Invariant: (in1 <= out2 + 1) /\ (in2 <= out1)

5. Producer/consumer problem
do true  do true 
produce pdata
<await(empty > 0)  empty= empty -1> <await(full > 0)  full = full -1 >
buf = pdata cdata = buf
<full = full + 1> <empty = empty + 1>
consume cdata
od od oc
Invariant: (in1 <= out2 + 1) /\ (in2 <= out1)
empty = out2 + 1 in1
full = out1 – in2

6. Producer/consumer problem
do true  do true 
produce pdata
P(empty) P(full)
buf = pdata cdata = buf
V(full) V(empty)
consume cdata
od od oc
Invariant: (in1 <= out2 + 1) /\ (in2 <= out1)
empty = out2 + 1 in1
full = out1 – in2

7. Producer/consumer problem
in1 = out1 = in2 = out2 = 0; co
Producer: Consumer1: Consumer2:
do true  do true  do true  produce pdata
in1 = in in2 = in in3 = in3 + 1
buf = pdata cdata = buf cdata = buf
out1 = out out2 = out out3 = out3 + 1
consume cdata consume cdata
od od od oc
Invariant:

8. Readers/writers problem
A reader and a writer cannot read and write respectively at the same time.
Two writers cannot write at the same time.

9. Readers/writers problemco
Reader[i]: Writer[j]:
do true  do true 
P(rw) P(rw)
read write
V(rw) V(rw)
od od oc

10. Readers/writers problemco
Reader[i]: Writer:
do true  do true 
nr = nr P(rw)
if nr == 1  P(rw) fi
write
read
nr = nr - 1
if (nr =0) V(rw) fi V(rw)
od od oc

11. Readers/writers problemco
Reader[i]: Writer:
do true  do true 
nr=nr+1 nw = nw + 1
read write
nr = nr nw = nw - 1
od od oc

12. Readers/writers problemco
Reader[i]: Writer:
do true  do true 
nr=nr+1 nw = nw + 1
read write
nr = nr nw = nw - 1
od od oc
BAD = (nr > 0 /\ nw > 0 ) \/ nw > 1
Invariant: not (BAD)
(nr = 0 \/ nw = 0) /\ nw <= 1

13. Readers/writers problemco
Reader[i]: Writer:
do true  do true 
inr = inr inw = inw + 1
read write
outr = outr outw = outw + 1
od od oc

14. Readers/writers problemco
Reader[i]: Writer:
do true  do true 
inr = inr inw = inw + 1
read write
outr = outr outw = outw + 1
od od oc
BAD =
Invariant:

15. Technique of Passing the baton
Solution may contain statements
F1 : <S>
F2 : < await (Bj)  Sj >
Introduce the following semaphores
e : for controlling access to evaluate the conditions
bj : for each guard Bj

16. Introduce a new counter dj for each guard Bj
Replace as follows:
F1:
P(e) Sj
SIGNAL

18. Readers/writers problemco
Reader[i]: Writer:
do true  do true 
<await(nw = 0)  nr=nr+1> <await(nw=0 /\ nr = 0)  nw = nw + 1>
read write
<nr = nr – 1> <nw = nw – 1>
od od oc

19. Readers/writers problemco
Reader[i]:
do true 
<await(nw = 0)  nr=nr+1>
read
<nr = nr – 1> od oc

20. Readers/writers problemco
Writer:
do true 
<await(nw=0 /\ nr = 0)  nw = nw + 1>
write
<nw = nw – 1> od oc

21. SIGNAL
if (nw = 0 and dr > 0)  dr = dr – 1; V(r)
elseif
(nw = 0 and nr = 0 and dw > 0) 
dw = dw – 1; V(w)
else V(e)

22. co Reader[i]: Writer: do true  do true  P(e) P(e)
if (nw > 0) dr++; V(e); P(r) fi if (nw=0 /\ nr = 0)  dw++; V(e); P(w) fi nr++; nw++;
SIGNAL SIGNAL
read write
nr--; nw--
SIGNAL SIGNAL
od od oc
SIGNAL
if (nw = 0 and dr > 0)  dr = dr – 1; V(r)
elseif
(nw = 0 and nr = 0 and dw > 0)  dw = dw – 1; V(w)
else V(e)

23. Weak Reader preference
If a reader A is reading and a reader B and a writer C is waiting then B is given preference.

24. Strong Reader preference
If a reader B and a writer C is waiting then B is given preference.

25. Strong Writer preference
If a reader B and a writer C is waiting then C is given preference.
SIGNAL:
If nw = 0 /\ dr > 0 /\ dw = 0  dr = dr – 1; V(r) []
nr = 0 /\ nw = 0 /\ dw > 0  dw = dw – 1; V(w);
[] else -> V(e) fi