1 / 25

CIS 720

CIS 720. Invariant Based Methodology. Producer/consumer problem. co Producer: Consumer: do true  do true  produce pdata buf = pdata cdata = buf; consume cdata

cecile
Télécharger la présentation

CIS 720

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CIS 720 Invariant Based Methodology

  2. Producer/consumer problem co Producer: Consumer: 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 = in1 + 1 in2 = in2 + 1 buf = pdata{in1 = out1+1}cdata = buf{in2 = out2+1} out1 = out1 + 1 out2 = out2 + 1 {in1=out1} consume cdata{in2 = out2} od od oc Invariant:

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

  5. Producer/consumer problem co Producer: Consumer: 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 co Producer: Consumer: 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 = in1 + 1 in2 = in2 + 1 in3 = in3 + 1 buf = pdata cdata = buf cdata = buf out1 = out1 + 1 out2 = out2 + 1 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 problem co Reader[i]: Writer[j]: do true  do true  P(rw) P(rw) read write V(rw) V(rw) od od oc

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

  12. Readers/writers problem co Reader[i]: Writer: do true  do true  nr=nr+1 nw = nw + 1 read write nr = nr - 1 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 problem co Reader[i]: Writer: do true  do true  inr = inr + 1 inw = inw + 1 read write outr = outr + 1 outw = outw + 1 od od oc

  14. Readers/writers problem co Reader[i]: Writer: do true  do true  inr = inr + 1 inw = inw + 1 read write outr = outr + 1 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

  17. Readers/writers problem co 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

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

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

  20. 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)

  21. 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) finr++; nw++; SIGNAL SIGNAL read write P(e) P(e) 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)

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

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

  24. 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

More Related