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CDA 5416 Computer System Verification HW Review

Computer Architecture A Quantitative Approach, Fifth Edition. CDA 5416 Computer System Verification HW Review. Instructor: Hao Zheng Department of Computer Science & Engineering University of South Florida Tampa, FL 33620 Email: haozheng@usf.edu Phone: (813)974-4757 Fax: (813)974-5456.

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CDA 5416 Computer System Verification HW Review

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  1. Computer Architecture A Quantitative Approach, Fifth Edition CDA 5416 Computer System Verification HW Review Instructor: Hao Zheng Department of Computer Science & Engineering University of South FloridaTampa, FL 33620Email: haozheng@usf.eduPhone: (813)974-4757Fax: (813)974-5456

  2. HW 4

  3. Problem 2

  4. Problem 3

  5. Problem 4

  6. HW 3

  7. HW 2

  8. Draw program graphs for twoprocesses with id = {0,1}. Show an execution of the composed program graph. Build a Promela model for this algorithm with threeprocesses.  Think about the type of channels that should be used. Format the relevant correctness requirement(s) using a separate process(es) and/or assertions in Promela.  Use the SPIN to check that the leader election model satisfies the correctness requirements.

  9. C0 (c0) P1 P0 C1 (c1) 0 0 /send(id0) /recv(c0) /send(c0) 1 1 /recv(m0) m0<id0 2 stop m0>id0/send(m0) m0=id0

  10. C0 P1 P0 C1 0 0 0 0 /<send(id0), recv(c1)> /<send(id1), recv(c0)> 0 1 0 0 1 1 1 0 /<send(id1), recv(c0)> /<send(id0), recv(c1)> 1 1 1 1

  11. C0 P1 P0 C1 /<send(id1), recv(c0)> /<send(id0), recv(c1)> 1 1 1 1 /<send(c0), recv(m0)> /<send(c1), recv(m1)> 1 2 1 0 0 2 1 1

  12. C0 P1 P0 C1 chanC1= [1] of {byte}; chan C2 = [1] of {byte};

  13. proctypeP0() { byte m0; C1 ! id0; do :: C0 ? m0; if :: m0<id0 -> goto stop :: m0>id0 -> C1 ! m0 fi od stop: } 0 /C1 ! id0 1 /C0 ? m0 m0<id0 2 stop m0>id0/ C1 ! m0 m0=id0

  14. HW 1

  15. byte n = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; } What are the largest and smallest values that n can be produced by the model upon termination?

  16. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } /* passed  active proctype checker1() {         if          :: stop==2 -> assert (n <= 20);         fi } */ /* failed */ active proctype checker2() {         if          :: stop==2 -> assert (n <= 19);         fi }

  17. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } /* passed active proctype checker4() {         if          :: stop==2 -> assert (n >= 2);         fi } */ /* failed */ active proctype checker4() {         if          :: stop==2 -> assert (n >= 3);         fi }

  18. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } 1: proc  1 (p:1) a1_3.pml:12 (state 3) [else]   2: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]   3: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)] n = 0; reg0=0, cnt0=0, reg1=1, cnt1=0   4: proc  0 (p:1) a1_3.pml:12 (state 3) [else]   5: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n] n = 0; reg0=0, cnt0=0, reg1=1, cnt1=0   6: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]   7: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 1; reg0=0, cnt0=0, reg1=1, cnt1=1   8: proc  1 (p:1) a1_3.pml:12 (state 3) [else]   9: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  10: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)] 11: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  12: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 2; reg0=0, cnt0=0, reg1=2, cnt1=2 spin -p -tmodel.pml /* replay error trace “model.pml.trail” */

  19. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 2; reg0=0, cnt0=0, reg1=2, cnt1=2 13: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  14: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  15: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  16: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  17: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 3; reg0=0, cnt0=0, reg1=3, cnt1=3 18: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  19: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  20: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  21: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  22: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 4; reg0=0, cnt0=0, reg1=4, cnt1=4

  20. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 4; reg0=0, cnt0=0, reg1=4, cnt1=4 23: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  24: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  25: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  26: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  27: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 5; reg0=0, cnt0=0, reg1=5, cnt1=5  28: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  29: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  30: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  31: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  32: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 6; reg0=0, cnt0=0, reg1=6, cnt1=6

  21. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 8; reg0=0, cnt0=0, reg1=8, cnt1=8 43: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  44: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  45: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  46: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  47: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]  48: proc  1 (p:1) a1_3.pml:12 (state 3) [else] n = 9; reg0=0, cnt0=0, reg1=9, cnt1=9  49: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  50: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg] n = 1; reg0=1, cnt0=0, reg1=9, cnt1=9  51: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  52: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)] n = 1; reg0=1, cnt0=0, reg1=2, cnt1=9

  22. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 1; reg0=1, cnt0=0, reg1=2, cnt1=9 53: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 1; reg0=1, cnt0=1, reg1=2, cnt1=9  54: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  55: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  56: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  57: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  58: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 2; reg0=2, cnt2=2, reg1=2, cnt1=9  59: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  60: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  61: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  62: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  63: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 3; reg0=3, cnt0=3, reg1=2, cnt1=9

  23. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 3; reg0=3, cnt0=3, reg1=2, cnt1=9 64: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  65: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  66: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  67: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  68: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 4; reg0=4, cnt0=4, reg1=2, cnt1=9  69: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  70: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  71: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  72: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  73: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 5; reg0=5, cnt0=5, reg1=2, cnt1=9

  24. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 5; reg0=5, cnt0=5, reg1=2, cnt1=9 74: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  75: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  76: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  77: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  78: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 6; reg0=6, cnt0=6, reg1=2, cnt1=9  79: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  80: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  81: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  82: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  83: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 7; reg0=7, cnt0=7, reg1=2, cnt1=9

  25. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 7; reg0=7, cnt0=7, reg1=2, cnt1=9 84: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  85: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  86: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  87: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  88: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 8; reg0=8, cnt0=8, reg1=2, cnt1=9 89: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  90: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  91: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  92: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  93: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 9; reg0=9, cnt0=9, reg1=2, cnt1=9

  26. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 9; reg0=9, cnt0=9, reg1=2, cnt1=9 94: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  95: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  96: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  97: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg] n = 10; reg0=10, cnt0=9, reg1=2, cnt1=9 98: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  99: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] 100: proc  1 (p:1) a1_3.pml:11 (state 1) [((cnt==10))] 101: proc  1 (p:1) a1_3.pml:18 (state 11) [stop = (stop+1)] n = 2; reg0=9, cnt0=9, reg1=2, cnt1=10; stop=1

  27. byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 2; reg0=9, cnt0=9, reg1=2, cnt1=10; stop=1 102: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] 103: proc  0 (p:1) a1_3.pml:11 (state 1) [((cnt==10))] 104: proc  0 (p:1) a1_3.pml:18 (state 11) [stop = (stop+1)] n = 2; reg0=9, cnt0=10, reg1=2, cnt1=10; stop=2 105: proc  2 (checker4:1) a1_3.pml:71 (state 1) [((stop==2))] 106: proc  2 (checker4:1) a1_3.pml:71 (state 2) [assert((n>=3))]

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