SERVER-35783 add integration test for plan cache oscillation

1 files changed, 122 insertions, 0 deletions

diff --git a/jstests/noPassthroughWithMongod/plan_cache_replanning.js b/jstests/noPassthroughWithMongod/plan_cache_replanning.js
new file mode 100644
index 00000000000..fa901b5e8d1
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+++ b/jstests/noPassthroughWithMongod/plan_cache_replanning.js
@@ -0,0 +1,122 @@
+/**
+ * This test will attempt to create a scenario where the plan cache entry for a given query shape
+ * oscillates. It achieves this by creating two indexes, A and B, on a collection, and interleaving
+ * queries which are "ideal" for index A with queries that are "ideal" for index B.
+*/
+(function() {
+    "use strict";
+
+    load('jstests/libs/analyze_plan.js');              // For getPlanStage().
+    load("jstests/libs/collection_drop_recreate.js");  // For assert[Drop|Create]Collection.
+
+    const coll = assertDropAndRecreateCollection(db, "plan_cache_replanning");
+
+    function getPlansForCacheEntry(query) {
+        let key = {query: query, sort: {}, projection: {}};
+        let res = coll.runCommand("planCacheListPlans", key);
+        assert.commandWorked(res, `planCacheListPlans(${tojson(key)}) failed`);
+        assert(res.hasOwnProperty("plans"),
+               `plans missing from planCacheListPlans(${tojson(key)}) failed`);
+
+        return res;
+    }
+
+    function planHasIxScanStageForKey(planStats, keyPattern) {
+        const stage = getPlanStage(planStats, "IXSCAN");
+        if (stage === null) {
+            return false;
+        }
+
+        return bsonWoCompare(keyPattern, stage.keyPattern) == 0;
+    }
+
+    const queryShape = {a: 1, b: 1};
+
+    // Carefully construct a collection so that some queries will do well with an {a: 1} index
+    // and others with a {b: 1} index.
+    for (let i = 1000; i < 1100; i++) {
+        assert.commandWorked(coll.insert({a: 1, b: i}));
+    }
+
+    for (let i = 1000; i < 1100; i++) {
+        assert.commandWorked(coll.insert({a: i, b: 2}));
+    }
+
+    // This query will be quick with {a: 1} index, and far slower {b: 1} index. With the {a: 1}
+    // index, the server should only need to examine one document. Using {b: 1}, it will have to
+    // scan through each document which has 2 as the value of the 'b' field.
+    const aIndexQuery = {a: 1099, b: 2};
+    // Opposite of 'aIndexQuery'. Should be quick if the {b: 1} index is used, and slower if the
+    // {a: 1} index is used.
+    const bIndexQuery = {a: 1, b: 1099};
+
+    assert.commandWorked(coll.createIndex({a: 1}));
+    assert.commandWorked(coll.createIndex({b: 1}));
+
+    // Run a query where the {b: 1} index will easily win.
+    assert.eq(1, coll.find(bIndexQuery).itcount());
+
+    // The plan cache should now hold an inactive entry.
+    let entry = getPlansForCacheEntry(queryShape);
+    let entryWorks = entry.works;
+    assert.eq(entry.isActive, false);
+    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
+
+    // Re-run the query. The inactive cache entry should be promoted to an active entry.
+    assert.eq(1, coll.find(bIndexQuery).itcount());
+    entry = getPlansForCacheEntry(queryShape);
+    assert.eq(entry.isActive, true);
+    assert.eq(entry.works, entryWorks);
+    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
+
+    // Now we will attempt to oscillate the cache entry by interleaving queries which should use
+    // the {a:1} and {b:1} index. When the plan using the {b: 1} index is in the cache, running a
+    // query which should use the {a: 1} index will perform very poorly, and trigger
+    // replanning (and vice versa).
+
+    // The {b: 1} plan is currently in the cache. Run the query which should use the {a: 1}
+    // index. The current cache entry will be deactivated, and then the cache entry for the {a: 1}
+    // will overwrite it (as active).
+    assert.eq(1, coll.find(aIndexQuery).itcount());
+    entry = getPlansForCacheEntry(queryShape);
+    assert.eq(entry.isActive, true);
+    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {a: 1}), true);
+
+    // Run the query which should use the {b: 1} index.
+    assert.eq(1, coll.find(bIndexQuery).itcount());
+    entry = getPlansForCacheEntry(queryShape);
+    assert.eq(entry.isActive, true);
+    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
+
+    // The {b: 1} plan is again in the cache. Run the query which should use the {a: 1}
+    // index.
+    assert.eq(1, coll.find(aIndexQuery).itcount());
+    entry = getPlansForCacheEntry(queryShape);
+    assert.eq(entry.isActive, true);
+    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {a: 1}), true);
+
+    // The {a: 1} plan is back in the cache. Run the query which would perform better on the plan
+    // using the {b: 1} index, and ensure that plan gets written to the cache.
+    assert.eq(1, coll.find(bIndexQuery).itcount());
+    entry = getPlansForCacheEntry(queryShape);
+    entryWorks = entry.works;
+    assert.eq(entry.isActive, true);
+    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
+
+    // Now run a plan that will perform poorly with both indices (it will be required to scan 500
+    // documents). This will result in replanning (and the cache entry being deactivated). However,
+    // the new plan will have a very high works value, and will not replace the existing cache
+    // entry. It will only bump the existing cache entry's works value.
+    for (let i = 0; i < 500; i++) {
+        assert.commandWorked(coll.insert({a: 3, b: 3}));
+    }
+    assert.eq(500, coll.find({a: 3, b: 3}).itcount());
+
+    // The cache entry should have been deactivated.
+    entry = getPlansForCacheEntry(queryShape);
+    assert.eq(entry.isActive, false);
+    assert.eq(planHasIxScanStageForKey(entry.plans[0].reason.stats, {b: 1}), true);
+
+    // The works value should have doubled.
+    assert.eq(entry.works, entryWorks * 2);
+})();